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
1
2
3
4
5
Fundamental Concepts . . . . . . .
Quantum Dynamics . . . . . . . . .
Theory of Angular Momentum . . .
Symmetry in Quantum Mechanics .
Approximation Methods . . . . . .
II Solutions
1
2
3
4
5
Fundamental Concepts . . . . . . .
Quantum Dynamics . . . . . . . . .
Theory of Angular Momentum . . .
Symmetry in Quantum Mechanics .
Approximation Methods . . . . . .
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23
2
CONTENTS
Part I
Problems
3
1. FUNDAMENTAL CONCEPTS
5
1 Fundamental Concepts
1.1 Consider a ket space spanned by the eigenkets fja0ig of a Hermitian operator A. There is no degeneracy.
(a) Prove that
Y
(A ; a0)
a0
is a null operator.
(b) What is the signicance of
Y (A ; a00)
0
00 ?
a00 6=a0 a ; a
(c) Illustrate (a) and (b) using A set equal to Sz of a spin 21 system.
1.2 A spin 12 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 Sx is measured. What is the probability of getting
+h=2?
(b) Evaluate the dispersion in Sx, that is,
h(Sx ; hSx i)2i:
(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
(hj + h j) (ji + 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 relation holds if the state in question satises
Aji = B ji
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
"
0 (x0 ; hxi)2 #
i
h
p
i
x
0
2
;
1
=
4
;
hx ji = (2d ) exp
h
4d2
satises the uncertainty relation
q
q
h(x)2i h(p)2i = h2 :
Prove that the requirement
hx0jxji = (imaginary number)hx0jpji
is indeed satised for such a Gaussian wave packet, in agreement
with (b).
1.4 (a) Let x and px be the coordinate and linear momentum in
one dimension. Evaluate the classical Poisson bracket
[x; F (px)]classical :
(b) Let x and px be the corresponding quantum-mechanical operators this time. Evaluate the commutator
x; exp iphxa :
(c) Using the result obtained in (b), prove that
ipxa exp h jx0i; (xjx0i = x0jx0i)
2. QUANTUM DYNAMICS
7
is an eigenstate of the coordinate operator x. What is the corresponding eigenvalue?
1.5 (a) Prove the following:
(i) hp0jxji = ih @p@ 0 hp0ji;
Z
@ (p0);
(ii) h jxji = dp0 (p0)ih @p
0 where (p0) = hp0 ji and (p0) = hp0j i are momentum-space wave
functions.
(b) What is the physical signicance of
ix exp 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
eB H = ; mc
Sz = !Sz ;
write the Heisenberg equations of motion for the time-dependent
operators Sx(t), Sy (t), and Sz (t). Solve them to obtain Sx;y;z as functions 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
2
H = 2p~m + V (~x):
By calculating [~x p~; H ] obtain
* +
d h~x ~pi = p2 ; h~x r
~ V i:
dt
m
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 ;hipa j0i:
You may use
2
!23 0
!1=21
0
h
x
hx0j0i = ;1=4x;0 1=2 exp 4; 21 x 5 ; @x0 m! A :
0
(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, dened
by
C (t) = hx(t)x(0)i;
where x(t) is the position operator in the Heisenberg picture. Evaluate 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 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)2i as a function of time using either picture.
2.7 A coherent state of a one-dimensional simple harmonic oscillator is dened to be an eigenstate of the (non-Hermitian) annihilation operator a:
aji = ji;
where is, in general, a complex number.
(a) Prove that
ji = e;jj =2eay j0i
is a normalized coherent state.
(b) Prove the minimum uncertainty relation for such a state.
(c) Write ji as
1
X
ji = f (n)jni:
2
n=0
Show that the distribution of jf (n)j2 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 ji remains coherent under timeevolution and calculate the time-evolved state j(t)i. (Hint: directly 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 ) =
Z1
0
dteiEt=hK (x; y; t; 0) = A
X sin(nrx) sin(nry)
E ; h22mr2 n2
n
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) = (x)
dx
where (x) is the (unit) step function, and (x) the Dirac delta
function. (Hint: study the eect on testfunctions.)
2.10 Derive the following expression
m! h(x2 + 2x2 ) cos(!T ) ; x x i
Scl = 2 sin(
0 T
T
!T ) 0
for the classical action for a harmonic oscillator moving from the
point x0 at t = 0 to the point xT at t = T .
2.11 The Lagrangian of the single harmonic oscillator is
L = 21 mx_ 2 ; 12 m!2x2
2. QUANTUM DYNAMICS
(a) Show that
11
hxbtbjxatai = exp iShcl G(0; tb; 0; ta)
where Scl is the action along the classical path xcl from (xa; ta) to
(xb; tb) and G is
G(0; tb; 0; ta) =
m (N2+1) 8
9
Z
N m
<i X
=
1
2 ; "m! 2y 2
lim
dy
:
:
:
dy
exp
(
y
;
y
)
1
N
j ;
: h
N !1
2ih"
2" j+1 j
2
j =0
ta .
where " = (tNb;+1)
(Hint: Let y(t) = x(t) ; xcl(t) be the new integration variable,
xcl(t) being the solution of the Euler-Lagrange equation.)
(b) Show that G can be written as
m (N2+1) Z
G = Nlim
dy1 : : : dyN exp(;nT n)
!1 2ih"
2 3
66 y..1 77
where n = 4 . 5 and nT is its transpose. Write the symmetric
yN
matrix .
(c) Show that
Z
dy1 : : : dyN
exp(;nT n) Z
dN ye;nT n
= p
N=2
det
[Hint: Diagonalize by an orhogonal matrix.]
(d) Let 2mih" N det detN0 pN . Dene j j matrices j0 that consist of the rst j rows and j columns of N0 and whose determinants
are pj . By expanding j0 +1 in minors show the following recursion
formula for the pj :
pj+1 = (2 ; "2!2)pj ; pj;1
j = 1; : : : ; N
(2.1)
12
(e) Let (t) = "pj for t = ta + j" and show that (2.1) implies that in
the limit " ! 0; (t) satises the equation
d2 = ;!2(t)
dt2
with initial conditions (t = ta) = 0; d(tdt=ta) = 1.
(f) Show that
(
)
im!
m!
2
2
hxbtbjxatai = 2ih sin(!T ) exp 2h sin(!T ) [(xb + xa) cos(!T ) ; 2xaxb]
where T = tb ; ta.
s
2.12 Show the composition property
Z
dx1Kf (x2; t2; x1; t1)Kf (x1; t1; x0; t0) = Kf (x2; t2; x0; t0)
where Kf (x1; t1; x0; t0) 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
h
e
[i; j ] = c "ijk Bk
where ~ m dt~x = ~p ; ecA~ and the relation
"
!#
2~x
~
d
1
d~
x
d~
x
d
~
~
~
m dt2 = dt = e E + 2c dt B ; B dt :
(b) Verify the continuity equation
@ + r
~ 0 ~j = 0
@t
2. QUANTUM DYNAMICS
with ~j given by
13
!
e
h
0
~
~ j j2:
~j =
=
(
r
)
;
A
m
mc
2.14 An electron moves in the presence of a uniform magnetic eld
in the z-direction (B~ = B z^).
(a) Evaluate
where
[x; y ];
y py ; eAc y :
x px ; eAc x ;
(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
!
2k 2
h
j
eB
j
h
1
E =
+
n+ ;
k;n
2m
mc
2
where hk is the continuous eigenvalue of the pz 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 cylinder 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 specied by
E.
(b) Discuss briey 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
! ;i3 ;
i
;
i
2
3
(1
=
2)
exp
exp
D (; ; ) = exp
2
=
e;i(+)=2 cos 2
ei(;)=2 sin 2
2
;e;i(;)=2 sin 2
ei(+)=2 cos 2
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 = mmax = j i is rotated
by an innitesimal angle
" about the y-axis. Without using the
(j )
explicit form of the dm0m 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
ml states?
(c) Suppose it is known somehow that (~x) is an energy eigenfunction 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 direction.
(a) Show explicitly that innitesimal rotations of %i(~x) are obtained
by acting with the operator
u~" = 1 ; i h~" (L~ + S~)
(3.1)
where L~ = hi r^ r~ . Determine S~ !
(b) Show that L~ and S~ commute.
(c) Show that S~ is a vector operator.
(d) Show that r~ %~(~x) = h1 (S~ ~p)%~ where p~ is the momentum operator.
2
3.5 We are to add angular momenta j1 = 1 and j2 = 1 to form
j = 2; 1; and 0 states. Using the ladder operator method express all
16
(nine) j; m eigenkets in terms of jj1j2; m1m2i. Write your answer as
jj = 1; m = 1i = p1 j+; 0i ; p1 j0; +i; : : : ;
2
2
where + and 0 stand for m1;2 = 1; 0; respectively.
(3.2)
3.6 (a) Construct a spherical tensor of rank 1 out of two dierent
vectors U~ = (Ux; Uy ; Uz ) and V~ = (Vx ; Vy ; Vz ). Explicitly write T(1)1;0 in
terms of Ux;y;z and Vx;y;z .
(b) Construct a spherical tensor of rank 2 out of two dierent
vectors U~ and V~ . Write down explicitly T(2)2;1;0 in terms of Ux;y;z
and Vx;y;z .
3.7 (a) Evaluate
j
X
m=;j
j)
jd(mm
0 ( )j2m
for any j (integer or half-integer); then check your answer for j = 21 .
(b) Prove, for any j ,
j
X
m=;j
m2jdm(j)0 m( )j2 = 12 j (j + 1) sin + m02 + 12 (3 cos2 ; 1):
[Hint: This can be proved in many ways. You may, for instance,
examine the rotational properties of Jz2 using the spherical (irreducible) tensor language.]
3.8 (a) Write xy, xz, and (x2 ; y2) as components of a spherical
(irreducible) tensor of rank 2.
4. SYMMETRY IN QUANTUM MECHANICS
17
(b) The expectation value
Q eh; j; m = j j(3z2 ; r2)j; j; m = j i
is known as the quadrupole moment. Evaluate
eh; j; m0j(x2 ; y2)j; j; m = j i;
(where m0 = j; j ; 1; j ; 2; : : : )in terms of Q and appropriate ClebschGordan coecients.
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 ei~p~x=h. Why does this not violate time-reversal
invariance?
4.2 Let (~p0) be the momentum-space wave function for state ji,
that is, (~p0) = h~p0 ji.Is the momentum-space wave function for the
time-reversed state ji given by (~p0, (;~p0), (~p0), or (;~p0)?
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 energybands in solids are described by the so called Bloch functions
n;k fulllling,
ika
n;k (x + a) = e
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,
X
n(x ; Ri)eikRi
n;k (x) =
Ri
where the sum is over all lattice vectors Ri. (In this simble one dimensional problem Ri = ia, but the construction generalizes easily
to three dimensions.).
The functions n are called Wannier functions, and are important in the tight-binding description of solids. Show that the Wannier functions are corresponding to dierent sites and/or dierent
bands are orthogonal, i:e: prove
Z
dx?m(x ; Ri)n(x ; Rj ) ij mn
Hint: Expand the n s in Bloch functions and use their orthonormality 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
hL~ i = 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
XX
l m
Flm(r)Ylm (; );
what kind of phase restrictions do we obtain on Flm(r)?
4.5 The Hamiltonian for a spin 1 system is given by
H = ASz2 + B (Sx2 ; Sy2):
5. APPROXIMATION METHODS
19
Solve this problem exactly to nd the normalized energy eigenstates 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 obtained transform under time reversal?
5 Approximation Methods
5.1 Consider an isotropic harmonic oscillator in two dimensions.
The Hamiltonian is given by
2
p2y m!2 2 2
p
x
H0 = 2m + 2m + 2 (x + y )
(a) What are the energies of the three lowest-lying states? Is there
any degeneracy?
(b) We now apply a perturbation
V = m!2xy
where is a dimensionless real number much smaller than unity.
Find the zeroth-order energy eigenket and the corresponding energy 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 H0 + V problem exactly. Compare with the perturbation results obtained in q(b). p
p
[You may use hn0jxjni = h=2m! ( n + 1 0 + n 0 ):]
n ;n+1
n ;n;1
5.2 A system that has three unperturbed states can be represented
by the perturbed Hamiltonian matrix
0
1
E1 0 a
B@ 0 E1 b CA
a b E2
20
where E2 > E1. The quantities a and b are to be regarded as perturbations that are of the same order and are small compared with
E2 ; E1. 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) = F0e;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
states?
q excited
p
p
0
[You may use hn jxjni = h=2m! ( n + 1n0;n+1 + nn0;n;1 ):]
5.4 Consider a composite system made up of two spin 21 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
4 H = 2 S~1 S~2:
h
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) = V0cos(kz ; !t):
Using time-dependent perturbation theory, obtain an expression
for the transition rate at which the electron is emitted with momentum p~. Show, in particular, how you may compute the angular
distribution of the ejected electron (in terms of and dened
with respect to the z-axis). Discuss briey the similarities and the
dierences between this problem and the (more realistic) photoelectric eect. (note: For the initial wave function use
Z 2
1
;Zr=a0 :
n=1;l=0(~x) = p
e
a0
3
If you have a normalization problem, the nal wave function may
be taken to be
1
f (~x) =
L
3
2
ei~p~x=h
with L very large, but you should be able to show that the observable eects are independent of L.)
22
Part II
Solutions
23
1. FUNDAMENTAL CONCEPTS
25
1 Fundamental Concepts
1.1 Consider a ket space spanned by the eigenkets fja0ig of a Hermitian operator A. There is no degeneracy.
(a) Prove that
Y
(A ; a0)
a0
is a null operator.
(b) What is the signicance of
Y (A ; a00)
0
00 ?
a00 6=a0 a ; a
(c) Illustrate (a) and (b) using A set equal to Sz of a spin 21 system.
(a) Assume that ji is an arbitrary state ket. Then
X Y
Y
Y
X
(A ; a0)ji = (A ; a0) ja00i h|a00{zji} = ca00 (A ; a0)ja00i
a0
a0
a00
a00
a0
ca00
X Y 00 0 00 a002fa0g
=
ca00 (a ; a )ja i = 0:
(1.1)
a00
a0
(b) Again for an arbitrary state ji we will have
2
3
2
3
1
00)
00) zX }| {
Y
Y
(
A
;
a
(
A
;
a
4
5
4
5 ja000iha000 ji
a0 ; a00 ji =
a0 ; a00
a00 6=a0
a000
X 000 Y (a000 ; a00) 000
ha ji
0
00 ja i =
a000
a00 =
6 a0 a ; a
X 000
ha jia000a0 ja000i = ha0jija0i )
a000
a00 6=a0
=
=
2
3
00)
Y
(
A
;
a
4
5 = ja0iha0j a0 :
0 ; a00
a
00
0
a 6=a
So it projects to the eigenket ja0i.
(1.2)
26
(c) It is Sz 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
Y
(Sz ; a0) = (Sz ; a01)
a0
"a0
#
h
h
= 2 (j+ih+j ; j;ih;j) ; 2 (j+ih+j + j;ih;j)
"
#
h
h
2 (j+ih+j ; j;ih;j) + 2 (j+ih+j + j;ih;j)
0
z
}|
{
= [;hj;ih;j][hj+ih+j] = ;h2j;i h;j+ih+j = 0; (1.3)
where we have used that j+ih+j + j;ih;j = 1.
For a0 = h=2 we have
Y (Sz ; a001) Sz + h2 1
Y (Sz ; a00)
0
00 = 00
00 = h=2 + h=2
a00 6=a0 a ; a
a 6=h=2 h=2 ; a
"
#
1
h
h
= h 2 (j+ih+j ; j;ih;j) + 2 (j+ih+j + j;ih;j)
= h1 hj+ih+j = j+ih+j:
(1.4)
Similarly for a0 = ;h=2 we have
Y (Sz ; a001)
Y (Sz ; a00)
Sz ; h2 1
=
=
0
00
00 ;h=2 ; h=2
a00 6=a0 a ; a
a00 6=;h=2 ;h=2 ; a
"
#
1
h
h
= ; h 2 (j+ih+j ; j;ih;j) ; 2 (j+ih+j + j;ih;j)
= ; h1 (;hj;ih;j) = j;ih;j:
(1.5)
1.2 A spin 12 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 Sx is measured. What is the probability of getting
+h=2?
1. FUNDAMENTAL CONCEPTS
27
(b) Evaluate the dispersion in Sx, that is,
h(Sx ; hSx i)2i:
(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 + Sx sin = # [(S-1.3.36),(S-1.4.18)]
"
#
h
h
= 2 (j+ih+j ; j;ih;j) cos + 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^ jS~ n^ ; +i = h =2jS~ n^ ; +i:
(1.8)
From (1.7) we have that
!
~S n^ = h cos sin :
(1.9)
2 sin ; cos The eigenvalues and eigenfuncions of this operator can be found if one solves
the secular equation
!
h
=
2
cos
;
h
=
2
sin
det(S~ n^ ; I ) = 0 ) det
h=2 sin ;h=2 cos ; = 0 )
2
2
2
; h4 cos2 + 2 ; h4 sin2 = 0 ) 2 ; h4 = 0 ) = h2 : (1.10)
!
a
~
Since the system is in the eigenstate jS n^; +i b we will have that
! !
! (
)
h cos sin a = h a ) a cos + b sin = a )
b
a sin ; b cos = b
2 sin ; cos 2 b
= a 2 sin2 2 = a tan :
b = a 1 ;sincos
(1.11)
2 sin 2 cos 2
2
28
But we want also the eigenstate jS~ n^ ; +i to be normalized, that is
a2 + b2 = 1 ) a2 + a2 tan2 2 = 1 ) a2 cos2 2 a2 sin2 2 = cos2 2
r 2
2
) a = cos 2 ) a = cos2 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 Sz operator by
(1.13)
jS~ n^ ; +i = cos 2 j+i + sin 2 j;i:
(a) From (S-1.4.17) we know that
jSx; +i = p12 j+i + p12 j;i:
(1.14)
So the propability of getting +h=2 when Sx is measured is given by
!
