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Title: Quantum Physics For All
Description: A Guide For All People Which Includes All Subjects Of Quantum Physics Simplified For Them In A Nice Presentable Way .
Description: A Guide For All People Which Includes All Subjects Of Quantum Physics Simplified For Them In A Nice Presentable Way .
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Quantum Mechanics
Richard Fitzpatrick
Professor of Physics
The University of Texas at Austin
Contents
1 Introduction
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2 Probability Theory
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5 Continuous Probability Distributions
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1 Introduction
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2 Wavefunctions
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3 Plane Waves
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4 Representation of Waves via Complex Functions
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5
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13
3
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15
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Heisenberg’s Uncertainty Principle
Schr¨dinger’s Equation
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4 Fundamentals of Quantum Mechanics
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4 Expectation Values and Variances
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5 Ehrenfest’s Theorem
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6 Operators
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7 Momentum Representation
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8 Heisenberg’s Uncertainty Principle
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9 Eigenstates and Eigenvalues
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10 Measurement
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11 Continuous Eigenvalues
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12 Stationary States
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1 Introduction
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2 Infinite Potential Well
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3 Square Potential Barrier
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4 WKB Approximation
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5 Cold Emission
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6 Alpha Decay
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7 Square Potential Well
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8 Simple Harmonic Oscillator
6 Multi-Particle Systems
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4 Two-Particle Systems
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5 Identical Particles
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1 Introduction
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2 Fundamental Concepts
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3 Particle in a Box
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4 Degenerate Electron Gases
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5 White-Dwarf Stars
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85
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1 Introduction
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2 Angular Momentum Operators
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3 Representation of Angular Momentum
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103
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9 Central Potentials
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4 Hydrogen Atom
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5 Rydberg Formula
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1 Introduction
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2 Spin Operators
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3 Spin Space
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4 Eigenstates of Sz and S2
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11 Addition of Angular Momentum
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4 Two Spin One-Half Particles
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1 Introduction
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2 Improved Notation
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3 Two-State System
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4 Non-Degenerate Perturbation Theory
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13 Time-Dependent Perturbation Theory
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175
4
QUANTUM MECHANICS
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14
Introduction
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Two-State System
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Perturbation Expansion
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Electromagnetic Radiation
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Spontaneous Emission
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2P → 1S Transitions in Hydrogen
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Forbidden Transitions
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1 Introduction
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2 Variational Principle
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3 Helium Atom
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4 Hydrogen Molecule Ion
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15 Scattering Theory
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6 Hard Sphere Scattering
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7 Low Energy Scattering
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8 Resonances
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1 Intended audience
These lecture notes outline a single semester course on non-relativistic quantum mechanics
which is primarily intended for upper-division undergraduate physics majors
...
In particular, prospective
students should be reasonably familiar with Newtonian dynamics, elementary classical
electromagnetism and special relativity, the physics and mathematics of waves (including the representation of waves via complex functions), basic probability theory, ordinary
and partial differential equations, linear algebra, vector algebra, and Fourier series and
transforms
...
2 Major Sources
The textbooks which I have consulted most frequently whilst developing course material
are:
The Principles of Quantum Mechanics, P
...
M
...
Quantum Mechanics, E
...
Introduction to the Quantum Theory, D
...
Modern Quantum Mechanics, J
...
Sakurai, (Benjamin/Cummings, Menlo Park CA, 1985)
...
Bohm, (Dover, New York NY, 1989)
...
L
...
Quantum Physics, S
...
Nonclassical Physics, R
...
Introduction to Quantum Mechanics, D
...
Griffiths, 2nd Edition, (Pearson Prentice Hall,
Upper Saddle River NJ, 2005)
...
3 Aim of Course
The aim of this course is to develop non-relativistic quantum mechanics as a complete
theory of microscopic dynamics, capable of making detailed predictions, with a minimum
of abstract mathematics
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4 Outline of Course
The first part of the course is devoted to an in-depth exploration of the basic principles
of quantum mechanics
...
e
...
We shall then proceed to investigate the rules of
quantum mechanics in a more systematic fashion in Chapter 4
...
Chapter 8 is devoted to the investigation of orbital angular momentum, and Chapter 9 to
the closely related subject of particle motion in a central potential
...
The second part of this course describes selected practical applications of quantum
mechanics
...
Time-dependent perturbation theory is employed to study radiative transitions in
the hydrogen atom in Chapter 13
...
Finally, Chapter 15 contains an introduction to quantum scattering
theory
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1 Introduction
This section is devoted to a brief, and fairly low level, introduction to a branch of mathematics known as probability theory
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2 What is Probability?
What is the scientific definition of probability? Well, let us consider an observation made on
a general system, S
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Suppose that we wish to find the probability of some general outcome, X
...
Mathematicians have a fancy name for a large group of similar systems
...
” So, let us consider
an ensemble, Σ, of similar systems, S
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We can write this
symbolically as
Ω(X)
,
(2
...
We can see that the probability P(X) must be a number
between 0 and 1
...
This is just another way of saying that there is no
chance of the outcome X
...
This is another way of saying that the
outcome X is bound to occur
...
3 Combining Probabilities
Consider two distinct possible outcomes, X and Y, of an observation made on the system
S, with probabilities of occurrence P(X) and P(Y), respectively
...
From the basic
definition of probability,
Ω(X | Y)
P(X | Y) = lim
,
(2
...
Now,
Ω(X | Y) = Ω(X) + Ω(Y)
(2
...
Thus,
P(X | Y) = P(X) + P(Y)
...
4)
So, the probability of the outcome X or the outcome Y is just the sum of the individual
probabilities of X and Y
...
It follows, from what has just been said, that the probability of throwing
either a one or a two is simply 1/6 + 1/6, which equals 1/3
...
Let us determine the probability of obtaining any of
these outcomes
...
But, this
quantity is also equal to the sum of the probabilities of all the individual outcomes, by
(2
...
e
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(2
...
This condition is equivalent to the self-evident statement that
an observation of a system must definitely result in one of its possible outcomes
...
Suppose that we make
an observation on a system picked at random from the ensemble, and then pick a second
system completely independently and make another observation
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The fancy
mathematical way of saying this is that the two observations are statistically independent
...
In order to determine
this probability, we have to form an ensemble of all of the possible pairs of systems which
we could choose from the ensemble Σ
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The number
of pairs of systems in this new ensemble is just the square of the number of systems in the
original ensemble, so
Ω(Σ ⊗ Σ) = Ω(Σ) Ω(Σ)
...
6)
Furthermore, the number of pairs of systems in the ensemble Σ ⊗ Σ which exhibit the
outcome X in the first system and Y in the second system is simply the product of the
number of systems which exhibit the outcome X and the number of systems which exhibit
the outcome Y in the original ensemble, so that
Ω(X ⊗ Y) = Ω(X) Ω(Y)
...
7)
Probability Theory
9
It follows from the basic definition of probability that
P(X ⊗ Y) =
Ω(X ⊗ Y)
= P(X) P(Y)
...
8)
Thus, the probability of obtaining the outcomes X and Y in two statistically independent
observations is the product of the individual probabilities of X and Y
...
2
...
We
could go to the central administration building and find out how many eighteen year-olds,
nineteen year-olds, etc
...
We would then write something like
Average Age ≃
N18 × 18 + N19 × 19 + N20 × 20 + · · ·
,
N18 + N19 + N20 · · ·
(2
...
Suppose that we were to pick
a student at random and then ask “What is the probability of this student being eighteen?”
From what we have already discussed, this probability is defined
P18 ≃
N18
Nstudents
,
(2
...
(Actually, this definition is only
accurate in the limit that Nstudents is very large
...
(2
...
So,
for a general variable u, which can take on any one of M possible values u1 , u2 , · · · , uM,
with corresponding probabilities P(u1 ), P(u2 ), · · · , P(uM ), the mean or average value of u,
which is denoted u , is defined as
M
u ≡
P(ui ) ui
...
12)
i=1
Suppose that f(u) is some function of u
...
Thus, f(u1 )
corresponds to u1 and occurs with the probability P(u1 ), and so on
...
i=1
(2
...
It follows that
M
f(u) + g(u) =
M
P(ui ) [f(ui ) + g(ui )] =
i=1
M
P(ui ) f(ui ) +
i=1
P(ui ) g(ui),
(2
...
(2
...
(2
...
But, how can
we characterize the scatter around the mean value? We could investigate the deviation of
u from its mean value u , which is denoted
(2
...
In fact, this is not a particularly interesting quantity, since its average is zero:
∆u = (u − u ) = u − u = 0
...
18)
This is another way of saying that the average deviation from the mean vanishes
...
The average value of this quantity,
M
(∆u)
2
P(ui ) (ui − u )2 ,
=
(2
...
The variance is a positive number, unless there is no scatter
at all in the distribution, so that all possible values of u correspond to the mean value u ,
in which case it is zero
...
20)
giving
(∆u)2 = u2 − u 2
...
21)
The variance of u is proportional to the square of the scatter of u around its mean value
...
22)
which is usually called the standard deviation of u
...
Probability Theory
11
2
...
In
general, we expect the probability that u takes on a value in the range u to u + du to be
directly proportional to du, in the limit that du → 0
...
23)
where P(u) is known as the probability density
...
5), (2
...
19)
generalize in a straightforward manner to give
∞
P(u) du,
(2
...
25)
P(u) (u − u )2 du = u2 − u 2 ,
1 =
(2
...
Exercises
1
...
The player then spins the drum,
aims at his/her head, and pulls the trigger
...
Suppose that the probability density for the speed s of a car on a road is given by
P(s) = A s exp −
s
,
s0
where 0 ≤ s ≤ ∞
...
More explicitly, P(s) ds gives the
probability that a car has a speed between s and s + ds
...
(b) What is the mean value of the speed?
(c) What is the “most probable” speed: i
...
, the speed for which the probability density has
a maximum?
(d) What is the probability that a car has a speed more than three times as large as the
mean value?
12
QUANTUM MECHANICS
3
...
e
...
Let P(t) be the probability of the atom not having decayed
at time t, given that it was created at time t = 0
...
What is the mean lifetime of the atom?
Wave-Particle Duality
13
3 Wave-Particle Duality
3
...
Waves are continuous and spatially extended, whereas particles are discrete and
have little or no spatial extent
...
In this chapter, we shall examine how wave-particle duality shapes the
general features of quantum mechanics
...
2 Wavefunctions
A wave is defined as a disturbance in some physical system which is periodic in both space
and time
...
g
...
1)
where x represents position, t represents time, and A, k, ω > 0
...
On the other hand, if we are considering
a light wave then ψ(x, t) might represent the wave’s transverse electric field
...
e
...
The function also oscillates between the minimum and
maximum values −1 and +1, respectively, as θ varies
...
1) is periodic in x with period λ = 2π/k: i
...
, ψ(x + λ, t) = ψ(x, t) for all x and t
...
e
...
Finally, the wavefunction oscillates between the minimum and maximum
values −A and +A, respectively, as x and t vary
...
Furthermore,
the quantity A is termed the wave amplitude, the quantity k the wavenumber, and the
quantity ω the wave angular frequency
...
The conventional wave frequency, in cycles per second (otherwise known as hertz), is ν =
1/T = ω/2π
...
1), is termed the phase
angle, and determines the exact positions of the wave maxima and minima at a given time
...
This follows
because the maxima of cos(θ) occur at θ = j 2π
...
It follows that the maximum, and, by implication,
the whole wave, propagates in the positive x-direction at the velocity ω/k
...
1: The solution of n · r = d is a plane
...
2)
is the wavefunction of a wave of amplitude A, wavenumber k, angular frequency ω, and
phase angle ϕ, which propagates in the negative x-direction at the velocity ω/k
...
3 Plane Waves
As we have just seen, a wave of amplitude A, wavenumber k, angular frequency ω, and
phase angle ϕ, propagating in the positive x-direction, is represented by the following
wavefunction:
ψ(x, t) = A cos(k x − ω t + ϕ)
...
3)
Now, the type of wave represented above is conventionally termed a one-dimensional plane
wave
...
Furthermore, it is a plane wave because the wave maxima, which
are located at
k x − ω t + ϕ = j 2π,
(3
...
These conclusions follow because Eq
...
4) can be re-written in the form
x = d,
(3
...
Moreover, as is well-known, (3
...
Wave-Particle Duality
15
The previous equation can also be written in the coordinate-free form
n · r = d,
(3
...
Since there is nothing
special about the x-direction, it follows that if n is re-interpreted as a unit vector pointing in an arbitrary direction then (3
...
As before, the plane is normal to n, and its distance of closest approach to the origin
is d
...
3
...
This observation allows us to write the three-dimensional equivalent to
the wavefunction (3
...
7)
where the constant vector k = (kx , ky , kz ) = k n is called the wavevector
...
It is threedimensional because its wavefunction, ψ(x, y, z, t), depends on all three Cartesian coordinates
...
8)
n · r = (j − ϕ/2π) λ + v t,
(3
...
Note that the wavenumber, k, is the magnitude of the
wavevector, k: i
...
, k ≡ |k|
...
(3
...
See Fig
...
2
...
3
...
e
...
Now, a real number, x (say), can take any value
in a continuum of different values lying between −∞ and +∞
...
It follows that
the square of a real number is a positive real number, whereas the square of an imaginary
number is a negative real number
...
10)
where x and y are real numbers
...
This is written mathematically as x = Re(z) and y = Im(z)
...
16
QUANTUM MECHANICS
k
λ
Figure 3
...
Now, just as we can visualize a real number as a point on an infinite straight-line, we
can visualize a complex number as a point in an infinite plane
...
e
...
This
idea is illustrated in Fig
...
3
...
This is
written mathematically as |z| = x2 + y2
...
The angle, θ = tan−1 (y/x), that the straight-line joining the representative point to the
origin subtends with the real axis is termed the argument of the corresponding complex
number, z
...
It follows from standard
trigonometry that x = r cos θ, and y = r sin θ
...
Complex numbers are often used to represent wavefunctions
...
11)
where φ is a real number
...
12)
where r = |z| and θ = arg(z) are real numbers
...
13)
where A is the wave amplitude, k the wavenumber, ω the angular frequency, and ϕ the
phase angle
...
14)
17
imaginary
Wave-Particle Duality
z
r
y
θ
x
real
Figure 3
...
where ψ0 is a complex constant
...
15)
where A is the modulus, and ϕ the argument, of ψ0
...
(3
...
(3
...
(3
...
13), can be represented
as the real part of a complex wavefunction of the form (3
...
For ease of notation, the
“take the real part” aspect of the above expression is usually omitted, and our general
one-dimension wavefunction is simply written
ψ(x, t) = ψ0 e i (k x−ω t)
...
18)
The main advantage of the complex representation, (3
...
13), is that the former enables us to combine the amplitude, A, and
18
QUANTUM MECHANICS
the phase angle, ϕ, of the wavefunction into a single complex amplitude, ψ0
...
19)
where k is the wavevector
...
5 Classical Light Waves
Consider a classical, monochromatic, linearly polarized, plane light wave, propagating
through a vacuum in the x-direction
...
Suppose that the wave is polarized such that this electric field oscillates in the y-direction
...
) Now, the electric field can be conveniently represented
in terms of a complex wavefunction:
¯
ψ(x, t) = ψ e i (k x−ω t)
...
20)
√
¯
Here, i = −1, k and ω are real parameters, and ψ is a complex wave amplitude
...
Suppose
that
¯
¯
ψ = |ψ| e i ϕ ,
(3
...
It follows that the physical electric field takes the form
¯
Ey (x, t) = Re[ψ(x, t)] = |ψ| cos(k x − ω t + ϕ),
(3
...
In addition, λ = 2π/k is the wavelength, and ν = ω/2π
the frequency (in hertz)
...
23)
ω = k c,
(3
...
Equations (3
...
24) yield
¯
¯
Ey (x, t) = |ψ| cos (k [x − (ω/k) t] + ϕ) = |ψ| cos (k [x − c t] + ϕ)
...
25)
Note that Ey depends on x and t only via the combination x − c t
...
(3
...
dt
(3
...
24), which determines the wave angular frequency as a function of the wavenumber, is generally termed a dispersion relation
...
(3
...
28)
vp = ,
k
which is known as the phase velocity
...
24) is effectively
saying that the phase velocity of a plane light wave propagating through a vacuum always
takes the fixed value c, irrespective of its wavelength or frequency
...
e
...
29)
ǫ0
where ǫ0 = 8
...
Hence, it follows from
Eqs
...
20) and (3
...
(3
...
This momentum
is directed along the wave’s direction of propagation, and is of density
G=
U
...
31)
3
...
The following facts regarding this effect can be established via careful observation
...
Second,
the current of photoelectrons, when it exists, is proportional to the intensity of the light
falling on the surface
...
These facts are inexplicable within
the framework of classical physics
...
According to this theory, light of fixed frequency ν consists of a
collection of indivisible discrete packages, called quanta,1 whose energy is
E = h ν
...
32)
20
QUANTUM MECHANICS
K
h
0
0
W/h
ν
Figure 3
...
Here, h = 6
...
Incidentally, h is called Planck’s constant, rather than Einstein’s constant, because Max
Planck first introduced the concept of the quantization of light, in 1900, whilst trying
to account for the electromagnetic spectrum of a black body (i
...
, a perfect emitter and
absorber of electromagnetic radiation)
...
In other words, the electrons have to acquire an energy W in order to be emitted from the
surface
...
Suppose that an electron absorbs a single quantum of light
...
If h ν is greater than W then the electron is emitted from the surface with
residual kinetic energy
K = h ν − W
...
33)
Otherwise, the electron remains trapped in the potential well, and is not emitted
...
Incidentally, we can
calculate Planck’s constant, and the work function of the metal, by simply plotting the
kinetic energy of the emitted photoelectrons as a function of the wave frequency, as shown
in Fig
...
4
...
Finally, the number of emitted electrons increases with the intensity of the
light because the more intense the light the larger the flux of light quanta onto the surface
...
Wave-Particle Duality
21
3
...
34)
where ¯ = h/2π = 1
...
Since classical light waves propagate at the fixed
h
velocity c, it stands to reason that photons must also move at this velocity
...
Hence, photons must be massless
...
c
(3
...
(3
...
It follows from the previous two equations that photons carry
momentum
p=¯k
h
(3
...
(3
...
3
...
Figure 3
...
The light from
the two slits is projected onto a screen a distance D behind them, where D ≫ d
...
Light from the first slit travels a distance x1 to get to this point,
whereas light from the second slit travels a slightly different distance x2
...
37)
∆x = x2 − x1 ≃ y,
D
provided d ≪ D
...
(3
...
38)
where ψ1 and ψ2 are the wavefunctions at the first and second slits, respectively
...
39)
22
QUANTUM MECHANICS
incoming wave
x1
y
double slits
x2
d
projection
screen
D
Figure 3
...
since the two slits are assumed to be illuminated by in-phase light waves of equal amplitude
...
This is reasonable provided D ≫ λ
...
e
...
Hence, it follows from Eq
...
30) that the light intensity on the screen a distance
y from the center-line is
I(y) ∝ |ψ(y, t)| 2
...
40)
Using Eqs
...
37)–(3
...
2D
(3
...
6 shows the characteristic interference pattern corresponding to the above expression
...
(3
...
9 Quantum Interference of Light
Let us now consider double-slit light interference from a quantum mechanical point of
view
...
6: Classical double-slit interference pattern
...
Hence, we expect the two slits in Fig
...
5 to be spraying
photons in all directions at the same rate
...
Let us also replace
the projection screen by a photographic film which records the position where it is struck
by each photon
...
3
...
Now, according to the above discussion, the interference pattern is built up one photon
at a time: i
...
, the pattern is not due to the interaction of different photons
...
Hence, the only way in which the classical interference pattern can
be reconstructed, after a great many photons have passed through the apparatus, is if
each photon has a greater probability of striking the film at points where the classical
interference pattern is bright, and a lesser probability of striking the film at points where
the interference pattern is dark
...
Let us call this number n(y)
...
Hence, in order to reconcile the classical
and quantum viewpoints, we need
Py (y) ≡ lim
N→∞
n(y)
∝ I(y) ∆y,
N
(3
...
(3
...
Here, Py (y) is the probability that a given photon strikes
24
QUANTUM MECHANICS
the film between y and y + ∆y
...
A
probability of 0 means that there is no chance of a photon striking the film between y and
y + ∆y, whereas a probability of 1 means that every photon is certain to strike the film in
this interval
...
In other words, the probability of a photon striking a
region of the film of width ∆y is directly proportional to this width
...
It is convenient to define a quantity known as the
probability density, P(y), which is such that the probability of a photon striking a region
of the film of infinitesimal width dy is Py (y) = P(y) dy
...
(3
...
However, according to Eq
...
40), I(y) ∝ |ψ(y, t)| 2
...
(3
...
Another way
of saying this is that the probability of a measurement of the photon’s distance from the
centerline, at the location of the film, yielding a result between y and y+dy is proportional
to |ψ(y, t)| 2 dy
...
If photons behaved classically then we
could, in principle, solve their equations of motion and predict exactly where each photon
was going to strike the film, given its initial position and velocity
...
In other words, we
can only reconcile the wave-like and particle-like properties of light in a statistical sense
...
In principle, each photon which passes through our apparatus is equally likely to pass
through one of the two slits
...
We know that we get an interference pattern in this
experiment
...
In this second experiment,
which is virtually identical to the first on the individual photon level, we know exactly
which slit each photon passed through
...
After
all, according to wave theory, it is impossible to obtain a two-slit interference pattern from
a single slit
...
It follows that, in the quantum mechanical version of the two-slit
interference experiment, we must think of each photon as essentially passing through both
slits simultaneously
...
10 Classical Particles
In this course, we are going to concentrate, almost exclusively, on the behaviour of nonrelativistic particles of non-zero mass (e
...
, electrons)
...
45)
m
and satisfy
p2
...
46)
E=
2m
3
...
For instance, it is possible to obtain a
double-slit interference pattern from a stream of mono-energetic electrons passing through
two closely spaced narrow slits
...
(3
...
It is found that
h
λ=
...
47)
p
The same relation is found for other types of particles
...
Note that the de Broglie wavelength is generally pretty
small
...
2 × 10−9 [E(eV)]−1/2 m,
(3
...
(An
electron accelerated from rest through a potential difference of 1000 V acquires an energy
of 1000 eV, and so on
...
9 × 10−11 [E(eV)]−1/2 m
...
49)
Given the smallness of the de Broglie wavelengths of common particles, it is actually
quite difficult to do particle interference experiments
...
Hence, particle interference experiments require either
very low energy particles (since λ ∝ E−1/2 ), or very closely spaced slits
...
Equation (3
...
50)
26
QUANTUM MECHANICS
which is exactly the same as the relation between momentum and wavenumber that we
obtained earlier for photons [see Eq
...
36)]
...
51)
where p is the particle’s vector momentum, and k its wavevector
...
Since the relation (3
...
34) between
energy and wave angular frequency should also apply to both photons and particles
...
52)
for particle waves, then Eqs
...
46) and (3
...
(3
...
k
(3
...
2m
(3
...
(3
...
Does this imply
that the dispersion relation (3
...
3
...
56)
where k = p/¯ > 0 and ω = E/¯ > 0
...
53)
...
56) represents a plane wave whose maxima and minima propagate
in the positive x-direction with the phase velocity vp = ω/k
...
From before, the most reasonable physical interpretation of the wavefunction is that
|ψ(x, t)| 2 is proportional to the probability density of finding the particle at position x at
¯
time t
...
56) is |ψ| 2 , which depends
on neither x nor t
...
Hence, the fact that the maxima
and minima of the wavefunction propagate at a phase velocity which does not correspond
to the classical particle velocity does not have any real physical consequences
...
e
...
e
...
ψ(x, t) =
(3
...
In writing the above expression, we are relying on the assumption that
particle waves are superposable: i
...
, it is possible to add two valid wave solutions to
form a third valid wave solution
...
As we shall see, in
Sect
...
15, this is indeed the case
...
Hence, the superposition (3
...
Now, there is a useful mathematical theorem, known as Fourier’s theorem, which states
that if
∞
1
¯
f(x) = √
f(k) e i k x dk,
(3
...
(3
...
We can use Fourier’s
¯
theorem to find the k-space function ψ(k) which generates any given x-space wavefunction
ψ(x) at a given time
...
4 (∆x) 2
(3
...
2 (∆x) 2
(3
...
3
...
It can be seen that a measurement of the particle’s position is most likely to yield
the value x0 , and very unlikely to yield a value which differs from x0 by more than 3 ∆x
...
60) is the wavefunction of a particle which is initially localized around x = x0 in
28
QUANTUM MECHANICS
Figure 3
...
some region whose width is of order ∆x
...
Now, according to Eq
...
57),
∞
¯
ψ(k) e i k x dk
...
62)
−∞
Hence, we can employ Fourier’s theorem to invert this expression to give
∞
¯
ψ(k) ∝
ψ(x, 0) e−i k x dx
...
63)
−∞
Making use of Eq
...
60), we obtain
∞
¯
ψ(k) ∝ e−i (k−k0 ) x0
exp −i (k − k0 ) (x − x0 ) −
−∞
(x − x0 )2
dx
...
64)
Changing the variable of integration to y = (x − x0 )/(2 ∆x), this reduces to
∞
¯
ψ(k) ∝ e−i k x0
exp −i β y − y2 dy,
(3
...
The above equation can be rearranged to give
∞
2
¯
ψ(k) ∝ e−i k x0 −β /4
2
e−(y−y0 ) dy,
−∞
(3
...
The integral now just reduces to a number, as can easily be seen by
making the change of variable z = y − y0
...
67)
where
1
...
68)
2 ∆x
Now, if |ψ(x)| 2 is proportional to the probability density of a measurement of the par¯
ticle’s position yielding the value x then it stands to reason that |ψ(k)| 2 is proportional to
the probability density of a measurement of the particle’s wavenumber yielding the value
k
...
According to Eq
...
67),
∆k =
¯
|ψ(k)| 2 ∝ exp −
(k − k0 ) 2
...
69)
Note that this probability distribution is a Gaussian in k-space
...
(3
...
3
...
Hence, a measurement of k is most likely to yield the value k0 , and very unlikely to yield
a value which differs from k0 by more than 3 ∆k
...
We have just seen that a Gaussian probability distribution of characteristic width ∆x in
x-space [see Eq
...
61)] transforms to a Gaussian probability distribution of characteristic
width ∆k in k-space [see Eq
...
69)], where
1
∆x ∆k =
...
70)
This illustrates an important property of wave packets
...
e
...
e
...
Conversely, if
we only combine plane waves whose wavenumbers differ by a small amount (i
...
, if ∆k
is small) then the resulting wave packet will be very extended in x-space (i
...
, ∆x will be
large)
...
13 Evolution of Wave Packets
We have seen, in Eq
...
60), how to write the wavefunction of a particle which is initially
localized in x-space
...
(3
...
71)
−∞
30
QUANTUM MECHANICS
where
(3
...
¯
The function ψ(k) is obtained by Fourier transforming the wavefunction at t = 0
...
(3
...
67)
...
(3
...
Thus, it is a reasonable approximation to Taylor expand φ(k) about k0
...
73)
where
(3
...
75)
′′
φ0
(3
...
77)
ω0 = ω(k0 ),
dω(k0 )
,
dk
d2 ω(k0 )
α =
...
78)
vg =
(3
...
(3
...
80)
−∞
where
β1 = 2 ∆k (x − x0 − vg t),
(3
...
(3
...
The above
expression can be rearranged to give
∞
ψ(x, t) ∝ e i (k0 x−ω0 t)−(1+i β2 ) β
2 /4
2
e−(1+i β2 ) (y−y0 ) dy,
−∞
(3
...
Again changing the variable of integration to
z = (1 + i β2 )1/2 (y − y0 ), we get
∞
ψ(x, t) ∝ (1 + i β2 )−1/2 e i (k0 x−ω0 t)−(1+i β2 ) β
2 /4
2
e−z dz
...
