Solutions to Problems in Sakurai s Quantum Mechanics P. Saltsidis B. Brinne

Prévia do material em texto

Solutions to Problems
in
Quantum Mechanics
P. Saltsidis, additions by B. Brinne
1995,1999
0
Most of the problems presented here are taken from the book
Sakurai, J. J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley,
1985.
Contents
I Problems 3
1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . 5
2 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Theory of Angular Momentum . . . . . . . . . . . . . . . . . . 14
4 Symmetry in Quantum Mechanics . . . . . . . . . . . . . . . . 17
5 Approximation Methods . . . . . . . . . . . . . . . . . . . . . 19
II Solutions 23
1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . 25
2 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 36
3 Theory of Angular Momentum . . . . . . . . . . . . . . . . . . 75
4 Symmetry in Quantum Mechanics . . . . . . . . . . . . . . . . 94
5 Approximation Methods . . . . . . . . . . . . . . . . . . . . . 103
1
2 CONTENTS
Part I
Problems
3
1. FUNDAMENTAL CONCEPTS 5
1 Fundamental Concepts
1.1 Consider a ket space spanned by the eigenkets fja
0
ig of a Her-
mitian operator A. There is no degeneracy.
(a) Prove that
Y
a
0
(A� a
0
)
is a null operator.
(b) What is the signi�cance of
Y
a
00
6=a
0
(A� a
00
)
a
0
� a
00
?
(c) Illustrate (a) and (b) using A set equal to S
z
of a spin
1
2
system.
1.2 A spin
1
2
system is known to be in an eigenstate of
~
S � n^ with
eigenvalue �h=2, where n^ is a unit vector lying in the xz-plane that
makes an angle 
 with the positive z-axis.
(a) Suppose S
x
is measured. What is the probability of getting
+�h=2?
(b) Evaluate the dispersion in S
x
, that is,
h(S
x
� hS
x
i)
2
i:
(For your own peace of mind check your answers for the special
cases 
 = 0, �=2, and �.)
1.3 (a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
(h�j + �
�
h�j) � (j�i + �j�i) � 0
for any complex number �; then choose � in such a way that the
preceding inequality reduces to the Schwarz inequility.
6
(b) Show that the equility sign in the generalized uncertainty re-
lation holds if the state in question satis�es
�Aj�i = ��Bj�i
with � purely imaginary.
(c) Explicit calculations using the usual rules of wave mechanics
show that the wave function for a Gaussian wave packet given by
hx
0
j�i = (2�d
2
)
�1=4
exp
"
ihpix
0
�h
�
(x
0
� hxi)
2
4d
2
#
satis�es the uncertainty relation
q
h(�x)
2
i
q
h(�p)
2
i =
�h
2
:
Prove that the requirement
hx
0
j�xj�i = (imaginary number)hx
0
j�pj�i
is indeed satis�ed for such a Gaussian wave packet, in agreement
with (b).
1.4 (a) Let x and p
x
be the coordinate and linear momentum in
one dimension. Evaluate the classical Poisson bracket
[x; F (p
x
)]
classical
:
(b) Let x and p
x
be the corresponding quantum-mechanical opera-
tors this time. Evaluate the commutator
�
x; exp
�
ip
x
a
�h
��
:
(c) Using the result obtained in (b), prove that
exp
�
ip
x
a
�h
�
jx
0
i; (xjx
0
i = x
0
jx
0
i)
2. QUANTUM DYNAMICS 7
is an eigenstate of the coordinate operator x. What is the corre-
sponding eigenvalue?
1.5 (a) Prove the following:
(i) hp
0
jxj�i = i�h
@
@p
0
hp
0
j�i;
(ii) h�jxj�i =
Z
dp
0
�
�
�
(p
0
)i�h
@
@p
0
�
�
(p
0
);
where �
�
(p
0
) = hp
0
j�i and �
�
(p
0
) = hp
0
j�i are momentum-space wave
functions.
(b) What is the physical signi�cance of
exp
�
ix�
�h
�
;
where x is the position operator and � is some number with the
dimension of momentum? Justify your answer.
2 Quantum Dynamics
2.1 Consider the spin-procession problem discussed in section 2.1
in Jackson. It can also be solved in the Heisenberg picture. Using
the Hamiltonian
H = �
�
eB
mc
�
S
z
= !S
z
;
write the Heisenberg equations of motion for the time-dependent
operators S
x
(t), S
y
(t), and S
z
(t). Solve them to obtain S
x;y;z
as func-
tions of time.
2.2 Let x(t) be the coordinate operator for a free particle in one
dimension in the Heisenberg picture. Evaluate
[x(t); x(0)] :
8
2.3 Consider a particle in three dimensions whose Hamiltonian is
given by
H =
~p
2
2m
+ V (~x):
By calculating [~x � ~p;H] obtain
d
dt
h~x � ~pi =
*
p
2
m
+
� h~x �
~
rV i:
To identify the preceding relation with the quantum-mechanical
analogue of the virial theorem it is essential that the left-hand side
vanish. Under what condition would this happen?
2.4 (a) Write down the wave function (in coordinate space) for the
state
exp
�
�ipa
�h
�
j0i:
You may use
hx
0
j0i = �
�1=4
x
�1=2
0
exp
2
4
�
1
2
 
x
0
x
0
!
2
3
5
;
0
@
x
0
�
 
�h
m!
!
1=2
1
A
:
(b) Obtain a simple expression that the probability that the state
is found in the ground state at t = 0. Does this probability change
for t > 0?
2.5 Consider a function, known as the correlation function, de�ned
by
C(t) = hx(t)x(0)i;
where x(t) is the position operator in the Heisenberg picture. Eval-
uate the correlation function explicitly for the ground state of a
one-dimensional simple harmonic oscillator.
2. QUANTUM DYNAMICS 9
2.6 Consider again a one-dimensional simple harmonic oscillator.
Do the following algebraically, that is, without using wave func-
tions.
(a) Construct a linear combination of j0i and j1i such that hxi is as
large as possible.
(b) Suppose the oscillator is in the state constructed in (a) at t = 0.
What is the state vector for t > 0 in the Schrodinger picture?
Evaluate the expectation value hxi as a function of time for t > 0
using (i) the Schrodinger picture and (ii) the Heisenberg picture.
(c) Evaluate h(�x)
2
i as a function of time using either picture.
2.7 A coherent state of a one-dimensional simple harmonic oscil-
lator is de�ned to be an eigenstate of the (non-Hermitian) annihi-
lation operator a:
aj�i = �j�i;
where � is, in general, a complex number.
(a) Prove that
j�i = e
�j�j
2
=2
e
�a
y
j0i
is a normalized coherent state.
(b) Prove the minimum uncertainty relation for such a state.
(c) Write j�i as
j�i =
1
X
n=0
f(n)jni:
Show that the distribution of jf(n)j
2
with respect to n is of the
Poisson form. Find the most probable value of n, hence of E.
(d) Show that a coherent state can also be obtained by applying
the translation (�nite-displacement) operator e
�ipl=�h
(where p is the
momentum operator, and l is the displacement distance) to the
ground state.
10
(e) Show that the coherent state j�i remains coherent under time-
evolution and calculate the time-evolved state j�(t)i. (Hint: di-
rectly apply the time-evolution operator.)
2.8 The quntum mechanical propagator, for a particle with mass
m, moving in a potential is given by:
K(x; y;E) =
Z
1
0
dte
iEt=�h
K(x; y; t; 0) = A
X
n
sin(nrx) sin(nry)
E �
�h
2
r
2
2m
n
2
where A is a constant.
(a) What is the potential?
(b) Determine the constant A in terms of the parameters describing
the system (such as m, r etc. ).
2.9 Prove the relation
d�(x)
dx
= �(x)
where �(x) is the (unit) step function,and �(x) the Dirac delta
function. (Hint: study the e�ect on testfunctions.)
2.10 Derive the following expression
S
cl
=
m!
2 sin(!T )
h
(x
2
0
+ 2x
2
T
) cos(!T )� x
0
x
T
i
for the classical action for a harmonic oscillator moving from the
point x
0
at t = 0 to the point x
T
at t = T .
2.11 The Lagrangian of the single harmonic oscillator is
L =
1
2
m _x
2
�
1
2
m!
2
x
2
2. QUANTUM DYNAMICS 11
(a) Show that
hx
b
t
b
jx
a
t
a
i = exp
�
iS
cl
�h
�
G(0; t
b
; 0; t
a
)
where S
cl
is the action along the classical path x
cl
from (x
a
; t
a
) to
(x
b
; t
b
) and G is
G(0; t
b
; 0; t
a
) =
lim
N!1
Z
dy
1
: : : dy
N
�
m
2�i�h"
�
(N+1)
2
exp
8
<
:
i
�h
N
X
j=0
�
m
2"
(y
j+1
� y
j
)
2
�
1
2
"m!
2
y
2
j
�
9
=
;
where " =
t
b
�t
a
(N+1)
.
(Hint: Let y(t) = x(t) � x
cl
(t) be the new integration variable,
x
cl
(t) being the solution of the Euler-Lagrange equation.)
(b) Show that G can be written as
G = lim
N!1
�
m
2�i�h"
�
(N+1)
2
Z
dy
1
: : : dy
N
exp(�n
T
�n)
where n =
2
6
6
4
y
1
.
.
.
y
N
3
7
7
5
and n
T
is its transpose. Write the symmetric
matrix �.
(c) Show that
Z
dy
1
: : : dy
N
exp(�n
T
�n) �
Z
d
N
ye
�n
T
�n
=
�
N=2
p
det�
[Hint: Diagonalize � by an orhogonal matrix.]
(d) Let
�
2i�h"
m
�
N
det� � det�
0
N
� p
N
. De�ne j � j matrices �
0
j
that con-
sist of the �rst j rows and j columns of �
0
N
and whose determinants
are p
j
. By expanding �
0
j+1
in minors show the following recursion
formula for the p
j
:
p
j+1
= (2 � "
2
!
2
)p
j
� p
j�1
j = 1; : : : ; N (2.1)
12
(e) Let �(t) = "p
j
for t = t
a
+ j" and show that (2.1) implies that in
the limit "! 0; �(t) satis�es the equation
d
2
�
dt
2
= �!
2
�(t)
with initial conditions �(t = t
a
) = 0;
d�(t=t
a
)
dt
= 1.
(f) Show that
hx
b
t
b
jx
a
t
a
i =
s
m!
2�i�h sin(!T )
exp
(
im!
2�h sin(!T )
[(x
2
b
+ x
2
a
) cos(!T )� 2x
a
x
b
]
)
where T = t
b
� t
a
.
2.12 Show the composition property
Z
dx
1
K
f
(x
2
; t
2
;x
1
; t
1
)K
f
(x
1
; t
1
;x
0
; t
0
) = K
f
(x
2
; t
2
;x
0
; t
0
)
where K
f
(x
1
; t
1
;x
0
; t
0
) is the free propagator (Sakurai 2.5.16), by
explicitly performing the integral (i.e. do not use completeness).
2.13 (a) Verify the relation
[�
i
;�
j
] =
 
i�he
c
!
"
ijk
B
k
where
~
� � m
~x
dt
= ~p�
e
~
A
c
and the relation
m
d
2
~x
dt
2
=
d
~
�
dt
= e
"
~
E +
1
2c
 
d~x
dt
�
~
B �
~
B �
d~x
dt
!#
:
(b) Verify the continuity equation
@�
@t
+
~
r
0
�
~
j = 0
2. QUANTUM DYNAMICS 13
with
~
j given by
~
j =
 