2
2
1
1
hSx; +jS~ n^ ; +i = p h+j + p h;j cos 2 j+i + sin 2 j;i 2
2
2
1
1
= p cos 2 + p sin 2 2
2
1
1
= 2 cos2 2 + 2 sin2 2 + cos 2 sin 2
= 12 + 12 sin = 12 (1 + sin ):
(1.15)
For = 0 which means that the system is in the jSz ; +i eigenstate we have
jhSx ; +jSz ; +ij2 = 21 (1) = 12 :
(1.16)
For = =2 which means that the system is in the jSx; +i eigenstate we
have
jhSx; +jSx; +ij2 = 1:
(1.17)
For = which means that the system is in the jSz ; ;i eigenstate we have
jhSx; +jSz ; ;ij2 = 21 (1) = 21 :
(1.18)
1. FUNDAMENTAL CONCEPTS
29
(b) We have that
h(Sx ; hSxi)2i = hSx2i ; (hSx i)2:
(1.19)
As we know
Sx = h2 (j+ih;j + j;ih+j) )
2
Sx2 = h4 (j+ih;j + j;ih+j) (j+ih;j + j;ih+j) )
2
2
h
h
2
Sx = 4 (|j+ih+j +
{z j;ih;j)} = 4 :
1
So
hSx i =
=
(hSxi)2 =
hSx2i =
=
(1.20)
h
cos 2 h+j + sin 2 h;j 2 (j+ih;j + j;ih+j) cos 2 j+i + sin 2 j;i
h cos sin + h sin cos = h sin )
2 2 2 2 2 2 2
h2 sin2 and
h2 4 cos 2 h+j + sin 2 h;j 4 cos 2 j+i + sin 2 j;i
h2 [cos2 + sin2 ] = h2 :
(1.21)
4
2
2 4
So substituting in (1.19) we will have
2
2
h(Sx ; hSxi)2i = h4 (1 ; sin2 ) = h4 cos2 :
(1.22)
and nally
2
h(Sx)2i=0;jSz ;+i = h4 ;
h(Sx)2i==2;jSx ;+i = 0;
2
h(Sx)2i=0;jSz ;;i = h4 :
(1.23)
(1.24)
(1.25)
30
1.3 (a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
(hj + h j) (ji + 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 relation holds if the state in question satises
Aji = B ji
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
"
0 (x0 ; hxi)2 #
i
h
p
i
x
0
2
;
1
=
4
;
hx ji = (2d ) exp
satises the uncertainty relation
q
h
4d2
q
h(x)2i
h(p)2i = h2 :
Prove that the requirement
hx0jxji = (imaginary number)hx0jpji
is indeed satised 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 = ji + j i where is a complex number,
we will have
(hj + h j) (ji + j i) 0 )
(1.27)
2
hji + hj i + h ji + jj h j i 0:
(1.28)
1. FUNDAMENTAL CONCEPTS
31
If we now choose = ;h ji=h j i the previous relation will be
hji ; hjhihjiji ; hjhihjiji + jhhjjiji 0 )
hjih j i jh jij2:
2
(1.29)
Notice that the equality sign in the last relation holds when
jci = ji + j i = 0 ) ji = ;j i
that is if ji and j i are colinear.
(1.30)
(b) The uncertainty relation is
h(A)2ih(B )2i 14 jh[A; B ]ij2 :
(1.31)
h(A)2ih(B )2i jhAB ij2:
(1.32)
To prove this relation we use the Schwarz inequality (1.29) for the vectors
ji = Ajai and j i = B jai which gives
The equality sign in this relation holds according to (1.30) when
Ajai = B jai:
(1.33)
On the other hand the right-hand side of (1.32) is
(1.34)
jhAB ij2 = 41 jh[A; B ]ij2 + 14 jhfA; B gij2
which means that the equality sign in the uncertainty relation (1.31) holds if
1 jhfA; B gij2 = 0 ) hfA; B gi = 0
4
) hajAB + B Ajai = 0 (1):33) haj(B )2jai + haj(B )2jai = 0
) ( + )haj(B )2jai = 0:
(1.35)
Thus the equality sign in the uncertainty relation holds when
Ajai = B jai
with purely imaginary.
(1.36)
32
(c) We have
hx0jxji hx0j(x ; hxi)ji = x0hx0ji ; hxihx0ji
= (x0 ; hxi)hx0ji:
(1.37)
On the other hand
But
hx0jpji hx0j(p ; hpi)ji
@ hx0ji ; hpihx0 ji
= ;ih @x
0
"
#
@ hx0ji = hx0ji @ ihpix0 ; (x0 ; hxi)2
0
@x0
@x
h
4d2 #
"
= hx0ji ihhpi ; 21d2 (x0 ; hxi)
(1.38)
(1.39)
So substituting in (1.38) we have
hx0jpji = hpihx0 ji + 2idh2 (x0 ; hxi) hx0ji ; hpihx0 ji
= 2idh2 (x0 ; hxi) hx0ji = 2idh2 hx0jxji )
2
;
i
2
d
0
hx jxji = h hx0jpji:
(1.40)
1.4 (a) Let x and px be the coordinate and linear momentum in
one dimension. Evaluate the classical Poisson bracket
[x; F (px)]classical :
(b) Let x and px be the corresponding quantum-mechanical operators this time. Evaluate the commutator
ipxa x; exp h :
(c) Using the result obtained in (b), prove that
ipxa exp h jx0i; (xjx0i = x0jx0i)
1. FUNDAMENTAL CONCEPTS
33
is an eigenstate of the coordinate operator x. What is the corresponding eigenvalue?
(a) We have
@x @F (px) ; @x @F (px)
[x; F (px)]classical @x
@px
@px @x
= @F (px) :
@px
(1.41)
(b) When x and px are treated as quantum-mechanical operators we have
" X
ipxa 1 (ia)n pn # X
1 1 (ia)n
x
n]
x; exp h
= x;
=
[
x;
p
n
n
x
n=0 h n!
n=0 n! h
;1
1 1 (ia)n nX
X
k [x; p ] pn;k;1
p
=
x x
x
n
n=0 n! h k=0
1 1 (ia)n
nX
;1
1 n (ia)n;1
X
X
n;1 (;a)
k
n
;
k
;
1
=
p
(
i
h
)
p
p
=
x
x
x
n
n
;
1
n=1 n! h
n=1 n! h
k=0
n;1
ipxa 1
X
= ;a (n ;1 1)! ia
p
=
;
a
exp
h x
h : (1.42)
n=1
(c) We have now
ipxa ipxa ipxa (b)
0
0
x exp h jx i = exp h xjx i ; a exp h jx0i
= x0 exp iphxa jx0i ; a exp iphxa jx0i
= (x0 ; a) exp iphxa jx0i:
(1.43)
So exp iphxa jx0i is an eigenstate of the operator x with eigenvalue x0 ; a.
So we can write
ipxa 0
(1.44)
jx ; ai = C exp h jx0i;
where C is a constant which due to normalization can be taken to be 1.
34
1.5 (a) Prove the following:
(i) hp0 jxji = ih @ 0 hp0ji;
Z @p
@ (p0);
(ii) h jxji = dp0 (p0)ih @p
0 where (p0) = hp0 ji and (p0) = hp0j i are momentum-space wave
functions.
(b) What is the physical signicance of
ix exp
h ;
where x is the position operator and is some number with the
dimension of momentum? Justify your answer.
(a) We have
(i)
hp0 jxji =
=
=
=
hp0 jxji =
(ii)
}|1 {
Z
hp0 jx dx0jx0ihx0j i = dx0hp0 jxjx0ihx0 ji
Z
Z
ip0 x0
dx0x0hp0jx0ihx0ji (S;1=:7:32) dx0x0Ae; h hx0ji
ip0x0 Z
Z
ip0 x0
A dx0 @p@ 0 e; h (ih)hx0ji = ih @p@ 0 dx0Ae; h hx0ji
Z
@
0
0
0
0
ih @p0 dx hp jx ihx ji = ih @p@ 0 hp0 ji )
ih @p@ 0 hp0ji:
(1.45)
zZ
Z
Z
h jxji = dp0 h jp0ihp0 jxji = dp0 (p0)ih @p@ 0 (p0);
(1.46)
where we have used (1.45) and that h jp0i = (p0) and hp0ji = (p0).
1. FUNDAMENTAL CONCEPTS
ix 35
(b) The operator exp h gives translation in momentum space. This can
be justied by calculating the following operator
" X
ix 1 1 ix n # X
1 1 i n
p; exp h
= p; n! h
= n! h [p; xn ]
n=0
n=0
n
1 1 i n X
X
xn;k [p; x]xk;1
=
n
!
h
n=1
k=1
n
1
n
1 1 i n
X 1 i X
X
n
;
1
=
(;ih)x = n! h n(;ih)xn;1
n
!
h
n=1
n=1
k=1
i 1
n
;
1
1 1 ix n
X
X
1
i
n
;
1
=
x (;ih) h = n! h
n=1 (n ; 1)! h
n=0
ix = exp h :
(1.47)
So when this commutator acts on an eigenstate jp0i of the momentum operator we will have
ix ix ix 0
p; exp h jp i = p exp h jp0i ; exp h p0jp0i )
ix ix ix exp h = p exp h jp0i ; p0 exp h jp0i )
(1.48)
p exp ixh jp0i = (p0 + ) exp ixh jp0i :
Thus we have that
exp ixh jp0i Ajp0 + 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
eB H = ; mc Sz = !Sz ;
write the Heisenberg equations of motion for the time-dependent
operators Sx(t), Sy (t), and Sz (t). Solve them to obtain Sx;y;z as functions of time.
Let us rst prove the following
[AS ; BS ] = CS ) [AH ; BH ] = CH :
(2.1)
Indeed we have
h
i
[AH ; BH ] = U yAS U ; U yBS U = U yAS BS U ; U yBS AS U
= U y [AS ; BS ] U = U yCS U = CH :
(2.2)
The Heisenberg equation of motion gives
dSx = 1 [S ; H ] = 1 [S ; !S ] (S;1=:4:20) ! (;ihS ) = ;!S ; (2.3)
y
y
dt
ih x
ih x z
ih
dSy = 1 [S ; H ] = 1 [S ; !S ] (S;1=:4:20) ! (ihS ) = !S ;
(2.4)
x
dt
ih y
ih y z
ih x
dSz = 1 [S ; H ] = 1 [S ; !S ] (S;1=:4:20) 0 ) S = constant: (2.5)
z
dt
ih z
ih z z
Dierentiating once more eqs. (2.3) and (2.4) we get
d2Sx = ;! dSy (2=:4) ;!2S ) S (t) = A cos !t + B sin !t ) S (0) = A
x
x
x
dt2
dt
d2Sy = ! dSx (2=:3) ;!2S ) S (t) = C cos !t + D sin !t ) S (0) = C:
y
y
y
dt2
dt
But on the other hand
dSx = ;!S )
y
dt
;A! sin !t + B! cos !t = ;C! cos !t ; D! sin !t )
A=D
C = ;B:
(2.6)
2. QUANTUM DYNAMICS
So, nally
37
Sx(t) = Sx(0) cos !t ; Sy (0) sin !t
Sy (t) = Sy (0) cos !t + Sx(0) sin !t
Sz (t) = Sz (0):
(2.7)
(2.8)
(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
2
H = 2pm :
(2.10)
This means that the Heisenberg equations of motion for the operators x and
p will be
"
#
@p(t) = 1 [p(t); H (t)] = 1 p(t); p2(t) = 0 )
@t
ih
ih
2m
p(t) = p(0)
(2.11)
"
#
2
@x(t) = 1 [x; H ] = 1 x(t); p (t) = 1 2p(t)ih = p(t) (2=:11) p(0) )
@t
ih
ih
2m
2mih
m
m
x(t) = mt p(0) + x(0):
(2.12)
Thus nally
ht : (2.13)
[x(t); x(0)] = mt p(0) + x(0); x(0) = mt [p(0); x(0)] = ; im
2.3 Consider a particle in three dimensions whose Hamiltonian is
given by
2
H = 2~pm + V (~x):
38
By calculating [~x p~; H ] obtain
* +
d h~x ~pi = p2 ; h~x r
~ V i:
dt
m
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 ]
3
"
# 2X
3
3 p2
2
X
p
~
[~x ~p; H ] = ~x ~p; 2m + V (~x) = 4 xipi; 2mj + V (~x)5
i=1
j =1
#
"
X
X
p2
(2.14)
=
xi; 2mj pi + xi [pi ; V (~x)] :
i
ij
The rst commutator in (2.14) will give
" 2#
p
xi; 2mj = 21m [xi; p2j ] = 21m (pj [xi; pj ] + [xi; pj ]pj ) = 21m (pj ihij + ihij pj )
= 21m 2ihij pj = imh ij pj :
(2.15)
The second commutator can be calculated if we Taylor expand
the function
P
V (~x) in terms of xi which means that we take V (~x) = n anxni with an
independent of xi. So
" X
# X
1
;1
X nX
[pi; V (~x)] = pi; anxni = an [pi ; xni] = an xki [pi ; xi] xni ;k;1
n=0
n
n
k=0
;1
X nX
X
@ X a xn
=
an (;ih)xni ;1 = ;ih annxni ;1 = ;ih @x
n i
i n
n
n
k=0
@ V (~x):
= ;ih @x
(2.16)
i
The right-hand side of (2.14) now becomes
X ih
X
@ V (~x)
[~x ~p; H ] =
ij pj pi + (;ih)xi
@xi
ij m
i
~ V (~x):
= imh p~2 ; ih~x r
(2.17)
2. QUANTUM DYNAMICS
The Heisenberg equation of motion gives
d ~x ~p = 1 [~x ~p; H ] (2=:17) ~p2 ; ~x r
~ V (~x) )
dt
i*h +
m
d h~x p~i = p2 ; h~x r
~ V i;
dt
m
39
(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 hnj~x p~jni = 1 hnj [~x ~p; H ] jni = 1 (E hnj~x p~jni ; E hnj~x ~pjni) = 0:
n
dt
ih
ih n
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
;ipa exp h j0i:
You may use
2
!23 0
!1=21
0
x
h
hx0j0i = ;1=4x;0 1=2 exp 4; 12 x 5 ; @x0 m! A :
0
(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
;ipa j; t = 0i = exp h j0i )
;ipa (Pr:1:4:c)
0
0
hx j; t = 0i = hx exp h j0i = hx0 ; aj0i
2
!23
0
x
;
a
5:
= ;1=4x;0 1=2 exp 4; 12 x
0
(2.19)
40
(b) This probability is given by the expression
;ipa 2
(2.20)
jh0j; t = 0ij = jhexp h j0ij2:
It is
;ipa ;ipa Z
0
0
0
hexp h j0i = dx h0jx ihx j exp h j0i
2
!2 3
Z
0
x
= dx0;1=4x;0 1=2 exp 4; 12 x 5 ;1=4x;0 1=2
0
2
3
!
2
0
exp 4; 12 x x; a 5
0
"
Z
02 02 2
#
1
0
;
1
=
2
;
1
0
= dx x0 exp ; 2 x + x + a ; 2ax
" 2x0
Z
2 a2 !#
1
2
a
a
= px dx0 exp ; 2x2 x02 ; 2x0 2 + 4 + 4
0
0
2! 1 p
2 !
a
a
= exp ; 4x2 px x0 = exp ; 4x2 : (2.21)
0
0
So
jh0j; t = 0ij2
For t > 0
jh0j; tij2
2!
a
= exp ; 2x2 :
0
0
(2.22)
iHt =
= jh0j exp ; h j; t = 0ij2
2
= e;iE0t=hh0j; t = 0i = jh0j; t = 0ij2:
(2.23)
jh0jU (t)j; t = 0ij2
2.5 Consider a function, known as the correlation function, dened
by
C (t) = hx(t)x(0)i;
(2.24)
where x(t) is the position operator in the Heisenberg picture. Evaluate 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
2 (t)
+ 12 m!2x2(t):
H = p2m
(2.25)
So the Heisenberg equations of motion will give
"
#
dx(t) = 1 [x(t); H ] = 1 x(t); p2(t) + 1 m!2x2(t)
dt
ih
ih
2m 2
1 hx(t); p2(t)i + 1 m!2 1 hx(t); x2(t)i
= 2mi
2
h
ih
(t)
(2.26)
= 22ihihm p(t) = pm
#
"
dp(t) = 1 [p(t); H ] = 1 p(t); p2 (t) + 1 m!2x2(t)
dt
ih
ih
2m 2
2h
i m!2
m!
2
= 2ih p(t); x (t) = 2ih [;2ihx(t)] = ;m!2x(t): (2.27)
Dierentiating once more the equations (2.26) and (2.27) we get
d2x(t) = 1 dp(t) (2=:27) ;!2x(t) ) x(t) = A cos !t + B sin !t ) x(0) = A
dt2
m dt
2
d p(t) = 1 dx(t) (2=:26) ;!2p(t) ) p(t) = C cos !t + D sin !t ) p(0) = C:
dt2
m dt
But on the other hand from (2.26) we have
dx(t) = p(t) )
dt
m
p
(0) cos !t + D sin !t )
;!x(0) sin !t + B! cos !t = m
m
p
(0)
B = m!
D = ;m!x(0):
So
(0) sin !t
x(t) = x(0) cos !t + pm!
and the correlation function will be
1 sin !t:
C (t) = hx(t)x(0)i (2=:29) hx2(0)i cos !t + hp(0)x(0)i m!
(2.28)
(2.29)
(2.30)
42
Since we are interested in the ground state the expectation values appearing
in the last relation will be
h (a + ay)(a + ay)j0i = h h0jaayj0i = h (2.31)
hx2(0)i = h0j 2m!
2m!
2m!
s
s
h h0j(ay ; a)(a + ay)j0i
hp(0)x(0)i = i m2h! 2m!
(2.32)
= ;i h2 h0jaayj0i = ;i h2 :
Thus
h cos !t ; i h sin !t = h e;i!t:
C (t) = 2m!
(2.33)
2m!
2m!
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)2i as a function of time using either picture.
(a) We want to nd a state ji = c0j0i + c1j1i such that hxi is as large as
possible. The state ji should be normalized. This means
q
jc0j2 + jc1j2 = 1 ) jc1j = 1 ; jc0j2:
(2.34)
We can write the constands c0 and c1 in the following form
c0 = jc0jei0
(2:34) i1 q
i
1
c1 = jc1je = e 1 ; jc0j2:
(2.35)
2. QUANTUM DYNAMICS
by
43
The average hxi in a one-dimensional simple harmonic oscillator is given
hxi = hjxji = (c0h0j + c1h1j) x (c0j0i + c1j1i)
= jc0j2hs0jxj0i + c0c1h0jxj1i + c1cs0h1jxj0i + jc1j2h1jxj1i
h h0ja + ayj0i + cc h h0ja + ayj1i
= jc0j2 2m!
0 1 2m!
s
h h1ja + ayj0i + jc j2 h h1ja + ayj1i
1
1 0 2m!
2m!
s
s
h (cc + cc ) = 2 h <(cc )
= 2m!
0 1
1 0
0 1
2
m!
s
q
h
= 2 2m! cos(1 ; 0)jc0j 1 ; jc0j2;
(2.36)
q h
where we have used that x = 2m!
(a + ay).
What we need is to nd the values of jc0j and 1 ; 0 that make the
average hxi as large as possible.
@ hxi = 0 ) q1 ; jc j2 ; q jc0j2 jc)
0 j6=1
1 ; jc0j2 ; jc0j2 = 0
0
2
@ jc0j
1 ; jc0j
) jc0j = p1
(2.37)
2
@ hxi = 0 ) ; sin( ; ) = 0 ) = + n; n 2 Z : (2.38)
1
0
1
0
@1
But for hxi maximum we want also
@ 2hxi (2.39)
@12 1=1max < 0 ) n = 2k; k 2 Z :
So we can write that
ji = ei0 p12 j0i + ei(0+2k) p12 j1i = ei0 p12 (j0i + j1i):
(2.40)
We can always take 0 = 0. Thus
ji = p12 (j0i + j1i):
(2.41)
+cc
s
44
(b) We have j; t0i = ji. So
j; t0; ti = U (t; t0 = 0)j; t0i = e;iHt=hji = p12 e;iE t=hj0i + p12 e;iE t=hj1i
0
1
= p1 e;i!t=2j0i + e;i!3t=2j1i = p1 e;i!t=2 j0i + e;i!tj1i :(2.42)
2
2
(i) In the Schrodinger picture
hxiS = h"; t0; tjxS j; t0; tiS
i!t=2
# " 1 ;i!t=2
#
1
i!