84)
−∞
The integral now just reduces to a number
...
4 (∆x) 2
,
(3
...
86)
Note that the above wavefunction is identical to our original wavefunction (3
...
This, justifies the approximation which we made earlier by Taylor expanding the phase
factor φ(k) about k = k0
...
(3
...
(3
...
Now, the most likely position of our particle coincides with the peak of the
distribution function
...
(3
...
89)
which is known as the group velocity
...
Now,
it follows from the dispersion relation (3
...
m
(3
...
(3
...
Hence, the dispersion relation (3
...
In fact, a plane wave is usually interpreted as a continuous stream
of particles propagating in the same direction as the wave
...
(3
...
Indeed, it follows from Eqs
...
53) and (3
...
¯
h
(3
...
e
...
Evidently, particle wave packets (for
freely moving particles) spread very rapidly
...
See Eqs
...
79) and
(3
...
This is why a functional relationship between ω and k is generally known as a
dispersion relation: i
...
, because it governs how wave packets disperse as time progresses
...
e
...
Now, the dispersion relation (3
...
It follows that light pulses propagate through a vacuum without spreading
...
Thus, both plane light waves
and light pulses propagate through a vacuum at the characteristic speed c = 3 × 108 m/s
...
53) for particle waves is not linear in k
...
3
...
However, as time progresses, the width of
the wave packet in x-space increases, whilst that of the wave packet in k-space stays the
same
...
(3
...
(3
...
] Hence, in general, we can say that
∆x ∆k >
∼
1
...
92)
Furthermore, we can think of ∆x and ∆k as characterizing our uncertainty regarding the
values of the particle’s position and wavenumber, respectively
...
Hence, an uncertainty in k of order ∆k translates to an
h
uncertainty in p of order ∆p = ¯ ∆k
...
2
(3
...
8: Heisenberg’s microscope
...
According to this principle, it is impossible to simultaneously measure the position
and momentum of a particle (exactly)
...
Note that the uncertainty
principle is a direct consequence of representing particles as waves
...
(3
...
79), and (3
...
m ∆x
(3
...
Since the initial uncertainty in the particle’s position is ∆x, it follows that the uncertainty
in its momentum is of order ¯ /∆x
...
Thus, if we imagine that parts of the wavefunction propagate at v0 + ∆v/2, and
h
others at v0 − ∆v/2, where v0 is the mean propagation velocity, then the wavefunction will
spread as time progresses
...
95)
m ∆x
which is identical to Eq
...
94)
...
Figure 3
...
Suppose that we try to image an electron using a simple optical system in which the objective lens is of diameter D and focal-length f
...
) It is a well-known result in optics that such a
system has a minimum angular resolving power of λ/D, where λ is the wavelength of the
light illuminating the electron
...
D
(3
...
97)
f
where α is the half-angle subtended by the lens at the electron
...
98)
2f
so
λ
...
99)
∆x ≃
2α
It follows that we can reduce the uncertainty in the electron’s position by minimizing the
ratio λ/α: i
...
, by using short wavelength radiation, and a wide-angle lens
...
According to quantum mechanics, the electron is imaged when it scatters an incoming
photon towards the objective lens
...
See Fig
...
8
...
Here, we are ignoring any wavelength shift of the photon on scattering—
i
...
, the magnitude of the k-vector is assumed to be the same before and after scattering
...
This
translates to a change in the photon’s x-component of momentum of ∆px = ¯ k (sin θ −
h
1)
...
However, θ can range all the way from −α to +α, and the scattered
photon will still be collected by the imaging system
...
(3
...
This is exactly the opposite of what we need to do to reduce the uncertainty in
the position
...
101)
which is essentially the uncertainty principle
...
First, it is impossible to measure any property of a microscopic dynamical system without
Wave-Particle Duality
35
disturbing the system somewhat
...
Hence, there is a limit to how small we can make the aforementioned disturbance
...
3
...
102)
−∞
¯
where ψ(k) is determined by ψ(x, 0), and
ω(k) =
¯ k2
h
...
103)
Now, it follows from Eq
...
102) that
∂ψ
=
∂x
and
∂2 ψ
=
∂x2
whereas
∂ψ
=
∂t
∞
¯
(i k) ψ(k) e i (k x−ω t) dk,
(3
...
105)
¯
(−i ω) ψ(k) e i (k x−ω t) dk
...
106)
−∞
∞
−∞
∞
−∞
Thus,
i
∂ψ
¯ ∂2 ψ
h
+
=
∂t
2 m ∂x2
∞
ω−
−∞
¯ k2 ¯
h
ψ(k) e i (k x−ω t) dk = 0,
2m
(3
...
103)
...
(3
...
Schr¨dinger’s equation is a linear, second-order, partial differential
o
o
equation which governs the time evolution of a particle wavefunction, and is generally
easier to solve than the integral equation (3
...
Of course, Eq
...
108) is only applicable to freely moving particles
...
It is plausible, from Eq
...
104), that we can identify k with the differential
operator −i ∂/∂x
...
(3
...
But, p = ¯ k
...
However, in the presence of a potential
V(x), the particle’s energy is written p2 /(2 m) + V
...
(3
...
∂t
2 m ∂x2
(3
...
16 Collapse of the Wave Function
Consider an extended wavefunction ψ(x, t)
...
If the wavefunction is extended then there is a wide range of
likely values that this measurement could give
...
We now know that the particle is located at x = x0
...
Thus,
immediately after the first measurement, a measurement of the particle’s position is certain to give the value x0 , and has no chance of giving any other value
...
This is illustrated in Fig
...
9
...
3
...
Thus, the second measurement must be
made reasonably quickly after the first, in order to guarantee that the same result will be
obtained
...
Namely,
that the wavefunction of a particle changes discontinuously (in time) whenever a measurement is made
...
First, there is a smooth evolution which is governed by
Schr¨dinger’s equation
...
Second, there
o
is a discontinuous evolution which takes place each time a measurement is made
...
A He-Ne laser emits radiation of wavelength λ = 633 nm
...
The ionization energy of a hydrogen atom in its ground state is Eion = 13
...
Calculate
Wave-Particle Duality
37
|ψ|2 →
BEFORE
x→
|ψ|2 →
AFTER
x0
x→
Figure 3
...
the frequency, wavelength, and wavenumber of the electromagnetic radiation which will just
ionize the atom
...
The maximum energy of photoelectrons from aluminium is 2
...
90 eV for radiation of wavelength 2580 A
...
4
...
29 × 10−9 V −1/2 m,
where V is measured in volts
...
If the atoms in a regular crystal are separated by 3×10−10 m demonstrate that an accelerating
voltage of about 1
...
6
...
What are the group and phase velocities of such
waves as functions of ν0 and λ?
7
...
Use
the uncertainty principle to show that electrons of energy 1 MeV could not be contained in
the nucleus before the decay
...
A particle of mass m has a wavefunction
ψ(x, t) = A exp[−a (m x2 /¯ + i t)],
h
where A and a are positive real constants
...
1 Introduction
The previous chapter serves as a useful introduction to many of the basic concepts of
quantum mechanics
...
For the sake of simplicity, we shall concentrate on one-dimensional systems
...
2 Schr¨ dinger’s Equation
o
Consider a dynamical system consisting of a single non-relativistic particle of mass m moving along the x-axis in some real potential V(x)
...
This wavefunction
evolves in time according to Schr¨dinger’s equation:
o
i¯
h
∂ψ
¯ 2 ∂2 ψ
h
=−
+ V(x) ψ
...
1)
The wavefunction is interpreted as follows: |ψ(x, t)| 2 is the probability density of a measurement of the particle’s displacement yielding the value x
...
(4
...
4
...
An outcome of a measurement which
has a probability 0 is an impossible outcome, whereas an outcome which has a probability
1 is a certain outcome
...
(4
...
(4
...
It follows that Px ∈ −∞:∞ = 1, or
∞
|ψ(x, t)| 2 dx = 1,
−∞
(4
...
For example, suppose that we wish to normalize the wavefunction of a Gaussian wave
packet, centered on x = x0 , and of characteristic width σ (see Sect
...
12): i
...
,
ψ(x) = ψ0 e−(x−x0 )
2 /(4 σ2 )
(4
...
In order to determine the normalization constant ψ0 , we simply substitute Eq
...
5) into
Eq
...
4), to obtain
∞
e−(x−x0 )
|ψ0 | 2
2 /(2 σ2 )
(4
...
−∞
√
Changing the variable of integration to y = (x − x0 )/( 2 σ), we get
√
|ψ0 | 2 2 σ
∞
2
e−y dy = 1
...
7)
−∞
However,
∞
2
e−y dy =
√
π,
(4
...
(2π σ2)1/2
(4
...
10)
where ϕ is an arbitrary real phase-angle
...
If this is not
o
the case then the probability interpretation of the wavefunction is untenable, since it does
not make sense for the probability that a measurement of x yields any possible outcome
(which is, manifestly, unity) to change in time
...
11)
−∞
for wavefunctions satisfying Schr¨dinger’s equation
...
(4
...
2
∂t
2m
∂x
¯
h
(4
...
14)
2
∂t
2m
∂x
¯
h
= A, and i∗ = −i]
...
−ψ
2 m ∂x
∂x
∂x
(4
...
12) and (4
...
(4
...
(4
...
(4
...
Hence, we conclude that all wavefunctions which are square-integrable [i
...
, are
such that the integral in Eq
...
4) converges] have the property that if the normalization
condition (4
...
It is also possible to demonstrate, via very similar analysis to the above, that
where Px ∈ a:b
dPx ∈ a:b
+ j(b, t) − j(a, t) = 0,
dt
is defined in Eq
...
2), and
j(x, t) =
i¯
h
∂ψ∗
∂ψ
ψ
− ψ∗
2m
∂x
∂x
(4
...
19)
is known as the probability current
...
Equation (4
...
According to this equation, the probability of a measurement of
x lying in the interval a to b evolves in time due to the difference between the flux of
probability into the interval [i
...
, j(a, t)], and that out of the interval [i
...
, j(b, t)]
...
Note, finally, that not all wavefunctions can be normalized according to the scheme set
out in Eq
...
4)
...
20)
is not square-integrable, and, thus, cannot be normalized
...
(4
...
42
QUANTUM MECHANICS
4
...
Suppose that we made a large number of
independent measurements of the displacement on an equally large number of identical
quantum systems
...
However, from the definition of probability, the mean of all these results is simply
∞
x |ψ| 2 dx
...
22)
−∞
Here, x is called the expectation value of x
...
f(x) =
(4
...
The degree of scatter is parameterized by the quantity
∞
σ2 =
x
−∞
(x − x ) 2 |ψ| 2 dx ≡ x2 − x 2 ,
(4
...
The square-root of this quantity, σx , is called the
standard deviation of x
...
For instance, consider the normalized Gaussian wave packet [see Eq
...
10)]
ψ(x) =
eiϕ
2
2
e−(x−x0 ) /(4 σ )
...
25)
The expectation value of x associated with this wavefunction is
x =√
∞
1
2π σ2
x e−(x−x0 )
2 /(2 σ2 )
dx
...
26)
2
(4
...
It follows that
x0
x =√
π
∞
e
−∞
−y2
√
2σ
dy + √
π
∞
y e−y dy
...
Hence, making
use of Eq
...
8), we obtain
x = x0
...
28)
Evidently, the expectation value of x for a Gaussian wave packet is equal to the most likely
value of x (i
...
, the value of x which maximizes |ψ| 2 )
...
25) is
σ2 = √
x
∞
1
(x − x0 ) 2 e−(x−x0 )
2π σ2
2 /(2 σ2 )
(4
...
−∞
√
Let y = (x − x0 )/( 2 σ)
...
However,
∞
2
y e
(4
...
31)
giving
2
σx = σ2
...
32)
This result is consistent with our earlier interpretation of σ as a measure of the spatial
extent of the wave packet (see Sect
...
12)
...
25) in the convenient form
ψ(x) =
eiϕ
2
2
e−(x− x ) /(4 σx )
...
33)
4
...
e
...
∂t
(4
...
35)
[this is just the differential form of Eq
...
18)], where j is the probability current defined
in Eq
...
19)
...
36)
−∞ ∂x
−∞
where we have integrated by parts
...
(4
...
37)
44
QUANTUM MECHANICS
where we have again integrated by parts
...
(4
...
39)
where we have integrated by parts
...
1), and
o
simplifying, we obtain
dp
=
dt
∞
−
−∞
∂|ψ| 2
¯ 2 ∂ ∂ψ∗ ∂ψ
h
+ V(x)
dx =
2 m ∂x ∂x ∂x
∂x
∞
V(x)
−∞
∂|ψ| 2
dx
...
40)
Integration by parts yields
dp
=−
dt
∞
−∞
dV
dV
|ψ| 2 dx = −
dx
dx
...
41)
Hence, according to Eqs
...
34) and (4
...
42)
p,
= −
dV
dx
...
43)
Evidently, the expectation values of displacement and momentum obey time evolution
equations which are analogous to those of classical mechanics
...
Suppose that the potential V(x) is slowly varying
...
Keeping terms up to second order, we obtain
dV( x ) dV 2 ( x )
1 dV 3 ( x )
dV(x)
=
+
(x − x ) +
(x − x ) 2
...
44)
Substitution of the above expansion into Eq
...
43) yields
2
dp
dV( x ) σx dV 3 ( x )
=−
−
,
dt
dx
2 dx3
(4
...
The final term on the righthand side of the above equation can be neglected when the spatial extent of the particle
Fundamentals of Quantum Mechanics
45
wavefunction, σx , is much smaller than the variation length-scale of the potential
...
(4
...
43) reduce to
m
dx
dt
dp
dt
=
p,
= −
dV( x )
...
46)
(4
...
Of course, if the spatial extent of the wavefunction is negligible then a measurement of x is almost certain to yield a result which lies
very close to x
...
This is an important result, since we know
that classical mechanics gives the correct answer in this limit
...
6 Operators
An operator, O (say), is a mathematical entity which transforms one function into another:
i
...
,
O(f(x)) → g(x)
...
48)
For instance, x is an operator, since x f(x) is a different function to f(x), and is fully specified once f(x) is given
...
Now,
x
df
d
=
(x f)
...
49)
This can also be written
d
d
=
x,
(4
...
The above expression illustrates an important
point: i
...
, in general, operators do not commute
...
g
...
(4
...
52)
where f is a general function, and c a general complex number
...
46
QUANTUM MECHANICS
Now, from Eqs
...
22) and (4
...
53)
−∞
∞
h
ψ∗ −i ¯
=
p
−∞
∂
ψ dx
...
54)
These expressions suggest a number of things
...
Second, displacement is represented by the algebraic operator x, and momentum
by the differential operator −i ¯ ∂/∂x: i
...
,
h
p ≡ −i ¯
h
∂
...
55)
Finally, the expectation value of some dynamical variable represented by the operator O(x)
is simply
∞
ψ∗ (x, t) O(x) ψ(x, t) dx
...
56)
−∞
Clearly, if an operator is to represent a dynamical variable which has physical significance then its expectation value must be real
...
57)
−∞
where O∗ is the complex conjugate of O
...
It is easily demonstrated that x and p are both Hermitian
...
(4
...
e
...
For a non-Hermitian operator, O (say), it is easily demonstrated that (O† )† = O,
and that the operator O + O† is Hermitian
...
Suppose that we wish to find the operator which corresponds to the classical dynamical
variable x p
...
However,
in quantum mechanics, we have already seen that x p = p x
...
However, (1/2) [x p + (x p)† ] is
Hermitian
...
Fundamentals of Quantum Mechanics
47
It is a reasonable guess that the operator corresponding to energy (which is called the
Hamiltonian, and conventionally denoted H) takes the form
H≡
p2
+ V(x)
...
59)
Note that H is Hermitian
...
(4
...
2 m ∂x2
(4
...
1), we have
o
−
¯ 2 ∂2
h
∂
+ V(x) = i ¯
h ,
2 m ∂x2
∂t
(4
...
∂t
Thus, the time-dependent Schr¨dinger equation can be written
o
H ≡ i¯
h
i¯
h
∂ψ
= H ψ
...
62)
(4
...
h
4
...
3
...
64)
ψ(x, t) e−i k x dx,
(4
...
However, p = ¯ k
...
66)
2π ¯ −∞
h
∞
1
h
√
φ(p, t) =
ψ(x, t) e−i p x/¯ dx,
(4
...
ψ(x, t) = √
48
QUANTUM MECHANICS
At this stage, it is convenient to introduce a useful function called the Dirac deltafunction
...
However, the
singularity at x = 0 is such that
∞
δ(x) dx = 1
...
68)
−∞
The delta-function is an example of what is known as a generalized function: i
...
, its value
is not well-defined at all x, but its integral is well-defined
...
(4
...
70)
−∞
where use has been made of Eq
...
68)
...
71)
−∞
which can also be thought of as an alternative definition of a delta-function
...
It follows from Eqs
...
67) and (4
...
2π ¯
h
(4
...
(4
...
(4
...
(4
...
The former scheme is
known as the momentum representation of quantum mechanics
...
Furthermore,
by analogy with Eq
...
56), the expectation value of some operator O(p) takes the form
∞
φ∗ (p, t) O(p) φ(p, t) dp
...
75)
Fundamentals of Quantum Mechanics
49
Consider momentum
...
76)
−∞ −∞ −∞
where use has been made of Eq
...
66)
...
(4
...
p =
(4
...
(4
...
−∞
(4
...
The above expression also strongly suggests [by comparison with Eq
...
22)] that |φ(p, t)| 2
can be interpreted as the probability density of a measurement of momentum yielding the
value p at time t
...
(4
...
e
...
(4
...
We can write
∞
x
ψ∗ (x, t) x ψ(x, t) dx
=
(4
...
∂p
Integration by parts yields
x =
1
2π ¯
h
∞
∞
∞
′
h
h
φ∗ (p ′ , t) e+i (p−p ) x/¯ i ¯
−∞ −∞ −∞
∂
φ(p, t) dx dp dp ′
...
81)
Hence, making use of Eqs
...
74) and (4
...
∂p
(4
...
83)
50
QUANTUM MECHANICS
in the momentum representation
...
We have
∞
1
2π ¯
h
ψ∗ (x, t) ψ(x, t) dx =
−∞
∞
∞
∞
′
h
φ∗ (p ′, t) φ(p, t) e+i (p−p ) x/¯ dx dp dp ′
...
84)
−∞ −∞ −∞
Thus, it follows from Eqs
...
71) and (4
...
|ψ(x, t)| dx =
−∞
(4
...
(4
...
(4
...
(4
...
The existence of the momentum representation illustrates an important point: i
...
,
that there are many different, but entirely equivalent, ways of mathematically formulating
quantum mechanics
...
4
...
A straightforward generalization of Eq
...
57) yields
∞
∞
ψ∗ (O ψ2 ) dx =
1
−∞
(O ψ1 )∗ ψ2 dx,
(4
...
Let f = (A − A ) ψ, where A(x) is an Hermitian operator, and ψ(x) a general wavefunction
...
−∞
(4
...
(4
...
88)
−∞
2
where σA is the variance of A [see Eq
...
24)]
...
89)
−∞
Now, there is a standard result in mathematics, known as the Schwartz inequality, which
states that
2
b
∗
f (x) g(x) dx
a
b
≤
b
|f(x)| 2 dx
a
|g(x)| 2 dx,
a
(4
...
Furthermore, if z is a complex number then
1
|z| = [Re(z)] + [Im(z)] ≥ [Im(z)] =
(z − z∗ )
2i
2
Hence, if z =
∞
f∗
−∞
2
2
2
2
...
91)
g dx then Eqs
...
88)–(4
...
(4
...
93)
−∞
where use has been made of Eq
...
86)
...
z=
(4
...
z =
(4
...
(4
...
96)
where
[A, B] ≡ A B − B A
...
97)
Equation (4
...
It states that if two dynamical variables are represented by the two Hermitian
operators A and B, and these operators do not commute (i
...
, A B = B A), then it is impossible to simultaneously (exactly) measure the two variables
...
For instance, displacement and momentum are represented (in real-space) by the operators x and p ≡ −i ¯ ∂/∂x, respectively
...
h
(4
...
99)
Thus,
σx σp ≥
52
QUANTUM MECHANICS
which can be recognized as the standard displacement-momentum uncertainty principle
(see Sect
...
14)
...
e
...
3
...
e
...
100)
φ(p) =
h
e−i p x0 /¯ −(p−p0 ) 2/4 σp
2
e
,
2 )1/4
(2π σp
(4
...
Energy and time are represented by the operators H ≡ i ¯ ∂/∂t and t, respectively
...
In fact,
[H, t] = i ¯ ,
h
(4
...
2
(4
...
104)
σE σt ≥
This can be written, somewhat less exactly, as
where ∆E and ∆t are the uncertainties in energy and time, respectively
...
For instance, suppose that a particle passes some fixed point on the x-axis
...
Thus, there is an uncertainty, ∆t, in the arrival time of the particle
...
e
...
Since a wave packet of finite extent is made up
of a combination of plane waves of different wavenumbers, and, hence, different frequencies, there will be an uncertainty ∆E in the particle’s energy which is proportional to the
range of frequencies of the plane waves making up the wave packet
...
To be more exact, if
ψ(t) is the wavefunction measured at the fixed point as a function of time, then we can
write
∞
1
h
χ(E) e−i E t/¯ dE
...
105)
ψ(t) = √
2π ¯ −∞
h
In other words, we can express ψ(t) as a linear combination of plane waves of definite
energy E
...
By Fourier’s theorem, we also have
χ(E) = √
1
2π ¯
h
∞
h
ψ(t) e+i E t/¯ dt
...
106)
Fundamentals of Quantum Mechanics
53
For instance, if ψ(t) is a Gaussian then it is easily shown that χ(E) is also a Gaussian: i
...
,
ψ(t) =
h
e−i E0 t/¯ −(t−t0 ) 2 /4 σt2
e
,
(2π σt2)1/4
(4
...
108)
where σE σt = ¯ /2
...
Conversely, non-Gaussian wave packets are characterized by σE σt >
h
¯ /2
...
9 Eigenstates and Eigenvalues
Consider a general real-space operator A(x)
...
However, there are certain special wavefunctions which are such that when A acts on them
the result is just a multiple of the original wavefunction
...
Thus, if
(4
...
Suppose that A is an Hermitian operator corresponding to some physical dynamical
variable
...
The expectation of value A in this
state is simply [see Eq
...
56)]
∞
∞
ψ∗
a
A =
ψ∗ ψa dx = a,
a
A ψa dx = a
−∞
(4
...
(4
...
4)
...
111)
−∞
−∞
so the variance of A is [cf
...
(4
...
(4
...
Thus, the eigenstate ψa is a state which is associated with a
unique value of the dynamical variable corresponding to A
...
54
QUANTUM MECHANICS
It is easily demonstrated that the eigenvalues of an Hermitian operator are all real
...
(4
...
(4
...
114)
(A ψ2 ) dx =
−∞
−∞
Hence, if ψ1 = ψ2 = ψa then
∞
∞
ψ∗ (A ψa ) dx =
a
−∞
−∞
which reduces to [see Eq
...
109)]
a = a∗ ,
(4
...
Two wavefunctions, ψ1 (x) and ψ2 (x), are said to be orthogonal if
∞
ψ∗ ψ2 dx = 0
...
116)
−∞
Consider two eigenstates of A, ψa and ψa ′ , which correspond to the two different eigenvalues a and a ′ , respectively
...
117)
A ψa ′ = a ′ ψa ′
...
118)
Multiplying the complex conjugate of the first equation by ψa ′ , and the second equation
by ψ∗ , and then integrating over all x, we obtain
a
∞
∞
(A ψa )∗ ψa ′ dx = a
−∞
∞
ψ∗ ψa ′ dx,
a
(4
...
a
ψ∗ (A ψa ′ ) dx = a ′
a
(4
...
(4
...
Hence, we can write
∞
′
ψ∗ ψa ′ dx = 0
...
121)
−∞
By assumption, a = a ′ , yielding
∞
ψ∗ ψa ′ dx = 0
...
122)
−∞
In other words, eigenstates of an Hermitian operator corresponding to different eigenvalues
are automatically orthogonal
...
Such eigenstates are termed degenerate
...
Note, however, that any linear combination
′
of ψa and ψa is also an eigenstate of A corresponding to the eigenvalue a
...
For instance, if ψa and ψa are properly normalized, and
∞
′
ψ∗ ψa dx = c,
a
(4
...
124)
is a properly normalized eigenstate of A, corresponding to the eigenvalue a, which is
orthogonal to ψa
...
Hence, we conclude that the eigenstates of an Hermitian operator
are, or can be chosen to be, mutually orthogonal
...
e
...
However, the proof is quite difficult, and we shall not attempt it here
...
125)
i
where the ci are complex weights, and the ψi are the properly normalized (and mutually
orthogonal) eigenstates of A: i
...
,
A ψi = ai ψi ,
(4
...
i
(4
...
It follows from Eqs
...
125) and (4
...
i
ci =
(4
...
(4
...
Moreover, if ψ is a properly normalized wavefunction then
Eqs
...
125) and (4
...
(4
...
10 Measurement
Suppose that A is an Hermitian operator corresponding to some dynamical variable
...
3
...
What sort of wavefunction, ψ, is
such that a measurement of A is bound to yield a certain result, a? Well, expressing ψ as
a linear combination of the eigenstates of A, we have
ψ=
(4
...
If a measurement of A
is bound to yield the result a then
A = a,
(4
...
(4
...
133)
|ci |2 ai2
...
134)
i
A2
=
i
Thus, Eq
...
132) gives
i
ai2 |ci |2 −
i
2
ai |ci |2 = 0
...
135)
Furthermore, the normalization condition yields
|ci |2 = 1
...
136)
i
For instance, suppose that there are only two eigenstates
...
(4
...
This result can easily be generalized to the
case where there are more than two eigenstates
...
In other words, the only states associated with definite values of A are the eigenstates
of A
...
Moreover, if a general wavefunction is expanded as a linear combination
Fundamentals of Quantum Mechanics
57
of the eigenstates of A, like in Eq
...
130), then it is clear from Eq
...
133), and the
general definition of a mean, that the probability of a measurement of A yielding the
eigenvalue ai is simply |ci |2 , where ci is the coefficient in front of the ith eigenstate in the
expansion
...
(4
...
e
...
Finally, if
a measurement of A results in the eigenvalue ai then immediately after the measurement
the system will be left in the eigenstate corresponding to ai
...
Under what circumstances is it possible to simultaneously measure these two
variables (exactly)? Well, the possible results of measurements of A and B are the eigenvalues of A and B, respectively
...
In fact, in order for A and
B to be simultaneously measurable under all circumstances, we need all of the eigenstates
of A to also be eigenstates of B, and vice versa, so that all states associated with unique
values of A are also associated with unique values of B, and vice versa
...
4
...
e
...
This suggests that the condition
for simultaneous measurement is that A and B should commute
...
It follows that
(A B − B A) ψi = (A B − B ai ) ψi = (A − ai ) B ψi = 0,
(4
...
(4
...
In other words, B ψi ∝ ψi , or
B ψi = bi ψi ,
(4
...
Hence, ψi is an eigenstate of B, and, thus, a
simultaneous eigenstate of A and B
...
4
...
Unfortunately, some
operators—most notably, x and p—possess eigenvalues which lie in a continuous range and
non-square-integrable eigenstates (in fact, these two properties go hand in hand)
...
58
QUANTUM MECHANICS
Let ψx (x, x ′ ) be the eigenstate of x corresponding to the eigenvalue x ′
...
141)
for all x
...
We can write
x δ(x − x ′ ) = x ′ δ(x − x ′ ),
(4
...
Evidently, ψx (x, x ′ ) is
proportional to δ(x − x ′ )
...