�h
m
!
=( 
�
~
r
0
 )�
�
e
mc
�
~
Aj j
2
:
2.14 An electron moves in the presence of a uniform magnetic �eld
in the z-direction (
~
B = Bz^).
(a) Evaluate
[�
x
;�
y
];
where
�
x
� p
x
�
eA
x
c
; �
y
� p
y
�
eA
y
c
:
(b) By comparing the Hamiltonian and the commutation relation
obtained in (a) with those of the one-dimensional oscillator problem
show how we can immediately write the energy eigenvalues as
E
k;n
=
�h
2
k
2
2m
+
 
jeBj�h
mc
!
�
n+
1
2
�
;
where �hk is the continuous eigenvalue of the p
z
operator and n is a
nonnegative integer including zero.
2.15 Consider a particle of mass m and charge q in an impenetrable
cylinder with radius R and height a. Along the axis of the cylin-
der runs a thin, impenetrable solenoid carrying a magnetic 
ux �.
Calculate the ground state energy and wavefunction.
2.16 A particle in one dimension (�1 < x < 1) is subjected to a
constant force derivable from
V = �x; (� > 0):
14
(a) Is the energy spectrum continuous or discrete? Write down an
approximate expression for the energy eigenfunction speci�ed by
E.
(b) Discuss brie
y what changes are needed if V is replaced be
V = �jxj:
3 Theory of Angular Momentum
3.1 Consider a sequence of Euler rotations represented by
D
(1=2)
(�; �; 
) = exp
�
�i�
3
�
2
�
exp
 
�i�
2
�
2
!
exp
�
�i�
3
2
�
=
 
e
�i(�+
)=2
cos
�
2
�e
�i(��
)=2
sin
�
2
e
i(��
)=2
sin
�
2
e
i(�+
)=2
cos
�
2
!
:
Because of the group properties of rotations, we expect that this
sequence of operations is equivalent to a single rotation about some
axis by an angle �. Find �.
3.2 An angular-momentum eigenstate jj;m = m
max
= ji is rotated
by an in�nitesimal angle " about the y-axis. Without using the
explicit form of the d
(j)
m
0
m
function, obtain an expression for the
probability for the new rotated state to be found in the original
state up to terms of order "
2
.
3.3 The wave function of a patricle subjected to a spherically
symmetrical potential V (r) is given by
 (~x) = (x+ y + 3z)f(r):
3. THEORY OF ANGULAR MOMENTUM 15
(a) Is an eigenfunction of
~
L? If so, what is the l-value? If
not, what are the possible values of l we may obtain when
~
L
2
is
measured?
(b)What are the probabilities for the particle to be found in various
m
l
states?
(c) Suppose it is known somehow that (~x) is an energy eigenfunc-
tion with eigenvalue E. Indicate how we may �nd V (r).
3.4 Consider a particle with an intrinsic angular momentum (or
spin) of one unit of �h. (One example of such a particle is the %-
meson). Quantum-mechanically, such a particle is described by a
ketvector j%i or in ~x representation a wave function
%
i
(~x) = h~x; ij%i
where j~x; ii correspond to a particle at ~x with spin in the i:th di-
rection.
(a) Show explicitly that in�nitesimal rotations of %
i
(~x) are obtained
by acting with the operator
u
~"
= 1� i
~"
�h
� (
~
L+
~
S) (3.1)
where
~
L =
�h
i
r^ �
~
r. Determine
~
S !
(b) Show that
~
L and
~
S commute.
(c) Show that
~
S is a vector operator.
(d) Show that
~
r� ~%(~x) =
1
�h
2
(
~
S � ~p)~% where ~p is the momentum oper-
ator.
3.5 We are to add angular momenta j
1
= 1 and j
2
= 1 to form
j = 2; 1; and 0 states. Using the ladder operator method express all
16
(nine) j;m eigenkets in terms of jj
1
j
2
;m
1
m
2
i. Write your answer as
jj = 1;m = 1i =
1
p
2
j+; 0i �
1
p
2
j0;+i; : : : ; (3.2)
where + and 0 stand for m
1;2
= 1; 0; respectively.
3.6 (a) Construct a spherical tensor of rank 1 out of two di�erent
vectors
~
U = (U
x
; U
y
; U
z
) and
~
V = (V
x
; V
y
; V
z
). Explicitly write T
(1)
�1;0
in
terms of U
x;y;z
and V
x;y;z
.
(b) Construct a sphericaltensor of rank 2 out of two di�erent
vectors
~
U and
~
V . Write down explicitly T
(2)
�2;�1;0
in terms of U
x;y;z
and V
x;y;z
.
3.7 (a) Evaluate
j
X
m=�j
jd
(j)
mm
0
(�)j
2
m
for any j (integer or half-integer); then check your answer for j =
1
2
.
(b) Prove, for any j,
j
X
m=�j
m
2
jd
(j)
m
0
m
(�)j
2
=
1
2
j(j + 1) sin � +m
0
2
+
1
2
(3 cos
2
� � 1):
[Hint: This can be proved in many ways. You may, for instance,
examine the rotational properties of J
2
z
using the spherical (irre-
ducible) tensor language.]
3.8 (a) Write xy, xz, and (x
2
� y
2
) as components of a spherical
(irreducible) tensor of rank 2.
4. SYMMETRY IN QUANTUM MECHANICS 17
(b) The expectation value
Q � eh�; j;m = jj(3z
2
� r
2
)j�; j;m = ji
is known as the quadrupole moment. Evaluate
eh�; j;m
0
j(x
2
� y
2
)j�; j;m = ji;
(where m
0
= j; j�1; j�2; : : : )in terms of Q and appropriate Clebsch-
Gordan coe�cients.
4 Symmetry in Quantum Mechanics
4.1 (a) Assuming that the Hamiltonian is invariant under time
reversal, prove that the wave function for a spinless nondegenerate
system at any given instant of time can always be chosen to be
real.
(b) The wave function for a plane-wave state at t = 0 is given by
a complex function e
i~p�~x=�h
. Why does this not violate time-reversal
invariance?
4.2 Let �(~p
0
) be the momentum-space wave function for state j�i,
that is, �(~p
0
) = h~p
0
j�i.Is the momentum-space wave function for the
time-reversed state �j�i given by �(~p
0
, �(�~p
0
), �
�
(~p
0
), or �
�
(�~p
0
)?
Justify your answer.
4.3 Read section 4.3 in Sakurai to refresh your knowledge of the
quantum mechanics of periodic potentials. You know that the en-
ergybands in solids are described by the so called Bloch functions
 
n;k
full�lling,
 
n;k
(x+ a) = e
ika
 
n;k
(x)
18
where a is the lattice constant, n labels the band, and the lattice
momentum k is restricted to the Brillouin zone [��=a; �=a].
Prove that any Bloch function can be written as,
 
n;k
(x) =
X
R
i
�
n
(x�R
i
)e
ikR
i
where the sum is over all lattice vectors R
i
. (In this simble one di-
mensional problem R
i
= ia, but the construction generalizes easily
to three dimensions.).
The functions �
n
are called Wannier functions, and are impor-
tant in the tight-binding description of solids. Show that the Wan-
nier functions are corresponding to di�erent sites and/or di�erent
bands are orthogonal, i:e: prove
Z
dx�
?
m
(x�R
i
)�
n
(x�R
j
) � �
ij
�
mn
Hint: Expand the �
n
s in Bloch functions and use their orthonor-
mality properties.
4.4 Suppose a spinless particle is bound to a �xed center by a
potential V (~x) so assymetrical that no energy level is degenerate.
Using the time-reversal invariance prove
h
~
Li = 0
for any energy eigenstate. (This is known as quenching of orbital
angular momemtum.) If the wave function of such a nondegenerate
eigenstate is expanded as
X
l
X
m
F
lm
(r)Y
m
l
(�; �);
what kind of phase restrictions do we obtain on F
lm
(r)?
4.5 The Hamiltonian for a spin 1 system is given by
H = AS
2
z
+B(S
2
x
� S
2
y
):
5. APPROXIMATION METHODS 19
Solve this problem exactly to �nd the normalized energy eigen-
states and eigenvalues. (A spin-dependent Hamiltonian of this kind
actually appears in crystal physics.) Is this Hamiltonian invariant
under time reversal? How do the normalized eigenstates you ob-
tained transform under time reversal?
5 Approximation Methods
5.1 Consider an isotropic harmonic oscillator in two dimensions.
The Hamiltonian is given by
H
0
=
p
2
x
2m
+
p
2
y
2m
+
m!
2
2
(x
2
+ y
2
)
(a) What are the energies of the three lowest-lying states? Is there
any degeneracy?
(b) We now apply a perturbation
V = �m!
2
xy
where � is a dimensionless real number much smaller than unity.
Find the zeroth-order energy eigenket and the corresponding en-
ergy to �rst order [that is the unperturbed energy obtained in (a)
plus the �rst-order energy shift] for each of the three lowest-lying
states.
(c) Solve the H
0
+V problem exactly. Compare with the perturba-
tion results obtained in (b).
[You may use hn
0
jxjni =
q
�h=2m!(
p
n + 1�
n
0
;n+1
+
p
n�
n
0
;n�1
):]
5.2 A system that has three unperturbed states can be represented
by the perturbed Hamiltonian matrix
0
B
@
E
1
0 a
0 E
1
b
a
�
b
�
E
2
1
C
A
20
where E
2
> E
1
. The quantities a and b are to be regarded as per-
turbations that are of the same order and are small compared with
E
2
� E
1
. Use the second-order nondegenerate perturbation theory
to calculate the perturbed eigenvalues. (Is this procedure correct?)
Then diagonalize the matrix to �nd the exact eigenvalues. Finally,
use the second-order degenerate perturbation theory. Compare
the three results obtained.
5.3 A one-dimensional harmonic oscillator is in its ground state
for t < 0. For t � 0 it is subjected to a time-dependent but spatially
uniform force (not potential!) in the x-direction,
F (t) = F
0
e
�t=�
(a) Using time-dependent perturbation theory to �rst order, obtain
the probability of �nding the oscillator in its �rst excited state for
t > 0). Show that the t ! 1 (� �nite) limit of your expression is
independent of time. Is this reasonable or surprising?
(b) Can we �nd higher excited states?
[You may use hn
0
jxjni =
q
�h=2m!(
p
n+ 1�
n
0
;n+1
+
p
n�
n
0
;n�1
):]
5.4 Consider a composite system made up of two spin
1
2
objects.
for t < 0, the Hamiltonian does not depend on spin and can be
taken to be zero by suitably adjusting the energy scale. For t > 0,
the Hamiltonian is given by
H =
�
4�
�h
2
�
~
S
1
�
~
S
2
:
Suppose the system is in j + �i for t � 0. Find, as a function of
time, the probability for being found in each of the following states
j++i, j+�i, j �+i, j � �i:
(a) By solving the problem exactly.
5. APPROXIMATION METHODS 21
(b) By solving the problem assuming the validity of �rst-order
time-dependent perturbation theory with H as a perturbation switched
on at t = 0. Under what condition does (b) give the correct results?
5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjected
to a time-dependent potential as follows:
V (~x; t) = V
0
cos(kz � !t):
Using time-dependent perturbation theory, obtain an expression
for the transition rate at which the electron is emitted with mo-
mentum ~p. Show, in particular, how you may compute the angular
distribution of the ejected electron (in terms of � and � de�ned
with respect to the z-axis). Discuss brie
y the similarities and the
di�erences between this problem and the (more realistic) photo-
electric e�ect. (note: For the initial wave function use
	