3
t=
2
;
i!
3
t=
2
= p e h0j + e h1j x p e
j0i + e
j1i
2
2
= 21 ei(!t=2;!3t=2)h0jxj1i + 12 ei(!3t=2;!t=2)h1jxj0i
s
s
s
h
h
h cos !t:
(2.43)
= 21 e;i!t 2m! + 21 ei!t 2m! = 2m!
(ii) In the Heisenberg picture we have from (2.29) that
(0) sin !t:
xH (t) = x(0) cos !t + pm!
So
hxiH = h"jxH ji
!"
#
#
1
1
1
p
(0)
1
= p h0j + p h1j x(0) cos !t + m! sin !t p j0i + p j1i
2
2
2
2
1 sin !th0jpj1i
= 21 cos !th0jxj1i + 12 cos !th1jxj0i + 12 m!
1 sin !th1jpj0i
+ 21 m!
s
s
s
h
h
1
= 21 2m! cos !t + 12 2m! cos !t + 2m! sin !t(;i) m2h!
s
1
+ 2m! sin !ti m2h!
s
h cos !t:
= 2m!
(2.44)
(c) It is known that
h(x)2i = hx2i ; hxi2
(2.45)
2. QUANTUM DYNAMICS
In the Schodinger picture we have
32
2s
h (a2 + ay2 + aay + aya);
h
2
y
x = 4 2m! (a + a )5 = 2m!
45
(2.46)
which means that
hxi2S = h"; t0; tjx2j; t0; tiS
i!t=2
# 2 " 1 ;i!t=2
#
1
i!
3
t=
2
;
i!
3
t=
2
= p e h0j + e h1j x p e
j0i + e
j1i
2
2
So
h1
i
i(!t=2;!t=2)h0jaayj0i + 1 ei(!3t=2;!3t=2)h1jaayj1i + 1 h1jayaj1i h
e
2
2
2
2m!
h 1 1 1 i h
h
(2.47)
= 2 + 2 2 + 2 2m! = 2m! :
=
h ; h cos2 !t = h sin2 !t:
h(x)2iS (2=:43) 2m!
2m!
2m!
In the Heisenberg picture
"
#2
p
(0)
x2H (t) = x(0) cos !t + m! sin !t
p2 (0) sin2 !t
= x2(0) cos2 !t + m
2! 2
p(0) cos !t sin !t + p(0)x(0) cos !t sin !t
+ x(0)
m!
m!
h (a2 + ay2 + aay + aya) cos2 !t
= 2m!
; 2mmh2!!2 (a2 + ay2 ; aay ; aya) sin2 !t
s
i
h! (a + ay)(ay ; a) sin 2!t
+ m! h4mm!
2
s
i hmh! (ay ; a)(a + ay) sin 2!t
+ m!
4m!
2
h
2
= 2m! (a2 + ay + aay + aya) cos2 !t
(2.48)
46
h (a2 + ay2 ; aay ; aya) sin2 !t + ih (ay2 ; a2) sin 2!t
; 2m!
2m!
h (aay + aya) + h a2 cos 2!t + h ay2 cos 2!t
= 2m!
2m!
2m!
i
h
+ 2m! (ay2 ; a2) sin 2!t;
(2.49)
which means that
hx2H iH = hjx2H"jiH
#
1
h
1
= 2m! p h0j + p h1j
2
2
"
#
1
1
p2 j0i + p2 j1i
h
i
aay + aya + a2 cos 2!t + ay2 cos 2!t + i(ay2 ; a2) sin 2!t
h hh0jaayj0i + h1jaayj1i + h1jayaj1ii
= 4m!
h [1 + 2 + 1] = h :
= 4m!
m!
(2.50)
So
h ; h cos2 !t = h sin2 !t:
h(x)2iH (2=:44) 2m!
2m!
2m!
(2.51)
2.7 A coherent state of a one-dimensional simple harmonic oscillator is dened to be an eigenstate of the (non-Hermitian) annihilation operator a:
aji = ji;
where is, in general, a complex number.
(a) Prove that
ji = e;jj =2eay j0i
is a normalized coherent state.
(b) Prove the minimum uncertainty relation for such a state.
2
2. QUANTUM DYNAMICS
(c) Write ji as
ji =
47
1
X
n=0
f (n)jni:
Show that the distribution of jf (n)j2 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 ji remains coherent under timeevolution and calculate the time-evolved state j(t)i. (Hint: directly apply the time-evolution operator.)
(a) We have
h
i
aji = e;jj2=2aeay j0i = e;jj2=2 a; eay j0i;
(2.52)
since aj0i = 0. The commutator is
# X
" X
1 1 h
1 1
i
h ay i
y
n
a; e
= a; n! (a ) = n! n a; (ay)n
n=0
n=0
1 1 X
n
1
n
h
i
X
n (ay)k;1 a; ay (ay)n;k = X 1 n X (ay)n;1
=
n=1 n! k=1
n=1 n! k=1
1 1
1
X
1 n (ay)n;1 = X
y)n = eay :
(
a
(2.53)
=
n=0 n!
n=1 (n ; 1)!
So from (2.52)
aji = e;jj2=2eay j0i = ji;
(2.54)
which means that ji is a coherent state. If it is normalized, it should satisfy
also hji = 1. Indeed
hji = h0jeae;jj eay j0i = e;jj h0jeaeay j0i
2
2
48
X 1 n m n ym
y)m j0i = pm!jmi]
(
)
h
0
j
a
j
(
a
)
j
0
i
[(
a
n;m n!m!
p p
n! m! ()n mhnjmi = e;jj2 X 1 (jj2)n
2 X
;j
j
= e
n;m n! m!
n n!
2
2
= e;jj ejj = 1:
(2.55)
= e;jj2
(b) According to problem (1.3) the state should satisfy the following relation
xji = cpji;
(2.56)
where x x ; hjxji, p p ; hjpji and c is a purely imaginary
number.
Since ji is a coherent state we have
aji = ji ) hjay = hj :
Using this relation we can write
s
s
h
h ( + ay)ji
xji = 2m! (a + ay)ji = 2m!
and
s
(2.57)
(2.58)
s
h
h (hjaji + hjayji)
hxi = hjxji = 2m! hj(a + ay)ji = 2m!
s
h ( + )
= 2m!
(2.59)
and so
s
h (ay ; )ji:
xji = (x ; hxi)ji = 2m!
(2.60)
q
Similarly for the momentum p = i m2h! (ay ; a) we have
s
p s mh! y
(2.61)
pji = i 2 (a ; a)ji = i m2h! (ay ; )ji
2. QUANTUM DYNAMICS
and
49
s
s
m
h
!
hpi = hjpji = i 2 hj(ay ; a)ji = i m2h! (hjayji ; hjaji)
s
= i m2h! ( ; )
(2.62)
and so
s
pji = (p ; hpi)ji = i m2h! (ay ; )ji )
s
(2.63)
(ay ; )ji = ;i m2h! pji:
So using the last relation in (2.60)
s
s
h
xji = 2m! (;i) m2h! pji =
i
; m!
pji (2.64)
| {z }
purely imaginary
and thus the minimum uncertainty condition is satised.
(c) The coherent state can be expressed as a superposition of energy eigenstates
1
1
X
X
ji = jnihnji = f (n)jni:
(2.65)
n=0
n=0
for the expansion coecients f (n) we have
f (n) = hnji = hnje;jj2=2eay j0i = e;jj2=2hnjeay j0i
1
1 1
X
2 =2 X 1
2 =2
y
m
;j
j
m hnj(ay)m j0i
;j
j
= e
hnj m! (a ) j0i = e
m=0 m!
m=0
1
p
X
= e;jj2=2 m1 ! m m!hnjmi = e;jj2=2 p1 n )
(2.66)
n!
m=0
2 n
jf (n)j2 = (jnj! ) exp(;jj2)
(2.67)
which means that the distribution of jf (n)j2 with respect to n is of the Poisson
type about some mean value n = jj2.
50
The most probable value of n is given by the maximum of the distribution
jf (n)j2 which can be found in the following way
jf (n + 1)j2 =
jf (n)j2
(jj2)n+1 exp(;jj2)
(n+1)!
(jj2 )n exp(;jj2 )
n!
2
j
j
= n+1 1
(2.68)
which means that the most probable value of n is jj2.
(d) We should check if the state exp (;ipl=h) j0i is an eigenstate of the annihilation operator a. We have
h
i
a exp (;ipl=h) j0i = a; e(;ipl=h) j0i
(2.69)
since aj0i = 0. For the commutator in the last relation we have
1 1 ;il !n X
n
1 1 ;il !n
h (;ipl=h)i
X
X
n] =
a; e
=
[
a;
p
pk;1 [a; p]pn;k
n
!
h
n
!
h
n=0
n=1
k=1
s
!
n
1
n
X 1 ;il X n;1 mh!
p i 2
=
n=1 n! h
k
=1
s
!X
!n;1
1
m
h
!
;
il
1
;
ilp
= i 2
h n=1 (n ; 1)! h
r m!
(2.70)
= l 2h e(;ipl=h);
where we have used that
s
s
m
h
!
[a; p] = i 2 [a; ay ; a] = i m2h! :
(2.71)
So substituting (2.70) in (2.69) we get
r m!
a [exp (;ipl=h) j0i] = l 2h [exp (;ipl=h) j0i]
(2.72)
which q
means that the state exp (;ipl=h) j0i is a coherent state with eigenvalue l m!
2h .
(e) Using the hint we have
1
X
j(t)i = U (t)ji = e;iHt=hji (2=:66) e;iHt=h e;jj2=2 p1n! n jni
n=0
2. QUANTUM DYNAMICS
51
1 ;it
1
:3:9) X
e;iEnt=he;jj2=2 p1 njni (2=
e h h!(n+ 2 )e;jj2=2 p1 n jni
n!
n!
n=0
n=0
1 X
n
=
e;i!t e;i!t=2e;jj2=2 p1 n jni
n!
n=0
1
;
i!t
X ;je;i!tj2=2 (e )n (2:66) ;i!t=2 ;i!t
;
i!t=
2
p jni = e je i
= e
e
(2.73)
n!
n=0
=
1
X
Thus
aj(t)i = e;i!t=2aje;i!ti = e;i!t e;i!t=2je;i!ti
= 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 ) =
Z1
0
dteiEt=hK (x; y; t; 0) = A
X sin(nrx) sin(nry)
E ; h22mr2 n2
n
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 ) =
Z1
Z01
Z01
dteiEt=hK (x; y; t; 0) dteiEt=hhxje;iHt=hjyi
Z1
0
dteiEt=hhx; tjy; 0i
X
dteiEt=h hxje;iHt=hjnihnjyi
n
Z01
X ;iEnt=h
iEt=
h
=
dte
e
hxjnihnjyi
0
nZ
1
X
=
n(x)n(y) ei(E;En)t=hdt
=
n
0
52
=
=
So
X
n
X
n
n(x)n(y) lim
"!0
"
;ih ei(E;En+i")t=h
E ; En + i"
n(x)n(y) E ;ihE
n
:
X
nry) )
n(x)n(y) E ;ihE = A sin(nrx)hsin(
2 r2
n
E ; 2m n2
n
n
s
2 2
n (x) = iAh sin(nrx); En = h2mr n2:
#1
0
(2.75)
X
(2.76)
For a one dimensional innite square well potential with size L the energy
eigenvalue En and eigenfunctions n(x) are given by
s
nx 2 2
2
h
n (x) = L sin L ; En = 2m L n2:
(2.77)
Comparing with (2.76) we get L = r ) L = r and
(
0 for 0 < x < r
V= 1
(2.78)
otherwise
while
A = 2r ) A = i 2hr :
(2.79)
ih 2.9 Prove the relation
d(x) = (x)
dx
where (x) is the (unit) step function, and (x) the Dirac delta
function. (Hint: study the eect on testfunctions.)
For an arbitrary test function f (x) we have
Z +1 d(x)
Z +1 d
Z +1 df (x)
f (x)dx = ;1 dx [(x)f (x)] dx ; ;1 (x) dx dx
;1 dx
2. QUANTUM DYNAMICS
53
Z
1 ; +1 df (x) dx
= (x)f (x) +;1
dx
0
+1
= x!lim
f (x) ; f (x) = f (0)
+1
0
Z +1
=
(x)f (x)dx )
;1
d(x) = (x):
dx
(2.80)
2.10 Derive the following expression
m! h(x2 + 2x2 ) cos(!T ) ; x x i
Scl = 2 sin(
0 T
T
!T ) 0
for the classical action for a harmonic oscillator moving from the
point x0 at t = 0 to the point xT at t = T .
The Lagrangian for the one dimensional harmonic oscillator is given by
L(x; x_ ) = 21 mx_ 2 ; 21 m!2x2:
From the Lagrange equation we have
@ L ; d @ L = 0 (2)
:81)
2x ; d (mx_ ) = 0 )
;
m!
@x dt @ x_
dt
2
x + ! x = 0:
(2.81)
(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) = x0 = A
(2.84)
x(t = T ) = xT = x0 cos !T + B sin !T ) B sin !T = xT ; x0 cos !T )
cos !T :
(2.85)
B = xT ;sinx0!T
54
So
cos !T sin !t
x(t) = x0 cos !t + xT ;sinx0!T
T sin !t ; x0 cos !T sin !t
= x0 cos !t sin !T + xsin
!T
sin !(T ; t) )
(2.86)
= xT sin !t +sinx0!T
! cos !(T ; t) :
x_ (t) = xT ! cos !t ;sinx0!T
(2.87)
With these at hand we have
ZT
ZT S =
dtL(x; x_ ) = dt 21 mx_ 2 ; 12 m!2x2
#
Z0 T " d 0
1
1
2
2
1
=
dt 2 m dt (xx_ ) ; 2 mxx ; 2 m! x
0
T
ZT
m
2
1
= ; 2 m dtx[x + ! x] + 2 xx_ 0
0
(2:82) m
= 2 [x(T )_x(T ) ; x(0)x_ (0)]
xT !
m
x
0!
=
!T )
2 sin !Th (xT cos !T ; x0) ; sin !T (xT ; x0 cos
i
m! x2 cos !T ; x x ; x x + x2 cos !T
=
0 T
0 T
0
2 sin !T h T
i
m!
= 2 sin !T (x2T + x20) cos !T ; 2x0xT :
(2.88)
2.11 The Lagrangian of the single harmonic oscillator is
L = 12 mx_ 2 ; 12 m!2x2
(a) Show that
hxb tbjxatai = exp iShcl G(0; tb ; 0; ta)
where Scl is the action along the classical path xcl from (xa; ta) to
(xb; tb) and G is
G(0; tb ; 0; ta) =
2. QUANTUM DYNAMICS
lim
N !1
Z
55
m (N2+1) 8
9
N m
<i X
=
1
2 ; "m! 2y 2
exp : h
dy1 : : : dyN 2ih"
(
y
;
y
)
j +1
j
j ;
2
j =0 2"
ta .
where " = (tNb;+1)
[Hint: Let y(t) = x(t) ; xcl(t) be the new integration variable,
xcl(t) being the solution of the Euler-Lagrange equation.]
(b) Show that G can be written as
m (N2+1) Z
dy1 : : : dyN exp(;nT n)
G = Nlim
!1 2ih"
2 3
66 y..1 77
where n = 4 . 5 and nT is its transpose. Write the symmetric
yN
matrix .
(c) Show that
Z
Z
N=2
dy1 : : : dyN exp(;nT n) dN ne;nT n = p
det
[Hint: Diagonalize by an orthogonal matrix.]
(d) Let 2mih" N det detN0 pN . Dene j j matrices j0 that consist of the rst j rows and j columns of N0 and whose determinants
are pj . By expanding j0 +1 in minors show the following recursion
formula for the pj :
pj+1 = (2 ; "2!2)pj ; pj;1
j = 1; : : : ; N
(2.89)
(e) Let (t) = "pj for t = ta + j" and show that (2.89) implies that in
the limit " ! 0; (t) satises the equation
d2 = ;!2(t)
dt2
with initial conditions (t = ta) = 0; d(dtt=ta) = 1.
56
(f) Show that
(
)
m!
im!
2
2
hxbtbjxatai = 2ih sin(!T ) exp 2h sin(!T ) [(xb + xa) cos(!T ) ; 2xaxb]
where T = tb ; ta.
s
(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
Z
hxbtbjxatai = Nlim
dx1dx2 : : : dxN hxb tbjxN tN ihxN tN jxN ;1tN ;1i : : : !1
hxi+1ti+1jxitii : : : hx1t1jxatai:
(2.91)
It is
hxi+1ti+1jxitii = hxi+1 je;iH (ti ;ti )=hjxii = hxi+1je;iH"=hjxii
1
" 1
= hxi+1 je;i h ( 2 mp + 2 m! x )jxii (since " is very small)
p
"1
= hxi+1 je;i h" m e;i h 2 m! x jxii
"1
"p
(2.92)
= e;i h 2 m! xi hxi+1 je;i h m jxii:
+1
2
2
2 2
2
2 2
2 2
2
2
For the second term in this last equation we have
Z
2
" p2
;
i
hxi+1 je h 2m jxii = dpi hxi+1je;i h" 2pm jpiihpi jxii
Z
p2
= dpi e;i h" 2mi hxi+1jpi ihpi jxii
Z
p2
= 21h dpi e;i h" 2im eipi(xi+1;xi )=h
h
i
Z
1
;i 2m" h p2i ;2pi m" (xi+1 ;xi )+ m"22 (xi+1 ;xi )2 ; m"22 (xi+1 ;xi )2
= 2h dpi e
2
i" m2 (xi+1 ;xi )2 Z
"
m
1
2
m
= 2h e h "2
dpie;i 2mh [pi;pi " (xi+1 ;xi)]
2. QUANTUM DYNAMICS
57
s
i" m2 (xi+1 ;xi )2 2
hm
1
= 2h e 2mh "2
i"
r m im
(2.93)
= 2hi" e 2h" (xi+1;xi)2 :
Substituting this in (2.92) we get
m 12 i m
hxi+1ti+1jxitii = 2ih" e h [ 2" (xi+1;xi )2; 12 "m!2 xi]
(2.94)
and this into (2.91):
i Z
hxbtbjxatai Dx exp h S [x] =
m (N2+1) 8
9
Z
N m
<i X
=
1
2 ; "m! 2 x2 :
lim
dx
:
:
:dx
exp
(
x
;
x
)
1
N
j
+1
j
j ;
: h
N !1
2ih"
2"
2
j =0
Let y(t) = x(t) ; xcl(t) ) x(t) = y(t) + xcl(t) ) x_ (t) = y_ (t) + x_ cl(t) with
boundary conditions y(ta) = y(tb) = 0: For this new integration variable we
have Dx = Dy and
Z tb
S [x] = S [y + xcl] = L(y + xcl; y_ + x_ cl)dt
ta
3
Z tb 2
2L 2L @
@
@
L
@
L
2
2
1
1
4L(xcl ; x_ cl) + y + y_ + 2 2 y + 2 2 y_ 5
=
@x
@ x_ xcl
@ x xcl
@ x_ xcl
ta
tb Z tb " xcl
!#
@ L ; d @ L y + Z tb h 1 my_ 2 ; 1 m!2y2i dt:
= Scl + @@Lx_ y +
2
ta @x dt @ x_ xcl
ta 2
ta
So
i
Z tb h
i i
2
2
2
1
1
hxb tbjxatai = Dy exp h Scl + h t 2 my_ ; 2 m! y dt
a
iScl = exp h G(0; tb; 0; ta)
(2.95)
Z
with
G(0; tb; 0; ta) =
m (N2+1) 8
9
Z
N m
=
<i X
1
2 ; "m! 2y 2 :
lim
dy
:
:
:
dy
(
y
;
y
)
exp
1
N
j ;
: h
N !1
2ih"
2" j+1 j
2
j =0
58
(b) For the argument of the exponential in the last relation we have
(y0 =0)
N m
1
iX
2
2
2
h j=0 2" (yj+1 ; yj ) ; 2 "m! yj =
N m
N 1
X
iX
i
=0)
2
2
2y y (yN +1
(
y
+
y
;
y
"m!