(4
...
(4
...
x
(4
...
127) satisfied by squareintegrable eigenstates
...
146)
−∞
where f(x) is a general function
...
147)
ψ∗ (x, x ′ ) ψ(x) dx
...
148)
−∞
where c(x ′) = ψ(x ′ ), or
∞
c(x ′) =
−∞
In other words, we can expand a general wavefunction ψ(x) as a linear combination of the
eigenstates, ψx (x, x ′ ), of the displacement operator
...
147) and (4
...
(4
...
128), respectively, for square-integrable eigenstates
...
4
...
Moreover, these probabilities are properly normalized provided ψ(x) is properly normalized [cf
...
(4
...
e
...
−∞
(4
...
e
...
3
...
Now, an eigenstate of the momentum operator p ≡ −i ¯ ∂/∂x corresponding to the
h
′
eigenvalue p satisfies
∂ψp (x, p ′)
= p ′ ψp (x, p ′)
...
150)
− i¯
h
∂x
It is evident that
ψp (x, p ′ ) ∝ e+i p
′
x/¯
h
(4
...
Now, we require ψp (x, p ′) to satisfy an analogous orthonormality condition to Eq
...
145):
i
...
,
∞
ψ∗ (x, p ′) ψp (x, p ′′) dx = δ(p ′ − p ′′ )
...
152)
−∞
Thus, it follows from Eq
...
74) that the constant of proportionality in Eq
...
151) should
be (2π ¯ )−1/2 : i
...
,
h
′
h
e+i p x/¯
ψp (x, p ) =
...
153)
Furthermore, according to Eqs
...
66) and (4
...
154)
ψ∗ (x, p ′) ψ(x) dx
...
155)
−∞
where c(p ′) = φ(p ′ ) [see Eq
...
67)], or
∞
c(p ′) =
−∞
In other words, we can expand a general wavefunction ψ(x) as a linear combination of
the eigenstates, ψp (x, p ′), of the momentum operator
...
154) and (4
...
(4
...
128), respectively, for square-integrable eigenstates
...
The probabilities are also
properly normalized provided ψ(x) is properly normalized [cf
...
(4
...
e
...
|φ(p )| dp =
−∞
(4
...
60
QUANTUM MECHANICS
4
...
(4
...
158)
where ψi (x) is a properly normalized stationary (i
...
, non-time-varying) wavefunction
...
Note that a stationary state is associated with a unique
value for the energy
...
1) yields the equation satisfied by the stationary wavefunction:
¯ 2 d2 ψi
h
= [V(x) − Ei ] ψi
...
159)
This is known as the time-independent Schr¨dinger equation
...
160)
where H is assumed not to be an explicit function of t
...
i
(4
...
162)
ψ(x, t) =
i
where
∞
ψ∗ (x) ψ(x, 0) dx
...
163)
−∞
Here, |ci | 2 is the probability that a measurement of the energy will yield the eigenvalue
Ei
...
The generalization of the above results to the case where H
has continuous eigenvalues is straightforward
...
(4
...
162) that the expectation value
and variance of A are time independent
...
Fundamentals of Quantum Mechanics
61
Exercises
˚
1
...
What is the subsequent spread in wavelengths of the no longer monochromatic
for 10
light?
2
...
Note that
1
2 a3
...
3
...
Show
that the classical expectation value of x is L/2, the expectation value of x2 is L2 /3, and the
√
standard deviation of x is L/ 12
...
Demonstrate that if a particle in a one-dimensional stationary state is bound then the expectation value of its momentum must be zero
...
Suppose that V(x) is complex
...
What does this tell us about a complex V(x)?
o
6
...
If
∞
−∞
ψ∗ ψ2 dx = c,
1
where c is real, find normalized linear combinations of ψ1 and ψ2 which are orthogonal to
(a) ψ1 , (b) ψ1 + ψ2
...
Demonstrate that p = −i ¯ ∂/∂x is an Hermitian operator
...
8
...
An operator B, corresponding to another
physical quantity β, has normalized eigenfunctions φ1 (x) and φ2 (x), with eigenvalues b1
and b2
...
α is measured and the value a1 is obtained
...
9
...
62
QUANTUM MECHANICS
10
...
It has eigenvalues a1 and a2 , corresponding to properly normalized
eigenfunctions
√
φ1 = (u1 + u2 )
2,
√
φ2 = (u1 − u2 )
2,
where u1 and u2 are properly normalized eigenfunctions of the Hamiltonian with eigenvalues
E1 and E2
...
2
2
¯
h
One-Dimensional Potentials
63
5 One-Dimensional Potentials
5
...
Since we are searching for
stationary solutions with unique energies, we can write the wavefunction in the form (see
Sect
...
12)
h
ψ(x, t) = ψ(x) e−i E t/¯ ,
(5
...
2
dx
¯
h
(5
...
Likewise, the solution must
be continuous, otherwise the probability current (4
...
5
...
otherwise
0
∞
(5
...
(5
...
Hence, ψ = 0 in the regions x ≤ 0
and x ≥ a
...
The boundary conditions on ψ in the region 0 < x < a
are
ψ(0) = ψ(a) = 0
...
4)
Furthermore, it follows from Eq
...
2) that ψ satisfies
d2 ψ
= −k2 ψ
dx2
(5
...
(5
...
It is easily demonstrated that there are no solutions
with E < 0 which are capable of satisfying the boundary conditions (5
...
k2 =
64
QUANTUM MECHANICS
The solution to Eq
...
5), subject to the boundary conditions (5
...
7)
where the An are arbitrary (real) constants, and
kn =
nπ
,
a
(5
...
Now, it can be seen from Eqs
...
6) and (5
...
e
...
En =
2 m a2
(5
...
This is a general
feature of bounded solutions: i
...
, solutions in which |ψ| → 0 as |x| → ∞
...
4
...
10)
ψn (x) ψm (x) dx = δnm
...
Hence,
2
x
sin n π
a
a
(5
...
Finally, again from Sect
...
12, the general time-dependent solution can be written as a
linear superposition of stationary solutions:
h
cn ψn (x) e−i En t/¯ ,
ψ(x, t) =
(5
...
(5
...
3 Square Potential Barrier
Consider a particle of mass m and energy E > 0 interacting with the simple square potential barrier
V0
for 0 ≤ x ≤ a
V(x) =
,
(5
...
In the regions to the left and to the right of the barrier, ψ(x) satisfies
d2 ψ
= −k2 ψ,
dx2
(5
...
(5
...
Let us adopt the following solution of the above equation to the left of the barrier (i
...
,
x < 0):
ψ(x) = e i k x + R e−i k x
...
16)
This solution consists of a plane wave of unit amplitude traveling to the right [since the
time-dependent wavefunction is multiplied by exp(−i ω t), where ω = E/¯ > 0], and
h
a plane wave of complex amplitude R traveling to the left
...
Hence, |R| 2 is the
probability of reflection
...
19) in
the region x < 0, which takes the form
jl = v (1 − |R| 2 ),
(5
...
h
Let us adopt the following solution to Eq
...
15) to the right of the barrier (i
...
x > a):
ψ(x) = T e i k x
...
18)
This solution consists of a plane wave of complex amplitude T traveling to the right
...
Hence,
|T | 2 is the probability of transmission
...
(5
...
(4
...
e
...
e
...
Hence, we must have jl = jr , or
|R| 2 + |T | 2 = 1
...
20)
In other words, the probabilities of reflection and transmission sum to unity, as must be
the case, since reflection and transmission are the only possible outcomes for a particle
incident on the barrier
...
e
...
21)
where
2 m (E − V0 )
...
22)
¯2
h
Let us, first of all, consider the case where E > V0
...
(5
...
23)
66
QUANTUM MECHANICS
where q = 2 m (E − V0 )/¯ 2
...
e
...
These boundary conditions ensure that the
probability current (4
...
Continuity of ψ and dψ/dx at the left edge of the barrier (i
...
, x = 0) yields
1 + R = A + B,
(5
...
(5
...
e
...
(5
...
27)
After considerable algebra, the above four equations yield
(k2 − q2 ) 2 sin2 (q a)
,
|R| =
4 k2 q2 + (k2 − q2 ) 2 sin2 (q a)
(5
...
4 k2 q2 + (k2 − q2 ) 2 sin2 (q a)
(5
...
20)
...
28) and (5
...
Now, according to classical physics, if a particle of energy E is incident on a potential
barrier of height V0 < E then the particle slows down as it passes through the barrier, but
is otherwise unaffected
...
The reflection and transmission probabilities obtained from Eqs
...
28) and (5
...
5
...
2
...
5
...
e
...
However, when V0 is of order E, there is a substantial
probability that the incident particle will be reflected by the barrier
...
It can also be seen, from Fig
...
2, that at certain barrier widths the probability of
reflection goes to zero
...
It is evident, from Eq
...
28), that these special barrier widths correspond to
q a = n π,
(5
...
In other words, the special barriers widths are integer multiples of
half the de Broglie wavelength of the particle inside the barrier
...
1: Transmission (solid-curve) and reflection (dashed-curve) probabilities for a square
potential barrier of width a = 1
...
Figure 5
...
75 E, as a function of
the ratio of the width of the barrier, a, to the free-space de Broglie wavelength, λ
...
Let us, now, consider the case E < V0
...
(5
...
31)
where q = 2 m (V0 − E)/¯ 2
...
e
...
32)
i k (1 − R) = q (A − B)
...
33)
Likewise, continuity of ψ and dψ/dx at the right edge of the barrier (i
...
, x = a) gives
A e q a + B e−q a = T e i k a ,
(5
...
(5
...
36)
4 k2 q2
...
37)
4 k2 q2 + (k2 + q2 ) 2 sinh2 (q a)
These expressions can also be obtained from Eqs
...
28) and (5
...
Note that Eqs
...
36) and (5
...
20)
...
36) and (5
...
Now, according to classical physics, if a particle of energy E is incident on a potential barrier of height V0 > E then the particle is reflected
...
The reflection and transmission probabilities obtained from Eqs
...
36) and (5
...
5
...
4
...
5
...
e
...
e
...
However, when V0
is of order E, there is a substantial probability that the incident particle will be transmitted
by the barrier
...
It can also be seen, from Fig
...
4, that the transmission probability decays exponentially as the width of the barrier increases
...
e
...
This phenomenon, which is inexplicable within the context of classical
physics, is called tunneling
...
3: Transmission (solid-curve) and reflection (dashed-curve) probabilities for a square
potential barrier of width a = 0
...
Figure 5
...
70
QUANTUM MECHANICS
5
...
The particle’s wavefunction satisfies
d2 ψ(x)
= −k2 (x) ψ(x),
2
dx
(5
...
¯2
h
Let us try a solution to Eq
...
38) of the form
k2 (x) =
(5
...
40)
0
where ψ0 is a complex constant
...
It follows that
h
dψ(x)
= i k(x) ψ(x),
dx
(5
...
42)
2
dx
where k ′ ≡ dk/dx
...
(5
...
42) reveals that Eq
...
40) represents an approximate solution to Eq
...
38) provided that the first term on its right-hand
side is negligible compared to the second
...
′|
|k
(5
...
Let us suppose that this is the case
...
(5
...
1 Similarly, Eq
...
40) is termed a WKB
solution
...
40), the probability density remains constant: i
...
,
|ψ(x)| 2 = |ψ0 | 2 ,
(5
...
e
...
Suppose, however, that the
1
After G
...
A
...
Brillouin
...
e
...
By definition, E < V(x) inside such a barrier, and k(x) is
consequently imaginary
...
The
WKB solution inside the barrier is written
x
|k(x ′ )| dx ′ ,
ψ(x) = ψ1 exp −
(5
...
(5
...
According to the WKB solution (5
...
e
...
47)
x1
where |ψ1 | 2 is the probability density at the left-hand side of the barrier (i
...
, x = x1 )
...
e
...
(5
...
Of course, in the region to the right of the barrier (i
...
, x > x2 ),
the probability density takes the constant value |ψ2 | 2
...
e
...
49)
x1
(see Sect
...
3)
...
Note that the criterion (5
...
Hence, the WKB approximation only applies
to situations in which there is very little chance of a particle tunneling through the potential
barrier in question
...
43) breaks down completely at
the edges of the barrier (i
...
, at x = x1 and x2 ), since k(x) = 0 at these points
...
(5
...
Hence, the above expression for the tunneling probability is a reasonable approximation
provided that the incident particle’s de Broglie wavelength is much smaller than the spatial
extent of the potential barrier
...
5: The potential barrier for an electron in a metal surface subject to an external
electric field
...
5 Cold Emission
Suppose that an unheated metal surface is subject to a large uniform external electric field
of strength E, which is directed such that it accelerates electrons away from the surface
...
3
...
Adopting a simple one-dimensional treatment of the problem, let the metal
lie at x < 0, and the surface at x = 0
...
Hence, the energy, E, say, of an electron just below the surface is
unaffected by the field
...
The electric field modifies this to V(x)−E = W−e E x
...
5
...
It can be seen, from Fig
...
5, that an electron just below the surface of the metal is
confined by a triangular potential barrier which extends from x = x1 to x2 , where x1 = 0
and x2 = W/e E
...
50)
¯
h
x1
or
|T | 2 = exp −
2
√
2m
¯
h
W/e E
0
√
W − e E x dx
...
51)
One-Dimensional Potentials
73
This reduces to
|T | 2 = exp −2
or
2
√ m1/2 W 3/2
2
¯ eE
h
|T | = exp −
4
1
1 − y dy ,
(5
...
3
¯ eE
h
(5
...
Note that the probability of
emission increases exponentially as the electric field-strength above the surface of the metal
increases
...
An STM consists of a very sharp conducting probe which is scanned over the surface of a metal (or any other solid conducting
medium)
...
Now,
the surface electric field-strength immediately below the probe tip is proportional to the
applied potential difference, and inversely proportional to the spacing between the tip and
the surface
...
The magnitude of this current is proportional to the tunneling probability
(5
...
It follows that the current is an extremely sensitive function of the surface electric
field-strength, and, hence, of the spacing between the tip and the surface (assuming that
the potential difference is held constant)
...
In fact, STMs are capable of
achieving sufficient resolution to image individual atoms
5
...
e
...
This process
is know as α-decay
...
Such a nucleus thus decays to produce a
daughter nucleus of charge-number Z1 = Z − 2 and mass-number A1 = A − 4, and an αparticle of charge-number Z2 = 2 and mass-number A2 = 4
...
Incidentally, nuclear radii are found to satisfy the empirical formula
1/3
R = 1
...
0 × 10−15 Z1 m
(5
...
In 1928, George Gamov proposed a very successful theory of α-decay, according to
which the α-particle moves freely inside the nucleus, and is emitted after tunneling through
the potential barrier between itself and the daughter nucleus
...
55)
4π ǫ0 r
74
QUANTUM MECHANICS
for r > R
...
56)
¯
h
r1
where r1 = R and r2 = Z1 Z2 e2 /(4π ǫ0 E)
...
The
above expression reduces to
|T | 2 = exp −2
√
Ec /E
2β
1
where
β=
Z1 Z2 e2 m R
4π ǫ0 ¯ 2
h
E
1
−
y Ec
1/2
dy ,
(5
...
74 Z1
(5
...
44 Z1 MeV
4π ǫ0 R
(5
...
Of course, E ≪ Ec
...
Hence
...
E
(5
...
61)
Now, the α-particle moves inside the nucleus with the characteristic velocity v =
2 E/m
...
62)
for a 1 MeV α-particle trapped inside a typical heavy nucleus of radius 10−14 m
...
If
each of these attempts has a probability |T | 2 of succeeding, then the probability of decay
per unit time is ν |T |2
...
(5
...
(5
...
It follows from the above formula that
τ=
ln 2
...
65)
Note that the half-life is independent of N(0)
...
66)
where
C1 = 28
...
67)
C2 = 1
...
68)
C3 = 1
...
(5
...
66)
...
5
...
9,
(5
...
60,
(5
...
61
...
72)
Note that these values are remarkably similar to those calculated above
...
7 Square Potential Well
Consider a particle of mass m and energy E interacting with the simple square potential
well
−V0
for −a/2 ≤ x ≤ a/2
V(x) =
,
(5
...
Now, if E > 0 then the particle is unbounded
...
As is easily demonstrated, the reflection and
76
QUANTUM MECHANICS
Figure 5
...
9 − 1
...
61 Z1/ E
...
Here, Z1 = Z − 2, where Z is the charge
number of the nucleus, and E the characteristic energy of the emitted α-particle in MeV
...
Data obtained from IAEA Nuclear Data Centre
...
(5
...
29), respectively, where
2mE
,
¯2
h
2 m (E + V0 )
=
...
74)
q2
(5
...
In this case, the particle is bounded (i
...
, |ψ|2 → 0 as
|x| → ∞)
...
73)?
Now, it is easily seen that independent solutions of Schr¨dinger’s equation (5
...
e
...
73) must be either totally symmetric [i
...
,
ψ(−x) = ψ(x)], or totally anti-symmetric [i
...
, ψ(−x) = −ψ(x)]
...
ψ→0
(5
...
In the region to the left of the
well (i
...
x < −a/2), the solution of Schr¨dinger’s equation which satisfies the boundary
o
condition ψ → 0 and x → −∞ is
ψ(x) = A e k x ,
(5
...
¯2
h
By symmetry, the solution in the region to the right of the well (i
...
, x > a/2) is
k2 =
ψ(x) = A e−k x
...
78)
(5
...
e
...
80)
where
2 m (V0 + E)
...
81)
¯2
h
Here, we have assumed that E > −V0
...
e
...
(5
...
83)
Let y = q a/2
...
m a2
(5
...
7: The curves tan y (solid) and λ − y2 /y (dashed), calculated for λ = 1
...
The
√
latter curve takes the value 0 when y > λ
...
(5
...
85)
with
λ=
V0
...
86)
√
Here, y must lie in the range 0 < y < λ: i
...
, E must lie in the range −V0 < E < 0
...
(5
...
Figure 5
...
In this case, the curves intersect twice, indicating the existence of two totally symmetric
bound states in the well
...
e
...
However, it is also
evident that there is always at least one totally symmetric bound state, no matter how
small λ becomes (i
...
, no matter how shallow the well becomes)
...
e
...
(5
...
This gives y = (2 j − 1) π/2, where j is a positive integer, or
q=
(2 j − 1) π
...
87)
These solutions are equivalent to the odd-n infinite square well solutions specified by
Eq
...
8)
...
8: The curves tan y (solid) and −y/ λ − y2 (dashed), calculated for λ = 1
...
For the case of a totally anti-symmetric bound state, similar analysis to the above yields
−
y
λ − y2
= tan y
...
88)
The solutions of this equation correspond to the intersection of the curve tan y with the
curve −y/ λ − y2
...
8 shows these two curves plotted for the same value of λ as
that used in Fig
...
7
...
It is, again, evident, from the figure,
that as λ increases (i
...
, as the well becomes deeper) there are more and more bound states
...
e
...
In other words, a very shallow potential
well always possesses a totally symmetric bound state, but does not generally possess a
totally anti-symmetric bound state
...
e
...
(5
...
This
gives y = j π, where j is a positive integer, or
q=
2jπ
...
89)
These solutions are equivalent to the even-n infinite square well solutions specified by
Eq
...
8)
...
8 Simple Harmonic Oscillator
The classical Hamiltonian of a simple harmonic oscillator is
p2
1
H=
+ K x2 ,
2m 2
(5
...
Assuming that the quantum mechanical Hamiltonian has the same form as the classical Hamiltonian, the timeindependent Schr¨dinger equation for a particle of mass m and energy E moving in a
o
simple harmonic potential becomes
2m
d2 ψ
= 2
2
dx
¯
h
1
K x2 − E ψ
...
91)
Let ω = K/m, where ω is the oscillator’s classical angular frequency of oscillation
...
92)
¯
h
and
2E
...
93)
ǫ=
¯ω
h
Equation (5
...
(5
...
e
...
Consider the behavior of the solution to Eq
...
94) in the limit |y| ≫ 1
...
2
dy
(5
...
96)
where A(y) is a relatively slowly varying function of y
...
This suggests that we should write
2
ψ(y) = h(y) e−y /2 ,
(5
...
Substituting Eq
...
97) into Eq
...
94), we obtain
dh
d2 h
−2y
+ (ǫ − 1) h = 0
...
98)
One-Dimensional Potentials
81
Let us attempt a power-law solution of the form
∞
ci yi
...
99)
i=0
Inserting this test solution into Eq
...
98), and equating the coefficients of yi , we obtain
the recursion relation
(2 i − ǫ + 1)
ci+2 =
ci
...
100)
(i + 1) (i + 2)
Consider the behavior of h(y) in the limit |y| → ∞
...
101)
ci+2 ≃ ci
...
j!
(5
...
This behavior is
unacceptable, since it does not satisfy the boundary condition ψ → 0 as |y| → ∞
...
99) terminate at some finite value of i
...
100), that
ǫ = 2 n + 1,
(5
...
Note that the number of terms in the power series
(5
...
Finally, using Eq
...
93), we obtain
E = (n + 1/2) ¯ ω,
h
(5
...
Hence, we conclude that a particle moving in a harmonic potential has quantized energy levels which are equally spaced
...
Furthermore, the lowest energy state (n = 0)
possesses the finite energy (1/2) ¯ ω
...
It is eash
ily demonstrated that the (normalized) wavefunction of the lowest energy state takes the
form
2
2
e−x /2 d
√ ,
(5
...
h
Let ψn (x) be an energy eigenstate of the harmonic oscillator corresponding to the eigenvalue
En = (n + 1/2) ¯ ω
...
106)
82
QUANTUM MECHANICS
Assuming that the ψn are properly normalized (and real), we have
∞
ψn ψm dx = δnm
...
107)
−∞
Now, Eq
...
94) can be written
d2
− 2 + y2 ψn = (2n + 1) ψn ,
dy
where x = d y, and d =
(5
...
It is helpful to define the operators
h
d
1
+y
...
109)
As is easily demonstrated, these operators satisfy the commutation relation
[a+ , a−] = −1
...
110)
Using these operators, Eq
...
108) can also be written in the forms
a+ a− ψn = n ψn ,
(5
...
(5
...
113)
n ψn−1
...
114)
We conclude that a+ and a− are raising and lowering operators, respectively, for the harmonic oscillator: i
...
, operating on the wavefunction with a+ causes the quantum number
n to increase by unity, and vice versa
...
115)
2
from which the result
H ψn = (n + 1/2) ¯ ω ψn = En ψn
h
(5
...
Finally, Eqs
...
107), (5
...
114) yield the useful expression
∞
d
ψm x ψn dx = √
2
−∞
=
∞
ψm (a+ + a− ) ψn dx
−∞
√
√
¯
h
m δm,n+1 + n δm,n−1
...
117)
One-Dimensional Potentials
83
Exercises
1
...
h
2
...
Its initial wavefunction is
ψ(x, 0) =
2/a sin(3π x/a)
...
What now is the subsequent time evolution? Calculate the probability of finding the particle
between 0 and a/2 as a function of time in each case
...
A particle of mass m is in the ground-state of an infinite one-dimensional square-well of
width a
...
The energy of the particle is now
measured
...
A stream of particles of mass m and energy E > 0 encounter a potential step of height
W(< E): i
...
, V(x) = 0 for x < 0 and V(x) = W for x > 0 with the particles incident from
−∞
...
h
h
5
...
Show that the fraction reflected is
R = β2 /(1 + β2 ),
where β = m α/¯ 2 k, and k2 = (2m/¯ 2 ) E
...
Two potential wells of width a are separated by a distance L ≫ a
...
Estimate the time required for the particle to tunnel to the
other well
...
Consider the half-infinite potential well
∞
V(x) =
−V0
0
x≤0
0
84
QUANTUM MECHANICS
where V0 > 0
...
h
8
...
Hint:
Consider the raising and lowering operators a± defined in Eq
...
109)
...
1 Introduction
In this chapter, we shall extend the single particle, one-dimensional formulation of nonrelativistic quantum mechanics introduced in the previous sections in order to investigate
one-dimensional chapters containing multiple particles
...
2 Fundamental Concepts
We have already seen that the instantaneous state of a system consisting of a single nonrelativistic particle, whose position coordinate is x, is fully specified by a complex wavefunction ψ(x, t)
...
The probability of finding
the particle between x and x + dx at time t is given by |ψ(x, t)|2 dx
...
1)
−∞
at all times
...
Consider a system containing N non-relativistic particles, labeled i = 1, N, moving in
one dimension
...
By analogy with the single-particle case, the instantaneous state of a multiparticle system is specified by a complex wavefunction ψ(x1 , x2 ,
...
The probability
of finding the first particle between x1 and x1 + dx1 , the second particle between x2 and
x2 + dx2 , etc
...
, xN , t)|2 dx1 dx2
...
It follows that the
wavefunction must satisfy the normalization condition
|ψ(x1 , x2 ,
...
dxN = 1
(6
...
xN space
...
4
...
By analh
ogy, in a multi-particle system, the position of the ith particle is represented by the algebraic operator xi , whereas the corresponding momentum is represented by the differential
operator
∂
...
3)
pi = −i ¯
h
∂xi
86
QUANTUM MECHANICS
Since the xi are independent variables (i
...
, ∂xi /∂xj = δij ), we conclude that the various
position and momentum operators satisfy the following commutation relations:
[xi , xj ] = 0,
(6
...
5)
[xi , pj ] = i ¯ δij
...
6)
Now, we know, from Sect
...
10, that two dynamical variables can only be (exactly) measured simultaneously if the operators which represent them in quantum mechanics commute with one another
...
Note, in particular, that a knowledge of the position or momentum of a given particle does
not in any way preclude a similar knowledge for a different particle
...
4)–(6
...
In this case, the different degrees of freedom correspond to the
different motions of the various particles making up the system
...
, xN , t) is the Hamiltonian of the system then the multi-particle
wavefunction ψ(x1 , x2 ,
...
(4
...
(6
...
e
...
4
...
, xN , t) = ψE (x1 , x2 ,
...
8)
where the stationary wavefunction ψE satisfies the time-independent Schr¨dinger equation
o
[see Eq
...
160)]
H ψE = E ψE
...
9)
Here, H is assumed not to be an explicit function of t
...
3 Non-Interacting Particles
In general, we expect the Hamiltonian of a multi-particle system to take the form
H(x1 , x2 ,
...
, xN , t)
...
10)
Here, the first term on the right-hand side represents the total kinetic energy of the system,
whereas the potential V specifies the nature of the interaction between the various particles
making up the system, as well as the interaction of the particles with any external forces
...
This implies that each
particle moves in a common potential: i
...
,
V(xi , t)
...
11)
Hi (xi , t),
(6
...
, xN , t) =
i=1,N
Hence, we can write
H(x1 , x2 ,
...
2 mi
(6
...
Here,
Hi represents the energy of the ith particle, and is completely unaffected by the energies
of the other particles
...
This immediately implies that the multi-particle wavefunction
ψ(x1 , x2 ,
...
e
...
, xN , t) = ψ1 (x1 , t) ψ2(x2 , t)
...
(6
...
This probability is completely unaffected by the positions of the other particles
...
(6
...
2) for the multi-particle wavefunction is automatically satisfied
...
14) illustrates an important point in quantum
mechanics: namely, that we can generally write the total wavefunction of a many degree
of freedom system as a product of different wavefunctions corresponding to each degree
of freedom
...
(6
...
14), the time-dependent Schr¨dinger equation (6
...
(6
...
9) also faco
torizes to give
Hi ψEi = Ei ψEi ,
(6
...
Hence, a
h
multi-particle state of definite energy E has a wavefunction of the form
h
ψ(x1 , x2 ,
...
, xN ) e−i E t/¯ ,
(6
...
, xN ) = ψE1 (x1 ) ψE2 (x2 )
...
19)
where
and
E=
Ei
...