n=1;l=0
(~x) =
1
p
�
�
Z
a
0
�
3
2
e
�Zr=a
0
:
If you have a normalization problem, the �nal wave function may
be taken to be
	
f
(~x) =
�
1
L
3
2
�
e
i~p�~x=�h
with L very large, but you should be able to show that the observ-
able e�ects are independent of L.)
22
Part II
Solutions23
1. FUNDAMENTAL CONCEPTS 25
1 Fundamental Concepts
1.1 Consider a ket space spanned by the eigenkets fja
0
ig of a Her-
mitian operator A. There is no degeneracy.
(a) Prove that
Y
a
0
(A� a
0
)
is a null operator.
(b) What is the signi�cance of
Y
a
00
6=a
0
(A� a
00
)
a
0
� a
00
?
(c) Illustrate (a) and (b) using A set equal to S
z
of a spin
1
2
system.
(a) Assume that j�i is an arbitrary state ket. Then
Y
a
0
(A� a
0
)j�i =
Y
a
0
(A� a
0
)
X
a
00
ja
00
i ha
00
j�i
| {z }
c
a
00
=
X
a
00
c
a
00
Y
a
0
(A� a
0
)ja
00
i
=
X
a
00
c
a
00
Y
a
0
(a
00
� a
0
)ja
00
i
a
00
2fa
0
g
= 0: (1.1)
(b) Again for an arbitrary state j�i we will have
2
4
Y
a
00
6=a
0
(A� a
00
)
a
0
� a
00
3
5
j�i =
2
4
Y
a
00
6=a
0
(A� a
00
)
a
0
� a
00
3
5
1
z }| {
X
a
000
ja
000
iha
000
j�i
=
X
a
000
ha
000
j�i
Y
a
00
6=a
0
(a
000
� a
00
)
a
0
� a
00
ja
000
i =
=
X
a
000
ha
000
j�i�
a
000
a
0
ja
000
i = ha
0
j�ija
0
i )
2
4
Y
a
00
6=a
0
(A� a
00
)
a
0
� a
00
3
5
= ja
0
iha
0
j � �
a
0
: (1.2)
So it projects to the eigenket ja
0
i.
26
(c) It is S
z
� �h=2(j+ih+j� j�ih�j). This operator has eigenkets j+i and j�i
with eigenvalues �h=2 and -�h=2 respectively. So
Y
a
0
(S
z
� a
0
) =
Y
a
0
(S
z
� a
0
1)
=
"
�h
2
(j+ih+j � j�ih�j)�
�h
2
(j+ih+j+ j�ih�j)
#
�
"
�h
2
(j+ih+j � j�ih�j) +
�h
2
(j+ih+j+ j�ih�j)
#
= [��hj�ih�j][�hj+ih+j] = ��h
2
j�i
0
z }| {
h�j+ih+j = 0; (1.3)
where we have used that j+ih+j+ j�ih�j = 1.
For a
0
= �h=2 we have
Y
a
00
6=a
0
(S
z
� a
00
)
a
0
� a
00
=
Y
a
00
6=�h=2
(S
z
� a
00
1)
�h=2 � a
00
=
S
z
+
�h
2
1
�h=2 + �h=2
=
1
�h
"
�h
2
(j+ih+j � j�ih�j) +
�h
2
(j+ih+j+ j�ih�j)
#
=
1
�h
�hj+ih+j = j+ih+j: (1.4)
Similarly for a
0
= ��h=2 we have
Y
a
00
6=a
0
(S
z
� a
00
)
a
0
� a
00
=
Y
a
00
6=��h=2
(S
z
� a
00
1)
��h=2 � a
00
=
S
z
�
�h
2
1
��h=2� �h=2
= �
1
�h
"
�h
2
(j+ih+j � j�ih�j)�
�h
2
(j+ih+j+ j�ih�j)
#
= �
1
�h
(��hj�ih�j) = j�ih�j: (1.5)
1.2 A spin
1
2
system is known to be in an eigenstate of
~
S � n^ with
eigenvalue �h=2, where n^ is a unit vector lying in the xz-plane that
makes an angle 
 with the positive z-axis.
(a) Suppose S
x
is measured. What is the probability of getting
+�h=2?
1. FUNDAMENTAL CONCEPTS 27
(b) Evaluate the dispersion in S
x
, that is,
h(S
x
� hS
x
i)
2
i:
(For your own peace of mind check your answers for the special
cases 
 = 0, �=2, and �.)
Since the unit vector n^ makes an angle 
 with the positive z-axis and is
lying in the xz-plane, it can be written in the following way
n^ = e^
z
cos 
 + e^
x
sin 
 (1.6)
So
~
S � n^ = S
z
cos 
 + S
x
sin 
 = [(S-1.3.36),(S-1.4.18)]
=
"
�h
2
(j+ih+j � j�ih�j)
#
cos 
 +
"
�h
2
(j+ih�j+ j�ih+j)
#
sin 
:(1.7)
Since the system is in an eigenstate of
~
S � n^ with eigenvalue �h=2 it has to
satisfay the following equation
~
S � n^j
~
S � n^; +i = �h=2j
~
S � n^; +i: (1.8)
From (1.7) we have that
~
S � n^
�
=
�h
2
 
cos 
 sin 
sin 
 � cos 
!
: (1.9)
The eigenvalues and eigenfuncions of this operator can be found if one solves
the secular equation
det(
~
S � n^� �I) = 0) det
 
�h=2 cos 
 � � �h=2 sin 
�h=2 sin 
 ��h=2 cos 
 � �
!
= 0)
�
�h
2
4
cos
2
 + �
2
�
�h
2
4
sin
2
 = 0) �
2
�
�h
2
4
= 0) � = �
�h
2
: (1.10)
Since the system is in the eigenstate j
~
S � n^; +i �
 
a
b
!
we will have that
�h
2
 
cos 
 sin 
sin 
 � cos 
! 
a
b
!
=
�h
2
 
a
b
!
)
(
a cos 
 + b sin 
 = a
a sin 
 � b cos 
 = b
)
)
b = a
1� cos 
sin 
= a
2 sin
2
2
2 sin
2
cos
2
= a tan
2
: (1.11)
28
But we want also the eigenstate j
~
S � n^; +i to be normalized, that is
a
2
+ b
2
= 1 ) a
2
+ a
2
tan
2
2
= 1) a
2
cos
2
2
a
2
sin
2
2
= cos
2
2
) a
2
= cos
2
2
) a = �
r
cos
2
2
= cos
2
; (1.12)
where the real positive convention has been used in the last step. This means
that the state in which the system is in, is given in terms of the eigenstates
of the S
z
operator by
j
~
S � n^; +i = cos
2
j+i+ sin
2
j�i: (1.13)
(a) From (S-1.4.17) we know that
jS
x
; +i =
1
p
2
j+i+
1
p
2
j�i: (1.14)
So the propability of getting +�h=2 when S
x
is measured is given by
�
�
�hS
x
; +j
~
S � n^; +i
�
�
�
2
=
�
�
�
�
�
 