=
j +1 yj ; yj yj +1 ) ;
i ij j
j
+1
j
h j=0 2"
h i;j=1 2
N
N
2 X
X
; 2"imh (2yiij yj ; yii;j+1yj ; yii+1;j yj ) ; i"m!
2h i;j=1 yiij yj :(2.96)
i;j =1
where the last step is written in such a form so that the matrix will be
symmetric. Thus we have
m (N2+1) Z
dy1 : : : dyN exp(;nT n)
(2.97)
G = Nlim
!1 2ih"
with
2
3
2
3
2 ;1 0 : : : 0 0
1 0 0 ::: 0 0
66 ;1 2 ;1 : : : 0 0 77
66 0 1 0 : : : 0 0 77
66 0 ;1 2 : : : 0 0 77
26
66 0 0 1 : : : 0 0 777
i"m!
m
6
7
= 2"ih 66 .. .. ..
:
... ... 77 + 2h 66 ... ... ...
... ... 77(2.98)
66 . . .
77
66
77
4 0 0 0 : : : 2 ;1 5
4 0 0 0 ::: 1 0 5
0 0 0 : : : ;1 2
0 0 0 ::: 0 1
(c) We can diagonalize by a unitary matrix U . Since is symmetric the
following will hold
= U yDU ) 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
= RT D R and det R = 1
(2.100)
whichZmeans that
Z
Z
T n
T RT Rn Rn=
N
;
n
N
;
n
d ne
= d ne
= dN e;T Z
Z
Z
2a
2a
2a
;
;
;
1
2
N
1
2
N
=
d1e
d2 e
: : : dN e
s s
s
N=2
N=2
q
p
= a a : : : a = QN =
det D
1
2
N
i=1 ai
N=2
= p
(2.101)
det 2. QUANTUM DYNAMICS
where ai are the diagonal elements of the matrix D .
(d) From (2.98) we have
!
2ih" N det =
m
82
3
21
>
2
;
1
0
:
:
:
0
0
>
66 ;1 2 ;1 : : : 0 0 77
66 0
>
>
6
7
>
<66 0 ;1 2 : : : 0 0 77 2 2 666 0
det >66 .. .. ..
... ... 77 ; " ! 66 ...
.
.
.
>
6
66
>
64 0 0 0 : : : 2 ;1 775
>
40
>
: 0 0 0 : : : ;1 2
0
det N0 pN :
59
0
1
0
...
0
0
0
0
1
...
0
0
39
::: 0 0 >
>
: : : 0 0 777>
>
=
: : : 0 0 77>
... ... 77> =
7>
>
: : : 1 0 75>
>
::: 0 1 ;
(2.102)
We dene j j matrices j0 that consist of the rst j rows and j columns of
N0 . So
2 2 ; "2!2 ;1 : : :
0
0
0 3
66 ;1 2 ; "2!2 : : :
0
0
0 777
66 0
;1 : : :
0
0
0 77
6
6
.
.
.
.
... 77 :
..
..
..
det j0 +1 = det 66 ..
77
66 0
2!2
0
:
:
:
2
;
"
;
1
0
77
64 0
2
2
0
: : : ;1 2 ; " !
;1 5
0
0
:::
0
;1 2 ; "2!2
From the above it is obvious that
det j0 +1 = (2 ; "2!2) det j0 ; det j0 ;1 )
pj+1 = (2 ; "2!2)pj ; pj;1 for j = 2; 3; : : : ; N
(2.103)
with p0 = 1 and p1 = 2 ; "2!2.
(e) We have
(t) (ta + j") "pj
) (ta + (j + 1)") = "pj+1 = (2 ; "2!2)"pj ; "pj;1
= 2(ta + j") ; "2!2(ta + j") ; (ta + (j ; 1)")
) (t + ") = 2(t) ; "2!2(t) ; (t ; "):
(2.104)
60
So
(t + ") ; (t) = (t) ; (t ; ") ; "2!2(t) )
; (t);"(t;") = ;!2(t) )
"
0(t) ; 0(t ; ")
2
2 (t) ) d = ;! 2 (t):
lim
=
;
!
(2.105)
"!0
"
dt2
(t+");(t)
"
From (c) we have also that
(ta) = "p0 ! 0
(2.106)
and
d (t ) = (ta + ") ; (ta) = "(p1 ; p0) = p ; p
1
0
dt a
"
"
= 2 ; "2!2 ; 1 ! 1:
The general solution to (2.105) is
(t) = A sin(!t + )
and from the boundary conditions (2.106) and (2.107) we have
(ta) = 0 ) A sin(!ta + ) = 0 ) = ;!ta + n n 2 Z
which gives that (t) = A sin !(t ; ta), while
d = A! cos(t ; t ) ) 0(t ) = A! (2)
:107)
a
a
dt
A! = 1 ) A = !1
Thus
(t) = sin !(!t ; ta) :
(f) Gathering all the previous results together we get
"
(N +1) N #1=2
m
p
G = Nlim
!1 2ih"
det
(2.107)
(2.108)
(2.109)
(2.110)
(2.111)
2. QUANTUM DYNAMICS
=
(d)
=
(2:111)
=
61
3;1=2
!N
m 1=2 2
2
i
h
"
4 lim "
5
N !1
2ih
m det
m 1=2 ;1=2 (e) m 1=2
lim "p
= 2ih [(tb)];1=2
N !1 N
2
i
h
s m!
(2.112)
2ih sin(!T ) :
So from (a)
iScl hxbtbjxatai = exp h G(0; tb ; 0; ta)
s m!
im! h
i
(2:88)
2 + x2 ) cos !T ; 2x x :
=
exp
(
x
b a
2ih sin(!T )
2h sin !T b a
2.12 Show the composition property
Z
dx1Kf (x2; t2; x1; t1)Kf (x1; t1; x0; t0) = Kf (x2; t2; x0; t0)
where Kf (x1; t1; x0; t0) is the free propagator (Sakurai 2.5.16), by
explicitly performing the integral (i.e. do not use completeness).
We have
Z
dx1Kf (x2; t2; x1; t1)Kf (x1; t1; x0; t0)
"
#
Z s
m
im
(
x
2 ; x1)2
= dx1 2ih(t ; t ) exp 2h(t ; t ) 2
1
2
"
# 1
s
2
m
im(x1 ; x0)
exp
2ih(t1 ; t0)
2h(t1 ; t0)
s
m
1
imx22 exp imx20 = 2i
exp
h (t2 "; t1)(t1 ; t0) 2h(t2 ; t1) 2h(t2 ; t1)
#
Z
im
im
im
im
2
2
dx1 exp 2h(t ; t ) x1 + 2h(t ; t ) x1 ; 2h(t ; t ) 2x1x2 ; 2h(t ; t ) 2x1x0
2
1
2
1
2
1
2
1
62
=
=
=
=
=
=
=
s
( " 2
2 #)
1
m
im
x
x
2
0
exp
+
2ih (t2 (; t1)("t1 ; t0)
2h (t2# ; t1) (t1 "; t0) #)
Z
im
1
1
im
x
x
2
0
2
dx1 exp 2h (t ; t ) + (t ; t ) x1 ; h x1 (t ; t ) + (t ; t )
2
1
2
1
0
s
( 1 "0 2
#)1
2
1
m
exp im x2 + x0
2ih (t2 (; t1)("t1 ; t0)
2h #(t2 ; t1) (t1 ; t0)
Z
t0
dx1 exp ;2imh (t ;tt2 ;
2
1)(t1 ; t0)
"
"
##)
2
h
t
im
x
2 ; t0
2 (t1 ; t0) + x0 (t2 ; t1)
2
x1 ; im (t ; t )(t ; t ) h x1
s 2 1 1 0 ( " 2 (t2 ; t1)(t1 ;2 t0) #)
1
m
im x2(t1 ; t0) + x0(t2 ; t1) exp
2h
(t2 ; t1)(t1 ; t0)
3
22Zih (t2 ;8t1)(t1"; t0)
#
"
#29
=
< ;m
t2 ; t0
x2(t1 ; t0) + x0(t2 ; t1) 5 4 dx1 exp :
x
1;
;
2ih (t2 ; t1)(t1 ; t0)
(t2 ; t0)
(
2)
im
1
[
x
2(t1 ; t0) + x0(t2 ; t1)]
exp ; 2h (t ; t )(t ; t )
(t2 ; t0)
2
1 1
0
v
s
(
u
u
1
2
i
h
(
t
;
t
)(
t
;
t
)
m
im
1
2
1
1
0
t
exp
2ih (t2 ; t1)(t1 ; t0)
m(t2 ; t0)
2h (t2 ; t1)(t1 ; t0) " 2
x2(t1 ; t0)(t2 ; t0) + x20(t2 ; t1)(t2 ; t0) ;
(t2 ; t0)
#)
2
2
2
x2(t1 ; t0) ; x0(t2 ; t1)2 ; 2x2x02(t1 ; t0)(t2 ; t1)
(t2 ; t0)
s
m
2ih(t2 ; t1) ( " 2
x2(t1 ; t0)(t2 ; t0 ; t1 + t0) + x20(t2 ; t1)(t2 ; t0 ; t2 + t1) ;
exp im
2h
#)(t2 ; t0)(t2 ; t1)(t1 ; t0)
2x2x02(t1 ; t0)(t2 ; t1)
(t ; t )(t ; t )(t ; t )
s 2 0m 2 1 1" im0(x ; x )2 #
2
0
exp
2ih(t ; t )
2h(t ; t )
2
0
Kf (x2; t2; x0; t0):
2
0
(2.113)
2. QUANTUM DYNAMICS
63
2.13 (a) Verify the relation
!
i
h
e
[i; j ] = c "ijk Bk
where ~ m dt~x = ~p ; ecA~ and the relation
"
!#
2~x
~
d
d
1
d~
x
d~
x
~
~
~
m dt2 = dt = e E + 2c dt B ; B dt :
(b) Verify the continuity equation
with ~j given by
(a) We have
@ + r
~ 0 ~j = 0
@t
!
e
h
0
~
~ 2
~j =
m =( r ) ; mc Aj j :
eAi
eA
j
[i; j ] = pi ; c ; pj ; c = ; ec [pi ; Aj ] ; ec [Ai; pj ]
!
i
h
e
@A
j ihe @Ai ihe @Aj @Ai
= c @x ; c @x = c @x ; @x
j
i
j
!i
= ihce "ijk Bk :
(2.114)
We have also that
2
3
2
3
dxi = 1 [x ; H ] = 1 4x ; ~ 2 + e5 = 1 4x ; ~ 2 5
dt
ih i
ih i 2m
ih i 2m
= ih12m f[xi; j ] j + j [xi; j ]g = ih21m f[xi; pj ]j + j [xi; pj ]g
= 22ihihm j ij = mi )
64
d2xi
dt2
=
=
(2:114)
=
=
=
m ddtx2i
2
m ddt~x2
2
=
=
2
3
"
#
1 dxi ; H = 1 4 ; ~ 2 + e5
ih dt
ihm i 2m
1 f[ ; ] + [ ; ]g + e [ ; ]
ih2m2 " i j j j i j #ihm i 1 ihe " B + ihe " B + e p ; eAi ; 2m2ih c ijk k j c ijk j k ihm i c
e (;" B + " B ) + e [p ; ]
2m2c ikj k j ijk j k ihm i
e m " xj B ; " B xj ; e @ )
2m2c "ijk dt k ! ikj k dt !m# @xi
~x B~ ; B~ ~x )
eEi + 2ec dt
i
"
!#dt i
~x B~ ; B~ ~x :
e E~ + 21c dt
(2.115)
dt
(b) The time-dependent Schrodinger equation is
0
12
~
@ hx0j; t ; ti = hx0jH j; t ; ti = hx0j 1 @~p ; eA A + ej; t ; ti
ih @t
0
0
0
2m
c
2
3 2
3
0
0
~
~
~ 0 ; eA(~x ) 5 4;ihr
~ 0 ; eA(~x ) 5 hx0j; t0; ti + e(~x0)hx0j; t0; ti
= 21m 4;ihr
c
c
"
#
2
e
e
e
1
2
0
0
0
0
0
0
2
0
~ r
~ + ihr
~ A~ (~x ) + ih A~ (~x ) r
~ + 2 A (~x ) (~x0; t)
= 2m ;h r
c
c
c
0
0
+e(~x ) (~x ; t)
~ 0 ~ 0
= 21m ;h2r02 (~x0; t)0 + ec ih r
A (~x ; t) + ec ihA~ (~x0) r~ 0 (~x0; t)
#
2
e
e
0
0
0
2
0
0
~ (~x ; t) + 2 A (~x ) (~x ; t) + e(~x0) (~x0; t)
+
ih c A~ (~x ) r
c
"
#
2
1
e
e
e
2
2
0
0
0
2
~
~
~
~
= 2m ;h r + c ih r A + 2ih c A r + c2 A + e : (2.116)
Multiplying the last equation by we get
@ =
ih @t
2. QUANTUM DYNAMICS
65
"
#
2
1 ;h2 r02 + e ih r
e
e
0
2
0
2
2
~ A~ j j + 2ih A~ r
~ + 2 A j j + ej j2:
2m
c
c
c
The complex conjugate of this eqution is
@ =
;ih @t
"
#
2
1 ;h2 r02 ; e ih r
e
e
~ 0 A~ j j2 ; 2ih A~ r
~ 0 + 2 A2j j2 + ej j2:
2m
c
c
c
Thus subtracting the last two equations we get
2 h
i
; 2hm r02 ; r02 e 2 e
0
~
~
~0 + r
~ 0 )
+ mc ih r A j j + mc ihA~ ( r
!
@
@
= ih @t + @t )
2
h
i e ~ 0 ~ 2 e ~ ~ 0 2
ih r A j j + mc ihA (r j j )
; 2hm r~ 0 r~ 0 ; r~ 0 + mc
= ih @ j j2 )
@t
h
i ~ 0 h ~ 2i
@ j j2 = ; h r
~ 0 =( r
~0 ) + e r
Aj j )
@t
m"
mc
#
@ j j2 + r
~ 0 h =( r
~ 0 ) ; e A~ j j2 = 0 )
@t
m
mc
@ + r
~ 0 ~j = 0
(2.117)
@t
~ 0 ) ; mce A~ j j2: and = j j2
with ~j = mh =( r
2.14 An electron moves in the presence of a uniform magnetic eld
in the z-direction (B~ = B z^).
(a) Evaluate
where
[x; y ];
x px ; eAc x ;
y py ; eAc y :
66
(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
!
2k2
j
eB
j
h
1
h
+
n+ ;
E =
k;n
2m
mc
2
where hk is the continuous eigenvalue of the pz operator and n is a
nonnegative integer including zero.
The magentic eld B~ = B z^ can be derived from a vector petential A~ (~x)
of the form
Bx ; A = 0:
Ax = ; By
;
A
(2.118)
y=
z
2
2
Thus we have
eA
eA
eBy
eBx
x
y (2:118)
[x; y ] = px ; c ; py ; c = px + 2c ; py ; 2c
eB [y; p ] = iheB + iheB
[
p
= ; eB
x ; x] +
2c
2c y
2c
2c
eB
= ih c :
(2.119)
(b) The Hamiltonian for this system is given by
0
1
~ 2 1 2 1 2 1 2
e
A
1
H = 2m @p~ ; c A = 2m x + 2m y + 2m pz = H1 + H2
(2.120)
where H1 21m 2x + 21m 2y and H2 21m p2z . Since
"
2 2 #
1
eBy
eBx
[H1; H2] = 4m2 px + 2c + ; py ; 2c ; p2z = 0 (2.121)
there exists a set of simultaneous eigenstates jk; ni of the operators H1 and
H2. So if hk is the continious eigegenvalue of the operator pz and jk; ni its
eigenstate we will have
2 2
2
H2jk; ni = 2pmz jk; ni = h2mk :
(2.122)
2. QUANTUM DYNAMICS
67
On the other hand H1 is similar to the Hamiltonian of the one-dimensional
oscillator problem which is given by
H = 21m p2 + 21 m!2x2
(2.123)
with [x; p] = ih. In order to use the eigenvalues of the harmonic oscillator
En = h ! n + 21 we should have the same commutator between the squared
operators in the Hamiltonian. From (a) we have
xc eB
[x; y ] = ih c ) [ eB ; y ] = ih:
(2.124)
So H1 can be written in the following form
2 j2
H1 21m 2x + 21m 2y = 21m 2y + 21m eBxc jeB
c2
!2 2
eB j xc :
(2.125)
= 21m 2y + 21 m jmc
eB
eB j to have
In this form it is obvious that we can replace ! with jmc
!
2k 2
h
j
eB
j
h
1
H jk; ni = H1jk; ni + H2jk; ni = 2m jk; ni + mc n + 2 jk; ni
" 2 2
!
#
h
k
j
eB
j
h
1
= 2m + mc n + 2 jk; ni:
(2.126)
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 cylinder runs a thin, impenetrable solenoid carrying a magnetic ux .
Calculate the ground state energy and wavefunction.
In the case where B~ = 0 the Schrodinger equation of motion in the
cylindrical coordinates is
h2 [r2 ] = 2E )
;
h @2 1 @ m 1 @2 @2 i
2
h
; m @2 + @ + 2 @2 + @z2 (~x) = 2E (~x)
(2.127)
68
If we write (; ; z) = ()R()Z (z) and k2 = 2mE
h2 we will have
2
2
()Z (z) d R2 + ()Z (z) 1 dR + R()Z2 (z) d 2
d
d
d
2
+R()() ddzZ2 + k2R()()Z (z) = 0 )
1 d2R + 1 dR + 1 d2 + 1 d2Z + k2 = 0(2.128)
R() d2 R() d 2() d2 Z (z) dz2
with initial conditions (a; ; z) = (R; ; z) = (; ; 0) = (; ; a) = 0.