20)
i=1,N
Clearly, for the case of non-interacting particles, the energy of the whole system is simply
the sum of the energies of the component particles
...
4 Two-Particle Systems
Consider a system consisting of two particles, mass m1 and m2 , interacting via the potential
V(x1 − x2 ) which only depends on the relative positions of the particles
...
(6
...
10), the Hamiltonian of the system is written
¯ 2 ∂2
h
¯ 2 ∂2
h
H(x1 , x2 ) = −
−
+ V(x1 − x2 )
...
21)
x ′ = x1 − x2
(6
...
23)
the position of the center of mass
...
∂x2
m1 + m2 ∂X ∂x
(6
...
25)
Hence, when expressed in terms of the new variables, x ′ and X, the Hamiltonian becomes
¯ 2 ∂2
h
¯ 2 ∂2
h
−
+ V(x ′ ),
H(x , X) = −
2
′2
2 M ∂X
2 µ ∂x
(6
...
27)
′
where
Multi-Particle Systems
89
is the total mass of the system, and
µ=
m1 m2
m1 + m2
(6
...
Note that the total momentum of the system can be written
P = −i ¯
h
∂
∂
+
∂x1 ∂x2
= −i ¯
h
∂
...
29)
The fact that the Hamiltonian (6
...
e
...
e
...
(6
...
7) also factorizes to give
o
i¯
h
∂ψx ′
¯ 2 ∂2 ψx ′
h
=−
+ V(x ′ ) ψx ′ ,
∂t
2 µ ∂x ′ 2
(6
...
i¯
h
∂t
2 M ∂X2
The above equation can be solved to give
ψX (X, t) = ψ0 e i (P
′
X/¯ −E ′ t/¯ )
h
h
(6
...
33)
where ψ0 , P ′ , and E ′ = P ′ 2 /2 M are constants
...
(6
...
30), and
(6
...
e
...
Suppose that we work in the centre of mass frame of the system, which is characterized
by P ′ = 0
...
In this case, we can write the wavefunction of the
system in the form ψ(x1 , x2 , t) = ψx ′ (x ′ , t) ψ0 ≡ ψ(x1 − x2 , t), where
∂ψ
¯ 2 ∂2 ψ
h
i¯
h
=−
+ V(x) ψ
...
34)
In other words, in the center of mass frame, two particles of mass m1 and m2 , moving
in the potential V(x1 − x2 ), are equivalent to a single particle of mass µ, moving in the
potential V(x), where x = x1 − x2
...
6
...
As before, the instantaneous state of the system is specified by the complex wavefunction ψ(x1 , x2 , t)
...
However, since the particles are identical, this must be the same as
the probability of finding the first particle between x2 and x2 +dx2 , and the second between
x1 and x1 + dx1 , at time t (since, in both cases, the physical outcome of the measurement
is exactly the same)
...
35)
ψ(x1 , x2 , t) = e i ϕ ψ(x2 , x1 , t),
(6
...
However, if we swap the labels on particles 1 and 2 (which are,
after all, arbitrary for identical particles), and repeat the argument, we also conclude that
ψ(x2 , x1 , t) = e i ϕ ψ(x1 , x2 , t)
...
37)
e 2 i ϕ = 1
...
38)
Hence,
The only solutions to the above equation are ϕ = 0 and ϕ = π
...
e
...
39)
ψ(x2 , x1 , t) = −ψ(x1 , x2 , t)
...
40)
or
The above argument can easily be extended to systems containing more than two identical
particles
...
Particles with wavefunctions
which are symmetric under label interchange are said to obey Bose-Einstein statistics, and
are called bosons—for instance, photons are bosons
...
Consider a system containing two identical and non-interacting bosons
...
The stationary wavefunction of the whole system is written
1
ψE boson (x1 , x2 ) = √ [ψ(x1 , Ea ) ψ(x2 , Eb) + ψ(x2 , Ea ) ψ(x1, Eb )] ,
2
(6
...
This expression automatically satisfies the symmetry requirement on the wavefunction
...
For a system consisting of two identical and non-interacting fermions, the stationary
wavefunction of the whole system takes the form
1
ψE fermion (x1 , x2 ) = √ [ψ(x1 , Ea ) ψ(x2 , Eb ) − ψ(x2 , Ea ) ψ(x1 , Eb )] ,
2
(6
...
Note that if Ea = Eb then the total wavefunction becomes zero everywhere
...
We
thus conclude that it is impossible for the two fermions in our system to occupy the same
single-particle stationary state
...
43)
ψE dist (x1 , x2 ) = ψ(x1 , Ea ) ψ(x2 , Eb )
...
It is easily demonstrated that if the particles are distinguishable then
(x1 − x2 )2 dist = x2 a + x2 b − 2 x a x b ,
(6
...
=
(6
...
46)
∞
x
ab
ψ∗ (x, Ea ) x ψ(x, Eb) dx
...
47)
−∞
Here, we have assumed that Ea = Eb , so that
∞
ψ∗ (x, Ea ) ψ(x, Eb) dx = 0
...
48)
−∞
Finally, for the case of two identical fermions, we obtain
(x1 − x2 )2
fermion
= (x1 − x2 )2
dist
+ 2| x
2
ab | ,
(6
...
46) shows that the symmetry requirement on the total wavefunction of two
identical bosons forces the particles to be, on average, closer together than two similar
distinguishable particles
...
(6
...
However, the strength of this effect
depends on square of the magnitude of x ab , which measures the overlap between the
wavefunctions ψ(x, Ea ) and ψ(x, Eb )
...
For a system containing N identical and non-interacting fermions, the anti-symmetric
stationary wavefunction of the system is written
ψ(x1 , E1 ) ψ(x2 , E1 )
...
ψ(xN , E2 )
1
ψE (x1 , x2 ,
...
...
...
...
...
N!
...
...
ψ(x1 , EN ) ψ(x2 , EN )
...
50)
This expression is known as the Slater determinant, and automatically satisfies the symmetry requirements on the wavefunction
...
, EN
...
e
...
We conclude that it is impossible for
any two identical fermions in a multi-particle system to occupy the same single-particle
stationary state
...
Exercises (N
...
Neglect spin in the following questions
...
Consider a system consisting of two non-interacting particles, and three one-particle states,
ψa (x), ψb (x), and ψc (x)
...
Consider two non-interacting particles, each of mass m, in a one-dimensional harmonic oscillator potential of classical oscillation frequency ω
...
3
...
What are the four lowest energies of the system? What are the degeneracies of these
energies if the two particles are (a) distinguishable, (b) indistinguishable bosons, or (c) indistingishable fermions?
4
...
Write
the properly normalized wavefunctions if the particles are (a) distinguishable, (b) indistinguishable bosons, or (c) indistinguishable fermions
...
1 Introduction
In this chapter, we shall extend our previous one-dimensional formulation of non-relativistic
quantum mechanics to produce a fully three-dimensional theory
...
2 Fundamental Concepts
We have seen that in one dimension the instantaneous state of a single non-relativistic
particle is fully specified by a complex wavefunction, ψ(x, t)
...
(7
...
2)
−∞
at all times
...
By analogy with the one-dimensional case, the
probability of finding the particle at time t between x and x + dx, between y and y + dx,
and between z and z + dz, is P(x, y, z, t) dx dy dz, where
P(x, y, z, t) = |ψ(x, y, z, t)|2
...
3)
As usual, this interpretation of the wavefunction only makes sense if the wavefunction is
normalized such that
∞
∞
∞
|ψ(x, y, z, t)|2 dx dy dz = 1
...
4)
−∞ −∞ −∞
This normalization constraint ensures that the probability of finding the particle anywhere
is space is always unity
...
4
...
5)
(7
...
Integrating Eq
...
5) over all space, and making
use of the fact that ψ → 0 as |x| → ∞ if ψ is to be square-integrable, we obtain
d
dt
∞
|ψ(x, t)|2 dx = 0
...
7)
−∞
In other words, if the wavefunction is initially normalized then it stays normalized as time
progresses
...
In three dimensions, by analogy with the one dimensional case, the probability conservation equation becomes
∂|ψ|2 ∂jx ∂jy ∂jz
+
+
+
= 0
...
8)
∂t
∂x
∂y
∂z
Here,
∂ψ∗
∂ψ
i¯
h
ψ
(7
...
10)
the flux of probability along the y-axis, etc
...
(7
...
(7
...
In one dimension, position is represented by the algebraic operator x, whereas momentum is represented by the differential operator −i ¯ ∂/∂x (see Sect
...
6)
...
∂z
px ≡ −i ¯
h
(7
...
13)
pz
(7
...
Since the xi are independent variables (i
...
,
∂xi /∂xj = δij ), we conclude that the various position and momentum operators satisfy the
Three-Dimensional Quantum Mechanics
95
following commutation relations:
[xi , xj ] = 0,
(7
...
16)
[xi , pj ] = i ¯ δij
...
17)
Now, we know, from Sect
...
10, that two dynamical variables can only be (exactly) measured simultaneously if the operators which represent them in quantum mechanics commute with one another
...
Note, however, that it is perfectly possible to simultaneously measure two different positions coordinates, or two different components of the momentum
...
15)–(7
...
6
...
In one dimension, the time evolution of the wavefunction is given by [see Eq
...
63)]
∂ψ
= H ψ,
(7
...
The same equation governs the time evolution of the wavefunction in three dimensions
...
19)
H=
2m
where V(x) is the potential energy
...
(7
...
(7
...
14) and (7
...
(4
...
∂t
2m
(7
...
22)
is known as the Laplacian
...
8) is
easily derivable from Eq
...
21)
...
(7
...
(7
...
4
...
24)
where the stationary wavefunction ψ(x, y, z) satisfies the three-dimensional version of the
time-independent Schr¨ndiger equation [see Eq
...
159)]:
o
∇2 ψ =
2m
(V − E) ψ,
¯2
h
(7
...
7
...
5
...
The particle’s stationary wavefunction, ψ(x, y, z), satisfies
∂2
∂2
∂2
2m
+ 2 + 2 ψ = − 2 E ψ,
∂x2 ∂y
∂z
¯
h
(7
...
The wavefunction satisfies the boundary condition that it
must be zero at the edges of the box
...
(7
...
Substituting (7
...
(7
...
28)
X
Y
Z
¯
h
where ′ denotes a derivative with respect to argument
...
29)
(7
...
31)
2
2
where kx , ky , and kz2 are spatial constants
...
The solutions to the above
Three-Dimensional Quantum Mechanics
97
equations which are properly normalized, and satisfy the boundary conditions, are [see
Eq
...
11)]
X(x) =
2
sin(kx x),
a
(7
...
33)
Z(z) =
2
sin(kz z),
a
(7
...
=
a
kx =
(7
...
36)
kz
(7
...
Thus, from Eqs
...
28)–(7
...
(5
...
(7
...
(7
...
4 Degenerate Electron Gases
Consider N electrons trapped in a cubic box of dimension a
...
According to Sect
...
3, the total energy of a system
consisting of many non-interacting particles is simply the sum of the single-particle energies of the individual particles
...
6
...
The exclusion principle states that no two electrons in our system can occupy the same single-particle energy
level
...
Thus, we conclude that
no two electrons in our system can have the same set of values of lx , ly , and lz
...
10)
...
Hence, when spin is taken
into account, we conclude that a maximum of two electrons (with different spin quantum
98
QUANTUM MECHANICS
numbers) can occupy a single-particle energy level corresponding to a particular set of values of lx , ly , and lz
...
(7
...
39), that the associated particle energy
2
2
is proportional to l2 = lx + ly + lz2
...
e
...
In this case, we would expect them to fill the lowest single-particle energy levels available
to them
...
Thus,
the energy levels are uniformly distributed in this space on a cubic lattice
...
This implies that the number
of energy levels per unit volume is also unity
...
Since we expect cold electrons to occupy the lowest energy levels available to them,
but only two electrons can occupy a given energy level, it follows that if the number of
electrons, N, is very large then the filled energy levels will be approximately distributed
in a sphere centered on the origin of quantum number space
...
It turns out that this is not
quite correct, because we have forgotten that the quantum numbers lx , ly , and lz can only
take positive values
...
The radius lF of the octant of filled energy levels in quantum number space can be
calculated by equating the number of energy levels it contains to the number of electrons,
N
...
(7
...
Thus,
lF =
3N
π
1/3
(7
...
According to Eq
...
38), the energy of the most energetic electrons—which is known as
the Fermi energy—is given by
2
π2 ¯ 2
h
lF π2 ¯ 2
h
=
EF =
2 me a2
2 m a2
3N
π
2/3
,
(7
...
This can also be written as
π2 ¯ 2
h
EF =
2 me
3n
π
2/3
,
(7
...
Note that the
Fermi energy only depends on the number density of the confined electrons
...
44)
since E ∝ l2 , and the energy levels are uniformly distributed in quantum number space
inside an octant of radius lF
...
Thus, if kB T ≪ EF then our original assumption that the electrons are cold
is valid
...
On the other hand, if
kB T ≫ EF then the electrons are hot, and are essentially governed by classical physics—
electrons in this state are termed non-degenerate
...
5
Hence, the gas pressure takes the form
P=−
(7
...
46)
since EF ∝ a−2 = V −2/3 [see Eq
...
42)]
...
Thus, a degenerate electron gas has a much higher pressure than that which
would be predicted by classical physics
...
Note that, according to
Eq
...
43), the mean spacing between degenerate electrons is
d ∼ n−1/3 ∼ √
h
h
∼ ∼ λ,
me E p
(7
...
Thus, an electron gas is non-degenerate when the
mean spacing between the electrons is much greater than the de Broglie wavelength, and
becomes degenerate as the mean spacing approaches the de Broglie wavelength
...
e
...
Indeed,
most metals are hard to compress as a direct consequence of the high degeneracy pressure of their conduction electrons
...
48)
Now, for a fixed number of electrons, P ∝ V −5/3 [see Eqs
...
42) and (7
...
Hence,
5
π3 ¯ 2
h
B= P=
3
9m
3n
π
5/3
...
49)
100
QUANTUM MECHANICS
For example, the number density of free electrons in magnesium is n ∼ 8
...
This
leads to the following estimate for the bulk modulus: B ∼ 6
...
The actual bulk
modulus is B = 4
...
7
...
Of course, the thermal energy of the star is generated
by nuclear reactions occurring deep inside its core
...
What is the ultimate fate of such a star?
A burnt-out star is basically a gas of electrons and ions
...
Eventually, the mean separation becomes of order the de Broglie wavelength of the electrons,
and the electron gas becomes degenerate
...
Now,
even at zero temperature, a degenerate electron gas exerts a substantial pressure, because
the Pauli exclusion principle prevents the mean electron separation from becoming significantly smaller than the typical de Broglie wavelength (see previous section)
...
Such stars are termed white-dwarfs
...
The total energy of a white-dwarf star can be written
(7
...
Let us assume, for the sake of
simplicity, that the density of the star is uniform
...
51)
5 R
where G is the gravitational constant, M is the stellar mass, and R is the stellar radius
...
52)
where N is the number of electrons, V the volume of the star, and me the electron mass
...
Thus,
M = 2 N mp,
(7
...
Equations (7
...
53) can be combined to give
E=
A
B
− ,
2
R
R
(7
...
5
B =
,
(7
...
56)
The equilibrium radius of the star, R∗ , is that which minimizes the total energy E
...
57)
B
which yields
R∗ =
¯2
h
(9π)2/3
...
58)
The above formula can also be written
M⊙
R∗
= 0
...
59)
where R⊙ = 7 × 105 km is the solar radius, and M⊙ = 2 × 1030 kg the solar mass
...
e
...
The first white-dwarf to be discovered (in 1862) was
the companion of Sirius
...
¯
Note from Eqs
...
52), (7
...
59) that E ∝ M4/3
...
Hence,
for a sufficiently massive white dwarf, the electrons can become relativistic
...
(7
...
It follows that electron degeneracy pressure is only able to halt the collapse
of a burnt-out star provided that the stellar mass does not exceed some critical value,
known as the Chandrasekhar limit, which turns out to be about 1
...
Stars whose mass exceeds the Chandrasekhar limit inevitably collapse to produce
extremely compact objects, such as neutron stars (which are held up by the degeneracy
pressure of their constituent neutrons), or black holes
...
Consider a particle of mass m moving in a three-dimensional isotropic harmonic oscillator
potential of force constant k
...
2
...
3
...
In other words, calculate the number of
electron states in the interval E to E + dE
...
4
...
5
...
5 × 1028 m−3 , calculate the
Fermi energy in electron volts, and the velocity of an electron whose kinetic energy is equal
to the Fermi energy
...
Obtain an expression for the Fermi energy (in eV) of an electron in a white dwarf star as a
function of the stellar mass (in solar masses)
...
1 Introduction
As is well-known, angular momentum plays a vitally important role in the classical description of three-dimensional motion
...
8
...
(8
...
2)
Ly = z px − x pz ,
(8
...
(8
...
) correspond to
the appropriate quantum mechanical position and momentum operators
...
2)–(8
...
(7
...
Moreover, given that the
xi and the pi are Hermitian operators, it is easily seen that the Li are also Hermitian
...
4
...
We, thus, conclude that Eqs
...
2)–(8
...
Let us now derive the commutation relations for the Li
...
5)
where use has been made of the definitions of the Li [see Eqs
...
2)–(8
...
15)–(7
...
There are two similar commutation
104
QUANTUM MECHANICS
relations: one for Ly and Lz , and one for Lz and Lx
...
6)
[Ly , Lz ] = i ¯ Lx ,
h
(8
...
h
(8
...
(8
...
(8
...
11)
where use has been made of Eqs
...
6)–(8
...
In other words, L2 commutes with Lx
...
Thus,
[L2, Lx ] = [L2 , Ly] = [L2 , Lz] = 0
...
12)
Recall, from Sect
...
10, that in order for two physical quantities to be (exactly) measured simultaneously, the operators which represent them in quantum mechanics must
commute with one another
...
6)–(8
...
12)
imply that we can only simultaneously measure the magnitude squared of the angular momentum vector, L2 , together with, at most, one of its Cartesian components
...
Finally, it is helpful to define the operators
L± = Lx ± i Ly
...
13)
Note that L+ and L− are not Hermitian operators, but are the Hermitian conjugates of one
another (see Sect
...
6): i
...
,
(L± )† = L∓ ,
(8
...
h
(8
...
16)
[L+ , L−] = 2 ¯ Lz
...
17)
[L+ , Lz] = [Lx , Lz] + i [Ly , Lz] = −i ¯ Ly − ¯ Lx = −¯ L+ ,
h
h
h
(8
...
h
(8
...
3 Representation of Angular Momentum
Now, we saw earlier, in Sect
...
2, that the operators, pi , which represent the Cartesian
components of linear momentum in quantum mechanics, can be represented as the spatial differential operators −i ¯ ∂/∂xi
...
It is most convenient to perform our investigation using conventional spherical polar
coordinates: i
...
, r, θ, and φ
...
20)
y = r sin θ sin φ,
(8
...
(8
...
23)
∂x
∂r
r
∂θ r sin θ ∂φ
∂
cos θ sin φ ∂
cos φ ∂
∂
= sin θ sin φ
+
+
,
(8
...
(8
...
2)–(8
...
9), and (8
...
12)–(7
...
(8
...
25), and a great deal of tedious algebra, we finally obtain
Lx = −i ¯ − sin φ
h
Ly = −i ¯ cos φ
h
Lz = −i ¯
h
∂
,
∂φ
∂
∂
,
− cos φ cot θ
∂θ
∂φ
∂
∂
,
− sin φ cot θ
∂θ
∂φ
(8
...
27)
(8
...
29)
and
L± = ¯ e±i φ ±
h
∂
∂
...
30)
We, thus, conclude that all of our angular momentum operators can be represented as differential operators involving the angular spherical coordinates, θ and φ, but not involving
the radial coordinate, r
...
4 Eigenstates of Angular Momentum
Let us find the simultaneous eigenstates of the angular momentum operators Lz and L2
...
Thus, we can write
Lz Yl,m (θ, φ) = m ¯ Yl,m (θ, φ),
h
(8
...
h
(8
...
Of course, we expect the Yl,m to be both mutually orthogonal and properly
normalized (see Sect
...
9), so that
Yl∗,m ′ (θ, φ) Yl,m (θ, φ) dΩ = δll ′ δmm ′ ,
′
(8
...
Now,
Lz (L+ Yl,m ) = (L+ Lz + [Lz , L+]) Yl,m = (L+ Lz + ¯ L+ ) Yl,m
h
= (m + 1) ¯ (L+ Yl,m ),
h
(8
...
(8
...
We, thus, conclude that when the operator L+
operates on an eigenstate of Lz corresponding to the eigenvalue m ¯ it converts it to an
h
eigenstate corresponding to the eigenvalue (m + 1) ¯
...
It is also easily demonstrated that
Lz (L− Yl,m ) = (m − 1) ¯ (L− Yl,m )
...
35)
Orbital Angular Momentum
107
In other words, when L− operates on an eigenstate of Lz corresponding to the eigenvalue
m ¯ it converts it to an eigenstate corresponding to the eigenvalue (m − 1) ¯
...
Writing
L+ Yl,m = c+ Yl,m+1 ,
l,m
(8
...
37)
L− L+ Yl,m = c+ c−
h2
l,m l,m+1 Yl,m = [l (l + 1) − m (m + 1)] ¯ Yl,m ,
(8
...
(8
...
Likewise,
−
L+ L− Yl,m = c+
h2
l,m−1 cl,m Yl,m = [l (l + 1) − m (m − 1)] ¯ Yl,m ,
(8
...
(8
...
It follows that
c+ c−
h2
l,m l,m+1 = [l (l + 1) − m (m + 1)] ¯ ,
(8
...
(8
...
h
l,m
(8
...
43)
L− Yl,m = [l (l + 1) − m (m − 1)]1/2 ¯ Yl,m−1
...
44)
Hence, we can write
8
...
(8
...
33) provided that
π
Θl∗′ ,m ′ (θ) Θl,m (θ) sin θ dθ = δll ′ ,
(8
...
(8
...
(8
...
It therefore follows
from Eqs
...
28), (8
...
45) that
− i¯
h
dΦm
= m ¯ Φm
...
48)
108
QUANTUM MECHANICS
The solution to this equation is
Φm (φ) ∼ e i m φ
...
49)
Here, the symbol ∼ just means that we are neglecting multiplicative constants
...
Otherwise, the probability density at a given point would not, in general, have a unique value, which does not
make physical sense
...
49) be single-valued:
i
...
, Φm (φ + 2 π) = Φm (φ) for all φ
...
In fact, m can only take integer values
...
[A more
h
rigorous argument is that Φm (φ) must be continuous in order to ensure that Lz is an Hermitian operator, since the proof of hermiticity involves an integration by parts in φ that
has canceling contributions from φ = 0 and φ = 2π
...
49) by making use of the orthonormality constraint (8
...
We obtain
eimφ
Φm (φ) = √
...
50)
This is the properly normalized eigenstate of Lz corresponding to the eigenvalue m ¯
...
6 Eigenvalues of L2
Consider the angular wavefunction ψ(θ, φ) = L+ Yl,m (θ, φ)
...
51)
since ψ∗ ψ ≡ |ψ|2 is a positive-definite real quantity
...
(4
...
14), we find that
(L+ Yl,m )∗ (L+ Yl,m ) dΩ =
∗
Yl,m (L+ )† (L+ Yl,m ) dΩ
=
∗
Yl,m L− L+ Yl,m dΩ ≥ 0
...
52)
It follows from Eqs
...
16), and (8
...
33) that
∗
Yl,m (L2 − Lz2 − ¯ Lz ) Yl,m dΩ =
h
∗
Yl,m ¯ 2 [l (l + 1) − m (m + 1)] Yl,m dΩ
h
= ¯ 2 [l (l + 1) − m (m + 1)]
h
∗
Yl,m Yl,m dΩ
= ¯ 2 [l (l + 1) − m (m + 1)] ≥ 0
...
53)
Orbital Angular Momentum
109
We, thus, obtain the constraint
l (l + 1) ≥ m (m + 1)
...
54)
Likewise, the inequality
(L− Yl,m )∗ (L− Yl,m ) dΩ =
∗
Yl,m L+ L− Yl,m dΩ ≥ 0
(8
...
(8
...
This is reasonable, from a physical
standpoint, since l (l + 1) ¯ 2 is supposed to represent the magnitude squared of something,
h
and should, therefore, only take non-negative values
...
54) and (8
...
(8
...
Well, if m can only take a restricted range of integer values then there must exist a
lowest possible value it can take
...
Suppose we act on this eigenstate with the lowering operator
L−
...
(8
...
However, no such state exists
...
Thus, we must have
L− Yl,m− = 0
...
58)
L2 = L+ L− + Lz2 − ¯ Lz
h
(8
...
60)
l (l + 1) Yl,m− = m− (m− − 1) Yl,m− ,
(8
...
(8
...
31), (8
...
58)
...
(8
...
(8
...
(8
...
We can now formulate the rules which determine the allowed values of the quantum numbers l and m
...
Once l is given, the quantum number m can take any integer value in the
range
− l, −l + 1, · · · 0, · · · , l − 1, l
...
65)
Thus, if l = 0 then m can only take the value 0, if l = 1 then m can take the values
−1, 0, +1, if l = 2 then m can take the values −2, −1, 0, +1, +2, and so on
...
7 Spherical Harmonics
The simultaneous eigenstates, Yl,m (θ, φ), of L2 and Lz are known as the spherical harmonics
...
Now, we know that
L+ Yl,l (θ, φ) = 0,
(8
...
Writing
Yl,l (θ, φ) = Θl,l (θ) e i l φ
(8
...
(8
...
49)], and making use of Eq
...
30), we obtain
¯ eiφ
h
∂
∂
Θl,l (θ) e i l φ = 0
...
68)
This equation yields
dΘl,l
− l cot θ Θl,l = 0
...
69)
Θl,l ∼ (sin θ)l
...
70)
Yl,l (θ, φ) ∼ (sin θ)l e i l φ
...
71)
Hence, we conclude that
Likewise, it is easy to demonstrate that
Yl,−l (θ, φ) ∼ (sin θ)l e−i l φ
...
72)
Orbital Angular Momentum
111
Once we know Yl,l , we can obtain Yl,l−1 by operating on Yl,l with the lowering operator
L−
...
73)
+ i cot θ
∂θ
∂φ
where use has been made of Eq
...
30)
...
dθ
Yl,l−1 ∼ e i (l−1) φ
(8
...
75)
where f(θ) is a general function
...
sin θ dθ
(8
...
sin θ dθ
(8
...
We get
Yl,l−2 ∼ L− Yl,l−1 ∼ e−i φ −
∂
∂
+ i cot θ
∂θ
∂φ
e i (l−1) φ
(sin θ)l−1
1 d
(sin θ)2 l ,
sin θ dθ
(8
...
sin θ dθ
(8
...
(8
...
(8
...
(8
...
(8
...
76), and (8
...
(8
...
1: The |Yl,m (θ, φ)| 2 plotted as a functions of θ
...
Here, m is assumed to be non-negative
...
(8
...
(8
...
77), and (8
...
(8
...
(8
...
86)
for m ≥ 0, where the Pl,m are known as associated Legendre polynomials, and are written
Pl,m (u) = (−1)
l+m
(l + m)! (1 − u2 )−m/2
(l − m)!
2l l!
for m ≥ 0
...
87)
l+m
(1 − u2 )l ,
(8
...
2: The |Yl,m (θ, φ)| 2 plotted as a functions of θ
...
for m ≥ 0
...
(8
...
e
...
90)
and also form a complete set
...
Finally, and most importantly, the spherical
harmonics are the simultaneous eigenstates of Lz and L2 corresponding to the eigenvalues
m ¯ and l (l + 1) ¯ 2 , respectively
...
91)
(8
...