1
p
2
h+j+
1
p
2
h�j
!
�
cos
2
j+i+ sin
2
j�i
�
�
�
�
�
�
2
=
�
�
�
�
�
1
p
2
cos
2
+
1
p
2
sin
2
�
�
�
�
�
2
=
1
2
cos
2
2
+
1
2
sin
2
2
+ cos
2
sin
2
=
1
2
+
1
2
sin 
 =
1
2
(1 + sin 
): (1.15)
For 
 = 0 which means that the system is in the jS
z
; +i eigenstate we have
jhS
x
; +jS
z
; +ij
2
=
1
2
(1) =
1
2
: (1.16)
For 
 = �=2 which means that the system is in the jS
x
; +i eigenstate we
have
jhS
x
; +jS
x
; +ij
2
= 1: (1.17)
For 
 = � which means that the system is in the jS
z
;�i eigenstate we have
jhS
x
; +jS
z
;�ij
2
=
1
2
(1) =
1
2
: (1.18)
1. FUNDAMENTAL CONCEPTS 29
(b) We have that
h(S
x
� hS
x
i)
2
i = hS
2
x
i � (hS
x
i)
2
: (1.19)
As we know
S
x
=
�h
2
(j+ih�j+ j�ih+j))
S
2
x
=
�h
2
4
(j+ih�j + j�ih+j) (j+ih�j+ j�ih+j))
S
2
x
=
�h
2
4
(j+ih+j + j�ih�j)
| {z }
1
=
�h
2
4
: (1.20)
So
hS
x
i =
�
cos
2
h+j + sin
2
h�j
�
�h
2
(j+ih�j+ j�ih+j)
�
cos
2
j+i+ sin
2
j�i
�
=
�h
2
cos
2
sin
2
+
�h
2
sin
2
cos
2
=
�h
2
sin 
 )
(hS
x
i)
2
=
�h
2
4
sin
2
 and
hS
2
x
i =
�
cos
2
h+j + sin
2
h�j
�
�h
2
4
�
cos
2
j+i+ sin
2
j�i
�
=
�h
2
4
[cos
2
2
+ sin
2
2
] =
�h
2
4
: (1.21)
So substituting in (1.19) we will have
h(S
x
� hS
x
i)
2
i =
�h
2
4
(1 � sin
2
) =
�h
2
4
cos
2
: (1.22)
and �nally
h(�S
x
)
2
i
=0;jS
z
;+i
=�h
2
4
; (1.23)
h(�S
x
)
2
i
=�=2;jS
x
;+i
= 0; (1.24)
h(�S
x
)
2
i
=0;jS
z
;�i
=
�h
2
4
: (1.25)
30
1.3 (a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
(h�j + �
�
h�j) � (j�i + �j�i) � 0
for any complex number �; then choose � in such a way that the
preceding inequality reduces to the Schwarz inequility.
(b) Show that the equility sign in the generalized uncertainty re-
lation holds if the state in question satis�es
�Aj�i = ��Bj�i
with � purely imaginary.
(c) Explicit calculations using the usual rules of wave mechanics
show that the wave function for a Gaussian wave packet given by
hx
0
j�i = (2�d
2
)
�1=4
exp
"
ihpix
0
�h
�
(x
0
� hxi)
2
4d
2
#
satis�es the uncertainty relation
q
h(�x)
2
i
q
h(�p)
2
i =
�h
2
:
Prove that the requirement
hx
0
j�xj�i = (imaginary number)hx
0
j�pj�i
is indeed satis�ed for such a Gaussian wave packet, in agreement
with (b).
(a) We know that for an arbitrary state jci the following relation holds
hcjci � 0: (1.26)
This means that if we choose jci = j�i+ �j�i where � is a complex number,
we will have
(h�j + �
�
h�j) � (j�i + �j�i) � 0) (1.27)
h�j�i + �h�j�i + �
�
h�j�i+ j�j
2
h�j�i � 0: (1.28)
1. FUNDAMENTAL CONCEPTS 31
If we now choose � = �h�j�i=h�j�i the previous relation will be
h�j�i �
h�j�ih�j�i
h�j�i
�
h�j�ih�j�i
h�j�i
+
jh�j�ij
2
h�j�i
� 0 )
h�j�ih�j�i � jh�j�ij
2
: (1.29)
Notice that the equality sign in the last relation holds when
jci = j�i + �j�i = 0) j�i = ��j�i (1.30)
that is if j�i and j�i are colinear.
(b) The uncertainty relation is
h(�A)
2
ih(�B)
2
i �
1
4
jh[A;B]ij
2
: (1.31)
To prove this relation we use the Schwarz inequality (1.29) for the vectors
j�i = �Ajai and j�i = �Bjai which gives
h(�A)
2
ih(�B)
2
i � jh�A�Bij
2
: (1.32)
The equality sign in this relation holds according to (1.30) when
�Ajai = ��Bjai: (1.33)
On the other hand the right-hand side of (1.32) is
jh�A�Bij
2
=
1
4
jh[A;B]ij
2
+
1
4
jhf�A;�Bgij
2
(1.34)
which means that the equality sign in the uncertainty relation (1.31) holds if
1
4
jhf�A;�Bgij
2
= 0) hf�A;�Bgi = 0
) haj�A�B +�B�Ajai = 0
(1:33)
) �
�
haj(�B)
2
jai+ �haj(�B)
2
jai = 0
) (� + �
�
)haj(�B)
2
jai = 0: (1.35)
Thus the equality sign in the uncertainty relation holds when
�Ajai = ��Bjai (1.36)
with � purely imaginary.
32
(c) We have
hx
0
j�xj�i � hx
0
j(x� hxi)j�i = x
0
hx
0
j�i � hxihx
0
j�i
= (x
0
� hxi)hx
0
j�i: (1.37)
On the other hand
hx
0
j�pj�i � hx
0
j(p� hpi)j�i
= �i�h
@
@x
0
hx
0
j�i � hpihx
0
j�i (1.38)
But
@
@x
0
hx
0
j�i = hx
0
j�i
@
@x
0
"
ihpix
0
�h
�
(x
0
� hxi)
2
4d
2
#
= hx
0
j�i
"
ihpi
�h
�
1
2d
2
(x
0
� hxi)
#
(1.39)
So substituting in (1.38) we have
hx
0
j�pj�i = hpihx
0
j�i +
i�h
2d
2
(x
0
� hxi) hx
0
j�i � hpihx
0
j�i
=
i�h
2d
2
(x
0
� hxi) hx
0
j�i =
i�h
2d
2
hx
0
j�xj�i )
hx
0
j�xj�i =
�i2d
2
�h
hx
0
j�pj�i: (1.40)
1.4 (a) Let x and p
x
be the coordinate and linear momentum in
one dimension. Evaluate the classical Poisson bracket
[x; F (p
x
)]
classical
:
(b) Let x and p
x
be the corresponding quantum-mechanical opera-
tors this time. Evaluate the commutator
�
x; exp
�
ip
x
a
�h
��
:
(c) Using the result obtained in (b), prove that
exp
�
ip
x
a
�h
�
jx
0
i; (xjx
0
i = x
0
jx
0
i)
1. FUNDAMENTAL CONCEPTS 33
is an eigenstate of the coordinate operator x. What is the corre-
sponding eigenvalue?
(a) We have
[x; F (p
x
)]
classical
�
@x
@x
@F (p
x
)
@p
x
�
@x
@p
x
@F (p
x
)
@x
=
@F (p
x
)
@p
x
: (1.41)
(b) When x and p
x
are treated as quantum-mechanical operators we have
�
x; exp
�
ip
x
a
�h
��
=
"
x;
1
X
n=0
(ia)
n
�h
n
p
n
x
n!
#
=
1
X
n=0
1
n!
(ia)
n
�h
n
[x; p
n
x
]
=
1
X
n=0
1
n!
(ia)
n
�h
n
n�1
X
k=0
p
k
x
[x; p
x
] p
n�k�1
x
=
1
X
n=1
1
n!
(ia)
n
�h
n
(i�h)
n�1
X
k=0
p
k
x
p
n�k�1
x
=
1
X
n=1
n
n!
(ia)
n�1
�h
n�1
p
n�1
x
(�a)
= �a
1
X
n=1
1
(n� 1)!
�
ia
�h
p
x
�
n�1
= �a exp
�
ip
x
a
�h
�
: (1.42)
(c) We have now
x
�
exp
�
ip
x
a
�h
��
jx
0
i
(b)
= exp
�
ip
x
a
�h
�
xjx
0
i � a exp
�
ip
x
a
�h
�
jx
0
i
= x
0
exp
�
ip
x
a
�h
�
jx
0
i � a exp
�
ip
x
a
�h
�
jx
0
i
= (x
0
� a) exp
�
ip
x
a
�h
�
jx
0
i: (1.43)
So exp
�
ip
x
a
�h
�
jx
0
i is an eigenstate of the operator x with eigenvalue x
0
� a.
So we can write
jx
0
� ai = C exp
�
ip
x
a
�h
�
jx
0
i; (1.44)
where C is a constant which due to normalization can be taken to be 1.
34
1.5 (a) Prove the following:
(i) hp
0
jxj�i = i�h
@
@p
0
hp
0
j�i;
(ii) h�jxj�i =
Z
dp
0
�
�
�
(p
0
)i�h
@
@p
0
�
�
(p
0
);
where �
�
(p
0
) = hp
0
j�i and �
�
(p
0
) = hp
0
j�i are momentum-space wave
functions.
(b) What is the physical signi�cance of
exp
�
ix�
�h
�
;
where x is the position operator and � is some number with the
dimension of momentum? Justify your answer.
(a) We have
(i)
hp
0
jxj�i = hp
0
jx
1
z }| {
Z
dx
0
jx
0
ihx
0
j�i =
Z
dx
0
hp
0
jxjx
0
ihx
0
j�i
=
Z
dx
0
x
0
hp
0
jx
0
ihx
0
j�i
(S�1:7:32)
=
Z
dx
0
x
0
Ae
�
ip
0
x
0
�h
hx
0
j�i
= A
Z
dx
0
@
@p
0
�
e
�
ip
0
x
0
�h
�
(i�h)hx
0
j�i = i�h
@
@p
0
�
Z
dx
0
Ae
�
ip
0
x
0
�h
hx
0
j�i
�
= i�h
@
@p
0
�
Z
dx
0
hp
0
jx
0
ihx
0
j�i
�
= i�h
@
@p
0
hp
0
j�i )
hp
0
jxj�i = i�h
@
@p
0
hp
0
j�i: (1.45)
(ii)
h�jxj�i =
Z
dp
0
h�jp
0
ihp
0
jxj�i =
Z
dp
0
�
�
�
(p
0
)i�h
@
@p
0
�
�
(p
0
); (1.46)
where we have used (1.45) and that h�jp
0
i = �
�
�
(p
0
) and hp
0
j�i = �
�
(p
0
).
1. FUNDAMENTAL CONCEPTS 35
(b) The operator exp
�
ix�
�h
�
gives translation in momentum space. This can
be justi�ed by calculating the following operator
�
p; exp
�
ix�
�h
��
=
"
p;
1
X
n=0
1
n!
�
ix�
�h
�
n
#
=
1
X
n=0
1
n!
�
i�
�h
�
n
[p; x
n
]
=
1
X
n=1
1
n!
�
i�
�h
�
n
n
X
k=1
x
n�k
[p; x]x
k�1
=
1X
n=1
1
n!
�
i�
�h
�
n
n
X
k=1
(�i�h)x
n�1
=
1
X
n=1
1
n!
�
i�
�h
�
n
n(�i�h)x
n�1
=
1
X
n=1
1
(n� 1)!
�
i�
�h
�
n�1
x
n�1
(�i�h)
�
i�
�h
�
= �
1
X
n=0
1
n!
�
ix�
�h
�
n
= �exp
�
ix�
�h
�
: (1.47)
So when this commutator acts on an eigenstate jp
0
i of the momentum oper-
ator we will have
�
p; exp
�
ix�
�h
��
jp
0
i = p
�
exp
�
ix�
�h
�
jp
0
i
�
�
�
exp
�
ix�
�h
��
p
0
jp
0
i )
�exp
�
ix�
�h
�
= p
�
exp
�
ix�
�h
�
jp
0
i
�
� p
0
�
exp
�
ix�
�h
��
jp
0
i )
p
�
exp
�
ix�
�h
�
jp
0
i
�
= (p
0
+ �)
�
exp
�
ix�
�h
�
jp
0
i
�
: (1.48)
Thus we have that
exp
�
ix�
�h
�
jp
0
i � Ajp
0
+ �i; (1.49)
where A is a constant which due to normalization can be taken to be 1.
36
2 Quantum Dynamics
2.1 Consider the spin-procession problem discussed in section 2.1
in Jackson. It can also be solved in the Heisenberg picture. Using
the Hamiltonian
H = �
�
eB
mc
�
S
z
= !S
z
;
write the Heisenberg equations of motion for the time-dependent
operators S
x
(t), S
y
(t), and S
z
(t). Solve them to obtain S
x;y;z
as func-
tions of time.
Let us �rst prove the following
[A
S
; B
S
] = C
S
) [A
H
; B
H
] = C
H
: (2.1)
Indeed we have
[A
H
; B
H
] =
h
U
y
A
S
U ;U
y
B
S
U
i
= U
y
A
S
B
S
U � U
y
B
S
A
S
U
= U
y
[A
S
; B
S
]U = U
y
C
S
U = C
H
: (2.2)
The Heisenberg equation of motion gives
dS
x
dt
=
1
i�h
[S
x
;H] =
1
i�h
[S
x
; !S
z
]
(S�1:4:20)
=
!
i�h
(�i�hS
y
) = �!S
y
; (2.3)
dS
y
dt
=
1
i�h
[S
y
;H] =
1
i�h
[S
y
; !S
z
]
(S�1:4:20)
=
!
i�h
(i�hS
x
) = !S
x
; (2.4)
dS
z
dt
=
1
i�h
[S
z
;H] =
1
i�h
[S
z
; !S
z
]
(S�1:4:20)
= 0) S
z
= constant: (2.5)
Di�erentiating once more eqs. (2.3) and (2.4) we get
d
2
S
x
dt
2
= �!
dS
y
dt
(2:4)
= �!
2
S
x
) S
x
(t) = A cos!t+B sin!t) S
x
(0) = A
d
2
S
y
dt
2
= !
dS
x
dt
(2:3)
= �!
2
S
y
) S
y
(t) = C cos!t+D sin!t) S
y
(0) = C:
But on the other hand
dS
x
dt
= �!S
y
)
�A! sin!t+B! cos !t = �C! cos!t�D! sin!t)
A = D C = �B: (2.6)
2. QUANTUM DYNAMICS 37
So, �nally
S
x
(t) = S
x
(0) cos!t� S
y
(0) sin!t (2.7)
S
y
(t) = S
y
(0) cos!t+ S
x
(0) sin!t (2.8)
S
z
(t) = S
z
(0): (2.9)
2.2 Let x(t) be the coordinate operator for a free particle in one
dimension in the Heisenberg picture. Evaluate
[x(t); x(0)] :
The Hamiltonian for a free particle in one dimension is given by
H =
p
2
2m
: (2.10)
This means that the Heisenberg equations of motion for the operators x and
p will be
@p(t)
@t
=
1
i�h
[p(t);H(t)] =
1
i�h
"
p(t);
p
2
(t)
2m
#
= 0)
p(t) = p(0) (2.11)
@x(t)
@t
=
1
i�h
[x;H] =
1
i�h
"
x(t);
p
2
(t)
2m
#
=
1
2mi�h
2p(t)i�h =
p(t)
m
(2:11)
=
p(0)
m
)
x(t) =
t
m
p(0) + x(0): (2.12)
Thus �nally
[x(t); x(0)] =
�
t
m
p(0) + x(0); x(0)
�
=
t
m
[p(0); x(0)] = �
i�ht
m
: (2.13)
2.3 Consider a particle in three dimensions whose Hamiltonian is
given by
H =
~p
2
2m
+ V (~x):
38
By calculating [~x � ~p;H] obtain
d
dt
h~x � ~pi =
*
p
2
m
+
� h~x �
~
rV i:
To identify the preceding relation with the quantum-mechanical
analogue of the virial theorem it is essential that the left-hand side
vanish. Under what condition would this happen?
Let us �rst calculate the commutator [~x � ~p;H]
[~x � ~p;H] =
"
~x � ~p;
~p
2
2m
+ V (~x)
#
=
2
4
3
X
i=1
x
i
p
i
;
3
X
j=1
p
2
j
2m
+ V (~x)
3
5
=
X
ij
"
x
i
;
p
2
j
2m
#
p
i
+
X
i
x
i
[p
i
; V (~x)] : (2.14)
The �rst commutator in (2.14) will give
"
x
i
;
p
2
j
2m
#
=
1
2m
[x
i
; p
2
j
] =
1
2m
(p
j
[x
i
; p
j
] + [x
i
; p
j
]p
j
) =
1
2m
(p
j
i�h�
ij
+ i�h�
ij
p
j
)
=
1
2m
2i�h�
ij
p
j
=
i�h
m
�
ij
p
j
: (2.15)
The second commutator can be calculated if we Taylor expand the function
V (~x) in terms of x
i
which means that we take V (~x) =
P
n
a
n
x
n
i
with a
n
independent of x
i
. So
[p
i
; V (~x)] =
"
p
i
;
1
X
n=0
a
n
x
n
i
#
=
X
n
a
n
[p
i
; x
n
i
] =
X
n
a
n
n�1
X
k=0
x
k
i
[p
i
; x
i
]x
n�k�1
i
=
X
n
a
n
n�1
X
k=0
(�i�h)x
n�1
i
= �i�h
X
n
a
n
nx
n�1
i
= �i�h
@
@x
i
X
n
a
n
x
n
i
= �i�h
@
@x
i
V (~x): (2.16)
The right-hand side of (2.14) now becomes
[~x � ~p;H] =
X
ij
i�h
m
�
ij
p
j
p
i
+
X
i
(�i�h)x
i
@
@x
i
V (~x)
=
i�h
m
~p
2
� i�h~x �
~
rV (~x): (2.17)
2. QUANTUM DYNAMICS 39
The Heisenberg equation of motion gives
d
dt
~x � ~p =
1
i�h
[~x � ~p;H]
(2:17)
=
~p
2
m
� ~x �
~
rV (~x))
d
dt
h~x � ~pi =
*
p
2
m
+
� h~x �
~
rV i; (2.18)
where in the last step we used the fact that the state kets in the Heisenberg
picture are independent of time.
The left-hand side of the last equation vanishes for a stationary state.
Indeed we have
d
dt
hnj~x � ~pjni =
1
i�h
hnj [~x � ~p;H] jni =
1
i�h
(E
n
hnj~x � ~pjni � E
n
hnj~x � ~pjni) = 0:
So to have the quantum-mechanical analogue of the virial theorem we can
take the expectation values with respect to a stationaru state.
2.4 (a) Write down the wave function (in coordinate space) for the
state
exp
�
�ipa
�h
�
j0i:
You may use
hx
0
j0i = �
�1=4
x
�1=2
0
exp
2
4
�
1
2
 