So
1 d2Z = ;l2 ) d2Z + l2Z (z) = 0 ) Z (z) = A eilz + B e;ilz (2.129)
1
1
Z (z) dz2
dz2
with
Z (0) = 0 ) A1 + B1 = 0 ) Z (z) = A1 eilz ; e;ilz = C sin lz
Z (a) = 0 ) C sin la = 0 ) la = n ) l = ln = n a n = 1; 2; : : :
So
Z (z) = C sin lnz
(2.130)
Now we will have
1 d2R + 1 dR + 1 d2 + k2 ; l2 = 0
R() d2 R() d 2() d2
2 2
1 d2 + 2(k2 ; l2) = 0
) R() ddR2 + R() dR
+
d () d2
2
(2.131)
) (1) dd2 = ;m2 ) () = eim:
with
( + 2) = () ) m 2 Z :
So the Schrodinger equation is reduced to
2 d2R + dR ; m2 + 2(k2 ; l2) = 0
R() d2 R() d
(2.132)
2. QUANTUM DYNAMICS
69
2 R 1 dR "
2#
d
m
2
2
) d2 + d + (k ; l ) ; 2 R() = 0
#
"
2
2R
d
1
dR
m
) d(pk2 ; l2)2 + pk2 ; l2 d(pk2 ; l2) + 1 ; (k2 ; l2)2 R() = 0
p
p
) R() = A3Jm( k2 ; l2) + B3Nm( k2 ; l2)
(2.133)
In the case at hand in which a ! 0 we should take B3 = 0 since Nm ! 1
when ! 0. From the other boundary condition we get
p
p
R(R) = 0 ) A3Jm(R k2 ; l2) = 0 ) R k2 ; l2 = m
(2.134)
where m is the -th zero of the m-th order Bessel function Jm. This means
that the energy eigenstates are given by the equation
2 2
2
p
m = R k2 ; l2 ) k2 ; l2 = Rm2 ) 2mE
;
n a = Rm2
2
h
2#
2 " 2
h
m
(2.135)
) E = 2m R2 + n a
while the corresponding eigenfunctions are given by
m )eim sin n z (
x
~
)
=
A
J
(
nm
c m
R
a
with n = 1; 2; : : : and m 2 Z .
Now suppose that B~ = B z^. We can then write
!
2!
B
a
A~ = 2 ^ = 2 ^:
(2.136)
(2.137)
The Schrodinger equation in the presence of the magnetic eld B~ can be
written as follows
2
3 2
3
~
~
1 4;ihr
e
A
(
x
~
)
e
A
(
x
~
)
~;
5 4 ~
5 (~x) = E (~x)
2m
c ;ihr ; c
!#
2 " @
h
@
1
@
ie
) ; 2m ^@ + z^ @z + ^ @ ; hc 2 "
!#
@
@
1
@
ie
^
^@ + z^ @z + @ ; hc 2 (~x) = E (~x):
(2.138)
70
Making now the transformation D @@ ; hiec 2 we get
# "
#
2 " @
@
1
@
@
1
h
; 2m ^@ + z^ @z + ^ D ^@ + z^ @z + ^ D (~x) = E (~x)
2 " @2
2#
h
1
@
1
@
2
) ; 2m @2 + @ + 2 D + @z2 (~x) = E (~x);
(2.139)
2
where D2 = @@ ; hiec 2 . Leting A = hec 2 we get
!
!
2
2
@
2
ie
@
@
@
2
2
2
D = @2 ; hc 2 @ ; A = @2 ; 2iA @ ; A : (2.140)
Following the same procedure we used before (i.e. (; ; z) = R()()Z (z))
we will get the same equations with the exception of
" 2
#
@ ; 2iA @ ; A2 = ;m2 ) d2 ; 2iA d + (m2 ; A2) = 0:
@2
@
d2
d
The solution to this equation is of the form el. So
l + (m2 ; A2)el = 0 ) l2 ; 2iAl + (m2 ; A2)
l2el ; 2iAle
q
2iA ;4A2 ; 4(m2 ; A2) 2iA 2im
=
= i(A m)
) l=
2
2
which means that
() = C2ei(Am):
(2.141)
But
( + 2) = () ) A m = m0 m0 2 Z
) m = (m0 ; A) m0 2 Z :
(2.142)
This means that the energy eigenfunctions will be
z m im0 (2.143)
nm (~x) = Ac Jm (
R )e sin n a
but now m is not an integer. As a result the energy of the ground state will
be
2#
2 " 2
h
m
(2.144)
E = 2m R2 + n a
2. QUANTUM DYNAMICS
71
where now m = m0 ; A is not zero in general but it corresponds to m0 2 Z
such that 0 m0 ; A < 1. Notice also that if we require the ground state to
be unchanged in the presence of B , we obtain ux quantization
0
m0 ; A = 0 ) hec 2 = m0 ) = 2me hc m0 2 Z :
(2.145)
2.16 A particle in one dimension (;1 < x < 1) is subjected to a
constant force derivable from
V = x; ( > 0):
(a) Is the energy spectrum continuous or discrete? Write down an
approximate expression for the energy eigenfunction specied by
E.
(b) Discuss briey what changes are needed if V is replaced be
V = jxj:
(a) In the case under construction there is only a continuous spectrum and
the eigenfunctions are non degenerate.
From the discussion on WKB approximation we had that for E > V (x)
i Z q
A
(
x
)
=
exp
2
m
[
E
;
V
(
x
)]
dx
I
[E ; V (x)]1=4
h
Zq
B
+ [E ; V (x)]1=4 exp ; hi
2m[E ; V (x)]dx
!
Z x0 q
c
1
= [E ; V (x)]1=4 sin h
2m[E ; V (x)]dx ; 4
x
p Z x0=E= !
1=2
c
2
m
E
= [E ; V (x)]1=4 sin h
; x dx ; 4
x
2 3
s
3
=
2
2m ; 5
= [E ; Vc(x)]1=4 sin 4; 32 E ; x
h
4
(2.146)
= [qc]11=4 sin 32 q3=2 + 4
72
h
i
1=3
where q = E ; x and = 2hm2 .
On the other hand when E < V (x)
!
Zx q
c
1
2
II (x) =
[x ; E ]1=4 exp " ; h x0=E= 2m(x ; E )dx
#
Zx q
c
1
2
= [x ; E ]1=4 exp ; h2m 0
2m(x ; E )d(2mx)
x =E=
= c31=4 exp ; 2 (;q)2=3 :
(2.147)
[;q]
3
We can nd an exact solution for this problem so we can compare with
the approximate solutions we got with the WKB method. We have
H ji = E ji ) hpjH ji = hpjE ji
2
) hpj 2pm + xji = E hpji
2
) 2pm (p) + ih dpd (p) = E(p)
2!
d
;
i
p
) dp (p) = h E ; 2m (p)
2!
d
(
p
)
;
i
p
) (p) = h E ; 2m dp
3!
p
;
i
) ln (p) = h Ep ; 6m + c1
"
!#
3
i
p
) E (p) = c exp h 6m ; Ep :
(2.148)
We also have
(E ; E 0 )
=
Z
Z
hE jE 0i = dphE jpihpjE 0 i = E (p)E0 (p)dp
Z
jcj2 dp exp hi (E ; E 0)p jcj22h(E ; E 0) )
(2.149)
c = p1 :
2h
(2:148)
=
So
"
!#
3
1
i
p
E (p) = p
exp h 6m ; Ep :
2h
(2.150)
2. QUANTUM DYNAMICS
73
These are the Hamiltonian eigenstates in momentum space. For the eigenfunctions in coordinate space we have
Z
Z ipx i p3 ;Ep
1
(2:150)
p dpe h e h 6m
(x) = dphxjpihpjE i =
2
h
" 3 #
Z
1
p
i
E
p dp exp i h6m ; h ; x p :
=
(2.151)
2h Using now the substitution
p3 = u3
p
)
(2.152)
u = (h2m
)1=3 h6m 3
we have
" 3
#
E 1=3 Z +1
iu
i
(
h
2
m
)
p
du exp 3 ; h ; x u(h2m)1=3
(x) =
2h ;1 "
#
Z +1
3
iu
= p
du exp 3 ; iuq ;
(2.153)
2 ;1
h
1=3
i
and q = E ; x . So
where = 2hm2
!
!
Z +1
Z +1
3
3
u
u
du cos 3 ; uq = p
cos
(x) = p
3 ; uq du
2 ;1
0
R +1 sin u3 ; uq du = 0. In terms of the Airy functions
since ;1
3
!
Z +1
3
u
1
cos 3 ; uq du
(2.154)
Ai(q) = p
0
we will have
(x) = p Ai(;q):
(2.155)
For large jqj, leading terms in the asymptotic series are as follows
2 1
p
Ai(q) 2 q1=4 exp ; 3 q3=2 ; q > 0
(2.156)
(2.157)
Ai(q) p(;1 q)1=4 sin 32 (;q)3=2 + 4 ; q < 0
74
Using these approximations in (2.155) we get
2
1
3
=
2
(q) p q1=4 sin 3 q + 4 ; for E > V (x)
2
1
3
=
2
p
exp ; 3 (;q) ; for E < V (x) (2.158)
(q) 2 (;q)1=4
as expected from the WKB approximation.
(b) When V = jxj we have bound states and therefore the energy spectrum is discrete. So in this case the energy eigenstates heve to satisfy the
consistency relation
Z x1 q
dx 2m[E ; jxj] = n + 21 h; n = 0; 1; 2; : : :
(2.159)
x2
The turning points are x1 = ; E and x2 = E . So
Z E= q
Z E= q
1
n + 2 h =
2m[E ; x]dx
dx 2m[E ; jxj] = 2
0
;E=
p Z E= E 1=2
= ;2 2m
; x d(;x)
0
p 2 E 3=2E= p 2 E 3=2
= ;2 2m 3 ; x = 2 2m 3 )
0
E 3=2
3 n + 12 h E [3 n + 21 h]2=3
p
=
) = 42=3(2m)1=3 )
4 2m
2 1 32=3
3 n + 2 h 5
p
En = 4
:
(2.160)
4 2m
3. THEORY OF ANGULAR MOMENTUM
75
3 Theory of Angular Momentum
3.1 Consider a sequence of Euler rotations represented by
! ;i3 ;
i
;
i
2
3
(1
=
2)
D (; ; ) = exp
exp
exp
2
=
e;i(+)=2 cos 2
ei(;)=2 sin 2
2
;e;i(;)=2 sin 2
ei(+)=2 cos 2
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 .
In the case of Euler angles we have
;i(+ )=2 cos ;e;i(; )=2 sin !
e
(1
=
2)
D (; ; ) = ei(;)=2 sin 2 ei(+)=2 cos 2
(3.1)
2
2
while the same rotation will be represented by
1
0 ;
n
)
sin
(S ;3:2:45) @ cos 2 ; inz sin 2 (;in
x
y
(1
=
2)
2 A : (3.2)
D (; n^ ) =
(;inx + ny ) sin 2 cos 2 + inz sin 2
Since these two operators must have the same eect, each matrix element
should be the same. That is
!
!
;
i
(
+
)
=
2
e
cos 2 = cos 2 ; inz sin 2
!
) cos 2 = cos ( +2 ) cos 2
) cos = 2 cos2 2 cos2 ( +2 ) ; 1
"
#
(
+
)
2
2
) = arccos 2 cos 2 cos 2 ; 1 :
(3.3)
3.2 An angular-momentum eigenstate jj; m = mmax = j i is rotated
by an innitesimal angle " about the y-axis. Without using the
76
explicit form of the d(mj)0m 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.
The rotated state is given by
jj; j iR =
R("; y^)jj; j i = d(j)(")jj; j i =
"
iJy " exp ; h jj; j i
#
iJ
y " (;i)2"2 2
= 1;
+
J jj; j i
(3.4)
h
2h2 y
up to terms of order "2. We can write Jy in terms of the ladder operators
)
J+ = Jx + iJy ) J = J+ ; J; :
(3.5)
y
J; = Jx ; iJy
2i
Subtitution of this in (3.4), gives
"
#
2
"
"
2
jj; j iR = 1 ; 2h (J+ ; J;) + 8h2 (J+ ; J; ) jj; j i
(3.6)
We know that for the ladder operators the following relations hold
q
J+jj; mi = h (j ; m)(j + m + 1)jj; m + 1i
(3.7)
q
J;jj; mi = h (j + m)(j ; m + 1)jj; m ; 1i
(3.8)
So
q
(J+ ; J;)jj; j i = ;J; jj; j i = ;h 2j jj; j ; 1i
(3.9)
q
(J+ ; J; )2jj; j i = ;h 2j (J+ ; J;)jj; j ; 1i
q
= ;h 2j (J+jj; j ; 1i ; J;jj; j ; 1i)
q
q q
= ;h 2j 2j jj; j i ; 2(2j ; 1)jj; j ; 2i
and from (3.6)
2
q
2 q
"
"
"
jj; j iR = jj; j i + 2 2j jj; j ; 1i ; 8 2j jj; j i + 8 2 j (2j ; 1)jj; j ; 2i
2 !
2q
q
"
= 1 ; 4 j jj; j i + 2" 2j jj; j ; 1i + "4 j (2j ; 1)jj; j ; 2i:
3. THEORY OF ANGULAR MOMENTUM
77
Thus the probability for the rotated state to be found in the original state
will be
2 !2
2
"
2
jhj; j jj; j iRj = 1 ; 4 j = 1 ; "2 j + O("4):
(3.10)
3.3 The wave function of a particle subjected to a spherically
symmetrical potential V (r) is given by
(~x) = (x + y + 3z)f (r):
(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
ml states?
(c) Suppose it is known somehow that (~x) is an energy eigenfunction with eigenvalue E . Indicate how we may nd V (r).
(a) We have
(~x) h~xj i = (x + y + 3z)f (r):
So
(3.11)
"
!#
2
1
@
1
@
@
(S ;3:6:15)
2
2
~
h~xjL j i = ;h sin2 @2 + sin @ sin @ (~x): (3.12)
If we write (~x) in terms of spherical coordinates (x = r sin cos ; y =
r sin sin ; z = r cos ) we will have
(~x) = rf (r) (sin cos + sin sin + 3 cos ) :
(3.13)
Then
1 @ 2 (~x) = rf (r) sin @ (cos ; sin ) = ; rf (r) (cos + sin ()3.14)
sin sin2 @2
sin2 @
78
and
!
1 @ sin @ (~x) = rf (r) @ h;3 sin2 + (cos + sin ) sin cos i =
sin @
@
sin @
i
rf (r) h;6 sin cos + (cos + sin )(cos2 ; sin2 )(3.15)
:
sin Substitution of (3.14) and (3.15) in (3.11) gives
1
2
2
2
2
~
h~xjL j i = ;h rf (r) ; sin (cos + sin )(1 ; cos + sin ) ; 6 cos = h2rf (r) sin1 2 sin2 (cos + sin ) + 6 cos = 2h2rf (r) [sin cos + sin sin + 3 cos ] = 2h2 (~x) )
L2 (~x) = 2h2 (~x) = 1(1 + 1)h2 (~x) = l(l + l)h2 (~x)
(3.16)
which means that (~x) is en eigenfunction of L~ 2 with eigenvalue l = 1.
(b) Since we already know that l = 1 we can try to write (~x) in terms of
the spherical harmonics Y1m (; ). We know that
s
s
s
3
3
z
Y10 = 4 cos = 4 r ) z = r 43 Y10
9 8
q q
Y1+1 = ;q 83 (x+riy) = < x = r q23 Y1;1 ; Y1+1 )
Y1;1 = 83 (x;riy) ; : y = ir 23 Y1;1 + Y1+1
So we can write
s
i
hp
(~x) = r 23 f (r) 3 2Y10 + Y1;1 ; Y1+1 + iY1+1 + iY1;1
s
hp
i
= 23 rf (r) 3 2Y10 + (1 + i)Y1;1 + (i ; 1)Y1+1 : (3.17)
But this means that the part of the state that depends on the values of m
can be written in the following way
hp
i
j im = N 3 2jl = 1; m = 0i + (1 + i)jl = 1; m = ;1i + (1 ; i)jl = 1; m = 1i
and if we want it normalized we will have
jN j2(18 + 2 + 2) = 1 ) N = p122 :
(3.18)
3. THEORY OF ANGULAR MOMENTUM
So
(c) If
9;
P (m = 0) = jhl = 1; m = 0j ij2 = 922 2 = 11
1;
P (m = +1) = jhl = 1; m = +1j ij2 = 222 = 11
1:
P (m = ;1) = jhl = 1; m == 1j ij2 = 222 = 11
79
(3.19)
(3.20)
(3.21)
E (~x) is an energy eigenfunction then it solves the Schrodinger equation
;h2 " @ 2 (~x) + 2 @ (~x) ; L2 (~x)# + V (r) (~x) =
E
2m @r2 E
r @r E
h2r2 E
E E (~x) "
#
2
2
d
2
d
2
;
h
m
) 2m Yl dr2 [rf (r)] + r dr [rf (r)] ; r2 [rf (r)] + V (r)rf (r)Ylm =
Erf (r)Ylm ) "
#
2 d
1
h
2
2
0
0
V (r) = E + rf (r) 2m dr [f (r) + rf (r)] + r [f (r) + rf (r)] ; r f (r) )
2
V (r) = E + rf1(r) 2hm [f 0(r) + f 0(r) + rf 00(r) + 2f 0(r)]] )
2 rf 00 (r) + 4f 0 (r)
h
V (r) = E + 2m
:
rf (r)
(3.22)
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 direction.
(a) Show explicitly that innitesimal rotations of %i(~x) are obtained
by acting with the operator
(3.23)
u~" = 1 ; i h~" (L~ + S~)
80
where L~ = hi r^ r~ . Determine S~ !
(b) Show that L~ and S~ commute.
(c) Show that S~ is a vector operator.
(d) Show that r~ %~(~x) = h1 (S~ ~p)%~ where ~p is the momentum operator.