93)
5
(3 cos2 θ − 1),
16π
(8
...
32π
(8
...
96)
The θ variation of these functions is illustrated in Figs
...
1 and 8
...
Exercises
2
1
...
Calculate Lx and Lx
...
Find the eigenvalues and eigenfunctions (in terms of the angles θ and φ) of Lx
...
Consider a beam of particles with l = 1
...
What
h
values will be obtained by a subsequent measurement of Lz , and with what probabilities?
Repeat the calculation for the cases in which the measurement of Lx yields the results 0 and
−¯
...
The Hamiltonian for an axially symmetric rotator is given by
H=
What are the eigenvalues of H?
2
2
Lx + Ly
L2
+ z
...
1 Introduction
In this chapter, we shall investigate the interaction of a non-relativistic particle of mass m
and energy E with various so-called central potentials, V(r), where r = x2 + y2 + z2 is
the radial distance from the origin
...
8
...
Thus, we shall
be searching for stationary wavefunctions, ψ(r, θ, φ), which satisfy the time-independent
Schr¨dinger equation (see Sect
...
12)
o
H ψ = E ψ,
where the Hamiltonian takes the standard non-relativistic form
p2
H=
+ V(r)
...
1)
(9
...
2 Derivation of Radial Equation
Now, we have seen that the Cartesian components of the momentum, p, can be represented
as (see Sect
...
2)
∂
(9
...
Likewise, it is easily
demonstrated, from the above expressions, and the basic definitions of the spherical polar
coordinates [see Eqs
...
20)–(8
...
(9
...
(8
...
(9
...
(9
...
The values of the various elements of this tensor are determined via a
simple rule:
if i, j, k not all different
0
1
if i, j, k are cyclic permutation of 1, 2, 3
ǫijk =
...
7)
−1
if i, j, k are anti-cyclic permutation of 1, 2, 3
116
QUANTUM MECHANICS
Thus, ǫ123 = ǫ231 = 1, ǫ321 = ǫ132 = −1, and ǫ112 = ǫ131 = 0, etc
...
6) also makes
use of the Einstein summation convention, according to which repeated indices are summed
(from 1 to 3)
...
Making use of this convention, as
well as Eq
...
7), it is easily seen that Eqs
...
5) and (9
...
Let us calculate the value of L2 using Eq
...
6)
...
Thus, we obtain
L2 = ǫijk xj pk ǫilm xl pm = ǫijk ǫilm xj pk xl pm
...
8)
Note that we are able to shift the position of ǫilm because its elements are just numbers,
and, therefore, commute with all of the xi and the pi
...
(9
...
if i and j different
(9
...
(9
...
9) that
L2 = xi pj xi pj − xi pj xj pi
...
11)
Here, we have made use of the fairly self-evident result that δij ai bj ≡ ai bi
...
We now need to rearrange the order of the terms on the right-hand side of Eq
...
11)
...
(7
...
h
(9
...
h
(9
...
(7
...
Next,
L2 = xi xi pj pj − xi pi (xj pj − [xj , pj ]) − 2 i ¯ xi pi
...
14)
Now, according to (9
...
h
(9
...
h
(9
...
h
(9
...
(9
...
This term originates from the lack of commutation between the xi and pi
operators in quantum mechanics
...
Equation (9
...
(9
...
(9
...
Hence, we obtain
r · p = r pr = −i ¯ r
h
p2 = −¯ 2
h
(9
...
r ∂r
∂r
r ∂r ¯ r
h
(9
...
(9
...
(9
...
Recall, from Sect
...
3, that Lz and L2 are represented as
differential operators which depend solely on the angular spherical polar coordinates, θ
and φ, and do not contain the radial polar coordinate, r
...
Moreover, L2 commutes both with itself, and with Lz (see Sect
...
2)
...
Now, according to Sect
...
10, if two operators commute with one another then they
possess simultaneous eigenstates
...
Now,
we have already found the simultaneous eigenstates of Lz and L2 —they are the spherical harmonics, Yl,m (θ, φ), discussed in Sect
...
7
...
This observation leads us to try the following
separable form for the stationary wavefunction:
ψ(r, θ, φ) = R(r) Yl,m (θ, φ)
...
22)
118
QUANTUM MECHANICS
It immediately follows, from (8
...
32), and the fact that Lz and L2 both obviously
commute with R(r), that
Lz ψ = m ¯ ψ,
h
(9
...
h
(9
...
8
...
Finally, making use of Eqs
...
1), (9
...
24), we obtain the following differential equation which determines the radial variation of the stationary wavefunction:
−
¯2
h
2m
d2
2 d
l (l + 1)
+
−
Rn,l + V Rn,l = E Rn,l
...
25)
Here, we have labeled the function R(r) by two quantum numbers, n and l
...
[Note that the azimuthal
quantum number, m, does not appear in the above equation, and, therefore, does not influence either the function R(r) or the energy, E
...
9
...
(9
...
Within this region, it is subject to the physical boundary conditions that it be well behaved (i
...
, squareintegrable) at r = 0, and that it be zero at r = a (see Sect
...
2)
...
27)
we deduce (see previous section) that the radial function Rn,l (r) satisfies
d2 Rn,l 2 dRn,l
l (l + 1)
+
+ k2 −
Rn,l = 0
2
dr
r dr
r2
in the region 0 ≤ r ≤ a, where
(9
...
(9
...
2
dz
z dz
z2
(9
...
1: The first few spherical Bessel functions
...
The two independent solutions to this well-known second-order differential equation are
called spherical Bessel functions,1 and can be written
jl (z) = z
1 d
−
z dz
l
yl (z) = −z
l
1 d
−
z dz
l
sin z
,
z
l
cos z
...
31)
(9
...
y1 (z) = − 2 −
z
z
j0 (z) =
(9
...
34)
(9
...
36)
These functions are also plotted in Fig
...
1
...
However, the yl (z) functions are badly behaved (i
...
, they are not square-integrable) at z = 0, whereas the jl (z)
1
M
...
A
...
10
...
120
QUANTUM MECHANICS
n=1
n=2
3
...
283
4
...
725
5
...
095
6
...
417
8
...
705
l=0
l=1
l=2
l=3
l=4
n=3
9
...
904
12
...
698
15
...
566
14
...
515
16
...
301
Table 9
...
functions are well behaved everywhere
...
In order to satisfy the boundary condition at r = a [i
...
, Rn,l (a) = 0], the
value of k must be chosen such that z = k a corresponds to one of the zeros of jl (z)
...
It follows that
k a = zn,l ,
(9
...
Hence, from (9
...
(9
...
1
...
We are now in a position to interpret the three quantum numbers—n, l, and m—which
determine the form of the wavefunction specified in Eq
...
27)
...
8,
the azimuthal quantum number m determines the number of nodes in the wavefunction
as the azimuthal angle φ varies between 0 and 2π
...
Likewise, the polar quantum number l
determines the number of nodes in the wavefunction as the polar angle θ varies between
0 and π
...
Finally, the
radial quantum number n determines the number of nodes in the wavefunction as the
radial variable r varies between 0 and a (not counting any nodes at r = 0 or r = a)
...
Note that,
for the case of an infinite potential well, the only restrictions on the values that the various
quantum numbers can take are that n must be a positive integer, l must be a non-negative
integer, and m must be an integer lying between −l and l
...
38) only depend on the values of the quantum numbers n and l
...
e
...
39)
0
when n = n ′
...
8), this ensures
that wavefunctions (9
...
Central Potentials
121
9
...
40)
where r is the position vector of the electron with respect to the proton
...
6
...
In the latter problem, a particle of mass
µ=
me mp
me + mp
(9
...
(9
...
Hence, in the following, we shall write neglect this difference entirely
...
43)
it follows from Sect
...
2 that the radial function Rn,l (r) satisfies
−
¯2
h
2 me
d2
2 d
l (l + 1)
e2
Rn,l −
+
−
+ E Rn,l = 0
...
44)
Let r = a z, with
a=
¯2
h
=
2 me (−E)
E0
a0 ,
E
(9
...
(9
...
58), respectively
...
The above
differential equation transforms to
d2
2 d
l (l + 1) ζ
+
−
+ − 1 Rn,l = 0,
2
dz
z dz
z2
z
(9
...
2 = 2
E
4π ǫ0 ¯
h
Suppose that Rn,l (r) = Z(r/a) exp(−r/a)/(r/a)
...
−2
−
+
2
dz
dz
z2
z
(9
...
48)
122
QUANTUM MECHANICS
We now need to solve the above differential equation in the domain z = 0 to z = ∞,
subject to the constraint that Rn,l (r) be square-integrable
...
Z(z) =
(9
...
(9
...
(9
...
(9
...
49) must terminate at small k, at some positive value of k,
otherwise Z(z) behaves unphysically as z → 0 [i
...
, it yields an Rn,l (r) that is not squareintegrable as r → 0]
...
There are two possibilities: kmin = −l or kmin = l + 1
...
Thus, we conclude that kmin = l + 1
...
e
...
For large values of z, the ratio of successive coefficients in the power series (9
...
52)
according to Eq
...
51)
...
53)
k!
k
which converges to exp(2 z)
...
It thus follows that Rn,l (r) ∼ Z(r/a) exp(−r/a)/(r/a) → exp(r/a)/(r/a) as r → ∞
...
The only way in which we can avoid this unphysical behaviour is if the power
series (9
...
According to the recursion relation
(9
...
54)
2
where n is an integer, and the last term in the series is cn zn
...
Finally, it is clear from Eqs
...
45), (9
...
54) that
E=
E0
n2
(9
...
6 eV,
=−
8π ǫ0 a0
2 (4π ǫ0)2 ¯ 2
h
(9
...
3 × 10−11 m
...
56)
(9
...
Note that |E0 | ∼ α2 me c2 , where
α = e2 /(4π ǫ0 ¯ c) ≃ 1/137 is the dimensionless fine-structure constant
...
We conclude that the wavefunction of a hydrogen atom takes the form
ψn,l,m (r, θ, φ) = Rn,l (r) Yl,m (θ, φ)
...
59)
Here, the Yl,m (θ, φ) are the spherical harmonics (see Sect 8
...
60)
2 dz
z
dz
z2
z
which varies as zl at small z
...
61)
where n is a positive integer, l a non-negative integer, and m an integer
...
e
...
62)
where dV is a volume element, and the integral is over all space
...
Moreover, we already know that the spherical
harmonics are orthonormal [see Eq
...
90)]: i
...
,
Yl∗,m ′ Yl,m dΩ = δll ′ δmm ′
...
63)
It, thus, follows that the radial wavefunction satisfies the orthonormality constraint
∞
R∗ ′ ,l Rn,l r2 dr = δnn ′
...
64)
124
QUANTUM MECHANICS
Figure 9
...
The solid, short-dashed, and
long-dashed curves correspond to n, l = 1, 0, and 2, 0, and 2, 1, respectively
...
exp −
3 a0
27 5 (3 a0)3/2 a0
R3,0 (r) =
(9
...
66)
(9
...
68)
(9
...
70)
These functions are illustrated in Figs
...
2 and 9
...
Given the (properly normalized) hydrogen wavefunction (9
...
71)
Central Potentials
125
Figure 9
...
The solid, short-dashed, and
long-dashed curves correspond to n, l = 3, 0, and 3, 1, and 3, 2, respectively
...
For instance, it can be demonstrated (after much tedious algebra) that
r2
r
1
r
2
a0 n2
[5 n2 + 1 − 3 l (l + 1)],
2
a0
[3 n2 − l (l + 1)],
=
2
1
=
,
2a
n 0
=
(9
...
73)
(9
...
75)
1
r3
=
1
...
76)
According to Eq
...
55), the energy levels of the bound-states of a hydrogen atom only
depend on the radial quantum number n
...
For a general central potential, V(r), the quantized energy levels of a
bound-state depend on both n and l (see Sect
...
3)
...
e
...
According to the inequality (9
...
e
...
Likewise, for a given value of n, there
126
QUANTUM MECHANICS
are n different allowed values of l (i
...
, 0, 1, · · · , n − 1)
...
e
...
Hence, the total number of
degenerate states corresponding to a given value of n is
1 + 3 + 5 + · · · + 2 (n − 1) + 1 = n2
...
77)
Thus, the ground-state (n = 1) is not degenerate, the first excited state (n = 2) is four-fold
degenerate, the second excited state (n = 3) is nine-fold degenerate, etc
...
10), the degeneracy of
the nth energy level becomes 2 n2
...
5 Rydberg Formula
An electron in a given stationary state of a hydrogen atom, characterized by the quantum
numbers n, l, and m, should, in principle, remain in that state indefinitely
...
g
...
Suppose that an electron in a hydrogen atom makes a transition from an initial state
whose radial quantum number is ni to a final state whose radial quantum number is nf
...
(9
...
2
nf
ni
(9
...
(3
...
Likewise, if ∆E is positive then the electron must absorb a photon
of energy ν = ∆E/h
...
79)
=R
2
λ
nf
ni
where
R=
−E0
me e4
=
= 1
...
3 ǫ2 ¯ 3 c
hc
(4π) 0 h
(9
...
Note that the emission spectrum of hydrogen is quantized: i
...
, a hydrogen atom can only emit photons with certain fixed set of wavelengths
...
This set of wavelengths constitutes the characteristic emission/absorption
spectrum of the hydrogen atom, and can be observed as “spectral lines” using a spectroscope
...
79) is known as the Rydberg formula
...
The Rydberg formula was actually discovered empirically in the nineteenth century
by spectroscopists, and was first explained theoretically by Bohr in 1913 using a primitive
Central Potentials
127
version of quantum mechanics
...
Transitions
to the first excited state (nf = 2) give rise to spectral lines in the visible band—this set of
lines is called the Balmer series
...
Exercises
1
...
Find the ground-state by solving the radial equation with l = 0
...
h
2
...
Solve the problem by separation of variables in spherical polar coordinates,
and, hence, determine the energy eigenvalues of the system
...
The normalized wavefunction for the ground-state of a hydrogen-like atom (neutral hydrogen, He+ , Li++ , etc
...
Show the following:
(a) A2 = β3 /π
...
h
(c) The energy is E = −Z2 E0 where E0 = (me /2 ¯ 2 ) (e2 /4π ǫ0 )2
...
(e) The expectation value of r is (3/2) (a0 /Z)
...
4
...
Suddenly the nucleus decays into a helium nucleus,
via the emission of a fast electron which leaves the atom without perturbing the extranuclear
electron, Find the probability that the resulting He+ ion will be left in an n = 1, l = 0 state
...
What is the probability that the
ion will be left in an l > 0 state?
5
...
6
...
Find a hydrogen
atom’s recoil energy when it emits a photon in an n = 2 to n = 1 transition
...
1 Introduction
Broadly speaking, a classical extended object (e
...
, the Earth) can possess two types of
angular momentum
...
g
...
The second type is due to the object’s internal motion—this is generally
known as spin angular momentum (since, for a rigid object, the internal motion consists of
spinning about an axis passing through the center of mass)
...
8), and spin angular momentum due to their internal motion
...
In fact, in quantum mechanics, it is best to think of spin angular momentum as a kind of intrinsic angular momentum possessed by particles
...
10
...
Thus, by analogy with Sect
...
2, we would
expect to be able to define three operators—Sx , Sy , and Sz —which represent the three
Cartesian components of spin angular momentum
...
(8
...
8)]
...
1)
[Sy, Sz ] = i ¯ Sx ,
h
(10
...
h
(10
...
(10
...
8
...
(10
...
4
...
By convention, we shall always choose to measure the z-component, Sz
...
(8
...
(10
...
e
...
(10
...
8
...
8)
2
(10
...
10)
[S− , Sz] = +¯ S−
...
11)
10
...
Unlike regular wavefunctions, spin wavefunctions do not exist in real space
...
Instead, we need to think of spin wavefunctions as existing in an
abstract (complex) vector space
...
Note that only the
directions of our vectors have any physical significance (just as only the shape of a regular
wavefunction has any physical significance)
...
Now, we expect the internal states of our particle to be superposable, since the superposability of states is one of the fundamental assumptions of quantum mechanics
...
Thus, if χ1 and χ2
are two vectors corresponding to two different internal states then c1 χ1 + c2 χ2 is another
vector corresponding to the state obtained by superposing c1 times state 1 with c2 times
state 2 (where c1 and c2 are complex numbers)
...
e
...
We now need to define the length of our vectors
...
Let the element of the second space which corresponds to the
element χ of the first space be called χ†
...
We shall assume that it is possible to combine χ and χ† in a
multiplicative fashion to generate a real positive-definite number which we interpret as
the length, or norm, of χ
...
Thus, we have
χ† χ ≥ 0
(10
...
We shall also assume that it is possible to combine unlike states in an analogous
multiplicative fashion to produce complex numbers
...
Two states χ and χ ′ are said to be mutually orthogonal, or
independent, if χ† χ ′ = 0
...
The dual of this state is (A χ)† ≡
χ† A† , where A† is the Hermitian conjugate of A (this is the definition of an Hermitian
conjugate in spin space)
...
(10
...
4
...
In order to ensure that these eigenvalues are all
real, A must be Hermitian: i
...
, A† = A (see Sect
...
9)
...
We can also normalize them such that they all have unit length
...
14)
χ† χa ′ = δaa ′
...
e
...
(10
...
10
...
4
...
Let these eigenstates take the form [see Eqs
...
31) and (8
...
16)
h
S2 χs,ms = s (s + 1) ¯ 2 χs,ms
...
17)
Now, it is easily demonstrated, from the commutation relations (10
...
11),
that
(10
...
(10
...
8
...
The eigenstates of Sz and S2 are assumed to be orthonormal:
i
...
,
′
′
χ† s χs ′ ,ms = δss ′ δms ms
...
20)
s,m
132
QUANTUM MECHANICS
Consider the wavefunction χ = S+ χs,ms
...
(10
...
21)
where use has been made of Eq
...
7)
...
9), (10
...
17), and (10
...
(10
...
(10
...
(10
...
Let ms min be the minimum possible value of ms
...
8
...
(10
...
h
(10
...
27)
s (s + 1) = ms min (ms min − 1)
...
28)
Now, from Eq
...
8),
Hence,
giving
Assuming that ms min < 0, this equation yields
ms min = −s
...
29)
Likewise, it is easily demonstrated that
ms max = +s
...
30)
S− χs,−s = S+ χs,s = 0
...
31)
Moreover,
Now, the raising operator S+ , acting upon χs,−s , converts it into some multiple of χs,−s+1
...
However,
this process cannot continue indefinitely, since there is a maximum possible value of ms
...
(10
...
If this is not the case then we will inevitably
obtain eigenstates of Sz corresponding to ms > s, which we have already demonstrated is
impossible
...
32)
where k is a positive integer
...
Up to now, our analysis has been very similar to that which
we used earlier to investigate orbital angular momentum (see Sect
...
Recall, that for orbital angular momentum the quantum number m, which is analogous to ms , is restricted
to take integer values (see Cha
...
5)
...
However, the origin of these restrictions is the representation of the orbital angular momentum operators as differential
operators in real space (see Sect
...
3)
...
Hence, we conclude that there is no reason
why the quantum number s cannot take half-integer, as well as integer, values
...
According to this theorem, all fermions possess half-integer spin (i
...
,
a half-integer value of s), whereas all bosons possess integer spin (i
...
, an integer value of
s)
...
In other words, electrons and protons are characterized by s = 1/2 and ms = ±1/2
...
5 Pauli Representation
Let us denote the two independent spin eigenstates of an electron as
χ± ≡ χ1/2,±1/2
...
33)
It thus follows, from Eqs
...
16) and (10
...
h
S2 χ± =
4
(10
...
35)
Note that χ+ corresponds to an electron whose spin angular momentum vector has a positive component along the z-axis
...
Likewise, χ− corresponds to an electron
whose spin points in the −z-direction (or whose spin is “down”)
...
36)
134
QUANTUM MECHANICS
and
χ† χ− = 0
...
37)
A general spin state can be represented as a linear combination of χ+ and χ− : i
...
,
(10
...
It is thus evident that electron spin space is two-dimensional
...
In the following,
we shall describe a particular representation of electron spin space due to Pauli
...
Let us attempt to represent a general spin state as a complex column vector in some
two-dimensional space: i
...
,
c+
χ≡
...
39)
c−
The corresponding dual vector is represented as a row vector: i
...
,
χ† ≡ (c∗ , c∗ )
...
40)
Furthermore, the product χ† χ is obtained according to the ordinary rules of matrix multiplication: i
...
,
χ† χ = (c∗ , c∗ )
+ −
c+
c−
= c∗ c+ + c∗ c− = |c+ |2 + |c− |2 ≥ 0
...
41)
Likewise, the product χ† χ ′ of two different spin states is also obtained from the rules of
matrix multiplication: i
...
,
χ† χ ′ = (c∗ , c∗ )
+ −
′
c+
′
c−
′
′
= c∗ c+ + c∗ c−
...
42)
Note that this particular representation of spin space is in complete accordance with the
discussion in Sect
...
3
...
A general spin operator A is represented as a 2 × 2 matrix which operates on a spinor:
i
...
,
A11 , A12
c+
Aχ ≡
...
43)
A21 , A22
c−
As is easily demonstrated, the Hermitian conjugate of A is represented by the transposed
complex conjugate of the matrix used to represent A: i
...
,
A† ≡
A∗ , A∗
11
21
A∗ , A∗
22
12
...
44)
Spin Angular Momentum
135
Let us represent the spin eigenstates χ+ and χ− as
χ+ ≡
1
0
,
(10
...
46)
and
respectively
...
36) and (10
...
It is convenient to write the spin operators Si (where i = 1, 2, 3
corresponds to x, y, z) as
¯
h
Si = σi
...
47)
2
Here, the σi are dimensionless 2 × 2 matrices
...
(10
...
3), the σi
satisfy the commutation relations
[σx , σy ] = 2 i σz,
(10
...
49)
[σz , σx ] = 2 i σy
...
50)
σz χ± = ±χ±
...
51)
Furthermore, Eq
...
34) yields
It is easily demonstrated, from the above expressions, that the σi are represented by the
following matrices:
σx ≡
0, 1
1, 0
σy ≡
0, −i
i, 0
,
(10
...
(10
...
52)
,
Incidentally, these matrices are generally known as the Pauli matrices
...
(10
...
(10
...
h
136
QUANTUM MECHANICS
10
...
The direction of the magnetic moment is conventionally taken to be normal to the
plane of the loop, in the sense given by a standard right-hand circulation rule
...
It is easily demonstrated that the electron’s orbital angular momentum L is related to the magnetic moment
µ of the loop via
e
µ=−
L,
(10
...
The above expression suggests that there may be a similar relationship between magnetic moment and spin angular momentum
...
58)
where g is called the gyromagnetic ratio
...
In fact,
g=2 1+
α
+ · · · = 2
...
59)
where α = e2 /(2 ǫ0 h c) ≃ 1/137 is the so-called fine-structure constant
...
Furthermore, the small corrections to the relativistic
result g = 2 come from quantum field theory
...
(10
...
61)
geB
...
62)
where
Ω=
Here, for the sake of simplicity, we are neglecting the electron’s translational degrees of
freedom
...
(4
...
63)
Spin Angular Momentum
137
where the spin state of the electron is characterized by the spinor χ
...
64)
χ=
c− (t)
where |c+ |2 + |c− |2 = 1
...
It follows from Eqs
...
47), (10
...
61), (10
...
64) that
c+
˙
c−
˙
Ω¯
h
2
1, 0
0, −1
c± = ∓i
˙
i¯
h
Ω
c±
...
Hence,
c+
c−
,
(10
...
66)
Let
c+ (0) = cos(α/2),
(10
...
(10
...
Solving Eq
...
66), subject to the initial conditions (10
...
68), we obtain
c+ (t) = cos(α/2) exp(−i Ω t/2),
(10
...
(10
...
By analogy with Eq
...
56), the expectation value of a
general spin operator A is simply
A = χ† A χ
...
71)
Hence, the expectation value of Sz is
Sz =
¯ ∗ ∗
h
(c , c )
2 + −
1, 0
0, −1
c+
c−
,
(10
...
73)
2
with the help of Eqs
...
69) and (10
...
Likewise, the expectation value of Sx is
Sz =
Sx =
¯ ∗ ∗
h
(c , c )
2 + −
0, 1
1, 0
c+
c−
,
(10
...
2
Sx =
(10
...
2
(10
...
(10
...
75), and (10
...
(10
...
Note, however,
that a measurement of Sx , Sy , or Sz will always yield either +¯ /2 or −¯ /2
...
Exercises
1
...
2
...
3
...
Demonstrate that a measurement of Sz yields ¯ /2 with probability
h
cos2 (θ/2), and −¯ /2 with probability sin2 (θ/2)
...
An electron is in the spin-state
χ=A
1 − 2i
2
in the Pauli representation
...
If a measurement of
Sz is made, what values will be obtained, and with what probabilities? What is the expectation value of Sz ? Repeat the above calculations for Sx and Sy
...
Consider a spin-1/2 system represented by the normalized spinor
χ=
cos α
sin α exp( i β)
in the Pauli representation, where α and β are real
...
An electron is at rest in an oscillating magnetic field
B = B0 cos(ω t) ez ,
where B0 and ω are real positive constants
...
(b) If the electron starts in the spin-up state with respect to the x-axis, determine the spinor
χ(t) which represents the state of the system in the Pauli representation at all subsequent times
...
(d) What is the minimum value of B0 required to force a complete flip in Sx ?
140
QUANTUM MECHANICS
Addition of Angular Momentum
141
11 Addition of Angular Momentum
11
...
As we have already seen, the electron’s motion
through space is parameterized by the three quantum numbers n, l, and m (see Sect
...
4)
...
Now, the quantum numbers l and m
specify the electron’s orbital angular momentum vector, L, (as much as it can be specified)
whereas the quantum numbers s and ms specify its spin angular momentum vector, S
...
2 General Principles
The three basic orbital angular momentum operators, Lx , Ly , and Lz , obey the commutation
relations (8
...
8), which can be written in the convenient vector form:
L × L = i ¯ L
...
1)
Likewise, the three basic spin angular momentum operators, Sx , Sy , and Sz , obey the
commutation relations (10
...
3), which can also be written in vector form: i
...
,
S × S = i ¯ S
...
2)
Now, since the orbital angular momentum operators are associated with the electron’s
motion through space, whilst the spin angular momentum operators are associated with
its internal motion, and these two types of motion are completely unrelated (i
...
, they
correspond to different degrees of freedom—see Sect
...
2), it is reasonable to suppose
that the two sets of operators commute with one another: i
...
,
[Li , Sj ] = 0,
(11
...
Let us now consider the electron’s total angular momentum vector
J = L + S
...
4)
We have
J × J = (L + S) × (L + S)
= L×L+S×S+L×S+S×L=L×L+S×S
= i¯ L+i¯ S
h
h
= i ¯ J
...
5)
142
QUANTUM MECHANICS
In other words,
J × J = i ¯ J
...
6)
It is thus evident that the three basic total angular momentum operators, Jx , Jy , and Jz , obey
analogous commutation relations to the corresponding orbital and spin angular momentum operators
...
For instance, it is only possible to simultaneously
measure the magnitude squared of the total angular momentum vector,
2
2
J2 = Jx + Jy + Jz2 ,
(11
...
By convention, we shall always choose to
measure Jz
...
8)
J2 ψj,mj = j (j + 1) ¯ 2 ψj,mj ,
h
(11
...
(11
...
11)
Now
which can also be written as
J2 = L2 + S2 + 2 Lz Sz + L+ S− + L− S+
...
12)
We know that the operator L2 commutes with itself, with all of the Cartesian components
of L (and, hence, with the raising and lowering operators L± ), and with all of the spin
angular momentum operators (see Sect
...
2)
...
(11
...
(11
...
8
...
It follows
that
[J2 , Lz] = 0
...