x
0
x
0
!
2
3
5
;
0
@
x
0
�
 
�h
m!
!
1=2
1
A
:
(b) Obtain a simple expression that the probability that the state
is found in the ground state at t = 0. Does this probability change
for t > 0?
(a) We have
j�; t = 0i = exp
�
�ipa
�h
�
j0i )
hx
0
j�; t = 0i = hx
0
exp
�
�ipa
�h
�
j0i
(Pr:1:4:c)
= hx
0
� aj0i
= �
�1=4
x
�1=2
0
exp
2
4
�
1
2
 
x
0
� a
x
0
!
2
3
5
: (2.19)
40
(b) This probability is given by the expression
jh0j�; t = 0ij
2
= jhexp
�
�ipa
�h
�
j0ij
2
: (2.20)
It is
hexp
�
�ipa
�h
�
j0i =
Z
dx
0
h0jx
0
ihx
0
j exp
�
�ipa
�h
�
j0i
=
Z
dx
0
�
�1=4
x
�1=2
0
exp
2
4
�
1
2
 
x
0
x0
!
2
3
5
�
�1=4
x
�1=2
0
� exp
2
4
�
1
2
 
x
0
� a
x
0
!
2
3
5
=
Z
dx
0
�
�1=2
x
�1
0
exp
"
�
1
2x
2
0
�
x
02
+ x
02
+ a
2
� 2ax
0
�
#
=
1
p
�x
0
Z
dx
0
exp
"
�
2
2x
2
0
 