2
(a) We have
j%i =
3 Z
X
i=1
j~x; iih~x; ij%i =
3 Z
X
i=1
j~x; ii%i(~x)d3x:
(3.24)
Under a rotation R we will have
3 Z
X
j%0i = U (R)j%i =
U (R) [j~xi jii] %i(~x)d3x
i=1
Z
3
3 Z
X
X
=
jR~xi jiiDil(1)(R)%l (~x)d3x det=R=1
j~x; iiDil(1)(R)%l (R;1~x)d3x
i=1
i=1
Z
3
X
=
j~x; ii%i0~x)d3x )
%i0(~x)
=
i=1
Dil(1)(R)%l(R;1~x) ) %~0(~x) = R~%(R;1~x):
(3.25)
Under an innitesimal rotation we will have
R(; n^ )~r = ~r + ~r = ~r + (^n ~r) = ~r + ~" ~r:
(3.26)
So
%~0(~x) = R()%~(R;1~x) = R()%~(~x ; ~" ~x)
= ~%(~x ; ~" ~x) + ~" %~(~x ; ~" ~x):
(3.27)
On the other hand
~ %~(~x) = ~%(~x) ; ~" (~x r
~ )%~(~x)
~%(~x ; ~" ~x) = ~%(~x) ; (~" ~x) r
(3.28)
= ~%(~x) ; hi ~" L~ %~(~x)
3. THEORY OF ANGULAR MOMENTUM
~ %~(~x) hr
~ %i (~x)i jii. Using this in (3.27) we get
where r
i
i
0
~
~
%~ (~x) = ~%(~x) ; h ~" L%~(~x) + ~" %~(~x) ; h ~" L%~(~x)
= ~%(~x) ; hi ~" L~ %~(~x) + ~" ~%(~x):
But
81
(3.29)
"~ %~ = "y %3 ; "z %2 e^x + "z %1 ; "x%1 e^y + "x%2 ; "y %1 e^z (3.30)
or in0matrix
1 form
0
10 1 1
1
0
%
0 ;"z "y
B@ %20 CA = B@ "z 0 ;"x CA B@ %%2 CA =
%30
;"y "x 0
%3
2 0
1
0
1
0
13 0 1 1
0 0 0
0 0 1
0 ;1 0
%
64"x B
C
B
C
B
C
7
B
@ 0 0 ;1 A + "y @ 0 0 0 A + "z @ 1 0 0 A5 @ %2 CA =
0 1 0
;1 0 0
0 0 0
%3
2 0
1
0
1
0
13 0 1 1
0
0
0
0
0
i
h
0
;
i
h
0
%
i
6
B
C
B
C
B
C
7
B
; h 4"x @ 0 0 ;ih A + "y @ 0 0 0 A + "z @ ih 0 0 A5 @ %2 CA
0 ih 0
;ih 0 0
0 0 0
%3
which means that
~" %~ = ; hi ~" S~~%(~x)
with (S)kl = ;ihkl.
Thus we will have that
i
0
~
~
%~ (~x) = U~"~%(~x) = 1 ; h ~" (L + S ) %~(~x) ) U~" = 1 ; hi ~" (L~ + S~): (3.31)
(b) From their denition it is obvious that L~ and S~ commute since L~ acts
only on the j~xi basis and S~ only on jii.
(c) S~ is a vector operator since
X
[Si; Sj ]km = [SiSj ; Sj Si]km = [(;ih)ikl(;ih)jlm ; (;ih)jkl(;ih)ilm]
i
Xh 2
=
h ikljml ; h2jkliml
X
= h2 (ij km ; imjk ; ij km + jmki)
X
= h2 (jmki ; imjk )
X
X
X
= h2 ijlkml = ihijl(;ihkml) = ihijl(Sl)km :
(3.32)
82
(d) It is
r~ %~(~x) = hi ~p %~(~x) = hi ilp %l(~x)jii = h12 (S)lm p%m jii
= 12 (S~ ~p)%~:
(3.33)
h
3.5 We are to add angular momenta j1 = 1 and j2 = 1 to form
j = 2; 1; and 0 states. Using the ladder operator method express all
(nine) j; m eigenkets in terms of jj1j2; m1m2i. Write your answer as
jj = 1; m = 1i = p12 j+; 0i ; p12 j0; +i; : : : ;
(3.34)
where + and 0 stand for m1;2 = 1; 0; respectively.
We want to add the angular momenta j1 = 1 and j2 = 1 to form j =
jj1 ; j2j; : : :; j1 + j2 = 0; 1; 2 states. Let us take rst the state j = 2, m = 2.
This state is related to jj1m1; j2m2i through the following equation
X
jj; mi =
hj1j2; m1m2jj1j2; jmijj1j2; m1m2i
(3.35)
m=m1 +m2
So setting j = 2, m = 2 in (3.35) we get
jj = 2; m = 2i = hj1j2; + + jj1j2; jmij + +i norm.
= j + +i (3.36)
If we apply the J; operator on this statet we will get
Jq
; jj = 2; m = 2i = (J1; + J2; )j + +i
) h (j + m)(j ; m + 1)jj = 2; m = 1i =
q
q
h (j + m )(j ; m + 1)j0+i + h (j2 + m2)(j2 ; m2 + 1)j + 0i
p 1 1 1 1p
p
) 4jj = 2; m = 1i = 2j0+i + 2j + 0i
) jj = 2; m = 1i = p12 j0+i + p12 j + 0i:
(3.37)
3. THEORY OF ANGULAR MOMENTUM
83
In the same way we have
J;jj = 2; m = 1i = p1 (J1; + J2;)j0+i + p1 (J1; + J2;)j + 0i )
2
2
i
i
h
hp
p
p
p
p
6jj = 2; m = 0i = p1 2j ; +i + 2j00i + p1 2j00i + 2j + ;i )
2
2
p
6jj = 2; m = 0i = 2sj00i + j ; +i + j + ;i )
(3.38)
jj = 2; m = 0i = 23 j00i + p16 j + ;i + p16 j ; +i
s
2 (J + J )j00i
3 1; 2;
+ p1 (J1; + J2;)j + ;i + p1 (J1; + J2;)j ; +i )
6
s 6h
i
p
p
p
p
p
6jj = 2; m = ;1i = 23 2j ; 0i + 2j0;i + p1 2j0;i + p1 2j ; 0i )
6
6
p
p
p
p
2
2
1
1
jj = 2; m = ;1i = 6 2j0;i + 6 2j ; 0i + 6 2j0;i + 6 2j ; 0i )
(3.39)
jj = 2; m = ;1i = p12 j0;i + p12 j ; 0i
J;jj = 2; m = 0i =
J;jj = 2; m = ;1i = p1 (J1; + J2;)j0;i + p1 (J1; + J2; )j ; 0i )
2
2
p
p
p
1
1
4jj = 2; m = ;2i = p 2j ; ;i + p 2j ; ;i )
2
2
jj = 2; m = ;2i = j ; ;i:
(3.40)
Now let us return to equation (3.35). If j = 1, m = 1 we will have
jj = 1; m = 1i = aj + 0i + bj0+i
(3.41)
This state should be orthogonal to all jj; mi states and in particular to jj =
2; m = 1i. So
hj = 2; m = 1jj = 1; m = 1i = 0 ) p12 a + p12 b = 0 )
a + b = 0 ) a = ;b
: (3.42)
84
In addition the state jj = 1; m = 1i should be normalized so
hj = 1; m = 1jj = 1; m = 1i = 1 ) jaj2 + jbj2 = 1 (3):42) 2jaj2 = 1 ) jaj = p12 :
By convention we take a to be real and positive so a = p12 and b = ; p12 .
That is
(3.43)
jj = 1; m = 1i = p12 j + 0i ; p12 j0+i:
Using the same procedure we used before
J;jj = 1; m = 1i = p1 (J1; + J2;)j + 0i ; p1 (J1; + J2;)j0+i )
2
2
i 1 hp
hp
p
p
p i
1
2jj = 1; m = 0i = p 2j00i + 2j + ;i ; p 2j ; +i + 2j00i )
2
2
1
1
(3.44)
jj = 1; m = 0i = p2 j + ;i ; p2 j ; +i
J;jj = 1; m = 0i = p1 (J1; + J2;)j + ;i ; p1 (J1; + J2;)j ; +i )
2
2
p
p
p
2jj = 1; m = ;1i = p1 2j0;i ; p1 2j ; 0i )
2
2
1
1
jj = 1; m = ;1i = p2 j0;i ; p2 j ; 0i:
(3.45)
Returning back to (3.35) we see that the state jj = 0; m = 0i can be written
as
jj = 0; m = 0i = c1j00i + c2j + ;i + c3j ; +i:
(3.46)
This state should be orthogonal to all states jj; mi and in particulat to jj =
2; m = 0i and to j = 1; m = 0i. So
s
hj = 2; m = 0jj = 0; m = 0i = 0 ) 23 c1 + p16 c2 + p16 c3
) 2c1 + c2 + c3 = 0
(3.47)
hj = 1; m = 0jj = 0; m = 0i = 0 ) p12 c2 ; p12 c3
) c2 = c3:
(3.48)
3. THEORY OF ANGULAR MOMENTUM
85
Using the last relation in (3.47), we get
2c1 + 2c2 = 0 ) c1 + c2 = 0 ) c1 = ;c2:
(3.49)
The state jj = 0; m = 0i should be normalized so
hj = 0; m = 0jj = 0; m = 0i = 1 ) jc1j2 + jc2j2 + jc3j2 = 1 ) 3jc2j2 = 1
) jc2j = p13 :
(3.50)
By convention we take c2 to be real and positive so c2 = c3 =
c1 = ; p13 . Thus
jj = 0; m = 0i = p13 j + ;i + p13 j ; +i ; p13 j00i:
p1
3
and
(3.51)
So gathering all the previous results together
jj = 2; m = 2i
jj = 2; m = 1i
jj = 2; m = 0i
jj = 2; m = ;1i
jj = 2; m = ;2i
jj = 1; m = 1i
jj = 1; m = 0i
jj = 1; m = ;1i
jj = 0; m = 0i
= j + +i
p1 j0+i + p1 j + 0i
= q
2
2
2
1
p
= 3 j00i + 6 j + ;i + p16 j ; +i
= p12 j0;i + p12 j ; 0i
= j ; ;i
= p12 j + 0i ; p12 j0+i
= p12 j + ;i ; p12 j ; +i
= p12 j0;i ; p12 j ; 0i
= p13 j + ;i + p13 j ; +i ; p13 j00i:
(3.52)
3.6 (a) Construct a spherical tensor of rank 1 out of two dierent
vectors U~ = (Ux; Uy ; Uz ) and V~ = (Vx ; Vy ; Vz ). Explicitly write T(1)1;0 in
terms of Ux;y;z and Vx;y;z .
(b) Construct a spherical tensor of rank 2 out of two dierent
vectors U~ and V~ . Write down explicitly T(2)2;1;0 in terms of Ux;y;z
and Vx;y;z .
86
(a) Since U~ and V~ are vector operators they will satisfy the following commutation relations
[Ui; Jj ] = ih"ijk Uk [Vi; Jj ] = ih"ijk Vk :
(3.53)
From the components of a vector operator we can construct a spherical tensor
of rank 1 in the following way. The dening properties of a spherical tensor
of rank 1 are the following
q
[Jz ; Uq(1)] = h qUq(1);
[J; Uq(1)] = h (1 q)(2 q)Uq(1)1 : (3.54)
It is
:54)
[Jz ; Uz ] (3=:53) 0hUz (3)
Uz = U0
p
[J+; U0] (3=:54) 2hU+1 = [J+; Uz ] = [Jx + iJy ; Uz ]
(3:53)
= ;ihUy + i(ih)Ux = ;h(Ux + iUy ) )
U+1 = ; p1 (Ux + iUy )
2
(3:54) p
2hU;1 = [J;; Uz ] = [Jx ; iJy ; Uz ]
[J;; U0] =
(3:53)
= ;ihUy ; i(ih)Ux = h(Ux ; iUy ) )
U;1 = p1 (Ux ; iUy )
2
So from the vector operators U~ and V~ we can construct spherical
with components
U0 = Uz
U+1 = ; p12 (Ux + iUy )
U;1 = p12 (Ux ; iUy )
V0 = Vz
V+1 = ; p12 (Vx + IVy )
V;1 = p12 (Vx ; iVy )
(3.55)
(3.56)
(3.57)
tensors
(3.58)
It is known (S-3.10.27) that if Xq(1k1 ) and Zq(2k2) are irreducible spherical tensors
of rank k1 and k2 respectively then we can construct a spherical tensor of
rank k
X
Tq(k) = hk1 k2; q1q2jk1k2 ; kqiXq(1k1)Zq(2k2)
(3.59)
q1 q 2
3. THEORY OF ANGULAR MOMENTUM
In this case we have
(1)
T+1
= h11; +10j11; 11iU+1 V0 + h11; 0 + 1j11; 11iU0 V+1
(3:52) 1
= p U+1V0 ; p1 U0V+1
2
2
1
= ; 2 (Ux + iUy )Vz + 21 Uz (Vx + iVy )
87
(3.60)
T0(1) = h11; 00j11; 10iU0 V0 + h11; ;1 + 1j11; 10iU;1 V+1
+h11; +1 ; 1j11; 10iU+1 V;1
(3:52)
= ; p1 U;1V+1 + p1 U+1V;1
2
2
1
= p 12 (Ux ; iUy )(Vx + iVy ) ; p1 21 (Ux + iUy )(Vx ; iVy )
2
2
1
= p [UxVx + iUxVy ; iUy Vx + Uy Vy ; UxVx + iUxVy ; iUy Vx ; Uy Vy ]
2 2
= pi (UxVy ; Uy Vx)
(3.61)
2
T;(1)1 = h11; ;10j11; 11iU;1 V0 + h11; 0 ; 1j11; 11iU0 V;1
(3:52)
= ; p1 U;1V0 + p1 U0V;1
2
2
= ; 21 (Ux ; iUy )Vz + 21 Uz (Vx ; iVy ):
(3.62)
(b) In the same manner we will have
(2)
T+2
= h11; +1 + 1j11; 2 + 2iU+1 V+1 (3=:52) U+1 V+1 = 12 (Ux + iUy )(Vx + iVy )
= ; 21 (UxVx ; Uy Vy + iUxVy + iUy Vx )
(3.63)
(2)
T+1
=
(3:52)
=
=
h11; 0 + 1j11; 2 + 1iU0 V+1 + h11; +10j11; 2 + 1iU+1 V0
p1 U0V+1 + p1 U+1V0
2
2
; 12 (Uz Vx + UxVz + iUz Vy + iUy Vz )
T0(2) = h11; 00j11; 20iU0 V0 + h11; ;1 + 1j11; 20iU;1 V+1
+h11; +1 ; 1j11; 20iU+1 V;1
(3.64)
88
(3:52)
=
=
=
=
T;(2)1 =
(3:52)
=
=
s
s
s
2U V + 1U V + 1U V
0 0
;1 +1
+1 ;1
3
6
6
s
s
s
2 U V ; 1 1 (U ; iU )(V + iV ) ; 1 p1 1 (U + iU )(V ; iV )
y x
y
y x
y
z z
62 x
6 22 x
s3 1 2U V ; 1 U V ; i U V + i U V
6 z z 2 x x 2 x y 2 y x
i
i
1
1
1
; 2 Uy Vy ; 2 UxVx + 2 UxVy ; 2 Uy Vx ; 2 Uy Vy
s
1 (2U V ; U V ; U V )
(3.65)
6 z z x x y y
h11; 0 ; 1j11; 2 ; 1iU0 V;1 + h11; ;10j11; 2 + 1iU;1 V0
p1 U0V;1 + p1 U;1 V0
2
2
1 (U V + U V ; iU V ; iU V )
x z
z y
y z
2 z x
(3.66)
T;(2)2 = h11; ;1 ; 1j11; 2 ; 2iU;1 V;1 (3=:52) U;1V;1 = 21 (Ux ; iUy )(Vx ; iVy )
= 12 (UxVx ; Uy Vy ; iUxVy ; iUy Vx):
(3.67)
3.7 (a) Evaluate
j
X
m=;j
j)
2
jd(mm
0 ( )j m
for any j (integer or half-integer); then check your answer for j = 21 .
(b) Prove, for any j ,
j
X
m=;j
m2jdm(j)0 m( )j2 = 12 j (j + 1) sin + m02 + 12 (3 cos2 ; 1):
[Hint: This can be proved in many ways. You may, for instance,
examine the rotational properties of Jz2 using the spherical (irreducible) tensor language.]
3. THEORY OF ANGULAR MOMENTUM
(a) We have
j
X
=
=
=
=
m=;j
j
X
m=;j
j
X
m=;j
j
X
m=;j
j
X
m=;j
89
j ) ( )j2m
jd(mm
0
mjhjmje;iJy =hjjm0ij2
mhjmje;iJy =hjjm0i hjmje;iJy =hjjm0i
mhjmje;iJy =hjjm0ihjm0jeiJy=hjjmi
hjm0jeiJy=hmjjmihjmje;iJy=hjjm0i
2 j
3
X
1
0
iJ
=
h
= h hjm je y Jz 4
jjmihjmj5 e;iJy =hjjm0i
m=;j
1
= h hjm0jeiJy=hJz e;iJy =hjjm0i
(3.68)
= h1 hjm0jD( ; e^y )Jz D( ; e^y )jjm0i:
But the momentum J~ is a vector operator so from (S-3.10.3) we will have
that
X
(3.69)
D ( ; e^y)Jz D( ; e^y ) = Rzj ( ; e^y )Jj :
j
On the other hand we know (S-3.1.5b) that
0
1
cos 0 sin R( ; e^y ) = B
(3.70)
@ 0 1 0 CA :
; sin 0 cos So
j
X
j ) ( )j2m = 1 [; sin hjm0jJ jjm0 i + cos hjm0jJ jjm0i]
jd(mm
0
x
z
h
m=;j
1
J
+ + J;
0
0
0
= h ; sin hjm j 2 jjm i + hm cos = m0 cos :
(3.71)
90
For j = 1=2 we know from (S-3.2.44) that
; sin !
cos
(1=2)
dmm0 ( ) = sin 2 cos 2 :
2
2
0
So for m = 1=2
1=2
X
jd(mj)1=2( )j2m = ; 12 sin2 2 + 12 cos2 2
m=;1=2
= 21 cos = m0 cos while for m0 = ;1=2
1=2
X
jd(mj)1=2( )j2m = ; 21 cos2 2 + 21 sin2 2
m=;1=2
= ; 12 cos = m0 cos :
(3.72)
(3.73)
(3.74)
(b) We have
j
X
=
=
=
=
m=;j
j
X
m=;j
j
X
m=;j
j
X
m=;j
j
X
m=;j
m2jd(mj)0 m( )j2
m2jhjm0je;iJy =hjjmij2
m2hjm0je;iJy =hjjmi hjm0je;iJy =hjjmi
m2hjm0je;iJy =hjjmihjmjeiJy=hjjm0i
hjm0je;iJy =hm2jjmihjmjeiJy =hjjm0i
2 j
3
X
1
= 2 hjm0je;iJy =hJz2 4
jjmihjmj5 eiJy =hjjm0i
h
m=;j
= 12 hjm0je;iJy =hJz2eiJy =hjjm0i
h
= h1 hjm0jD( ; e^y )Jz2Dy( ; e^y)jjm0i:
(3.75)
3. THEORY OF ANGULAR MOMENTUM
From (3.65) we know that
T0(2) =
s
91
1 (3J 2 ; J 2)
6 z
(3.76)
where T0(2) is the 0-component of a second rank tensor. So
p
Jz2 = 36 T0(2) + 31 J 2
(3.77)
and since D(R)J 2Dy(R) = J 2D(R)Dy (R) = J 2 we will have
Pj
2 (j )
2
m=;j m jdmm0 ( )j =
s
1 1 hjm0jJ 2jjm0i + 2 1 hjm0jD( ; e^ )J 2Dy( ; e^ )jjm0(3.78)
i:
y z
y
3 h2
h3 3
We know that for a spherical tensor (S-3.10.22b)
k
X
(
k
)
y
D(R)Tq D (R) =
Dq(k0q)(R)Tq(0k)
(3.79)
q0 =;k
which means in our case that
hjm0jD( ; e^y )Jz2Dy( ; e^y )jjm0i = hjm0j
=
2
X
2
X
q0 =;2
(2)
Tq(2)
0 Dq0 0 ( ; e^y )jjm0i
q0 =;2
Dq(2)00 ( ; e^y)hjm0jTq(2)0 jjm0i:
(3.80)
But we know from the Wigner-Eckart theorem that hjm0jTq(2)
06=0 jjm0i = 0. So
j
X
m=;j
j)
2
m2jd(mm
0 ( )j
s
1
1
2
= 2 h j (j + 1) + 2 23 D00(2)( ; e^y)hjm0jT0(2)jjm0i
3h
h
1
1
(2)
= 3 j (j + 1) + 2 d00 ( )hjm0jJz2 ; 13 J 2jjm0i
2 1
1
1
(2)
0
= 3 j (j + 1) + 2 d00 ( ) m ; 3 j (j + 1)
92
2 1
1
1
2
0
= 3 j (j + 1) + 2 (3 cos ; 1) m ; 3 j (j + 1)
02
1
1
1
m
2
= ; j (j + 1) cos + j (j + 1) + j (j + 1) + (3 cos2 ; 1)
2
6
3
2
2
2
0
2
1
1
(3.81)
= 2 j (j + 1) sin + m 2 (3 cos ; 1)
1
2
where we have used d(2)
00 ( ) = P2 (cos ) = 2 (3 cos ; 1).