15)
Likewise, we can also show that
[J2 , Sz] = 0
...
16)
Addition of Angular Momentum
143
Finally, we have
(11
...
Recalling that only commuting operators correspond to physical quantities which can be
simultaneously measured (see Sect
...
10), it follows, from the above discussion, that there
are two alternative sets of physical variables associated with angular momentum which we
can measure simultaneously
...
The second set correspond to the operators L2 , S2 , J2 , and Jz
...
In addition, we can
either choose to measure the z-components of the orbital and spin angular momentum
vectors, or the magnitude squared of the total angular momentum vector
...
18)
(1)
(1)
(11
...
20)
h
Lz ψl,s;m,ms = m ¯ ψl,s;m,ms ,
(1)
(1)
(11
...
It is easily seen that
(1)
(1)
(1)
h
Jz ψl,s;m,ms = (Lz + Sz ) ψl,s;m,ms = (m + ms ) ¯ ψl,s;m,ms
(1)
(11
...
h
Hence,
(11
...
In other words, the quantum numbers controlling the z-components of the various angular
momentum vectors can simply be added algebraically
...
24)
(2)
(2)
(11
...
h
(11
...
27)
144
QUANTUM MECHANICS
11
...
8
...
If the electron is also in an eigenstate of S2 and Sz then the quantum numbers
s and ms take the values 1/2 and ±1/2, respectively, and the internal state of the electron
is specified by the spinors χ± (see Sect
...
5)
...
(11
...
(1)
Since the eigenstates ψl,1/2;m,±1/2 are (presumably) orthonormal, and form a complete
(2)
set, we can express the eigenstates ψl,1/2;j,mj as linear combinations of them
...
29)
where α and β are, as yet, unknown coefficients
...
(11
...
Assuming that the ψ(2) eigenstates are properly normalized, we have
α2 + β2 = 1
...
30)
Now, it follows from Eq
...
26) that
(2)
(2)
J2 ψl,1/2;j,m+1/2 = j (j + 1) ¯ 2 ψl,1/2;j,m+1/2 ,
h
(11
...
(11
...
(11
...
(11
...
29), we can write
(2)
ψl,1/2;j,m+1/2 = α Yl,m χ+ + β Yl,m+1 χ−
...
33)
Recall, from Eqs
...
43) and (8
...
34)
L− Yl,m = [l (l + 1) − m (m − 1)]1/2 ¯ Yl,m−1
...
35)
By analogy, when the spin raising and lowering operators, S± , act on a general spinor,
χs,ms , we obtain
h
S+ χs,ms = [s (s + 1) − ms (ms + 1)]1/2 ¯ χs,ms +1 ,
(11
...
(11
...
e
...
38)
and
S± χ∓ = ¯ χ±
...
39)
It follows from Eqs
...
32) and (11
...
39) that
J2 Yl,m χ+ = [l (l + 1) + 3/4 + m] ¯ 2 Yl,m χ+
h
+[l (l + 1) − m (m + 1)]1/2 ¯ 2 Yl,m+1 χ− ,
h
(11
...
h
(11
...
(11
...
33) yield
(x − m) α − [l (l + 1) − m (m + 1)]1/2 β = 0,
−[l (l + 1) − m (m + 1)]
1/2
α + (x + m + 1) β = 0,
(11
...
43)
where
(11
...
Equations (11
...
43) can be solved to give
x (x + 1) = l (l + 1),
(11
...
β
x−m
(11
...
Once x is specified, Eqs
...
30) and (11
...
We
obtain
(2)
ψl+1/2,m+1/2
=
l+m+1
2l+ 1
=
l−m
2l+ 1
1/2
l−m
+
2l+ 1
1/2
l+m+1
−
2l+ 1
1/2
(1)
ψm,1/2
(1)
(11
...
48)
ψm+1,−1/2 ,
and
(2)
ψl−1/2,m+1/2
1/2
(1)
ψm,1/2
ψm+1,−1/2
...
1: Clebsch-Gordon coefficients for adding spin one-half to spin l
...
e
...
The above equations can easily be inverted to give the ψ(1) eigenstates
in terms of the ψ(2) eigenstates:
(1)
ψm,1/2
(1)
ψm+1,−1/2
=
l+m+1
2l+ 1
=
l−m
2l+ 1
1/2
l−m
+
2l+ 1
1/2
l+m+1
−
2l+ 1
1/2
(2)
ψl+1/2,m+1/2
1/2
(2)
ψl+1/2,m+1/2
(2)
(11
...
50)
ψl−1/2,m+1/2 ,
ψl−1/2,m+1/2
...
(11
...
50) is neatly summarized in Table 11
...
For
instance, Eq
...
47) is obtained by reading the first row of this table, whereas Eq
...
50)
is obtained by reading the second column
...
As an example, let us consider the l = 1 states of a hydrogen atom
...
Since m can take the values −1, 0, 1, whereas ms can
m,m
(1)
(1)
(1)
take the values ±1/2, there are clearly six such states: i
...
, ψ1,±1/2 , ψ0,±1/2 , and ψ−1,±1/2
...
Since l = 1 and s = 1/2 can be
combined together to form either j = 3/2 or j = 1/2 (see earlier), there are also six such
(2)
(2)
(2)
states: i
...
, ψ3/2,±3/2 , ψ3/2,±1/2 , and ψ1/2,±1/2
...
1, the various different
eigenstates are interrelated as follows:
(2)
(1)
(11
...
52)
1 (1)
ψ
−
3 0,1/2
(2)
ψ3/2,1/2 =
2 (1)
ψ
,
3 1,−1/2
(11
...
54)
1 (1)
ψ
+
3 −1,1/2
(2)
ψ1/2,−1/2 =
2 (1)
ψ
,
3 0,−1/2
(11
...
56)
Addition of Angular Momentum
147
−1, −1/2 −1, 1/2 0, −1/2 0, 1/2 1, −1/2 1, 1/2 m, ms
3/2, −3/2
3/2, −1/2
1/2, −1/2
3/2, 1/2
1/2, 1/2
3/2, 3/2
j, mj
1
√
1/3
√
2/3
√
2/3
√
−
1/3
√
2/3
√
1/3
√
1/3
√
−
2/3
1
Table 11
...
Only non-zero
coefficients are shown
...
57)
2 (2)
ψ
+
3 3/2,1/2
1 (2)
ψ
,
3 1/2,1/2
(11
...
59)
1 (2)
ψ
+
3 3/2,−1/2
2 (2)
ψ
,
3 1/2,−1/2
(11
...
e
...
(11
...
Suppose that we make such a measurement, and obtain the result j = 3/2, mj = 1/2
...
It thus follows from Eq
...
52) that
a subsequent measurement of Lz and Sz will yield m = 0, ms = 1/2 with probability 2/3,
and m = 1, ms = −1/2 with probability 1/3
...
(11
...
59) is neatly summed up in Table 11
...
Note that each row and column of this table has unit norm, and also that the different
rows and different columns are mutually orthogonal
...
11
...
Suppose that the system does
not possess any orbital angular momentum
...
61)
S = S1 + S2
be the total spin angular momentum operator
...
Let the quantum numbers associated with measurements of S1 , S1z , S2 ,
S2z , S2 , and Sz be s1 , ms1 , s2 , ms2 , s, and ms , respectively
...
62)
2
S2 χ(1),s2 ;ms
s1
1
,ms2
= s2 (s2 + 1) ¯ 2 χs1 ,s2 ;ms
h (1)
1
,ms2 ,
(11
...
64)
S2z χ(1),s2 ;ms
s1
1
,ms2
= ms2 ¯ χs1 ,s2 ;ms
h (1)
1
,ms2 ,
(11
...
1
(11
...
67)
2
h (2)
S2 χ(2),s2 ;s,ms = s2 (s2 + 1) ¯ 2 χs1 ,s2 ;s,ms ,
s1
(11
...
69)
Sz χ(2),s2 ;s,ms = ms ¯ χs1 ,s2 ;s,ms
...
70)
Of course, since both particles have spin one-half, s1 = s2 = 1/2, and s1z , s2z = ±1/2
...
(11
...
By
analogy, when spin one-half is added to spin one-half then the possible values of the total
spin quantum number are s = 1/2 ± 1/2
...
To be more exact, there are three possible s = 1 states (corresponding to ms = −1,
0, 1), and one possible s = 0 state (corresponding to ms = 0)
...
The Clebsch-Gordon coefficients for adding spin one-half to spin one-half can easily be
inferred from Table 11
...
3
...
2 ,
(11
...
3: Clebsch-Gordon coefficients for adding spin one-half to spin one-half
...
1
(2)
(1)
(1)
χ1,0 = √ χ−1/2,1/2 + χ1/2,−1/2 ,
2
(2)
(1)
(11
...
74)
χ1,1 = χ1/2,1/2 ,
(2)
where χ(2) s is shorthand for χs1 ,s2 ;s,ms , etc
...
2
(11
...
An electron in a hydrogen atom occupies the combined spin and position state
R2,1
1/3 Y1,0 χ+ +
2/3 Y1,1 χ−
...
(c) Same for S2
...
(e) Same for J2
...
(g) What is the probability density for finding the electron at r, θ, φ?
(h) What is the probability density for finding the electron in the spin up state (with respect
to the z-axis) at radius r?
2
...
Calculate the potential energy for the neutron-proton
system:
150
QUANTUM MECHANICS
(a) In the spin singlet state
...
3
...
(a) If a measurement of the spin of one of the electrons shows that it is in the state with
Sz = ¯ /2, what is the probability that a measurement of the z-component of the spin of
h
the other electron yields Sz = ¯ /2?
h
(b) If a measurement of the spin of one of the electrons shows that it is in the state with
Sy = ¯ /2, what is the probability that a measurement of the x-component of the spin
h
of the other electron yields Sx = −¯ /2?
h
Finally, if electron 1 is in a spin state described by cos α1 χ+ + sin α1 e i β1 χ− , and electron
2 is in a spin state described by cos α2 χ+ + sin α2 e i β2 χ− , what is the probability that the
two-electron spin state is a triplet state?
Time-Independent Perturbation Theory
151
12 Time-Independent Perturbation Theory
12
...
The Hamiltonian of a quantum
mechanical system is written
H = H0 + H1
...
1)
Here, H0 is a simple Hamiltonian whose eigenvalues and eigenstates are known exactly
...
However, H1 can, in some sense (which we shall specify more precisely later
on), be regarded as being small compared to H0
...
Incidentally, in this chapter, we shall only discuss so-called time-independent perturbation theory, in which the modification to the Hamiltonian, H1 , has no explicit dependence
on time
...
12
...
Let the ψi be a complete set of eigenstates of the Hamiltonian, H, corresponding to the
eigenvalues Ei : i
...
,
H ψi = Ei ψi
...
2)
Now, we expect the ψi to be orthonormal (see Sect
...
9)
...
(12
...
7), the above expression generalizes to
∞
∞
∞
ψ∗ ψj dx dy dz = δij
...
4)
−∞ −∞ −∞
Finally, if the ψi are spinors (see Cha
...
i
(12
...
We can represent all of the above possibilities by writing
ψi |ψj ≡ i|j = δij
...
6)
152
QUANTUM MECHANICS
Here, the term in angle brackets represents the integrals in Eqs
...
3) and (12
...
5) in spinspace
...
e
...
Expanding a general wavefunction, ψa , in terms of the energy eigenstates, ψi , we
obtain
ψa =
ci ψi
...
7)
i
In one dimension, the expansion coefficients take the form (see Sect
...
9)
∞
ψ∗ ψa dx,
i
ci =
(12
...
i
ci =
(12
...
i
(12
...
(12
...
7) thus becomes
ψa =
i
ψi |ψa ψi ≡
i|a ψi
...
12)
i
Incidentally, it follows that
i|a
∗
= a|i
...
13)
Finally, if A is a general operator, and the wavefunction ψa is expanded in the manner
shown in Eq
...
7), then the expectation value of A is written (see Sect
...
9)
c∗ cj Aij
...
14)
i,j
Here, the Aij are unsurprisingly known as the matrix elements of A
...
15)
−∞
whereas in three dimensions we get
∞
∞
∞
ψ∗ A ψj dx dy dz
...
16)
Time-Independent Perturbation Theory
153
Finally, if ψ is a spinor then we have
Aij = ψ† A ψj
...
17)
We can represent all of the above possibilities by writing
Aij = ψi |A|ψj ≡ i|A|j
...
18)
The expansion (12
...
(12
...
(4
...
(12
...
(12
...
21)
where the ψi are a complete set of eigenstates, and 1 is the identity operator
...
3 Two-State System
Consider the simplest possible non-trivial quantum mechanical system
...
e
...
22)
H0 ψ2 = E2 ψ2
...
23)
It is assumed that these states, and their associated eigenvalues, are known
...
Let us now try to solve the modified energy eigenvalue problem
(H0 + H1 ) ψE = E ψE
...
24)
We can, in fact, solve this problem exactly
...
(12
...
(12
...
26)
It follows from (12
...
Equations (12
...
23), (12
...
26), and the orthonormality
condition
i|j = δij ,
(12
...
28)
where
e11 =
1|H1 |1 ,
(12
...
30)
e12 =
1|H1 |2 = 2|H1 |1 ∗
...
31)
Here, use has been made of the fact that H1 is an Hermitian operator
...
32)
e11 = e22 = 0
...
(12
...
|E1 − E2 |
...
33)
(12
...
(12
...
36)
|e12 |2
= E2 −
+ ···
...
37)
′
E1 = E1 +
′
E2
Note that H1 causes the upper eigenvalue to rise, and the lower to fall
...
E1 − E2
′
ψ1 = ψ1 +
(12
...
39)
Time-Independent Perturbation Theory
155
Thus, the modified energy eigenstates consist of one of the unperturbed eigenstates, plus
a slight admixture of the other
...
This suggests that the condition for the validity of the perturbation method as a whole is
|e12 | ≪ |E1 − E2 |
...
40)
In other words, when we say that H1 needs to be small compared to H0 , what we are really
saying is that the above inequality must be satisfied
...
4 Non-Degenerate Perturbation Theory
Let us now generalize our perturbation analysis to deal with systems possessing more
than two energy eigenstates
...
41)
where n runs from 1 to N
...
42)
m|n = δnm ,
and to form a complete set
...
(12
...
44)
where m can take any value from 1 to N
...
45)
k
where k runs from 1 to N
...
46)
k=m
where
emk = m|H1 |k
...
47)
Let us now develop our perturbation expansion
...
48)
156
QUANTUM MECHANICS
for all m = k, where ǫ ≪ 1 is our expansion parameter
...
49)
for all m
...
50)
and
n|E
= 1,
(12
...
52)
for m = n
...
(12
...
We find that
(Em − En ) m|E + emn ≃ 0,
(12
...
(12
...
(12
...
(12
...
En − Ek
(12
...
57)
Incidentally, it is easily demonstrated that the modified eigenstates remain orthonormal to
O(ǫ2 )
...
5 Quadratic Stark Effect
Suppose that a hydrogen atom is subject to a uniform external electric field, of magnitude
|E|, directed along the z-axis
...
Namely, the unperturbed Hamiltonian,
e2
p2
−
,
H0 =
2 me 4πǫ0 r
(12
...
(12
...
Hence,
the energy eigenstates of the unperturbed Hamiltonian are characterized by three quantum
numbers—the radial quantum number n, and the two angular quantum numbers l and m
(see Cha
...
Let us denote these states as the ψnlm , and let their corresponding energy
eigenvalues be the Enlm
...
Enlm − En ′ l ′ m ′
n ′ ,l ′ ,m ′ =n,l,m
(12
...
The sum on the right-hand side of the above equation seems very complicated
...
This follows because the
matrix elements n, l, m|z|n ′ , l ′, m ′ are zero for virtually all choices of the two sets of
quantum number, n, l, m and n ′ , l ′ , m ′
...
These rules are usually referred to as the selection rules for the problem in hand
...
(8
...
61)
it follows that [see Eqs
...
15)–(7
...
(12
...
63)
since ψnlm is, by definition, an eigenstate of Lz corresponding to the eigenvalue m ¯
...
Let us now determine the selection rule for l
...
64)
158
QUANTUM MECHANICS
= Lx [Lx , z] + [Lx , z] Lx + Ly [Ly , z] + [Ly , z] Ly
= i ¯ (−Lx y − y Lx + Ly x + x Ly )
h
= 2 i ¯ (Ly x − Lx y + i ¯ z)
h
h
= 2 i ¯ (Ly x − y Lx ) = 2 i ¯ (x Ly − Lx y),
h
h
(12
...
(7
...
17), (8
...
4), and (8
...
Thus,
[L2 , [L2, z]] = 2 i ¯ L2 , Ly x − Lx y + i ¯ z
h
h
= 2 i ¯ Ly [L2, x] − Lx [L2 , y] + i ¯ [L2, z]
h
h
= −4 ¯ 2 Ly (y Lz − Ly z) + 4 ¯ 2 Lx (Lx z − x Lz )
h
h
−2 ¯ 2 (L2 z − z L2),
h
(12
...
h
(12
...
(8
...
4) that
Lx x + Ly y + Lz z = 0
...
68)
[L2, [L2, z]] = 2 ¯ 2 (L2 z + z L2 )
...
69)
Hence, we obtain
Finally, the above expression expands to give
L4 z − 2 L2 z L2 + z L4 − 2 ¯ 2 (L2 z + z L2) = 0
...
70)
Equation (12
...
h
(12
...
72)
which reduces to
(l + l ′ + 2) (l + l ′ ) (l − l ′ + 1) (l − l ′ − 1) n, l, m|z|n ′, l ′ , m = 0
...
73)
Time-Independent Perturbation Theory
159
According to the above formula, the matrix element n, l, m|z|n ′ , l ′ , m vanishes unless
l = l ′ = 0 or l ′ = l ± 1
...
] Recall, however, from Cha
...
It, therefore, follows, from symmetry, that
the matrix element n, l, m|z|n ′ , l ′, m is zero when l = l ′ = 0
...
(12
...
64) and (12
...
(12
...
Enlm − En ′ l ′ m
′ ,l ′ =l±1
n
(12
...
(12
...
Only those terms which vary quadratically
with the field-strength survive
...
Now, the electric
polarizability of an atom is defined in terms of the energy-shift of the atomic state as
follows:
1
∆E = − α |E|2
...
76)
2
Hence, we can write
αnlm
| n, l, m|z|n ′ , l ′ , m |2
...
77)
Unfortunately, there is one fairly obvious problem with Eq
...
75)
...
e
...
Clearly, our perturbation method breaks down completely in this
situation
...
(12
...
77) are only applicable to cases
where the coupled eigenstates are non-degenerate
...
Now, the
unperturbed eigenstates of a hydrogen atom have energies which only depend on the radial quantum number n (see Cha
...
It follows that we can only apply the above results
to the n = 1 eigenstate (since for n > 1 there will be coupling to degenerate eigenstates
with the same value of n but different values of l)
...
e
...
En00 − E100
(12
...
The sum in the above expression can
be evaluated approximately by noting that (see Sect
...
4)
En00 = −
where
a0 =
e2
,
8π ǫ0 a0 n2
(12
...
80)
is the Bohr radius
...
81)
| 1, 0, 0|z|n, 1, 0 |2
...
82)
En00 − E100 ≥ E200 − E100 =
which implies that
α<
16
4πǫ0 a0
3
n>1
However, [see Eq
...
21)]
| 1, 0, 0|z|n, 1, 0 |2 =
n>1
1, 0, 0|z|n, 1, 0 n, 1, 0|z|1, 0, 0
n>1
1, 0, 0|z|n ′, l ′ , m ′ n ′, l ′ , m ′ |z|1, 0, 0
=
n ′ ,l ′ ,m ′
=
1, 0, 0|z2|1, 0, 0 =
1
1, 0, 0|r2|1, 0, 0 ,
3
(12
...
Finally, it
follows from Eq
...
72) that
2
1, 0, 0|r2|1, 0, 0 = 3 a0
...
84)
Hence, we conclude that
16
3
3
4πǫ0 a0 ≃ 5
...
(12
...
5 4πǫ0 a0
...
86)
2
α<
12
...
e
...
We can write
H0 ψnlm = En ψnlm ,
(12
...
Making use of the selection rules (12
...
74), non-degenerate
perturbation theory yields the following expressions for the perturbed energy levels and
eigenstates [see Eqs
...
56) and (12
...
88)
(12
...
(12
...
e
...
These particular matrix elements give rise to singular factors
1/(En − En ) in the summations
...
Of course, there is no n = 1 state with l ′ = 1
...
Unfortunately, if n > 1 then there are
multiple coupled states corresponding to the eigenvalue En
...
e
...
91)
n, l ′, m|H1 |n, l, m = λnl δll ′
...
(12
...
89) would reduce to zero
...
Fortunately, we can always redefine
the unperturbed eigenstates corresponding to the eigenvalue En in such a manner that
Eq
...
91) is satisfied
...
Let us define Nn new states which are linear combinations of our Nn
original degenerate eigenstates:
(1)
n, k, m|n, l(1) , m ψnkm
...
92)
k=1,Nn
Note that these new states are also degenerate energy eigenstates of the unperturbed
(1)
Hamiltonian, H0 , corresponding to the eigenvalue En
...
e
...
Thus,
(1)
(1)
H1 ψnlm = λnl ψnlm
...
93)
162
QUANTUM MECHANICS
(1)
The ψnlm are also chosen so as to be orthonormal: i
...
,
n, l ′(1) , m|n, l(1) , m = δll ′
...
94)
n, l ′(1) , m|H1 |n, l(1) , m = λnl δll ′
...
95)
It follows that
Thus, if we use the new eigenstates, instead of the old ones, then we can employ Eqs
...
88)
and (12
...
The only remaining difficulty
is to determine the new eigenstates in terms of the original ones
...
(12
...
96)
l=1,Nn
where 1 denotes the identity operator in the sub-space of all coupled unperturbed eigenstates corresponding to the eigenvalue En
...
93) can be transformed into a straightforward matrix equation:
n, l ′ , m|H1 |n, l ′′ , m n, l ′′, m|n, l(1) , m = λnl n, l ′, m|n, l(1) , m
...
97)
l ′′ =1,Nn
This can be written more transparently as
U x = λ x,
(12
...
(12
...
(12
...
The normalized
eigenvectors specify the weights of the new eigenstates in terms of the original eigenstates:
i
...
,
(xnl )k = n, k, m|n, l(1) , m ,
(12
...
In our new scheme, Eqs
...
88) and (12
...
101)
en ′ l ′ nl
ψn ′ l ′ m
...
102)
′
Enl = En + λnl +
and
(1) ′
(1)
ψnlm = ψnlm +
There are no singular terms in these expressions, since the summations are over n ′ = n:
i
...
, they specifically exclude the problematic, degenerate, unperturbed energy eigenstates
corresponding to the eigenvalue En
...
98)
...
7 Linear Stark Effect
Returning to the Stark effect, let us examine the effect of an external electric field on the
energy levels of the n = 2 states of a hydrogen atom
...
All of these states possess the same unperturbed energy, E200 = −e2 /(32π ǫ0 a0 )
...
103)
H1 = e |E| z
...
e
...
Hence, non-degenerate perturbation theory breaks
down when applied to these two states
...
In order to apply perturbation theory to the ψ200 and ψ210 states, we have to solve the
matrix eigenvalue equation
U x = λ x,
(12
...
Thus,
U = e |E|
0
2, 1, 0|z|2, 0, 0
2, 0, 0|z|2, 1, 0
0
,
(12
...
Here, we have
again made use of the selection rules, which tell us that the matrix element of z between
two hydrogen atom states is zero unless the states possess l quantum numbers which differ
by unity
...
(12
...
The corresponding normalized eigenvectors are
√
1/ 2
x1 = √ ,
(12
...
x2 =
(12
...
=
2
ψ1 =
(12
...
110)
164
QUANTUM MECHANICS
In the absence of an external electric field, both of these states possess the same energy,
E200
...
111)
∆E2 = −3 e a0 |E|
...
112)
Thus, in the presence of an electric field, the energies of states 1 and 2 are shifted upwards
and downwards, respectively, by an amount 3 e a0 |E|
...
Note that the energy shifts are linear
in the electric field-strength, so this effect—which is known as the linear Stark effect—is
much larger than the quadratic effect described in Sect
...
5
...
Of course, to
second-order the energies of these states are shifted by an amount which depends on the
square of the electric field-strength (see Sect
...
5)
...
8 Fine Structure of Hydrogen
According to special relativity, the kinetic energy (i
...
, the difference between the total
energy and the rest mass energy) of a particle of rest mass m and momentum p is
T=
p2 c2 + m2 c4 − m c2
...
113)
In the non-relativistic limit p ≪ m c, we can expand the square-root in the above expression to give
p2
1 p 2
p 4
T=
1−
...
114)
+O
2m
4 mc
mc
Hence,
p2
p4
T≃
−
...
115)
2 m 8 m3 c2
Of course, we recognize the first term on the right-hand side of this equation as the standard non-relativistic expression for the kinetic energy
...
Let us consider the effect of this type of correction
on the energy levels of a hydrogen atom
...
(12
...
3
8 me c2
(12
...
12
...
3
8 me c2
(12
...
118)
where V = −e2 /(4πǫ0 r)
...
(9
...
75) that
∆Enlm
1
e2
E 2 + 2 En
= −
2 me c2 n
4πǫ0
1
e2
+
n2 a0
4πǫ0
2
1
...
119)
1
...
120)
Finally, making use of Eqs
...
55), (9
...
58), the above expression reduces to
∆Enlm = En
α2
n2
3
n
,
−
l + 1/2 4
(12
...
122)
4πǫ0 ¯ c
h
137
is the dimensionless fine structure constant
...
It
turns out that this is not the case for l = 0 states
...
Note, also, that we are able to use
non-degenerate perturbation theory in the above calculation, using the ψnlm eigenstates,
because the perturbing Hamiltonian commutes with both L2 and Lz
...
Hence, all coupled
states have different n quantum numbers, and therefore have different energies
...
123)
due to the charge on the nucleus
...
124)
B=− 2
...
(10
...
59)]
µ=−
e
S
me
(12
...
We, therefore, expect an additional contribution to
the Hamiltonian of a hydrogen atom of the form [see Eq
...
60)]
H1 = −µ · B
e2
= −
v×r·S
4πǫ0 me c2 r3
=
e2
L · S,
2
4πǫ0 me c2 r3
(12
...
This effect is known as
spin-orbit coupling
...
Hence, the true spin-orbit
correction to the Hamiltonian is
H1 =
e2
L · S
...
127)
Let us now apply perturbation theory to the hydrogen atom, using the above expression as
the perturbing Hamiltonian
...
128)
is the total angular momentum of the system
...
129)
giving
1 2
(J − L2 − S2 )
...
130)
2
Recall, from Sect
...
2, that whilst J2 commutes with both L2 and S2 , it does not commute
with either Lz or Sz
...
127) also commutes
with both L2 and S2 , but does not commute with either Lz or Sz
...
58) and the perturbing Hamiltonian (12
...
11
...
It is important to know this since, according to Sect
...
6, we can only
safely apply perturbation theory to the simultaneous eigenstates of the unperturbed and
perturbing Hamiltonians
...
11
...
131)
Time-Independent Perturbation Theory
167
(2)
(2)
S2 ψl,s;j,mj = s (s + 1) ¯ 2 ψl,s;j,mj ,
h
(2)
(12
...
133)
(2)
Jz ψl,s;j,mj = mj ¯ ψl,s;j,mj
...
134)
According to standard first-order perturbation theory, the energy-shift induced in such a
state by spin-orbit coupling is given by
∆El,1/2;j,mj =
l, 1/2; j, mj|H1 |l, 1/2; j, mj
e2
J2 − L2 − S2
=
1, 1/2; j, mj
l, 1/2; j, mj
2
16π ǫ0 me c2
r3
=
e2 ¯ 2
h
[j (j + 1) − l (l + 1) − 3/4]
2
16π ǫ0 me c2
1
...