x
02
� 2x
0
a
2
+
a
2
4
+
a
2
4
!#
= exp
 
�
a
2
4x
2
0
!
1
p
�x
0
p
�x
0
= exp
 
�
a
2
4x
2
0
!
: (2.21)
So
jh0j�; t = 0ij
2
= exp
 
�
a
2
2x
2
0
!
: (2.22)
For t > 0
jh0j�; tij
2
= jh0jU(t)j�; t = 0ij
2
= jh0j exp
�
�
iHt
�h
�
j�; t = 0ij
2
=
�
�
�e
�iE
0
t=�h
h0j�; t = 0i
�
�
�
2
= jh0j�; t = 0ij
2
: (2.23)
2.5 Consider a function, known as the correlation function, de�ned
by
C(t) = hx(t)x(0)i; (2.24)
where x(t) is the position operator in the Heisenberg picture. Eval-
uate the correlation function explicitly for the ground state of a
one-dimensional simple harmonic oscillator.
2. QUANTUM DYNAMICS 41
The Hamiltonian for a one-dimensional harmonic oscillator is given by
H =
p
2
(t)
2m
+
1
2
m!
2
x
2
(t): (2.25)
So the Heisenberg equations of motion will give
dx(t)
dt
=
1
i�h
[x(t);H] =
1
i�h
"
x(t);
p
2
(t)
2m
+
1
2
m!
2
x
2
(t)
#
=
1
2mi�h
h
x(t); p
2
(t)
i
+
1
2
m!
2
1
i�h
h
x(t); x
2
(t)
i
=
2i�h
2i�hm
p(t) =
p(t)
m
(2.26)
dp(t)
dt
=
1
i�h
[p(t);H] =
1
i�h
"
p(t);
p
2
(t)
2m
+
1
2
m!
2
x
2
(t)
#
=
m!
2
2i�h
h
p(t); x
2
(t)
i
=
m!
2
2i�h
[�2i�hx(t)] = �m!
2
x(t): (2.27)
Di�erentiating once more the equations (2.26) and (2.27) we get
d
2
x(t)
dt
2
=
1
m
dp(t)
dt
(2:27)
= �!
2
x(t)) x(t) = A cos!t+B sin!t) x(0) = A
d
2
p(t)
dt
2
=
1
m
dx(t)
dt
(2:26)
= �!
2
p(t)) p(t) = C cos!t+D sin!t) p(0) = C:
But on the other hand from (2.26) we have
dx(t)
dt
=
p(t)
m
)
�!x(0) sin !t+B! cos!t =
p(0)
m
cos!t+
D
m
sin!t)
B =
p(0)
m!
D = �m!x(0): (2.28)
So
x(t) = x(0) cos!t+
p(0)
m!
sin!t (2.29)
and the correlation function will be
C(t) = hx(t)x(0)i
(2:29)
= hx
2
(0)i cos !t+ hp(0)x(0)i
1
m!
sin!t: (2.30)
42
Since we are interested in the ground state the expectation values appearing
in the last relation will be
hx
2
(0)i = h0j
�h
2m!
(a+ a
y
)(a+ a
y
)j0i =
�h
2m!
h0jaa
y
j0i =
�h
2m!
(2.31)
hp(0)x(0)i = i
s
m�h!
2
s
�h
2m!
h0j(a
y
� a)(a+ a
y
)j0i
= �i
�h
2
h0jaa
y
j0i = �i
�h
2
: (2.32)
Thus
C(t) =
�h
2m!
cos!t� i
�h
2m!
sin!t =
�h
2m!
e
�i!t
: (2.33)
2.6 Consider a one-dimensional simple harmonic oscillator. Do the
following algebraically, that is, without using wave functions.
(a) Construct a linear combination of j0i and j1i such that hxi is as
large as possible.
(b) Suppose the oscillator is in the state constructed in (a) at t = 0.
What is the state vector for t > 0 in the Schrodinger picture?
Evaluate the expectation value hxi as a function of time for t > 0
using (i) the Schrodinger picture and (ii) the Heisenberg picture.
(c) Evaluate h(�x)
2
i as a function of time using either picture.
(a) We want to �nd a state j�i = c
0
j0i + c
1
j1i such that hxi is as large as
possible. The state j�i should be normalized. This means
jc
0
j
2
+ jc
1
j
2
= 1) jc
1
j =
q
1 � jc
0
j
2
: (2.34)
We can write the constands c
0
and c
1
in the following form
c
0
= jc
0
je
i�
0
c
1
= jc
1
je
i�
1
(2:34)
= e
i�
1
q
1� jc
0
j
2
: (2.35)
2. QUANTUM DYNAMICS 43
The average hxi in a one-dimensional simple harmonic oscillator is given
by
hxi = h�jxj�i = (c
�
0
h0j + c
�
1
h1j) x (c
0
j0i + c
1
j1i)
= jc
0
j
2
h0jxj0i + c
�
0
c
1
h0jxj1i + c
�
1
c
0
h1jxj0i + jc
1
j
2
h1jxj1i
= jc
0
j
2
s
�h
2m!
h0ja+ a
y
j0i+ c
�
0
c
1
s
�h
2m!
h0ja + a
y
j1i
+c
�
1
c
0
s
�h
2m!
h1ja + a
y
j0i + jc
1
j
2
s
�h
2m!
h1ja + a
y
j1i
=
s
�h
2m!
(c
�
0
c
1
+ c
�
1
c
0
) = 2
s
�h
2m!
<(c
�
0
c
1
)
= 2
s
�h
2m!
cos(�
1
� �
0
)jc
0
j
q
1 � jc
0
j
2
; (2.36)
where we have used that x =
q
�h
2m!
(a+ a
y
).
What we need is to �nd the values of jc
0
j and �
1
� �
0
that make the
average hxi as large as possible.
@hxi
@jc
0
j
= 0 )
q
1 � jc
0
j
2
�
jc
0
j
2
q
1 � jc
0
j
2
jc
0
j6=1
) 1 � jc
0
j
2
� jc
0
j
2
= 0
) jc
0
j =
1
p
2
(2.37)
@hxi
@�
1
= 0 ) � sin(�
1
� �
0
) = 0) �
1
= �
0
+ n�; n 2 Z: (2.38)
But for hxi maximum we want also
@
2
hxi
@�
2
1
�
�
�
�
�
�
1
=�
1max
< 0) n = 2k; k 2 Z: (2.39)
So we can write that
j�i = e
i�
0
1
p
2
j0i + e
i(�
0
+2k�)
1
p
2
j1i = e
i�
0
1
p
2
(j0i + j1i): (2.40)
We can always take �
0
= 0. Thus
j�i =
1
p
2
(j0i + j1i): (2.41)
44
(b) We have j�; t
0
i = j�i. So
j�; t
0
; ti = U(t; t
0
= 0)j�; t
0
i = e
�iHt=�h
j�i =
1
p
2
e
�iE
0
t=�h
j0i +
1
p
2
e
�iE
1
t=�h
j1i
=
1
p
2
�
e
�i!t=2
j0i+ e
�i!3t=2
j1i
�
=
1
p
2
e
�i!t=2
�
j0i+ e
�i!t
j1i
�
:(2.42)
(i) In the Schrodinger picture
hxi
S
= h�; t
0
; tjx
S
j�; t
0
; ti
S
=
"
1
p
2
�
e
i!t=2
h0j+ e
i!3t=2
h1j
�
#
x
"
1
p
2
�
e
�i!t=2
j0i + e
�i!3t=2
j1i
�
#
=
1
2
e
i(!t=2�!3t=2)
h0jxj1i +
1
2
e
i(!3t=2�!t=2)
h1jxj0i
=
1
2
e
�i!t
s
�h
2m!
+
1
2
e
i!t
s
�h
2m!
=
s
�h
2m!
cos !t: (2.43)
(ii) In the Heisenberg picture we have from (2.29) that
x
H
(t) = x(0) cos!t+
p(0)
m!
sin!t:
So
hxi
H
= h�jx
H
j�i
=
"
1
p
2
h0j +
1
p
2
h1j
# 
x(0) cos!t+
p(0)
m!
sin!t
!"
1
p
2
j0i+
1
p
2
j1i
#
=
1
2
cos !th0jxj1i+
1
2
cos !th1jxj0i +
1
2
1
m!
sin !th0jpj1i
+
1
2
1
m!
sin!th1jpj0i
=
1
2
s
�h
2m!
cos!t+
1
2
s
�h
2m!
cos!t+
1
2m!
sin!t(�i)
s
m�h!
2
+
1
2m!
sin!ti
s
m�h!
2
=
s
�h
2m!
cos!t: (2.44)
(c) It is known that
h(�x)
2
i = hx
2
i � hxi
2
(2.45)
2. QUANTUM DYNAMICS 45
In the Schodinger picture we have
x
2
=
2
4
s
�h
2m!
(a+ a
y
)
3
5
2
=
�h
2m!
(a
2
+ a
y
2
+ aa
y
+ a
y
a); (2.46)
which means that
hxi
2
S= h�; t
0
; tjx
2
j�; t
0
; ti
S
=
"
1
p
2
�
e
i!t=2
h0j + e
i!3t=2
h1j
�
#
x
2
"
1
p
2
�
e
�i!t=2
j0i+ e
�i!3t=2
j1i
�
#
=
h
1
2
e
i(!t=2�!t=2)
h0jaa
y
j0i +
1
2
e
i(!3t=2�!3t=2)
h1jaa
y
j1i +
1
2
h1ja
y
aj1i
i
�h
2m!
=
h
1
2
+ 2
1
2
+
1
2
i
�h
2m!
=
�h
2m!
: (2.47)
So
h(�x)
2
i
S
(2:43)
=
�h
2m!
�
�h
2m!
cos
2
!t =
�h
2m!
sin
2
!t: (2.48)
In the Heisenberg picture
x
2
H
(t) =
"
x(0) cos!t+
p(0)
m!
sin!t
#
2
= x
2
(0) cos
2
!t+
p
2
(0)
m
2
!
2
sin
2
!t
+
x(0)p(0)
m!
cos!t sin!t+
p(0)x(0)
m!
cos !t sin!t
=
�h
2m!
(a
2
+ a
y
2
+ aa
y
+ a
y
a) cos
2
!t
�
m�h!
2m
2
!
2
(a
2
+ a
y
2
� aa
y
� a
y
a) sin
2
!t
+
i
m!
s
�hm�h!
4m!
(a+ a
y
)(a
y
� a)
sin 2!t
2
+
i
m!
s
�hm�h!
4m!
(a
y
� a)(a+ a
y
)
sin 2!t
2
=
�h
2m!
(a
2
+ a
y
2
+ aa
y
+ a
y
a) cos
2
!t
46
�
�h
2m!
(a
2
+ a
y
2
� aa
y
� a
y
a) sin
2
!t+
i�h
2m!
(a
y
2
� a
2
) sin 2!t
=
�h
2m!
(aa
y
+ a
y
a) +
�h
2m!
a
2
cos 2!t+
�h
2m!
a
y
2
cos 2!t
+
i�h
2m!
(a
y
2
� a
2
) sin 2!t; (2.49)
which means that
hx
2
H
i
H
= h�jx
2
H
j�i
H
=
�h
2m!
"
1
p
2
h0j+
1
p
2
h1j
#
�
h
aa
y
+ a
y
a+ a
2
cos 2!t+ a
y
2
cos 2!t+ i(a
y
2
� a
2
) sin 2!t
i
�
"
1
p
2
j0i +
1
p
2
j1i
#
=
�h
4m!
h
h0jaa
y
j0i+ h1jaa
y
j1i + h1ja
y
aj1i
i
=
�h
4m!
[1 + 2 + 1] =
�h
m!
: (2.50)
So
h(�x)
2
i
H
(2:44)
=
�h
2m!
�
�h
2m!
cos
2
!t =
�h
2m!
sin
2
!t: (2.51)
2.7 A coherent state of a one-dimensional simple harmonic oscil-
lator is de�ned to be an eigenstate of the (non-Hermitian) annihi-
lation operator a:
aj�i = �j�i;
where � is, in general, a complex number.
(a) Prove that
j�i = e
�j�j
2
=2
e
�a
y
j0i
is a normalized coherent state.
(b) Prove the minimum uncertainty relation for such a state.
2. QUANTUM DYNAMICS 47
(c) Write j�i as
j�i =
1
X
n=0
f(n)jni:
Show that the distribution of jf(n)j
2
with respect to n is of the
Poisson form. Find the most probable value of n, hence of E.
(d) Show that a coherent state can also be obtained by applying
the translation (�nite-displacement) operator e
�ipl=�h
(where p is the
momentum operator, and l is the displacement distance) to the
ground state.
(e) Show that the coherent state j�i remains coherent under time-
evolution and calculate the time-evolved state j�(t)i. (Hint: di-
rectly apply the time-evolution operator.)
(a) We have
aj�i = e
�j�j
2
=2
ae
�a
y
j0i = e
�j�j
2
=2
h
a; e
�a
y
i
j0i; (2.52)
since aj0i = 0. The commutator is
h
a; e
�a
y
i
=
"
a;
1
X
n=0
1
n!
(�a
y
)
n
#
=
1
X
n=0
1
n!
�
n
h
a; (a
y
)
n
i
=
1
X
n=1
1
n!
�
n
n
X
k=1
(a
y
)
k�1
h
a; a
y
i
(a
y
)
n�k
=
1
X
n=1
1
n!
�
n
n
X
k=1
(a
y
)
n�1
=
1
X
n=1
1
(n� 1)!
�
n
(a
y
)
n�1
= �
1
X
n=0
1
n!
(�a
y
)
n
= �e
�a
y
: (2.53)
So from (2.52)
aj�i = e
�j�j
2
=2
�e
�a
y
j0i = �j�i; (2.54)
which means that j�i is a coherent state. If it is normalized, it should satisfy
also h�j�i = 1. Indeed
h�j�i = h0je
�
�
a
e
�j�j
2
e
�a
y
j0i = e
�j�j
2
h0je
�
�
a
e
�a
y
j0i
48
= e
�j�j
2
X
n;m
1
n!m!
(�
�
)
n
�
m
h0ja
n
j(a
y
)
m
j0i [(a
y
)
m
j0i =
p
m!jmi]
= e
�j�j
2
X
n;m
p
n!
n!
p
m!
m!
(�
�
)
n
�
m
hnjmi = e
�j�j
2
X
n
1
n!
(j�j
2
)
n
= e
�j�j
2
e
j�j
2
= 1: (2.55)
(b) According to problem (1.3) the state should satisfy the following relation
�xj�i = c�pj�i; (2.56)
where �x � x � h�jxj�i, �p � p � h�jpj�i and c is a purely imaginary
number.
Since j�i is a coherent state we have
aj�i = �j�i ) h�ja
y
= h�j�
�
: (2.57)
Using this relation we can write
xj�i =
s
�h
2m!
(a+ a
y
)j�i =
s
�h
2m!
(�+ a
y
)j�i (2.58)
and
hxi = h�jxj�i =
s
�h
2m!
h�j(a+ a
y
)j�i =
s
�h
2m!
(h�jaj�i + h�ja
y
j�i)
=
s
�h
2m!
(� + �
�
) (2.59)
and so
�xj�i = (x� hxi)j�i =
s
�h
2m!
(a
y
� �
�
)j�i: (2.60)
Similarly for the momentum p = i
q
m�h!
2
(a
y
� a) we have
pj�i =
p
i
s
m�h!
2
(a
y
� a)j�i = i
s
m�h!
2
(a
y
� �)j�i (2.61)
2. QUANTUM DYNAMICS 49
and
hpi = h�jpj�i = i
s
m�h!
2
h�j(a
y
� a)j�i = i
s
m�h!
2
(h�ja
y
j�i � h�jaj�i)
= i
s
m�h!
2
(�
�
� �) (2.62)
and so
�pj�i = (p� hpi)j�i = i
s
m�h!
2
(a
y
� �
�
)j�i )
(a
y
� �
�
)j�i = �i
s
2
m�h!
�pj�i: (2.63)
So using the last relation in (2.60)
�xj�i =
s
�h
2m!
(�i)
s
2
m�h!
�pj�i = �
i
m!
| {z }
purely imaginary
�pj�i (2.64)
and thus the minimum uncertainty condition is satis�ed.
(c) The coherent state can be expressed as a superposition of energy eigen-
states
j�i =
1
X
n=0
jnihnj�i =
1
X
n=0
f(n)jni: (2.65)
for the expansion coe�cients f(n) we have
f(n) = hnj�i = hnje
�j�j
2
=2
e
�a
y
j0i = e
�j�j
2
=2
hnje
�a
y
j0i
= e
�j�j
2
=2
hnj
1
X
m=0
1
m!
(�a
y
)
m
j0i = e
�j�j
2
=2
1
X
m=0
1
m!
�
m
hnj(a
y
)
m
j0i
= e
�j�j
2
=2
1
X
m=0
1
m!
�
m
p
m!hnjmi = e
�j�j
2
=2
1
p
n!
�
n
) (2.66)
jf(n)j
2
=
(j�j
2
)
n
n!
exp(�j�j
2
) (2.67)
whichmeans that the distribution of jf(n)j
2
with respect to n is of the Poisson
type about some mean value n = j�j
2
.
50
The most probable value of n is given by the maximumof the distribution
jf(n)j
2
which can be found in the following way
jf(n + 1)j
2
jf(n)j
2
=
(j�j
2
)
n+1
(n+1)!
exp(�j�j
2
)
(j�j
2
)
n
n!
exp(�j�j
2
)
=
j�j
2
n+ 1
� 1 (2.68)
which means that the most probable value of n is j�j
2
.
(d) We should check if the state exp (�ipl=�h) j0i is an eigenstate of the an-
nihilation operator a. We have
a exp (�ipl=�h) j0i =
h
a; e
(�ipl=�h)
i
j0i (2.69)
since aj0i = 0. For the commutator in the last relation we have
h
a; e
(�ipl=�h)
i
=
1
X
n=0
1
n!
 