3.8 (a) Write xy, xz, and (x2 ; y2) as components of a spherical
(irreducible) tensor of rank 2.
(b) The expectation value
Q eh; j; m = j j(3z2 ; r2)j; j; m = j i
is known as the quadrupole moment. Evaluate
eh; j; m0j(x2 ; y2)j; j; m = j i;
(where m0 = j; j ; 1; j ; 2; : : : )in terms of Q and appropriate ClebschGordan coecients.
(a) Using the relations (3.63-3.67) we can nd that in the case where U~ =
V~ = ~x the components of a spherical tensor of rank 2 will be
(2)
T+2
= 12 (x2 ; y2) + ixy
T;(2)2 = 21 (x2 ; y2) ; ixy
(2) = ;(xz + izy )
T+1
T;(2)1 = xz ; izy
(3.82)
q
q
(2)
1
1
2
2
2
2
2
T0 = 6 (2z ; x ; y ) = 6 (3z ; r )
So from the above we have
(2) ; T (2)
(2) ; T (2)
2 2 (2) (2)
T
T
+2
;
2
;
1
+1 : (3.83)
x ; y = T+2 + T;2 ; xy =
;
xz
=
2i
2
(b) We have
Q = eh; j; m = j j(3z2 ; r2)j; j; m = j i
3. THEORY OF ANGULAR MOMENTUM
(3:82)
=
)
93
p
(2) kj i p
h
j
k
T
(W:;E:)
(2)
p
6eh; j; m = j jT0 j; j; m = j i = hj 2; j 0jj 2; jj i 2j + 1 6e
hj kT (2)kj i =
p
pQ hj 2; j20jjj+2;1jj i :
6e
(3.84)
So
e h; j; m0j(x2 ; y2)j; j; m = j i
(3:83)
(2)
= eh; j; m0jT+2
j; j; m = j i + eh; j; m0jT;(2)2 j; j; m = j i
z
}|0
{
(2)
(2)
= e hj 2; j 2jj 2; jm0i hjpk2Tj +k1j i + em0;j;2hj 2; j ; 2jj 2; jj ; 2i hjpk2Tj +k1j i
(3:84) Q hj 2; j; ;2jj 2; j; j ; 2i
= p hj 2; j; 0jj 2; j; j i m0;j;2:
(3.85)
6
94
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 ei~p~x=h. Why does this not violate time-reversal
invariance?
(a) Suppose that jni in a nondegenerate energy eigenstate. Then
H jni = H jni = En jni ) jni = ei jni
) jn; t0 = 0; ti = e;itH=hjni = e;itEn=hjni =
2E t
E t
eitEZn=hjni = ei( hn +) jni = ei( hn +)jn;Zt0 = 0; ti ) d3xj~xih~xj jn; t0 = 0; ti = ei( 2Ehn t +) d3xj~xih~xj jn; t0 = 0; ti
Z
Z
3
) d xh~xjn; t0 = 0; ti j~xi = d3xei( 2Ehn t +)h~xjn; t0 = 0; tij~xi
) n(~x; t) = ei( Ehn t +) n(~x; t):
2
(4.1)
the wave function will be
So if we choose at any instant of time = ; 2Ehnt
real.
(b) In the case of a free particle the Schrodinger equation is
p2 jni = E jni ) ; h2 r
~
2m
2m n (x) = En(x)
) n (x) = Aei~p~x=h + Be;i~p~x=h
(4.2)
The wave functions n(2x) = e;i~p~x=h and 0n (x) = ei~p~x=h correspond to the
same eigenvalue E = 2pm and so there is degeneracy since these correspond
to dierent state kets j~pi and j ; p~i. So we cannot apply the previous result.
4.2 Let (~p0) be the momentum-space wave function for state ji,
that is, (~p0) = h~p0ji.Is the momentum-space wave function for the
4. SYMMETRY IN QUANTUM MECHANICS
95
time-reversed state ji given by (~p0), (;~p0), (~p0 ), or (;~p0)?
Justify your answer.
In the momentum space we have
Z
Z
ji = d3p0h~p0 jij~p0i ) ji = d3p0(~p0)j~p0i
Z
Z
3
0
0
0
) ji = d p [h~p jij~p i] = d3p0h~p0 jij~p0i:
(4.3)
For the momentum it is natural to require
hj~pji = ;h~j~pj~ i )
h~ j~p;1j~i ) ~p;1 = ;~p
(4.4)
So
~pj~p0i (4=:4) ;~pj~p0i ) j~p0i = j ; ~p0i
up to a phase factor. So nally
Z
Z
ji = d3p0h~p0ji j ; p~0i = d3p0h;~p0 jij~p0i
) h~p0jji = ~(~p0) = h;~p0ji = (;~p0):
(4.5)
(4.6)
4.3 Read section 4.3 in Sakurai to refresh your knowledge of the
quantum mechanics of periodic potentials. You know that the energybands in solids are described by the so called Bloch functions
n;k fulllling,
ika
n;k (x + a) = e
n;k (x)
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,
X
n(x ; Ri)eikRi
n;k (x) =
Ri
96
where the sum is over all lattice vectors Ri. (In this simble one dimensional problem Ri = ia, but the construction generalizes easily
to three dimensions.).
The functions n are called Wannier functions, and are important in the tight-binding description of solids. Show that the Wannier functions are corresponding to dierent sites and/or dierent
bands are orthogonal, i:e: prove
Z
dx?m(x ; Ri)n(x ; Rj ) ij mn
Hint: Expand the n s in Bloch functions and use their orthonormality properties.
The dening property of a Bloch function
n;k (x + a) = e
ika
n;k (x)
is
n;k (x):
(4.7)
We can show that the functions PRi n(x ; Ri)eikRi satisfy the same relation
X
X
n(x + a ; Ri)eikRi = n [x ; (Ri ; a)]eik(Ri;a)eika
Ri
Ri
Ri ;a=Rj ika X
= e
n(x ; Rj )eikRj
(4.8)
Rj
which means that it is a Bloch function
X
n(x ; Ri )eikRi :
n;k (x) =
(4.9)
Ri
The last relation gives the Bloch functions in terms of Wannier functions.
To nd the expansion of a Wannier function in terms of Bloch functions we
multiply this relation by e;ikRj and integrate over k.
X
n(x ; Ri)eikRi
n;k (x) =
Ri
Z =a
Z =a
X
)
dke;ikRj n;k (x) = n (x ; Ri)
eik(Ri;Rj )dk (4.10)
;=a
Ri
;=a
4. SYMMETRY IN QUANTUM MECHANICS
But
97
ik(Ri;Rj ) =a
(Ri ; Rj )]
e
eik(Ri;Rj )dk = i(R ; R ) = 2 sin [=a
Ri ; Rj
;=a
i
j ;=a
= ij 2a
(4.11)
where in the last step we used that Ri ; Rj = na, with n 2 Z . So
Z =a
X
dke;ikRj n;k (x) = n(x ; Ri)ij 2a
;=a
R
Z =a i
a
) n(x ; Ri) = 2 ;=a e;ikRi n;k (x)dk
(4.12)
So using the orthonormality properties of the Bloch functions
Z
dxm(x ; Ri )n(x ; Rj )
Z Z Z a2
ikRi (x)e;ik0 Rj
0
=
e
n;k0 (x)dkdk dx
m;k
2
(2)
Z Z a2
Z
0 Rj
ikR
;
ik
0
i
=
m;k (x) n;k0 (x)dxdkdk
(2)2 e
Z Z a2
ikRi ;ik0 Rj (k ; k 0 )dkdk 0
e
=
mn
2
(2)
Z =a
2
ik(Ri ;Rj ) dk = a :
= a 2 mn
e
(4.13)
(2)
2 mn ij
Z =a
;=a
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
hL~ i = 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
XX
l m
Flm(r)Ylm(; );
98
what kind of phase restrictions do we obtain on Flm(r)?
Since the Hamiltonian is invariant under time reversal
H = H:
(4.14)
So if jni is an energy eigenstate with eigenvalue En we will have
H jni = H jni = Enjni:
(4.15)
If there is no degeneracy jni and jni can dier at most by a phase factor.
Hence
jn~ i jni = eijni:
(4.16)
For the angular-momentum operator we have from (S-4.4.53)
hnjL~ jni = ;hn~ jL~ jn~ i (4=:16) ;hnjL~ jni )
hnjL~ jni = 0
:
We have
Z
Z
jni = d3xh~xjni j~xi
Z
= d3xh~xjnij~xi (4=:16) eijni )
hx~0jjni = hx~0jni = ei hx~0jni:
(4.18)
So if we use h~xjni = Pl Pm Flm(r)Ylm(; )
X X
Flm(r)Ylm(; ) = ei Flm(r)Ylm(; )
ml
ml
X
(S ;4:4:57) X m
;
m
)
Flm(r)(;1) Yl (; ) = ei Flm(r)Ylm(; )
ml
Zml 0 X
Z
X
m
m
;
m
)
Yl0
Flm(r)(;1) Yl (; )d
= ei Ylm0 0 Flm(r)Ylm (; )d
ml
X ml m
X
)
Flm(r)(;1) m0;;m l0l = ei Flm(r)m0;m l0l
)
ml
Fl0;;m0 (r)(;1);m0
d3xj~xih~xjni =
(4.17)
ml
i
0
0
= e Fl m (r) ) Fl0;;m0 (r) = (;1)m0 Fl0m0 (r)ei :
(4.19)
4. SYMMETRY IN QUANTUM MECHANICS
99
4.5 The Hamiltonian for a spin 1 system is given by
H = ASz2 + B (Sx2 ; Sy2):
Solve this problem exactly to nd the normalized energy eigenstates 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 obtained transform under time reversal?
For a spin 1 system l = 1 and m = ;1; 0; +1. For the operator Sz we
have
Sz jl; mi = hmjl; mi ) hlnjSz jl; mi = hmhnjmi ) (Sz )nm = hmnm (4.20)
So
0
1
0
1
1
0
0
1
0
0
Sz = h B
@ 0 0 0 CA ) Sz2 = h2 B@ 0 0 0 CA
0 0 ;1
0 0 1
For the operator Sx we have
S+ + S; j1; mi = 1 S j1; mi + 1 S j1; mi !
Sxjl; mi =
2 +
2 ;
2
h1; njSxj1; mi = 12 h1; njS+ j1; mi + 12 h1; njS; j1; mi
q
(S ;3:5:39) 1 q
1 h (1 + m)(2 ; m)
=
h
(1
;
m
)(2
+
m
)
+
n;m+1 2
n;m;1 :
2
So
1 0 0 0 01
0 p
0
2
0
p
p
Sx = h2 B
@ 0 0 2 CA + h2 B@ 2 p0 0 CA
0 0 0
0 2 0
p
0
1
p0 2 p0
= h2 B
@ 2 p0 2 CA )
0 2 0
0
1
01
1
0 12
2 2 0 2
2
h
(4.21)
Sx2 = 4 B
@ 0 4 0 CA = h2 B@ 0 1 0 CA :
1
1
2 0 2
2 0 2
100
In the same manner for the operator Sy = S+2;iS; we nd
1
0 0 p
2
0
p
p
Sx = 2hi B
0
2C
A)
@; 2 p
0 ; 2 0
0
1
0 1
1 1
0
;
;
2
0
2
2
2
2
Sx2 = ; h4 B
@ 0 ;4 0 CA = h 2 B@ 0 1 0 CA :
; 21 0 12
2 0 ;2
Thus the Hamiltonian can be represented by the matrix
0
1
A
0
B
H = h2 B
@ 0 0 0 CA :
B 0 A
(4.22)
(4.23)
To nd the energy eigenvalues we have to solve the secular equation
0 2
2 1
A
h
;
0
B
h
det(H ; I ) = 0 ) det B
; 0 CA = 0
@ 0
B h2 0 Ah2 ; h
i
) (Ah2 ; )2(;) + (B h2)2 = 0 ) (Ah2 ; )2 ; (B h2)2 = 0
) (Ah2 ; ; B h2)(Ah2 ; + B h2) = 0
) 1 = 0; 2 = h2(A + B ); 3 = h2(A ; B ):
(4.24)
To nd the eigenstate jnc i that corresponds to the eigenvalue c we have to
solve the following equation
0
10 1
0 1
A
0
B
a
aC
B
C
B
C
B
2
h @ 0 0 0 A @ b A = c @ b A :
(4.25)
B 0 A
c
c
For 1 = 0
0
10 1
(
A 0 B
a
+ cB = 0
C
B
C
B
h 2 @ 0 0 0 A @ b A = 0 ) aA
aB + cA = 0
c
B 0 A
(
(
B
a
=
;
c
A
) ;c B2 + cA = 0 ) ac == 00
A
(4.26)
4. SYMMETRY IN QUANTUM MECHANICS
So
0 1
0 1
0
0
B b C norm: B 1 C
jn0i = @ A = @ A )
0
0
jn0i = j10i:
In the same way for = h 2(A + B )
0
10 1
0
A
0
B
a
B@ 0 0 0 CA B@ b C
A = (A + B ) B@
B 0 A
c
(
) ab == 0c
So
So
(4.27)
1 8
aC >
< aA + cB = a(A + B )
b A)>
0 = b(A + B )
: aB + cA = c(A + B )
c
(4.28)
0 1
0 1
c
1
1
norm: p B 0 C
B
C
0
jnA+B i = @ A = 2 @ A )
c
1
jnA+B i = p12 j1; +1i + p12 j1; ;1i:
For = h2(A ; B ) we have
0
10 1
0
A 0 B
a
B@ 0 0 0 CA B@ b CA = (A ; B ) B@
B 0 A
c
(
) a =b =;0c
101
(4.29)
1 8
>
a
< aA + cB = a(A ; B )
C
b A)>
0 = b(A ; B )
: aB + cA = c(A ; B )
c
0 1
0 1
c
1
norm: p1 B
B
C
jnA+B i = @ 0 A =
@ 0 CA )
2 ;1
;c
jnA;B i = p12 j1; +1i ; p12 j1; ;1i:
(4.30)
(4.31)
102
Now we are going to check if the Hamiltonian is invariant under time reversal
H ;1 = ASz2;1 + B (Sx2;1 ; Sy2;1)
= ASz ;1Sz ;1 + B (Sx;1Sx;1 ; Sy ;1Sy ;1)
= ASz2 + B (Sx2 ; Sy2) = H:
(4.32)
To nd the transformation of the eigenstates under time reversal we use the
relation (S-4.4.58)
jl; mi = (;1)mjl; ;mi:
(4.33)
So
jn0i = j10i (4=:33) j10i
= jn0i
jnA+B i =
p1 j1; +1i + p1 j1; ;1i
2
2
= ; p1 j1; ;1i ; p1 j1; +1i
2
2
= ;jnA+B i
(4:33)
jnA;B i =
p1 j1; +1i ; p1 j1; ;1i
2
2
= ; p1 j1; ;1i + p1 j1; +1i
2
2
= jnA;B i:
(4:33)
(4.34)
(4.35)
(4.36)
(4.37)
(4.38)
5. APPROXIMATION METHODS
103
5 Approximation Methods
5.1 Consider an isotropic harmonic oscillator in two dimensions.
The Hamiltonian is given by
2
p2y m!2 2 2
p
x
H0 = 2m + 2m + 2 (x + y )
(a) What are the energies of the three lowest-lying states? Is there
any degeneracy?
(b) We now apply a perturbation
V = m!2xy
where is a dimensionless real number much smaller than unity.
Find the zeroth-order energy eigenket and the corresponding energy 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 H0 + V problem exactly. Compare with the perturbation results obtained in q(b). p
p
[You may use hn0jxjni = h=2m! ( n + 1 0 + n 0 ):]
n ;n+1
n ;n;1
Dene step operators:
r m!
ipx );
ax 2h (x + m!
r m!
ipx );
y
ax 2h (x ; m!
r m!
ipy );
ay 2h (y + m!
r
(y ; ipy ):
ayy m!
2h
m!
From the fundamental commutation relations we can see that
[ax; ayx] = [ay ; ayy ] = 1:
(5.1)
104
Dening the number operators
Nx ayxax;
Ny ayy ay
we nd
N Nx + Ny = hH!0 ; 1 )
H0 = h !(N + 1):
(5.2)
I.e. energy eigenkets are also eigenkets of N :
Nx j m; n i = m j m; n i;
Ny j m; n i = n j m; n i )
N j m; n i = (m + n) j m; n i
so that
(5.3)
H0 j m; n i = Em;n j m; n i = h!(m + n + 1) j m; n i:
(a) The lowest lying states are
state
degeneracy
E0;0 = h !
1
E1;0 = E0;1 = 2h!
2
E2;0 = E0;2 = E1;1 = 3h!
3
(b) Apply the perturbation V = m!2xy.
Full problem:
(H0 + V ) j l i = E j l i
Unperturbed problem: H0 j l0 i = E 0 j l0 i
Expand the energy levels and the eigenkets as
E = E 0 + 1 + 2 + : : :
j l i = j l0 i + j l1 i + : : :
(5.4)
so that the full problem becomes
h
i
h
i
(E 0 ; H0 ) j l0 i + j l1 i + : : : = (V ; 1 ; 2 : : :) j l0 i + j l1 i + : : : :
5. APPROXIMATION METHODS
105
To 1'st order:
(E 0 ; H0) j l1 i = (V ; 1) j l0 i:
(5.5)
Multiply with h l0 j to nd
h l0 j E 0 ; H0 j l1 i = 0 = h l0 j V ; 1 j l0 i )
1h l0 j l0 i = 1 = h l0 j V j l0 i
(5.6)
In the degenerate case this does not work since we're not using the right basis
kets. Wind back to (5.5) and multiply it with another degenerate basis ket
h m0 j E 0 ; H0 j l1 i = 0 = h m0 j V ; 1 j l0 i )
1h m0 j l0 i = h m0 j V j l0 i:
(5.7)
Now, h m0 j l0 i is not necessarily kl since only states corresponding to dierent eigenvalues have to be orthogonal!