135)
Here, we have made use of the fact that s = 1/2 for an electron
...
(9
...
136)
2 c2 a 3
16π ǫ0 me
l (l + 1/2) (l + 1) n3
0
where n is the radial quantum number
...
(9
...
57), and
(9
...
137)
where α is the fine structure constant
...
(12
...
We can add these
two corrections together (making use of the fact that j = l ± 1/2 for a hydrogen atom—see
Sect
...
3) to obtain a net energy-shift of
∆El,1/2;j,mj = En
α2
n2
3
n
...
138)
This modification of the energy levels of a hydrogen atom due to a combination of relativity
and spin-orbit coupling is known as fine structure
...
Let us examine the effect of the fine structure energy-shift (12
...
For n = 1, in the absence of fine structure, there are two degenerate 1S1/2 states
...
(12
...
1: Effect of the fine structure energy-shift on the n = 1, 2 and 3 states of a hydrogen
atom
...
the same
...
For n = 2, in the absence of fine structure, there are two 2S1/2 states, two 2P1/2 states,
and four 2P3/2 states, all of which are degenerate
...
(12
...
Hence, fine structure
does not break the degeneracy of the 2S1/2 and 2P1/2 states of hydrogen, but does break
the degeneracy of these states relative to the 2P3/2 states
...
According to Eq
...
138), fine structure breaks these states into three groups: the 3S1/2 and
3P1/2 states, the 3P3/2 and 3D3/2 states, and the 3D5/2 states
...
12
...
Note, finally, that although expression (12
...
121) it, somewhat fortuitously, gives rise to an
expression (12
...
Time-Independent Perturbation Theory
169
12
...
The modification to the Hamiltonian of the system is
(12
...
140)
2 me
is the total electron magnetic moment, including both orbital and spin contributions [see
Eqs
...
57)–(10
...
Thus,
eB
H1 =
(Lz + 2 Sz)
...
141)
2 me
Suppose that the applied magnetic field is much weaker than the atom’s internal magnetic field (12
...
Since the magnitude of the internal field is about 25 tesla, this is a
fairly reasonable assumption
...
Of course, these states are the simultaneous eigenstates of L2 , S2 , J2 ,
and Jz (see previous section)
...
142)
since Jz = Lz + Sz
...
(11
...
48),
(2)
ψj,mj =
j + mj
2l+1
1/2
(1)
ψmj −1/2,1/2 +
j − mj
2l+1
1/2
(1)
ψmj +1/2,−1/2
(12
...
144)
when j = l − 1/2
...
In particular,
(1)
Sz ψm,±1/2 = ±
¯ (1)
h
ψ
...
145)
It follows from Eqs
...
143)–(12
...
146)
170
QUANTUM MECHANICS
when j = l ± 1/2
...
147)
e¯
h
= 5
...
148)
∆El,1/2;j,mj = µB B mj 1 ±
where the ± signs correspond to j = l ± 1/2
...
Of course, the quantum number mj takes values differing
by unity in the range −j to j
...
(12
...
(12
...
2l+ 1
(12
...
Likewise, the four degenerate 2S1/2
and 2P1/2 states are split by (2/3) µB B, whereas the four degenerate 2P3/2 states are split
(2)
by (4/3) µB B
...
12
...
Note, finally, that since the ψl,mj are not
simultaneous eigenstates of the unperturbed and perturbing Hamiltonians, Eqs
...
149)
and (12
...
However, as long as the external magnetic field is much weaker than the internal magnetic field, these expectation values are almost identical to the actual measured
values of the energy-shifts
...
10
Hyperfine Structure
The proton in a hydrogen atom is a spin one-half charged particle, and therefore possesses
a magnetic moment
...
(10
...
151)
where µp is the proton magnetic moment, Sp is the proton spin, and the proton gyromagnetic ratio gp is found experimentally to take that value 5
...
Note that the magnetic
Time-Independent Perturbation Theory
171
(4/3)ǫ
2P3/2
2S1/2
(4/3)ǫ
(4/3)ǫ
2P1/2
(2/3)ǫ
(2/3)ǫ
(2/3)ǫ
1S1/2
2ǫ
unperturbed + fine structure
+ Zeeman
Figure 12
...
Here, ǫ = µB B
...
moment of a proton is much smaller (by a factor of order me /mp ) than that of an electron
...
152)
where er = r/r
...
All
magnetic field-lines generated by the loop must pass through the loop
...
Now, the Hamiltonian of the electron in the
magnetic field generated by the proton is simply
H1 = −µe · B,
(12
...
(12
...
(10
...
59)], and Se the
electron spin
...
8π mp me
r3
3 mp me
(12
...
According to standard first-order perturbation theory, the energy-shift induced by spinspin coupling between the proton and the electron is the expectation value of the perturbing Hamiltonian
...
3 mp me
(12
...
Moreover, it is easily demonstrated that |ψ000 (0)|2 =
3
1/(π a0 )
...
(12
...
158)
be the total spin
...
2
(12
...
However, both the proton and
the electron are spin one-half particles
...
11
...
In fact, there are three spin 1 states, known as triplet states, and a
2
2
single spin 0 state, known as the singlet state
...
The eigenvalue of S2 is 0 for the singlet state, and 2 ¯ 2 for the triplet states
...
160)
Sp · Se = − ¯ 2
4
for the singlet state, and
1
Sp · Se = ¯ 2
h
(12
...
It follows, from the above analysis, that spin-spin coupling breaks the degeneracy of the
two 1S1/2 states in hydrogen, lifting the energy of the triplet configuration, and lowering
that of the singlet
...
The net energy difference
between the singlet and the triplet states is
∆E =
me 2
8
gp
α E0 = 5
...
162)
Time-Independent Perturbation Theory
173
where E0 = 13
...
Note that the hyperfine
energy-shift is much smaller, by a factor me /mp , than a typical fine structure energy-shift
...
1 cm
...
163)
This is the wavelength of the radiation emitted by a hydrogen atom which is collisionally
excited from the singlet to the triplet state, and then decays back to the lower energy
singlet state
...
174
QUANTUM MECHANICS
Time-Dependent Perturbation Theory
175
13 Time-Dependent Perturbation Theory
13
...
(13
...
However, H1 now represents a small time-dependent external
perturbation
...
(13
...
4
...
However, the presence
of a small time-dependent perturbation can, in principle, give rise to a finite probability
that if the system is initially in some eigenstate ψn of the unperturbed Hamiltonian then
it is found in some other eigenstate at a subsequent time (since ψn is no longer an exact
eigenstate of the total Hamiltonian)
...
Let us
investigate this effect
...
2 Preliminary Analysis
Suppose that at t = 0 the state of the system is represented by
ψ(0) =
cm ψm ,
(13
...
Thus, the initial state is some linear superposition of
the unperturbed energy eigenstates
...
4
...
h
ψ(t) =
(13
...
5)
since the unperturbed eigenstates are assummed to be orthonormal: i
...
,
n|m = δnm
...
6)
176
QUANTUM MECHANICS
Clearly, with H1 = 0, the probability of finding the system in state ψn at time t is exactly
the same as the probability of finding the system in this state at the initial time, t = 0
...
Thus, we can
write
ψ(t) =
cm (t) exp (−i Em t/¯ ) ψm ,
h
(13
...
Here, we have carefully separated the fast phase oscillation of the
eigenstates, which depends on the unperturbed Hamiltonian, from the slow variation of
the amplitudes cn (t), which depends entirely on the perturbation (i
...
, cn is constant in
time if H1 = 0)
...
(13
...
The time-dependent Schr¨dinger equation [see Eq
...
63)] yields
o
i¯
h
∂ψ(t)
= H(t) ψ(t) = [H0 + H1 (t)] ψ(t)
...
8)
Now, it follows from Eq
...
7) that
cm exp (−i Em t/¯ ) (Em + H1 ) ψm
...
9)
dcm
+ cm Em exp (−i Em t/¯ ) ψm ,
h
dt
(13
...
According to Eq
...
8), we can equate the right-hand
sides of the previous two equations to obtain
i¯
h
m
dcm
exp (−i Em t/¯ ) ψm =
h
dt
cm exp (−i Em t/¯ ) H1 ψm
...
11)
m
Projecting out the component of the above equation which is proportional to ψn , using
Eq
...
6), we obtain
i¯
h
dcn (t)
=
dt
Hnm (t) exp ( i ωnm t) cm (t),
(13
...
13)
En − Em
...
14)
¯
h
Suppose that there are N linearly independent eigenstates of the unperturbed Hamiltonian
...
(13
...
Note that Eqs
...
12) are exact—we have
made no approximations at this stage
...
Instead, we have to obtain approximate solutions via suitable
expansions in small quantities
...
e
...
(13
...
This solution is of great practical importance
...
3 Two-State System
Consider a system in which the time-independent Hamiltonian possesses two eigenstates,
denoted
H0 ψ1 = E1 ψ1 ,
(13
...
(13
...
e
...
(13
...
e
...
18)
where γ and ω are real
...
For a two-state system, Eq
...
12) reduces to
dc1
= γ exp [+i (ω − ω21 ) t] c2 ,
dt
dc2
i
= γ exp [−i (ω − ω21 ) t] c1 ,
dt
i
(13
...
20)
where ω21 = (E2 −E1 )/¯
...
e
...
2
dt
dt
(13
...
(13
...
Let us
search for a solution in which the system is certain to be in state 1 (and, thus, has no
chance of being in state 2) at time t = 0
...
It is easily demonstrated that the appropriate solutions to (13
...
20)
are
c2 (t) =
−i (ω − ω21 ) t
−i γ
exp
sin(Ω t),
Ω
2
(13
...
23)
where
Ω=
γ2 + (ω − ω21 )2 /4
...
24)
Now, the probability of finding the system in state 1 at time t is simply P1 (t) = |c1 (t)|2
...
It
follows that
(13
...
2 /4
+ (ω − ω21 )
(13
...
Equation (13
...
At resonance, when
the oscillation frequency of the perturbation, ω, matches the frequency ω21 , we find that
P1 (t) = cos2 (γ t),
2
P2 (t) = sin (γ t)
...
27)
(13
...
After a time interval
π/(2 γ) it is certain to be in state 2
...
Thus, the system periodically flip-flops between states 1 and
2 under the influence of the time-dependent perturbation
...
The absorption-emission cycle also takes place away from the resonance, when ω =
ω21
...
This means
that the maximum value of P2 (t) is no longer unity, nor is the minimum of P1 (t) zero
...
Thus, if the applied frequency differs from
the resonant frequency by substantially more than 2 γ then the probability of the system
jumping from state 1 to state 2 is always very small
...
Clearly, the weaker the perturbation
(i
...
, the smaller γ becomes), the narrower the resonance
...
4 Spin Magnetic Resonance
Consider a system consisting of a spin one-half particle with no orbital angular momentum
(e
...
, a bound electron) placed in a uniform z-directed magnetic field, and then subject to
Time-Dependent Perturbation Theory
179
a small time-dependent magnetic field rotating in the x-y plane at the angular frequency
ω
...
29)
where B0 and B1 are constants, with B1 ≪ B0
...
In
this system, the electric component of the wave has no effect
...
30)
(13
...
(13
...
(12
...
The eigenstates of the unperturbed Hamiltonian are the “spin up” and “spin down”
states, denoted χ+ and χ− , respectively
...
10)
...
(13
...
34)
where S+ and S− are the conventional raising and lowering operators for spin angular
momentum (see Sect
...
It follows that
+|H1 |+ = −|H1 |− = 0,
(13
...
(13
...
γ → −
4m
(13
...
38)
(13
...
40)
180
QUANTUM MECHANICS
The resonant frequency, ω21 , is simply the spin precession frequency in a uniform magnetic
field of strength B0 (see Sect
...
6)
...
If we now apply a magnetic perturbation rotating at the resonant
frequency then, according to the analysis of the previous subsection, the system undergoes
a succession of spin flips, χ+ ↔ χ− , in addition to the spin precession
...
The width of
the resonance (in frequency) is determined by the strength of the oscillating magnetic
perturbation
...
By determining the resonant frequency (i
...
, the frequency
at which the particles absorb energy from the oscillating field), it is possible to determine
the gyromagnetic ratio (assuming that the mass is known)
...
5 Perturbation Expansion
Let us recall the analysis of Sect
...
2
...
Thus, H0 ψn = En ψn , where the
En are the unperturbed energy levels, and n|m = δnm
...
41)
n
where ωn = En /¯
...
42)
m
where Hnm (t) = n|H1 (t)|m and ωnm = (En − Em )/¯
...
43)
(assuing that, initially, n |cn |2 = 1)
...
Equation (13
...
(13
...
(13
...
Now, according to (13
...
Hence, the zeroth-order solution
is simply
c(0) (t) = δni
...
45)
n
The first-order solution is obtained, via iteration, by substituting the zeroth-order solution
into the right-hand side of Eq
...
42)
...
46)
m
subject to the boundary condition c(1) (0) = 0
...
(13
...
(13
...
49)
Hfi (t ′) exp( i ωfi t ′ ) dt ′
...
(13
...
6 Harmonic Perturbations
Consider a (Hermitian) perturbation which oscillates sinusoidally in time
...
Such a perturbation takes the form
H1 (t) = V exp( i ω t) + V † exp(−i ω t),
(13
...
It follows from Eqs
...
48) and (13
...
52)
0
where
Vfi =
f|V|i ,
(13
...
(13
...
1: The functions sinc(x) (dashed curve) and sinc2 (x) (solid curve)
...
Integration with respect to t ′ yields
cf (t) = −
it
(Vfi exp [ i (ω + ωfi ) t/2] sinc [(ω + ωfi ) t/2]
¯
h
†
+Vfi exp [−i (ω − ωfi ) t/2] sinc [(ω − ωfi ) t/2] ,
(13
...
(13
...
13
...
Thus, the first and second terms on the right-hand side of
Eq
...
55) are only non-negligible when
sinc x ≡
|ω + ωfi | <
∼
2π
,
t
(13
...
58)
t
respectively
...
Eventually, when t ≫ 2π/|ωfi |, these two ranges become strongly non-overlapping
...
¯
h
(13
...
13
...
It follows that the above expression exhibits a resonant response
∼
to the applied perturbation at the frequencies ω = ±ωfi
...
At each of the resonances (i
...
, at ω =
±ωfi ), the transition probability Pi→f (t) varies as t2 [since sinh(0) = 1]
...
28), for the two-state system, in the limit
γ t ≪ 1 (recall that our perturbative solution is only valid as long as Pi→f ≪ 1)
...
h
(13
...
This process is
h
known as stimulated emission
...
h
(13
...
This
h
process is known as absorption
...
Hence, we can write the transition
probabilities for both processes separately
...
59), the transition probability
for stimulated emission is
t2 †
stm
Pi→f (t) = 2 |Vif | 2 sinc2 [(ω − ωif ) t/2] ,
(13
...
Likewise, the
transition probability for absorption is
abs
Pi→f (t)
t2 † 2
= 2 |Vfi | sinc2 [(ω − ωfi ) t/2]
...
63)
13
...
e
...
The unperturbed Hamiltonian of the system is
H0 =
p2
+ V0 (r)
...
64)
Now, the standard classical prescription for obtaining the Hamiltonian of a particle of
charge q in the presence of an electromagnetic field is
p → p + q A,
H → H − q φ,
(13
...
66)
184
QUANTUM MECHANICS
where A(r) is the vector potential, and φ(r) the scalar potential
...
67)
(13
...
This prescription also works in quantum mechanics
...
69)
where A and φ are functions of the position operators
...
(13
...
71)
provided that we adopt the gauge ∇·A = 0
...
2 me
me
2 me
(13
...
In this case,
φ = 0,
(13
...
74)
where k is the wavevector (note that ω = k c), and ǫ a unit vector which specifies the
direction of polarization (i
...
, the direction of E)
...
The Hamiltonian
becomes
H = H0 + H1 (t),
(13
...
76)
(13
...
The perturbing Hamiltonian can be written
H1 = −
e A0 ǫ·p
[exp( i k·r − i ωt) + exp(−i k·r + i ωt)]
...
78)
Time-Dependent Perturbation Theory
185
This has the same form as Eq
...
51), provided that
V† = −
e A0 ǫ·p
exp( i k·r )
...
79)
It follows from Eqs
...
53), (13
...
79) that the transition probability for
radiation induced absorption is
abs
Pi→f (t) =
t2 e2 |A0 |2
| f|ǫ·p exp( i k·r)|i | 2 sinc2 [(ω − ωfi ) t/2]
...
80)
Now, the mean energy density of an electromagnetic wave is
u=
1
2
ǫ0 |E0 |2 |B0 |2
+
2
2 µ0
=
1
ǫ0 |E0 |2 ,
2
(13
...
It thus follows that
abs
Pi→f (t) =
t 2 e2
| f|ǫ·p exp( i k·r)|i | 2 u sinc2 [(ω − ωfi ) t/2]
...
82)
Thus, not surprisingly, the transition probability for radiation induced absorption (or stimulated emission) is directly proportional to the energy density of the incident radiation
...
We can write
∞
u=
ρ(ω) dω,
(13
...
Equation (13
...
(13
...
e
...
This follows because it is permissible to add the intensities of incoherent radiation,
whereas we must always add the amplitudes of coherent radiation
...
13
...
85)
−∞
the above equation reduces to
abs
Pi→f (t) =
π e2 ρ(ωfi )
| f|ǫ·p exp( i k·r)|i | 2 t
...
86)
186
QUANTUM MECHANICS
Note that in integrating over the frequencies of the incoherent radiation we have transformed a transition probability which is basically proportional to t2 [see Eq
...
82)] to
one which is proportional to t
...
However, the result that
wabs ≡
i→f
abs
dPi→f
π e2 ρ(ωfi )
| f|ǫ·p exp( i k·r)|i | 2
=
2
2 ω2
dt
ǫ0 ¯ me fi
h
(13
...
Here, wabs is the transition probability per unit time
i→f
interval, otherwise known as the transition rate
...
2)
abs
abs
abs
Pi→f (t + dt) − Pi→f (t) = 1 − Pi→f (t) wabs dt :
i→f
(13
...
e
...
It follows that
i→f
abs
dPi→f
abs
+ wabs Pi→f = wabs ,
i→f
i→f
dt
(13
...
The above equation can be solved to give
abs
Pi→f (t) = 1 − exp −wabs t
...
90)
abs
This result is consistent with Eq
...
86) provided wabs t ≪ 1: i
...
, provided that Pi→f ≪ 1
...
91)
where the corresponding transition rate is written
wstm
i→f
π e2 ρ(ωif )
=
| i|ǫ·p exp( i k·r)|f | 2
...
92)
13
...
Thus,
exp( i k·r) = 1 + i k·r + · · · ,
(13
...
This approach is known as the electric dipole
approximation
...
(13
...
95)
so
me
f|[r, H0]|i = i me ωfi f|r|i
...
96)
¯
h
Thus, our previous expressions for the transition rates for radiation induced absorption
and stimulated emission reduce to
π
2
(13
...
98)
i→f
2 |ǫ·dif | ρ(ωif ),
ǫ0 ¯
h
f|p|i = −i
respectively
...
99)
is the effective electric dipole moment of the atom when making a transition from state i
to state f
...
97) and (13
...
Actually, we are more
interested in the transition rates induced by unpolarized isotropic radiation
...
(13
...
98) over all possible polarizations and propagation
directions of the wave
...
It follows that the vector ǫ, which specifies the direction of
wave polarization, must lie in the x-y plane, since it has to be orthogonal to k
...
100)
k = (0, 0, k),
dif = (dif sin θ, 0, dif cos θ),
ǫ = (cos φ, sin φ, 0),
(13
...
102)
which implies that
|ǫ·dif | 2 = d2 sin2 θ cos2 φ
...
103)
We must now average the above quantity over all possible values of θ and φ
...
104)
188
QUANTUM MECHANICS
where dΩ = sin θ dθ dφ, and the integral is taken over all solid angle
...
(13
...
(13
...
107)
wstm
i→f
(13
...
13
...
This process is known as absorption when the energy of the final state exceeds that
of the initial state, and stimulated emission when the energy of the final state is less than
that of the initial state
...
On the
other hand, it should be possible for such an atom to spontaneously jump into an state
with a lower energy via the emission of a photon whose energy is equal to the difference
between the energies of the initial and final states
...
It is possible to derive the rate of spontaneous emission between two atomic states
from a knowledge of the corresponding absorption and stimulated emission rates using
a famous thermodynamic argument due to Einstein
...
Let the system
have attained thermal equilibrium
...
109)
where kB is the Boltzmann constant
...
Consider two atomic states, labeled i and f, with Ei > Ef
...
This
Time-Dependent Perturbation Theory
189
means that, irrespective of any other atomic states, the rate at which atoms in the ensemble leave state i due to transitions to state f is exactly balanced by the rate at which atoms
enter state i due to transitions from state f
...
e
...
110)
where wspn is the rate of spontaneous emission (for a single atom) between states i and
i→f
f, and Ni is the number of atoms in the ensemble in state i
...
111)
f→i
where Nf is the number of atoms in the ensemble in state f
...
In thermal equilibrium, we have Wi→f = Wf→i , which gives
wspn =
i→f
Nf abs
wf→i − wstm
...
112)
According to Eqs
...
107) and (13
...
Ni
3 ǫ0 ¯
h
(13
...
This implies that
exp(−Ef /kB T )
Nf
=
= exp( ¯ ωif /kB T )
...
114)
Thus, it follows from Eq
...
109), (13
...
114) that the rate of spontaneous
emission between states i and f takes the form
wspn =
i→f
3
2
ωif dif
...
115)
Note, that, although the above result has been derived for an atom in a radiation-filled
cavity, it remains correct even in the absence of radiation
...
116)
wstm =
i→f
3 2
ωif dif
1
,
3 exp(¯ ω /k T ) − 1
3π ǫ0 ¯ c
h
h if B
(13
...
Let us estimate the typical value of the spontaneous emission rate for a hydrogen atom
...
(9
...
We also expect ωif to be of order |E0 |/¯ , where E0 is the energy of the groundh
state [see Eq
...
57)]
...
(13
...
118)
where α = e2 /(4π ǫ0 ¯ c) ≃ 1/137 is the fine-structure constant
...
e
...
13
...
This is indeed the case
...
10
Radiation from a Harmonic Oscillator
Consider an electron in a one-dimensional harmonic oscillator potential aligned along the
x-axis
...
5
...
119)
where ω0 is the frequency of the corresponding classical oscillator
...
Let the ψn (x) be the (real) properly normalized
unperturbed eigenstates of the system
...
e
...
In principle, the
electron can decay to a lower energy state via the spontaneous emission of a photon of
the appropriate frequency
...
Now, according to Eq
...
115),
the system can only make a spontaneous transition from an energy state corresponding to
the quantum number n to one corresponding to the quantum number n ′ if the associated
electric dipole moment
∞
(dx )
n,n ′
′
= n|e x|n = e
ψn (x) x ψn ′ (x) dx
(13
...
However, according to Eq
...
117),
∞
ψn x ψn ′ dx =
−∞
√
√
¯
h
n δn,n ′ +1 + n ′ δn,n ′ −1
...
121)
Since we are dealing with emission, we must have n > n ′
...
2 me ω0
(13
...
Thus, the frequency of the
photon emitted when the nth excited state decays is
ωn,n−1 =
En − En−1
= ω0
...
123)
Hence, we conclude that, no matter which state decays, the emitted photon always has the
same frequency as the classical oscillator
...
(13
...
3π ǫ0 ¯ c3
h
(13
...
6π ǫ0 me c3
(13
...
h
6π ǫ0 me c3
(13
...
127)
where E is the oscillator energy
...
The only difference
is the factor (1/2) ¯ ω0 in Eq
...
126)—this is needed to ensure that the ground-state of
h
the quantum oscillator does not radiate
...
11
Selection Rules
Let us now consider spontaneous transitions between the different energy levels of a hydrogen atom
...
77) does not contain any spin operators,
we can neglect electron spin in our analysis
...
9
...
According to Eqs
...
106) and (13
...
128)
192
QUANTUM MECHANICS
is non-zero
...
12
...
It turns out that the proof that
this matrix element is zero unless l ′ = l ± 1 can, via a trivial modification, also be used to
demonstrate that n, l, m|x|n ′ , l ′ , m ′ and n, l, m|y|n ′ , l ′ , m ′ are also zero unless l ′ = l±1
...
(13
...
h
(13
...
131)
and
n, l, m|[Lz, x− ] + ¯ x− |n ′ , l ′ , m ′ = ¯ (m − m ′ + 1) n, l, m|x− |n ′ , l ′ , m ′ = 0
...
132)
Clearly, n, l, m|x+ |n ′ , l ′ , m ′ is zero unless m ′ = m − 1, and n, l, m|x− |n ′ , l ′, m ′ is zero
unless m ′ = m + 1
...
Hence, we conclude
that n, l, m|x|n ′ , l ′ , m ′ and n, l, m|y|n ′ , l ′ , m ′ are only non-zero if m ′ = m ± 1
...
(13
...
134)
These are termed the selection rules for electric dipole transitions (i
...
, transitions calculated
using the electric dipole approximation)
...
Hence, we have the additional selection rule that ms = ms
...
12
2P → 1S Transitions in Hydrogen
Let us calculate the rate of spontaneous emission between the first excited state (i
...
,
n = 2) and the ground-state (i
...
, n ′ = 1) of a hydrogen atom
...
Hence, in order to satisfy the selection rules (13
...
134), the excited state must have the quantum numbers l = 1 and m = 0, ±1
...
Note, incidentally,
that a spontaneous transition from a 2S to a 1S state is forbidden by our selection rules
...
9
...
135)
Time-Dependent Perturbation Theory
193
where the radial functions Rn,l are given in Sect
...
4, and the spherical harmonics Yl,m are
given in Sect
...
7
...
136)
35
27
1, 0, 0|y|2, 1, ±1 = i 5 a0 ,
(13
...
138)
2 5 a0 ,
3
where a0 is the Bohr radius specified in Eq
...
58)
...
If follows from Eq
...
128) that
the modulus squared of the dipole moment for the 2P → 1S transition takes the same
value
215
(13
...
Clearly, the transition rate is independent of the quantum number m
...
Now, the energy of the eigenstate of the hydrogen atom characterized by the quantum
numbers n, l, m is E = E0 /n2 , where the ground-state energy E0 is specified in Eq
...
57)
...
140)
¯ ω = E0 /4 − E0 = − E0 = 10
...
h
4
This corresponds to a wavelength of 1
...
Finally, according to Eq
...
115), the 2P → 1S transition rate is written
1, 0, 0|x|2, 1, ±1
w2P→1S
which reduces to
= ±
ω 3 d2
=
,
3π ǫ0 ¯ c3
h
(13
...
27 × 108 s−1
(13
...
(13
...
140)
...
Hence, the mean life-time of a hydrogen 2P state is
w2P→1S =
τ2P = (w2P→1S )−1 = 1
...
(13
...
(13
...
This natural line-width is of order
∆λ ∆E2P
∼
∼ 4 × 10−8
...
145)
λ
¯ω
h
194
13
...
12
...
In fact,
these states are divided into two groups with slightly different energies
...
The remaining two states are characterized by j = 1/2, and are
thus called the 2P1/2 states
...
In fact, the energy difference is
∆E = −
α2
E0 = 4
...
16
(13
...
4 × 10−6
...
147)
Note that this splitting is much greater than the natural line-width estimated in Eq
...
145),
so there really are two spectral lines
...
It follows
that the transition rate is independent of the z-component of total angular momentum
quantum number mj = m + ms
...
Hence, we
expect the 2P3/2 → 1S and 2P1/2 → 1S transition rates to be the same
...