�il
�h
!
n
[a; p
n
] =
1
X
n=1
1
n!
 
�il
�h
!
n
n
X
k=1
p
k�1
[a; p]p
n�k
=
1
X
n=1
1
n!
 
�il�h
!
n
n
X
k=1
p
n�1
i
s
m�h!
2
= i
s
m�h!
2
 
�il
�h
!
1
X
n=1
1
(n � 1)!
 
�ilp
�h
!
n�1
= l
r
m!
2�h
e
(�ipl=�h)
; (2.70)
where we have used that
[a; p] = i
s
m�h!
2
[a; a
y
� a] = i
s
m�h!
2
: (2.71)
So substituting (2.70) in (2.69) we get
a [exp (�ipl=�h) j0i] = l
r
m!
2�h
[exp (�ipl=�h) j0i] (2.72)
which means that the state exp (�ipl=�h) j0i is a coherent state with eigen-
value l
q
m!
2�h
.
(e) Using the hint we have
j�(t)i = U(t)j�i = e
�iHt=�h
j�i
(2:66)
= e
�iHt=�h
1
X
n=0
e
�j�j
2
=2
1
p
n!
�
n
jni
2. QUANTUM DYNAMICS 51
=
1
X
n=0
e
�iE
n
t=�h
e
�j�j
2
=2
1
p
n!
�
n
jni
(2:3:9)
=
1
X
n=0
e
�it
�h
�h!(n+
1
2
)
e
�j�j
2
=2
1
p
n!
�
n
jni
=
1
X
n=0
�
e
�i!t
�
n
e
�i!t=2
e
�j�j
2
=2
1
p
n!
�
n
jni
= e
�i!t=2
1
X
n=0
e
�j�e
�i!t
j
2
=2
(�e
�i!t
)
n
p
n!
jni
(2:66)
= e
�i!t=2
j�e
�i!t
i (2.73)
Thus
aj�(t)i = e
�i!t=2
aj�e
�i!t
i = �e
�i!t
e
�i!t=2
j�e
�i!t
i
= �e
�i!t
j�(t)i: (2.74)
2.8 The quntum mechanical propagator, for a particle with mass
m, moving in a potential is given by:
K(x; y;E) =
Z
1
0
dte
iEt=�h
K(x; y; t; 0) = A
X
n
sin(nrx) sin(nry)
E �
�h
2
r
2
2m
n
2
where A is a constant.
(a) What is the potential?
(b) Determine the constant A in terms of the parameters describing
the system (such as m, r etc. ).
We have
K(x; y;E) �
Z
1
0
dte
iEt=�h
K(x; y; t; 0) �
Z
1
0
dte
iEt=�h
hx; tjy; 0i
=
Z
1
0
dte
iEt=�h
hxje
�iHt=�h
jyi
=
Z
1
0
dte
iEt=�h
X
n
hxje
�iHt=�h
jnihnjyi
=
Z
1
0
dte
iEt=�h
X
n
e
�iE
n
t=�h
hxjnihnjyi
=
X
n
�
n
(x)�
�
n
(y)
Z
1
0
e
i(E�E
n
)t=�h
dt
52
=
X
n
�
n
(x)�
�
n
(y) lim
"!0
"
�i�h
E � E
n
+ i"
e
i(E�E
n
+i")t=�h
#
1
0
=
X
n
�
n
(x)�
�
n
(y)
i�h
E � E
n
: (2.75)
So
X
n
�
n
(x)�
�
n
(y)
i�h
E �E
n
= A
X
n
sin(nrx) sin(nry)
E �
�h
2
r
2
2m
n
2
)
�
n
(x) =
s
A
i�h
sin(nrx); E
n
=
�h
2
r
2
2m
n
2
: (2.76)
For a one dimensional in�nite square well potential with size L the energy
eigenvalue E
n
and eigenfunctions �
n
(x) are given by
�
n
(x) =
s
2
L
sin
�
n�x
L
�
; E
n
=
�h
2
2m
�
�
L
�
2
n
2
: (2.77)
Comparing with (2.76) we get
�
L
= r) L =
�
r
and
V =
(
0 for 0 < x <
�
r
1 otherwise
(2.78)
while
A
i�h
=
2r
�
) A = i
2�hr
�
: (2.79)
2.9 Prove the relation
d�(x)
dx
= �(x)
where �(x) is the (unit) step function, and �(x) the Dirac delta
function. (Hint: study the e�ect on testfunctions.)
For an arbitrary test function f(x) we have
Z
+1
�1
d�(x)
dx
f(x)dx =
Z
+1
�1
d
dx
[�(x)f(x)]dx �
Z
+1
�1
�(x)
df(x)
dx
dx
2. QUANTUM DYNAMICS 53
= �(x)f(x)
�
�
�
�
�
+1
�1
�
Z
+1
0
df(x)
dx
dx
= lim
x!+1
f(x)� f(x)
�
�
�
�
+1
0
= f(0)
=
Z
+1
�1
�(x)f(x)dx)
d�(x)
dx
= �(x): (2.80)
2.10 Derive the following expression
S
cl
=
m!
2 sin(!T )
h
(x
2
0
+ 2x
2
T
) cos(!T )� x
0
x
T
i
for the classical action for a harmonic oscillator moving from the
point x
0
at t = 0 to the point x
T
at t = T .
The Lagrangian for the one dimensional harmonic oscillator is given by
L(x; _x) =
1
2
m _x
2
�
1
2
m!
2
x
2
: (2.81)
From the Lagrange equation we have
@L
@x
�
d
dt
@L
@ _x
= 0
(2:81)
) �m!
2
x�
d
dt
(m _x) = 0)
x+ !
2
x = 0: (2.82)
which is the equation of motion for the system. This can be solved to give
x(t) = A cos!t+B sin!t (2.83)
with boundary conditions
x(t = 0) = x
0
= A (2.84)
x(t = T ) = x
T
= x
0
cos!T +B sin!T ) B sin!T = x
T
� x
0
cos!T )
B =
x
T
� x
0
cos!T
sin!T
: (2.85)
54
So
x(t) = x
0
cos!t+
x
T
� x
0
cos!T
sin!T
sin!t
=
x
0
cos!t sin!T + x
T
sin!t� x
0
cos!T sin!t
sin!T
=
x
T
sin!t+ x
0
sin!(T � t)
sin!T
) (2.86)
_x(t) =
x
T
! cos!t� x
0
! cos!(T � t)
sin!T
: (2.87)
With these at hand we have
S =
Z
T
0
dtL(x; _x) =
Z
T
0
dt
�
1
2
m _x
2
�
1
2
m!
2
x
2
�
=
Z
T
0
dt
"
1
2
m
d
dt
(x _x)�
1
2
mxx�
1
2
m!
2
x
2
#
= �
1
2
m
Z
T
0
dtx[x+ !
2
x] +
m
2
x _x
�
�
�
�
T
0
(2:82)
=
m
2
[x(T ) _x(T )� x(0) _x(0)]
=
m
2
�
x
T
!
sin!T
(x
T
cos!T � x
0
)�
x
0
!
sin!T
(x
T
� x
0
cos!T )
�
=
m!
2 sin!T
h
x
2
T
cos!T � x
0
x
T
� x
0
x
T
+ x
2
0
cos!T
i
=
m!
2 sin!T
h
(x
2
T
+ x
2
0
) cos!T � 2x
0
x
T
i
: (2.88)
2.11 The Lagrangian of the single harmonic oscillator is
L =
1
2
m _x
2
�
1
2
m!
2
x
2
(a) Show that
hx
b
t
b
jx
a
t
a
i = exp
�
iS
cl
�h
�
G(0; t
b
; 0; t
a
)
where S
cl
is the action along the classical path x
cl
from (x
a
; t
a
) to
(x
b
; t
b
) and G is
G(0; t
b
; 0; t
a
) =
2. QUANTUM DYNAMICS 55
lim
N!1
Z
dy
1
: : : dy
N
�
m
2�i�h"
�
(N+1)
2
exp
8
<
:
i
�h
N
X
j=0
�
m
2"
(y
j+1
� y
j
)
2
�
1
2
"m!
2
y
2
j
�
9
=
;
where " =
t
b
�t
a
(N+1)
.
[Hint: Let y(t) = x(t) � x
cl
(t) be the new integration variable,
x
cl
(t) being the solution of the Euler-Lagrange equation.]
(b) Show that G can be written as
G = lim
N!1
�
m
2�i�h"
�
(N+1)
2
Z
dy
1
: : : dy
N
exp(�n
T
�n)
where n =
2
6
6
4
y
1
.
.
.
y
N
3
7
7
5
and n
T
is its transpose. Write the symmetric
matrix �.
(c) Show that
Z
dy
1
: : : dy
N
exp(�n
T
�n) �
Z
d
N
ne
�n
T
�n
=
�
N=2
p
det�
[Hint: Diagonalize � by an orthogonal matrix.]
(d) Let
�
2i�h"
m
�
N
det� � det�
0
N
� p
N
. De�ne j � j matrices �
0
j
that con-
sist of the �rst j rows and j columns of �
0
N
and whose determinants
are p
j
. By expanding �
0
j+1
in minors show the following recursion
formula for the p
j
:
p
j+1
= (2 � "
2
!
2
)p
j
� p
j�1
j = 1; : : : ; N (2.89)
(e) Let �(t) = "p
j
for t = t
a
+ j" and show that (2.89) implies that in
the limit "! 0; �(t) satis�es the equation
d
2
�
dt
2
= �!
2
�(t)
with initial conditions�(t = t
a
) = 0;
d�(t=t
a
)
dt
= 1.
56
(f) Show that
hx
b
t
b
jx
a
t
a
i =
s
m!
2�i�h sin(!T )
exp
(
im!
2�h sin(!T )
[(x
2
b
+ x
2
a
) cos(!T )� 2x
a
x
b
]
)
where T = t
b
� t
a
.
(a) Because at any given point the position kets in the Heisenberg picture
form a complete set, it is legitimate to insert the identity operator written
as
Z
dxjxtihxtj = 1 (2.90)
So
hx
b
t
b
jx
a
t
a
i = lim
N!1
Z
dx
1
dx
2
: : : dx
N
hx
b
t
b
jx
N
t
N
ihx
N
t
N
jx
N�1
t
N�1
i : : :�
hx
i+1
t
i+1
jx
i
t
i
i : : : hx
1
t
1
jx
a
t
a
i: (2.91)
It is
hx
i+1
t
i+1
jx
i
t
i
i = hx
i+1
je
�iH(t
i+1
�t
i
)=�h
jx
i
i = hx
i+1
je
�iH"=�h
jx
i
i
= hx
i+1
je
�i
"
�h
(
1
2
mp
2
+
1
2
m!
2
x
2
)
jx
i
i (since " is very small)
= hx
i+1
je
�i
"
�h
p
2
2m
e
�i
"
�h
1
2
m!
2
x
2
jx
i
i
= e
�i
"
�h
1
2
m!
2
x
2
i
hx
i+1
je
�i
"
�h
p
2
2m
jx
i
i: (2.92)
For the second term in this last equation we have
hx
i+1
je
�i
"
�h
p
2
2m
jx
i
i =
Z
dp
i
hx
i+1
je
�i
"
�h
p
2
2m
jp
i
ihp
i
jx
i
i
=
Z
dp
i
e
�i
"
�h
p
2
i
2m
hx
i+1
jp
i
ihp
i
jx
i
i
=
1
2��h
Z
dp
i
e
�i
"
�h
p
i
2
2m
e
ip
i
(x
i+1
�x
i
)=�h
=
1
2��h
Z
dp
i
e
�i
"
2m�h
h
p
2
i
�2p
i
m
"
(x
i+1
�x
i
)+
m
2
"
2
(x
i+1
�x
i
)
2
�
m
2
"
2
(x
i+1
�x
i
)
2
i
=
1
2��h
e
i"
2m�h
m
2
"
2
(x
i+1
�x
i
)
2
Z
dp
i
e
�i
"
2m�h
[
p
i
�p
i
m
"
(x
i+1
�x
i
)
]
2
2. QUANTUM DYNAMICS 57
=
1
2��h
e
i"
2m�h
m
2
"
2
(x
i+1
�x
i
)
2
s
�2�hm
i"
=
r
m
2��hi"
e
im
2�h"
(x
i+1
�x
i
)
2
: (2.93)
Substituting this in (2.92) we get
hx
i+1
t
i+1
jx
i
t
i
i =
�
m
2�i�h"
�
1
2
e
i
�h
[
m
2"
(x
i+1
�x
i
)
2
�
1
2
"m!
2
x
i
]
(2.94)
and this into (2.91):
hx
b
t
b
jx
a
t
a
i �
Z
Dx exp
�
i
�h
S[x]
�
=
lim
N!1
Z
dx
1
: : : dx
N
�
m
2�i�h"
�
(N+1)
2
exp
8
<
:
i
�h
N
X
j=0
�
m
2"
(x
j+1
� x
j
)
2
�
1
2
"m!
2
x
2
j
�
9
=
;
:
Let y(t) = x(t)� x
cl
(t)) x(t) = y(t)+ x
cl
(t)) _x(t) = _y(t) + _x
cl
(t) with
boundary conditions y(t
a
) = y(t
b
) = 0: For this new integration variable we
have Dx = Dy and
S[x] = S[y + x
cl
] =
Z
t
b
t
a
L(y + x
cl
; _y + _x
cl
)dt
=
Z
t
b
t
a
2
4
L(x
cl
; _x
cl
) +
@L
@x
�
�
�
�
�
x
cl
y +
@L
@ _x
�
�
�
�
�
x
cl
_y +
1
2
@
2
L
@
2
x
�
�
�
�
�
x
cl
y
2
+
1
2
@
2
L
@ _x
2
�
�
�
�
�
x
cl
_y
2
3
5
= S
cl
+
@L
@ _x
y
�
�
�
�
�
t
b
t
a
+
Z
t
b
t
a
"
@L
@x
�
d
dt
 
@L
@ _x
!#
�
�
�
�
�
x
cl
y +
Z
t
b
t
a
h
1
2
m _y
2
�
1
2
m!
2
y
2
i
dt:
So
hx
b
t
b
jx
a
t
a
i =
Z
Dy exp
�
i
�h
S
cl
+
i
�h
Z
t
b
t
a
h
1
2
m _y
2
�
1
2
m!
2
y
2
i
dt
�
= exp
�
iS
cl
�h
�
G(0; t
b
; 0; t
a
) (2.95)
with
G(0; t
b
; 0; t
a
) =
lim
N!1
Z
dy
1
: : : dy
N
�
m
2�i�h"
�
(N+1)
2
exp
8
<
:
i
�h
N
X
j=0
�
m
2"
(y
j+1
� y
j
)
2
�
1
2
"m!
2
y
2
j
�
9
=
;
:
58
(b) For the argument of the exponential in the last relation we have
i
�h
N
X
j=0
�
m
2"
(y
j+1
� y
j
)
2
�
1
2
"m!
2
y
2
j
�
(y
0
=0)
=
i
�h
N
X
j=0
m
2"
(y
2
j+1
+ y
2
j
� y
j+1
y
j
� y
j
y
j+1
)�
i
�h
N
X
i;j=1
1
2
"m!
2
y
i
�
ij
y
j
(y
N+1
=0)
=
�
m
2"i�h
N
X
i;j=1
(2y
i
�
ij
y
j
� y
i
�
i;j+1
y
j
� y
i
�
i+1;j
y
j
)�
i"m!
2
2�h
N
X
i;j=1
y
i
�
ij
y
j
:(2.96)
where the last step is written in such a form so that the matrix � will be
symmetric. Thus we have
G = lim
N!1
�
m
2�i�h"
�
(N+1)
2
Z
dy
1
: : : dy
N
exp(�n
T
�n) (2.97)
with
� =
m
2"i�h
2
6
6
6
6
6
6
6
6
6
4
2 �1 0 : : : 0 0
�1 2 �1 : : : 0 0
0 �1 2 : : : 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 : : : 2 �1
0 0 0 : : : �1 2
3
7
7
7
7
7
7
7
7
7
5
+
i"m!
2
2�h
2
6
6
6
6
6
6
6
6
6
4
1 0 0 : : : 0 0
0 1 0 : : : 0 0
0 0 1 : : : 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 : : : 1 0
0 0 0 : : : 0 1
3
7
7
7
7
7
7
7
7
7
5
:(2.98)
(c) We can diagonalize � by a unitary matrix U . Since � is symmetric the
following will hold
� = U
y
�
D
U ) �
T
= U
T
�
D
(U
y
)
T
= U
T
�
D
U
�
= �) U = U
�
: (2.99)
So we can diagonalize � by an orthogonal matrix R. So
� = R
T
�
D
R and detR = 1 (2.100)
which means that
Z
d
N
ne
�n
T
�n
=
Z
d
N
ne
�n
T
R
T
�Rn
Rn=�
=
Z
d
N
�e
��
T
��
=
�
Z
d�
1
e
��
2
1
a
1
� �
Z
d�
2
e
��
2
2
a
2
�
: : :
�
Z
d�
N
e
��
2
N
a
N
�
=
s
�
a
1
s
�
a
2
: : :
s
�
a
N
=
�
N=2
q
Q
N
i=1
a
i
=
�
N=2
p
det�
D
=
�
N=2
p
det�
(2.101)
2. QUANTUM DYNAMICS 59
where a
i
are the diagonal elements of the matrix �
D
.
(d) From (2.98) we have
 
2i�h"
m
!
N
det� =
det
8
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
:
2
6
6
6
6
6
6
6
6
6
4
2 �1 0 : : : 0 0
�1 2 �1 : : : 0 0
0 �1 2 : : : 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 : : : 2 �1
0 0 0 : : : �1 2
3
7
7
7
7
7
7
7
7
7
5
� "
2
!
2
2
6
6
6
6
6
6
6
6
6
4
1 0 0 : : : 0 0
0 1 0 : : : 0 0
0 0 1 : : : 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 : : : 1 0
0 0 0 : : : 0 1
3