If these states are equally populated—which we
would certainly expect to be the case in thermal equilibrium, since they have almost the
same energies—and since they decay to the 1S state at the same rate, it stands to reason
that the spectral line associated with the 2P3/2 → 1S transition is twice as bright as that
associated with the 2P1/2 → 1S transition
...
14
Forbidden Transitions
Atomic transitions which are forbidden by the electric dipole selection rules (13
...
134) are unsurprisingly known as forbidden transitions
...
13
...
However, this matrix element is only an approximation to the true matrix element
for radiative transitions, which takes the form f|ǫ·p exp( i k·r)|i
...
(13
...
In
fact, in Sect
...
9, we calculated that the typical rate of an electric dipole transition is
wi→f ∼ α3 ωif
...
149)
Since the transition rate is proportional to the square of the radiative matrix element, it
is clear that the transition rate for a forbidden transition enabled by the residual matrix
element (13
...
Estimating r as the Bohr radius, and k as the wavenumber of a typical spectral line of
hydrogen, it is easily demonstrated that
wi→f ∼ α5 ωif
(13
...
Of course, there are some transitions (in particular, the 2S → 1S
transition) for which the true radiative matrix element f|ǫ· p exp( i k· r)|i is zero
...
Finally, it is fairly obvious that excited states which decay via forbidden transitions
have much longer life-times than those which decay via electric dipole transitions
...
196
QUANTUM MECHANICS
Variational Methods
197
14 Variational Methods
14
...
9
...
Unfortunately, it is not possible to find exact
solutions of Schr¨dinger’s equation for atoms more complicated than hydrogen, or for
o
molecules
...
Most
of the methods which have been developed for finding such solutions employ the so-called
variational principle discussed below
...
2 Variational Principle
Suppose that we wish to solve the time-independent Schr¨dinger equation
o
H ψ = E ψ,
(14
...
Let ψ be
a normalized trial solution to the above equation
...
e
...
(14
...
Let us prove the variational principle
...
e
...
(14
...
4)
Furthermore, let
so that ψ0 is the ground-state, ψ1 the first excited state, etc
...
e
...
(14
...
6)
198
QUANTUM MECHANICS
where
|cn | 2 = 1
...
7)
n
Now, the expectation value of H, calculated with ψ, takes the form
ψ|H|ψ
=
n
m
n,m
∗
cn cm Em ψn |ψm =
=
∗
cn cm ψn |H|ψm
=
cm ψm
cn ψn H
n
En |cn | 2 ,
(14
...
(14
...
5)
...
ψ|H|ψ = |c0 | 2 E0 +
(14
...
(14
...
(14
...
ψ|H|ψ = E0 +
(14
...
4)]
...
(14
...
If ψ is a normalized trial wavefunction which is orthogonal to ψ0 (i
...
, ψ|ψ0 = 0)
then, by repeating the above analysis, we can easily demonstrate that
ψ|H|ψ ≥ E1
...
13)
Thus, by varying ψ until the expectation value of H is minimized, we can obtain an approximation to the wavefunction and energy of the first excited state
...
Note,
however, that the errors are clearly cumulative in this method, so that any approximations
to highly excited states are unlikely to be very accurate
...
Variational Methods
199
14
...
Let us
attempt to calculate its ground-state energy
...
The Hamiltonian of the system thus takes
the form
¯2
h
e2
2
2
1
2
2
H=−
∇1 + ∇2 −
,
(14
...
The terms in the above expression
represent the kinetic energy of the first electron, the kinetic energy of the second electron,
the electrostatic attraction between the nucleus and the first electron, the electrostatic
attraction between the nucleus and the second electron, and the electrostatic repulsion between the two electrons, respectively
...
Indeed, if this term is neglected then we can write
(14
...
2 me 1,2 4π ǫ0 r1,2
(14
...
In this case, we would expect the wavefunction to be separable: i
...
,
ψ(r1 , r2) = ψ1 (r1 ) ψ2 (r2 )
...
17)
Hψ = Eψ
(14
...
19)
E = E1 + E2
...
20)
Hence, Schr¨dinger’s equation
o
reduces to
where
Of course, Eq
...
19) is the Schr¨dinger equation of a hydrogen atom whose nuclear
o
charge is +2 e, instead of +e
...
9
...
21)
ψ2 (r2 ) = ψ0 (r2 ),
(14
...
a0
(14
...
(9
...
Note that ψ0 is properly normalized
...
24)
where E0 = −13
...
(9
...
Thus, our
crude estimate for the ground-state energy of helium becomes
E = 4 E0 + 4 E0 = 8 E0 = −108
...
(14
...
98 eV
...
Fortunately,
however, we can use the variational principle to estimate this contribution
...
Thus,
ψ(r1 , r2) = ψ0 (r1 ) ψ0 (r2 ) =
8
2 [r1 + r2 ]
exp −
...
26)
The expectation value of the Hamiltonian (14
...
27)
where
Vee = ψ
e2
e2
ψ =
4π ǫ0 |r2 − r1 |
4π ǫ0
|ψ(r1 , r2)| 2 3
d r1 d3 r2
...
28)
The variation principle only guarantees that (14
...
In reality, we hope that it will give a reasonably accurate estimate of this
energy
...
(9
...
26) and (14
...
29)
where ^1,2 = 2 r1,2 /a0
...
30)
Vee = − 2
2
2
π
r + r − 2 r1 r2 cos θ
1
2
where θ is the angle subtended between vectors r1 and r2
...
31)
Variational Methods
201
where
e−2 r1
I(r2 ) =
2
2
r1 + r2 − 2 r1 r2 cos θ
d3 r1
...
32)
Our first task is to evaluate the function I(r2 )
...
It follows that
θ = θ1
...
33)
which trivially reduces to
π
∞
I(r2 ) = 2π
0
0
e−2 r1
2
r1
+
2
r2
− 2 r1 r2 cos θ1
2
r1 dr1 sin θ1 dθ1
...
34)
Making the substitution µ = cos θ1 , we can see that
π
0
1
1
2
r1
+
2
r2
dµ
sin θ1 dθ1 =
− 2 r1 r2 cos θ1
−1
+
...
35)
for r1 > r2
,
for r1 < r2
2
r1
(14
...
+
(14
...
r2
(14
...
39)
(14
...
31)
reduces to
16 E0 ∞ −2 r2
2
Vee = −
e
I(r2 ) r2 dr2 ,
(14
...
2
(14
...
27), our estimate for the ground-state energy of helium is
H = 8 E0 −
11
5
E0 =
E0 = −74
...
2
2
(14
...
We can actually refine our estimate further
...
26) essentially
treats the two electrons as non-interacting particles
...
Hence, a better
trial wavefunction might be
ψ(r1 , r2) =
Z3
Z [r1 + r2 ]
exp −
,
3
π a0
a0
(14
...
Let us recalculate
the ground-state energy of helium as a function of Z, using the above trial wavefunction,
and then minimize the result with respect to Z
...
We can rewrite the expression (14
...
45)
where
¯2
h
Z e2
2
H1,2 (Z) = −
∇ −
2 me 1,2 4π ǫ0 r1,2
is the Hamiltonian of a hydrogen atom with nuclear charge +Z e,
Vee =
e2
1
4π ǫ0 |r2 − r1 |
(14
...
47)
is the electron-electron repulsion term, and
U(Z) =
e2
4π ǫ0
[Z − 2] [Z − 2]
+
...
48)
It follows that
H (Z) = 2 E0 (Z) + Vee (Z) + U (Z),
(14
...
44) [actually, all we need to do is to make the
substitution a0 → (2/Z) a0 ], and
U (Z) = 2 (Z − 2)
e2
4π ǫ0
1
...
50)
Here, 1/r is the expectation value of 1/r calculated for a hydrogen atom with nuclear
charge +Z e
...
(9
...
(14
...
52)
since E0 = −e2 /(8π ǫ0 a0 )
...
4
4
(14
...
= −4 Z +
dZ
4
(14
...
69
...
55)
16
The fact that Z < 2 confirms our earlier conjecture that the electrons partially shield the
nuclear charge from one another
...
5 eV
...
56)
H (1
...
43) [recall that the correct
result is −78
...
Obviously, we could get even closer to the correct value of the helium ground-state
energy by using a more complicated trial wavefunction with more adjustable parameters
...
6)
...
Our spatial wavefunction (14
...
This means that the spinor must be anti-symmetric
...
1: The hydrogen molecule ion
...
11
...
e
...
Hence, the
ground-state of helium has overall electron spin zero
...
4 Hydrogen Molecule Ion
The hydrogen molecule ion consists of an electron orbiting about two protons, and is the
simplest imaginable molecule
...
e
...
According to the variation principle, we can deduce
that the H+ ion has a bound state if we can find any trial wavefunction for which the total
2
Hamiltonian of the system has an expectation value less than that of a hydrogen atom and
a free proton
...
In fact, let them lie on the
z-axis, with the first at the origin, and the second at z = R (see Fig
...
1)
...
This is reasonable, since the electron
moves far more rapidly than the protons
...
57)
Variational Methods
205
as our trial wavefunction, where
ψ0 (r) = √
1
3/2
π a0
e−r/a0
(14
...
14
...
Obviously, this is a very simplistic wavefunction, since it is just a linear combination of
hydrogen ground-state wavefunctions centered on each proton
...
Our first task is to normalize our trial wavefunction
...
(14
...
57), A = I−1/2 , where
I=
|ψ0 (r1 )|2 + |ψ0 (r2 )|2 ± 2 ψ0 (r1 ) ψ(r2) d3 r
...
60)
It follows that
I = 2 (1 ± J),
with
J=
ψ0 (r1 ) ψ0 (r2 ) d3 r
...
61)
(14
...
Now, it is easily seen
that r1 = r and r2 = (r2 + R2 − 2 r R cos θ)1/2
...
63)
0
where X = R/a0
...
Let y = (x2 +
X2 − 2 x X cos θ)1/2
...
64)
|x−X|
1
e−(x+X) (1 + x + X) − e−|x−X| (1 + |x − X|)
...
65)
206
QUANTUM MECHANICS
which evaluates to
J = e−X 1 + X +
X3
...
66)
Now, the Hamiltonian of the electron is written
H=−
¯2
h
e2
∇2 −
2 me
4π ǫ0
1
1
+
...
67)
Note, however, that
−
¯2
h
e2
ψ0 (r1,2 ) = E0 ψ0 (r1,2 ),
∇2 −
2 me
4π ǫ0 r1,2
(14
...
It follows that
H ψ± = A −
¯2
h
e2
∇2 −
2 me
4π ǫ0
e2
4π ǫ0
= E0 ψ − A
1
1
+
r1 r2
[ψ0 (r1 ) ± ψ0 (r2 )]
ψ0 (r1 ) ψ0 (r2 )
...
69)
Hence,
H = E0 + 4 A2 (D ± E) E0 ,
(14
...
r1
D =
ψ0 (r1 )
E =
(14
...
72)
Now,
∞
π
0
0
D=2
e−2 x
x2 dx sin θ dθ,
(x2 + X2 − 2 x X cos θ)1/2
which reduces to
4
D=
X
X
∞
e
0
(14
...
74)
X
giving
1
1 − [1 + X] e−2 X
...
75)
exp −x − (x2 + X2 − 2 x X cos θ)1/2 x dx sin θ dθ,
(14
...
77)
X
yielding
E = (1 + X) e−X
...
78)
Our expression for the expectation value of the electron Hamiltonian is
H = 1+2
(D ± E)
E0 ,
(1 ± J)
(14
...
(14
...
75), and
(14
...
In order to obtain the total energy of the molecule, we must add to
this the potential energy of the two protons
...
Hence, we can write
Etotal = H +
Etotal = −F± (R/a0 ) E0 ,
(14
...
81)
where E0 is the hydrogen ground-state energy, and
F± (X) = −1 +
2 (1 + X) e−2 X ± (1 − 2 X2 /3) e−X
...
82)
The functions F+ (X) and F− (X) are both plotted in Fig
...
2
...
e
...
It follows from Eq
...
81) that a bound state corresponds to
F± < −1
...
(14
...
This is hardly surprising, since the even
wavefunction maximizes the electron probability density between the two protons, thereby
reducing their mutual electrostatic repulsion
...
The binding energy of the H+ ion is defined as the difference
2
between its energy and that of a hydrogen atom and a free proton: i
...
,
Ebind = Etotal − E0 = −(F+ + 1) E0
...
83)
According to the variational principle, the binding energy is less than or equal to the
minimum binding energy which can be inferred from Fig
...
2
...
5 and F+ ≃ −1
...
Thus, our estimates for the separation between the
two protons, and the binding energy, for the H+ ion are R = 2
...
33 × 10−10 m
2
and Ebind = 0
...
77 eV, respectively
...
06 × 10−10 m, and Ebind = −2
...
Clearly, our estimates are not
particularly accurate
...
2
208
QUANTUM MECHANICS
Figure 14
...
Scattering Theory
209
15 Scattering Theory
15
...
Firstly, from the study of spectroscopic lines, and, secondly, from scattering experiments
...
Let us now examine the quantum theory of scattering
...
2 Fundamentals
Consider time-independent, energy conserving scattering in which the Hamiltonian of the
system is written
H = H0 + V(r),
(15
...
2)
H0 =
2m
2m
is the Hamiltonian of a free particle of mass m, and V(r) the scattering potential
...
Let
√
(15
...
Of
h
course,
H0 ψ0 = E ψ0 ,
(15
...
Schr¨dinger’s equation for the scattering
h
o
problem is
(H0 + V) ψ = E ψ,
(15
...
The above equation can be rearranged to give
(∇2 + k2 ) ψ =
2m
V ψ
...
6)
Now,
(∇2 + k2 ) u(r) = ρ(r)
is known as the Helmholtz equation
...
4π |r − r ′ |
(15
...
11
...
(15
...
Hence, Eq
...
6) can be inverted, subject
to the boundary condition ψ → ψ0 as V → 0, to give
e i k |r−r |
V(r ′ ) ψ(r ′) d3 r ′
...
9)
Let us calculate the value of the wavefunction ψ(r) well outside the scattering region
...
10)
to first-order in r ′ /r, where ^/r is a unit vector which points from the scattering region to
r
the observation point
...
This is the wavevector for particles
r
with the same energy as the incoming particles (i
...
, k ′ = k) which propagate from the
scattering region to the observation point
...
9) reduces to
ψ(r) ≃
where
f(k, k ′) = −
√
n e i k·r +
m
√
2π n ¯ 2
h
eikr
f(k, k ′) ,
r
′
′
e−i k ·r V(r ′ ) ψ(r ′) d3 r ′
...
11)
(15
...
(15
...
The differential scattering cross-section dσ/dΩ is defined as the number of particles per
unit time scattered into an element of solid angle dΩ, divided by the incident particle flux
...
7
...
e
...
(15
...
14)
where v = ¯ k/m is the velocity of the incident particles
...
15)
where v ′ = ¯ k ′ /m is the velocity of the scattered particles
...
16)
dσ
= |f(k, k ′)|2
...
17)
which yields
Scattering Theory
211
Thus, |f(k, k ′ )|2 gives the differential cross-section for particles with incident velocity v =
¯ k/m to be scattered such that their final velocities are directed into a range of solid
h
angles dΩ about v ′ = ¯ k ′ /m
...
15
...
17) is not particularly useful, as it stands, because the quantity f(k, k ′) depends on the, as yet, unknown wavefunction ψ(r) [see Eq
...
12)]
...
In this case, it is reasonable to suppose that
the total wavefunction, ψ(r), does not differ substantially from the incident wavefunction, ψ0 (r)
...
(15
...
This procedure is called the Born approximation
...
′
′
(15
...
For a spherically symmetric potential,
f(k ′ , k) ≃ −
m
2π ¯ 2
h
exp( i q r ′ cos θ ′ ) V(r ′ ) r ′ 2 dr ′ sin θ ′ dθ ′ dφ ′ ,
(15
...
(15
...
It is easily
demonstrated that
q ≡ |k − k ′ | = 2 k sin(θ/2),
(15
...
In other words, θ is the
scattering angle
...
Consider scattering by a Yukawa potential
V(r) =
V0 exp(−µ r)
,
µr
(15
...
It follows from
Eq
...
20) that
2 m V0
1
f(θ) = − 2
,
(15
...
(15
...
25)
2
dΩ
[2 k2 (1 − cos θ) + µ2] 2
¯ µ
h
given that
q2 = 4 k2 sin2 (θ/2) = 2 k2 (1 − cos θ)
...
26)
exp(−µ r ′ ) sin(q r ′ ) dr ′ =
The Yukawa potential reduces to the familiar Coulomb potential as µ → 0, provided
that V0 /µ → Z Z ′ e2 /4π ǫ0
...
16 k4 sin4 (θ/2)
(15
...
28)
where E = p2 /2 m is the kinetic energy of the incident particles
...
(15
...
The Born approximation is valid provided that ψ(r) is not too different from ψ0 (r) in
the scattering region
...
(15
...
29)
V(r ′ ) d3 r ′ ≪ 1
...
At low energies, (i
...
, k ≪ µ) we can
replace exp( i k r ′) by unity, giving
2 m |V0 |
≪1
(15
...
The condition for the Yukawa
potential to develop a bound state is
2 m |V0 |
≥ 2
...
31)
where V0 is negative
...
In the high-k limit, Eq
...
29) yields
2 m |V0 |
≪ 1
...
32)
This inequality becomes progressively easier to satisfy as k increases, implying that the
Born approximation is more accurate at high incident particle energies
...
4 Partial Waves
We can assume, without loss of generality, that the incident wavefunction is characterized
by a wavevector k which is aligned parallel to the z-axis
...
The direction of k ′ is specified by the polar angle θ (i
...
,
the angle subtended between the two wavevectors), and an azimuthal angle φ about the
z-axis
...
20) and (15
...
e
...
e
...
33)
f(θ, φ) = f(θ)
...
34)
nor the large-r form of the total wavefunction,
ψ(r) =
√
n exp( i k r cos θ) +
exp( i k r) f(θ)
,
r
(15
...
Outside the range of the scattering potential, both ψ0 (r) and ψ(r) satisfy the free space
Schr¨dinger equation
o
(∇2 + k2 ) ψ = 0
...
36)
What is the most general solution to this equation in spherical polar coordinates which
does not depend on the azimuthal angle φ? Separation of variables yields
ψ(r, θ) =
Rl (r) Pl(cos θ),
(15
...
The Legendre functions are related to the spherical harmonics, introduced in Cha
...
2l+1
Pl (cos θ) =
(15
...
36) and (15
...
(15
...
9
...
Recall that
r2
jl (z) = zl −
yl (z) = −z
l
1 d
z dz
1 d
−
z dz
l
sin z
,
z
l
cos z
...
40)
(15
...
The asymptotic behaviour of these functions in the limit z → ∞ is
sin(z − l π/2)
,
z
cos(z − l π/2)
yl (z) → −
...
42)
(15
...
44)
l
where the al are constants
...
The Legendre functions are orthonormal,
1
Pn (µ) Pm (µ) dµ =
−1
δnm
,
n + 1/2
(15
...
al jl (k r) = (l + 1/2)
(15
...
47)
−1
where l = 0, 1, 2, · · · [see M
...
A
...
10
...
14]
...
48)
giving
ψ0 (r) =
√
n exp( i k r cos θ) =
√
i l (2 l + 1) jl(k r) Pl (cos θ)
...
49)
l
The above expression tells us how to decompose the incident plane-wave into a series of
spherical waves
...
The most general expression for the total wavefunction outside the scattering region is
√
ψ(r) = n
[Al jl (k r) + Bl yl (k r)] Pl (cos θ),
(15
...
Note that the yl (k r) functions are allowed to appear
in this expansion, because its region of validity does not include the origin
...
51)
Scattering Theory
215
where use has been made of Eqs
...
42) and (15
...
The above expression can also be
written
√
sin(k r − l π/2 + δl )
ψ(r) ≃ n
Pl (cos θ),
(15
...
Note that Al = Cl cos δl and Bl = −Cl sin δl
...
52) yields
ψ(r) ≃
√
n
e i (k r−l π/2+δl ) − e−i (k r−l π/2+δl )
Pl (cos θ),
2ikr
Cl
l
(15
...
What is the source of the
incoming waves? Obviously, they must be part of the large-r asymptotic expansion of the
incident wavefunction
...
(15
...
49) that
ψ0 (r) ≃
√
i l (2l + 1)
n
l
e i (k r−l π/2) − e−i (k r−l π/2)
Pl (cos θ)
2ikr
(15
...
Now, Eqs
...
34) and (15
...
n
r
(15
...
This implies that
the coefficients of the incoming spherical waves in the large-r expansions of ψ(r) and ψ0 (r)
must be the same
...
(15
...
54) that
Cl = (2 l + 1) exp[ i (δl + l π/2)]
...
56)
Thus, Eqs
...
53)–(15
...
k
(15
...
e
...
Now, the differential scattering cross-section dσ/dΩ is simply the modulus squared of
the scattering amplitude f(θ) [see Eq
...
17)]
...
58)
216
QUANTUM MECHANICS
where µ = cos θ
...
59)
l
where use has been made of Eq
...
45)
...
5 Determination of Phase-Shifts
Let us now consider how the phase-shifts δl in Eq
...
57) can be evaluated
...
In the region r > a, the wavefunction ψ(r) satisfies the free-space
Schr¨dinger equation (15
...
The most general solution which is consistent with no
o
incoming spherical-waves is
ψ(r) =
√
∞
n
l=0
il (2 l + 1) Rl (r) Pl (cos θ),
(15
...
(15
...
The logarithmic derivative of
the lth radial wavefunction, Rl (r), just outside the range of the potential is given by
βl+ = k a
cos δl jl′ (k a) − sin δl yl′ (k a)
,
cos δl jl (k a) − sin δl yl (k a)
(15
...
The above equation can be inverted to give
tan δl =
k a jl′ (k a) − βl+ jl (k a)
...
63)
Thus, the problem of determining the phase-shift δl is equivalent to that of obtaining βl+
...
2
dr
r2
¯
h
(15
...
65)
(15
...
67)
ensures that the radial wavefunction is well-behaved at the origin
...
(15
...
(15
...
(15
...
15
...
Consider scattering by a
hard sphere, for which the potential is infinite for r < a, and zero for r > a
...
Thus,
for all l
...
63) thus gives
βl− = βl+ = ∞,
tan δl =
jl (k a)
...
70)
(15
...
Equation (15
...
72)
− cos(k a)/ka
where use has been made of Eqs
...
40) and (15
...
It follows that
δ0 = −k a
...
73)
The S-wave radial wave function is [see Eq
...
61)]
[cos(k a) sin(k r) − sin(k a) cos(k r)]
kr
sin[k (r − a)]
= exp(−i k a)
...
74)
The corresponding radial wavefunction for the incident wave takes the form [see Eq
...
49)]
sin(k r)
˜
...
75)
218
QUANTUM MECHANICS
Thus, the actual l = 0 radial wavefunction is similar to the incident l = 0 wavefunction,
except that it is phase-shifted by k a
...
Low energy implies
that k a ≪ 1
...
76)
(15
...
It follows that
tan δl =
−(k a)2 l+1
...
78)
It is clear that we can neglect δl , with l > 0, with respect to δ0
...
e
...
It
follows from Eqs
...
17), (15
...
73) that
sin2 k a
dσ
=
≃ a2
2
dΩ
k
(15
...
Note that the total cross-section
σtotal =
dσ
dΩ = 4π a2
dΩ
(15
...
e
...
However, low energy scattering implies relatively
long wavelengths, so we would not expect to obtain the classical result in this limit
...
At high energies, all partial waves up to lmax =
k a contribute significantly to the scattering cross-section
...
(15
...
(15
...
Thus,
ka
2π
(2 l + 1) ≃ 2π a2
...
82)
σtotal ≃
k2
l=0
This is twice the classical result, which is somewhat surprizing, since we might expect
to obtain the classical result in the short wavelength limit
...
However, in order to
produce a “shadow” behind the sphere, there must also be some scattering in the forward
Scattering Theory
219
direction in order to produce destructive interference with the incident plane-wave
...
The effective cross-section associated
with this bright spot is π a2 which, when combined with the cross-section for classical
reflection, π a2, gives the actual cross-section of 2π a2
...
7 Low Energy Scattering
In general, at low energies (i
...
, when 1/k is much larger than the range of the potential)
partial waves with l > 0 make a negligible contribution to the scattering cross-section
...
As a specific example, let us consider scattering by a finite potential well, characterized
by V = V0 for r < a, and V = 0 for r ≥ a
...
The potential is
repulsive for V0 > 0, and attractive for V0 < 0
...
(15
...
83)
where use has been made of Eqs
...
40) and (15
...
The inside wavefunction follows
from Eq
...
66)
...
84)
R0 (r) = B
r
where use has been made of the boundary condition (15
...
Here, B is a constant, and
¯ 2 k′ 2
h
E − V0 =
...
85)
Note that Eq
...
84) only applies when E > V0
...
86)
where
¯ 2 κ2
h
...
87)
tan(k a + δ0 ) =
k
tan(k ′ a)
k′
(15
...
89)
for E > V0 , and
220
QUANTUM MECHANICS
for E < V0
...
Suppose that |V0 | ≫ E (i
...
, the
depth of the potential well is much larger than the energy of the incident particles), so
that k ′ ≫ k
...
(15
...
k′
(15
...
δ0 ≃ k a
k′ a
(15
...
(15
...
(15
...
93)
so for sufficiently small values of k a,
k′ a ≃
2 m |V0 | a2
...
94)
It follows that the total (S-wave) scattering cross-section is independent of the energy of
the incident particles (provided that this energy is sufficiently small)
...
g
...
49) at which δ0 → π, and the scattering
cross-section (15
...
In reality,
the cross-section is not exactly zero, because of contributions from l > 0 partial waves
...
It follows that there are certain
values of V0 and k which give rise to almost perfect transmission of the incident wave
...
15
...
Suppose that the quantity 2 m |V0 | a2 /¯ 2 is slightly less than π/2
...
(15
...
In
this case, tan(k ′ a) becomes infinite, so we can no longer assume that the right-hand side
of Eq
...
88) is small
...
(15
...
This implies that
σtotal =
1
4π
sin2 δ0 = 4π a2
...
95)
Note that the cross-section now depends on the energy
...
(15
...
The origin of this rather strange behaviour is quite simple
...
96)
is equivalent to the condition that a spherical well of depth V0 possesses a bound state at
zero energy
...
In this situation,
an incident particle would like to form a bound state in the potential well
...
Nevertheless, this
sort of resonance scattering is best understood as the capture of an incident particle to form
a metastable bound state, and the subsequent decay of the bound state and release of the
particle
...
We have seen that there is a resonant effect when the phase-shift of the S-wave takes
the value π/2
...
Suppose that δl attains the value π/2 at the incident energy E0 , so that
δl (E0 ) =
π
...
97)
Let us expand cot δl in the vicinity of the resonant energy:
cot δl (E) = cot δl (E0 ) +
= −
1 dδl
sin2 δl dE
d cot δl
dE
E=E0
E=E0
(E − E0 ) + · · ·
(E − E0 ) + · · ·
...
98)
Defining
dδl (E)
dE
we obtain
=
E=E0
2
,
Γ
2
cot δl (E) = − (E − E0 ) + · · ·
...
99)
(15
...
(15
...
(15
...
k2
(E − E0 )2 + Γ 2 /4
(15
...
The variation of the partial cross-section σl with
the incident energy has the form of a classical resonance curve
...
We can interpret the Breit-Wigner formula as describing the
absorption of an incident particle to form a metastable state, of energy E0 , and lifetime
τ = ¯ /Γ
Title: Quantum Physics For All
Description: A Guide For All People Which Includes All Subjects Of Quantum Physics Simplified For Them In A Nice Presentable Way .
Description: A Guide For All People Which Includes All Subjects Of Quantum Physics Simplified For Them In A Nice Presentable Way .