Insert a 1:
X 0
h m j V j k0 ih k0 j l0 i = 1h m0 j l0 i:
k 2D
This is the eigenvalue equation which gives the correct zeroth order eigenvectors!
Let us use all this:
1. The ground state is non-degenerate )
100 = h 0; 0 j V j 0; 0 i = m!2h 0; 0 j xy j 0; 0 i h 0; 0 j (ax +ayx)(ay +ayy ) j 0; 0 i = 0
2. First excited state is degenerate j 1; 0 i, j 0; 1 i. We need the matrix
elements h 1; 0 j V j 1; 0 i, h 1; 0 j V j 0; 1 i, h 0; 1 j V j 1; 0 i, h 0; 1 j V j 0; 1 i.
h (a +ay )(a +ay ) = h! (a a +ay a +a ay +ay ay )
V = m!2xy = m!2 2m!
x x y y
2 xy xy x y x y
and
p
ax j m; n i = m j m ; 1; n i
p
ayx j m; n i = m + 1 j m + 1; n i etc:
106
Together this gives
V10;10 = V01;01 = 0;
V10;01 = h2! h 1; 0 j ax ayy j 0; 1 i = h2! ;
V01;10 = h2! h 1; 0 j ayx ay j 0; 1 i = h2! :
The V -matrix becomes
(5.8)
!
h! 0 1
2 1 0
and so the eigenvalues (= 1) are
1 = h! :
2
To get the eigenvectors we solve
!
!
!
0 1
x = x
1 0
y
y
and get
j i+ = p12 ( j 0; 1 i + j 1; 0 i); E+ = h!(2 + 2 );
j i; = p12 ( j 0; 1 i ; j 1; 0 i); E; = h !(2 ; 2 ): (5.9)
3. The second excited state is also degenerate j 2; 0 i, j 1; 1 i, j 0; 2 i, so
we need the corresponding 9 matrix elements. However the only nonvanishing ones are:
V11;20 = V20;11 = V11;02 = V02;11 = ph!
(5.10)
2
p
(where the 2 came from going from level 1 to 2 in either of the oscillators) and thus to get the eigenvalues we evaluate
0
1
;
1 0
0 = det B
@ 1 ; 1 CA = ;(2 ; 1) + = (2 ; 2)
0 1 ;
5. APPROXIMATION METHODS
107
which means that the eigenvalues are f0; h!g. By the same method
as above we get the eigenvectors
p
j i+ = 21 ( j 2; 0 i + 2 j 1; 1 i + j 0; 2 i);
j i0 = p12 (; j 2; 0 i + j 0; 2 i);
p
j i; = 21 ( j 2; 0 i ; 2 j 1; 1 i + j 0; 2 i);
E+ = h !(3 + );
E0 = 3h!;
E; = h !(3 ; ):
(c) To solve the problem exactly we will make a variable change. The potential is
h
i
m!2 21 (x2 + y2) + xy =
#
"
1
2
2
2
2
2
= m! 4 ((x + y) + (x ; y) ) + 4 (x + y) ; (x ; y) ) : (5.11)
Now it is natural to introduce
p0x p1 (p0x + p0y );
x0 p1 (x + y);
2
2
1
1
y0 p (x ; y);
p0y p (p0x ; p0y ):
(5.12)
2
2
Note: [x0; p0x] = [y0; p0y ] = ih, so that (x0, p0x ) and (y0, p0y ) are canonically
conjugate.
In these new variables the problem takes the form
2
02
02
H = 21m (p0x2 + p0y2) + m!
2 [(1 + )x + (1 ; )y ]:
p
p
So we get one oscillator with !x0 = ! 1 + and another with !y0 = ! 1 ; .
The energy levels are:
E0;0 = h!;
p
E1;0 = h ! + h!x0 = h !(1 + 1 + ) =
= h!(1 + 1 + 12 + : : :) = h !(2 + 21 ) + O(2);
108
E0;1
E2;0
E1;1
E0;2
=
=
=
=
h ! + h!y0 = : : : = h !(2 ; 21 ) + O(2);
h! + 2h!x0 = : : : = h!(3 + ) + O(2);
h! + h!x0 + h!y0 = : : : = 3h! + O(2);
h! + 2h!y0 = : : : = h!(3 ; ) + O(2):
(5.13)
So rst order perturbation theory worked!
5.2 A system that has three unperturbed states can be represented
by the perturbed Hamiltonian matrix
0
1
E
1 0 a
B@ 0 E1 b CA
a b E2
where E2 > E1. The quantities a and b are to be regarded as per-
turbations that are of the same order and are small compared with
E2 ; E1. 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.
(a) First, nd the exact result by diagonalizing the Hamiltonian:
0
a E1 ; E
0 = 0
E1 ; E
b =
a
E2 ; E h b
i
= (E1 ; E ) (E1 ; E )(E2 ; E ) ; jbj2 + a [0 ; a(E1 ; E )] =
= (E1 ; E )2(E2 ; E ) ; (E1 ; E )(jbj2 + jaj2):
(5.14)
So, E = E1 or (E1 ; E )(E2 ; E ) ; (jbj2 + jaj2) = 0 i.e.
E 2 ; (E1 + E2)Es+ E1E2 ; (jaj2 + jbj2) = 0 )
2
E = E1 +2 E2 E1 +2 E2 ; E1E2 + jaj2 + jbj2 =
5. APPROXIMATION METHODS
s
2
E
+
E
= 1 2 E1 ; E2 + jaj2 + jbj2:
109
(5.15)
2
2
Since jaj2 + jbj2 is small we can expand the square root and write the three
energy levels as:
E = E1 ;
2
E
1 + E2 E1 ; E2
2
2
2
1
E =
1 + 2 (jaj + jbj )( E ; E ) + : : : =
2 + 2
1
2
2
2
+ jbj ;
= E1 + jaEj ;
1 E2
2 + jbj2
E = E1 +2 E2 ; E1 ;2 E2 (: : :) = E2 ; jaEj ;
:
1 E2
(5.16)
(b) Non degenerate perturbation theory to 2'nd order. The basis we use is
0 1
0 1
0 1
1
0
0
j 1 i = B@ 0 CA ;
j 2 i = B@ 1 CA ;
j 3 i = B@ 0 CA :
0
0
1
0
1
0
0
a
The matrix elements of the perturbation V = B
@ 0 0 b CA are
a b 0
h 1 j V j 3 i = a; h 2 j V j 3 i = b; h 1 j V j 2 i = h k j V j k i = 0:
Since (1)
k = h k j V j k i = 0 1'st order gives nothing. But the 2'nd order
shifts are
X jVk1 j2
jh 3 j V j 1 ij2 = jaj2 ;
=
0
0
E1 ; E2
E1 ; E2
k6=1 E1 ; Ek
2
2
2
X jVk2 j
jh
3
j
V
j
2
ij
j
b
j
=
0
0 = E1 ; E2 = E1 ; E2 ;
k6=2 E2 ; Ek
X jVk3 j2
jaj2 + jbj2 = ; jaj2 + jbj2 :
=
=
0
0
E2 ; E1 E2 ; E1
E1 ; E2
k6=3 E3 ; Ek
(2)
1 =
(2)
2
(2)
3
(5.17)
110
The unperturbed problem has two (degenerate) states j 1 i and j 2 i with
energy E1, and one (non-degenerate) state j 3 i with energy E2. Using nondegenerate perturbation theory we expect only the correction to E2 (i.e. (2)
3 )
to give the correct result, and indeed this turns out to be the case.
(c) To nd the correct energy shifts for the two degenerate states we have
to use degenerate perturbation
theory. The V -matrix for the degenerate
!
0
0
subspace is 0 0 , so 1'st order pert.thy. will again give nothing. We have
to go to 2'nd order. The problem we want to solve is (H0 + V ) j l i = E j l i
using the expansion
j l i = j l0 i + j l1 i + : : : E = E 0 + (1) + (2) + : : :
(5.18)
where H0 j l0 i = E 0 j l0 i. Note that the superscript index in a bra or ket denotes which order it has in the perturbation expansion. Dierent solutions to
the full problem are denoted by dierent l's. Since the (sub-) problem we are
now solving is 2-dimensional we expect to nd two solutions corresponding
to l = 1; 2. Inserting the expansions in (5.18) leaves us with
h
i
(E 0 ; H0 ) j l0 i + j l1 i + : : : =
h
i
(V ; (1) ; (2) : : :) j l0 i + j l1 i + : : : :
(5.19)
At rst order in the perturbation this says:
(E 0 ; H0) j l1 i = (V ; (1)) j l0 i;
where of course (1) = 0 as noted above. Multiply this from the left with a
bra h k0 j from outside the deg. subspace
h k0 j E 0 ; H0 j l1 i = h k0 j V j l0 i
0
X 0 0
(5.20)
) j l1 i = j k Eih0k;jEV j l i :
k6=D
k
This expression for j l1 i we will use in the 2'nd order equation from (5.19)
(E 0 ; H0) j l2 i = V j l1 i ; (2) j l0 i:
To get rid of the left hand side, multiply with a degenerate bra h m0 j
(H0 j m0 i = E 0 j m0 i)
h m0 j E 0 ; H0 j l2 i = 0 = h m0 j V j l1 i ; (2)h m0 j l0 i:
5. APPROXIMATION METHODS
111
Inserting the expression (5.20) for j l1 i we get
X h m0 j V j k0 ih k0 j V j l0 i
= (2)h m0 j l0 i:
0 ; Ek
E
k6=D
To make this look like an eigenvalue equation we have to insert a 1:
X X h m0 j V j k0 ih k0 j V j n0 i 0 0
h n j l i = (2)h m0 j l0 i:
0 ; Ek
E
n2D k6=D
Maybe it looks more familiar in matrix form
X
Mmn xn = (2)xm
n2D
where
X h m0 j V j k0 ih k0 j V j n0 i
;
E 0 ; Ek
k6=D
= 0 h m j l0 i
Mmn =
xm
are expressed in the basis dened by j l0 i. Evaluate M in the degenerate
subspace basis D = f j 1 i; j 2 ig
2
M11 = EV13;V31E 0 = E ja;j E ; M12 = EV13;V32E 0 = E ab; E ;
1
1
2
1
1
2
3
3
2
2
M21 = EV23;V31E 0 = E a;bE ; M22 = EjV;23jE 0 = E jb;j E :
1
3
1
2
1
3
1
2
With this explicit expression for M , solve the eigenvalue equation (dene
= (2)(E1 ; E2), and take out a common factor E1;1 E2 )
2 ;
!
j
a
j
ab
0 = det
ab jbj2 ; =
= (jaj2 ; )(jbj2 ; ) ; jaj2jbj2 =
= 2 ; (jaj2 + jbj2)
) = 0; jaj2 + jbj2
2 + jbj2
j
a
j
(2)
(2)
) 1 = 0 2 = E ; E :
(5.21)
1
2
112
From before we knew the non-degenerate energy shift, and now we see that
degenerate perturbation theory leads to the correct shifts for the other two
levels. Everything is as we would have expected.
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) = F0e;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
q excited
p states?0
0
[You may use hn jxjni = h=2m! ( n + 1n ;n+1 + pnn0;n;1 ):]
(a) The problem is dened by
2
2
H0 = 2pm + m!22x V (t) = ;F0xe;t= (F = ; @V
@x )
At t = 0 the system is in its ground state j ; 0 i = j 0 i. We want to calculate
X
j ; t i = cn (t)e;Ent=h j n i
En0
n
= h!(n + 12 )
where we get cn (t) from its di. eqn. (S. 5.5.15):
@ c (t) = X V ei!nmtc (t)
ih @t
n
nm
m
m
Vnm = h n j V j m i
!nm = En ;h Em = !(n ; m)
We need the matrix elements Vnm
Vnm = h n j ; F0sxe;t= j m i = ;F0e;t= h n j x j m i =
p
h (pm
= ;F0e;t= 2m!
n;m;1 + m + 1n;m+1 ):
(5.22)
5. APPROXIMATION METHODS
113
Put it back into (5.22)
s
@
h pn + 1e;i!tc (t) + pnei!tc (t) :
;
t=
ih @t cn(t) = ;F0e
n+1
n;1
2m!
(1)
Perturbation theory means expanding cn(t) = c(0)
n + cn + : : :, and to zeroth
order this is
@ c(0)(t) = 0 ) c(0) = n0
n
@t n
To rst order we get
1 Z t dt0 X V (t0)ei!nmt0 c(0) =
c(1)
(
t
)
=
n
m
ih 0 m nm
s
h Z t dt0e;t= pn + 1e;i!t0 c(0) (t) + pnei!t0 c(0) (t)
= ; Fih0 2m!
n+1
n;1
0
We get one non-vanishing term for n = 1, i.e. at rst order in perturbation
theory with the H.O. in the ground state at t = 0 there is just one non-zero
expansion coecient
s
h Z t dt0ei!t0;t0 = p1
F
0
c(1)
(
t
)
=
;
1;1;0 =
1
ih s 2m! 0
#t
"
F
1 )t0
h
1
0
(
i!
;
= ; ih 2m! i! ; 1 e 0
s
h 1 1 ; e(i!; 1 )t
= Fih0 2m!
i! ; 1
and
X
;iE1 t
;iEn t
(1)
j ; t i = c(1)
n (t)e h j n i = c1 (t)e h j 1 i:
n
The probability of nding the H.O. in j 1 i is
As t ! 1
2
jh 1 j ; t ij2 = jc(1)
1 (t)j :
s
F
h 1
0
c(1)
1 ! ih 2m! i! ; 1 = const:
This is of course reasonable since applying a static force means that the
system asymptotically nds a new equilibrium.
114
(b) As remarked earlier there are no other non-vanishing cn's at rst order,
so no higher excited states can be found. However, going to higher order in
perturbation theory such states will be excited.
5.4 Consider a composite system made up of two spin 21 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
4 H = 2 S~1 S~2:
h
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.
(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?
(a) The basis we are using is of course j S1z ; S2z i. Expand the interaction
potential in this basis:
S~1 S~2 = S1x"S2x + S1y S2y + S1z S2z = fin this basisg
2
= h4 ( j + ih ; j + j ; ih + j )1 ( j + ih ; j + j ; ih + j )2 +
+ i2(; j + ih ; j + j ; ih + j )1 (; j + ih ; j + j ; ih +
# j )2 +
+ ( j + ih + j ; j ; ih ; j )1 ( j + ih + j ; j ; ih ; j )2 =
2"
h
= 4 j + + ih ; ; j + j + ; ih ; + j +
+ j ; + ih + ; j + j ; ; ih + + j +
2
+ i ( j + + ih ; ; j ; j + ; ih ; + j +
; j ; + ih + ; j + j ; ; ih + + j ) +
5. APPROXIMATION METHODS
115
+ j + + ih + + j ; j + ; ih + ; j + #
; j ; + ih ; + j + j ; ; ih ; ; j =
In matrix form this is (using j 1 i = j + + i j 2 i = j + ; i
j 3 i = j ; + i j 4 i = j ; ; i)
01 0 0 01
B 0 ;1 2 0 CC
H = B
B@ 0 2 ;1 0 CA :
(5.23)
0 0 0 1
This basis is nice to use, since even though the problem is 4-dimensional we
get a 2-dimensional matrix to diagonalize. Lucky us! (Of course this luck is
due to the rotational invariance of the problem.)
Now diagonalize the 2 2 matrix to nd the eigenvalues and eigenkets
!
;
1
;
2
2
2
0 = det
2
;1 ; = (;1 ; ) ; 4 = + 2 ; 3
) = 1; ;3
=1:
!
!
x = x
y
y
) ;x + 2y = x ) x = y = p12
;1 2
2 ;1
= ;3 :
!
!
!
x = ;3 x
y
y
) ;x + 2y = ;3x ) x = ;y = p12
So, the complete spectrum is:
8
1
>
< j + + i; j ; ; i; p2 ( j + ; i + j ; + i with energy
>
: p1 ( j + ; i ; j ; + i
with energy ; 3
;1 2
2 ;1
2
!
116
This was a cumbersome but straightforward way to calculate the spectrum.
A smarter way would have been to use S~ = S~1 + S~2 to nd
S~ 2 = S 2 = S~12 + S~22 + 2S~1 S~2 ) S~1 S~2 = 21 S~ 2 ; S~12 ; S~22
We know that S~12 = S~22 = h2 21 21 + 1 = 3h42 so
2!
3
h
2
1
S~1 S~2 = 2 S ; 2
Also, we know that two spin 21 systems add up to one triplet (spin 1) and one
singlet (spin 0), i.e.
S = 1 (3 states) ) S~1 S~2 = 21 (h21(1 + 1) ; 3h22 ) = 14 h2
: (5.24)
2
2
3
h
3
1
~
~
S = 0 (1 state) ) S1 S2 = 2 (; 2 ) = ; 4 h
~ ~
Since H = 4
h2 S1 S2 we get
2
E(spin=1) = 42 14h = ;
h
4
h2 = ;3:
E(spin=0) = 2 ;3
h 4
n
(5.25)
From Clebsch-Gordan odecomposition we know that j + + i; j ; ; i;
p1 ( j + ; i + j ; + i) are spin 1, and p1 ( j + ; i ; j ; + i) is spin 0!
2
2
Let's get back on track and nd the dynamics. In the new basis H is diagonal
and time-independent, so we can use the simple form of tthe time-evolution
operator:
i
U (t; t0) = exp ; h H (t ; t0) :
The initial state was j + ; i. In the new basis
n
j 1 i = j + + i; j 2 i = j ; ; i; j 3 i = p12 ( j + ; i + j ; + i);
o
j 4 i = p12 ( j + ; i ; j ; + i)
5. APPROXIMATION METHODS
the initial state is
117
j + ; i = p12 ( j 3 i + j 4 i):
Acting with U (t; 0) on that we get
j ; t i = p1 exp ; hi Ht ( j 3 i + j 4 i) =
2 3i 1
i
= p exp ; t j 3 i + exp t j 4 i =
h
h
"2 = exp ;iht p1 ( j + ; i + j ; + i)+
2
#
3it 1
+exp h p ( j + ; i ; j ; + i) =
2
h ;i!t 3i!t
i
1
= 2 (e + e ) j + ; i + (e;i!t + e3i!t) j ; + i
where
(5.26)
! h :
The probability to nd the system in the state j i is as usual jh j ; t ij2
8
>
h + + j ; t i = h ; ; j ; t i = 0
>
>
<
jh + ; j ; t ij2 = 14 (2 + e4i!t + e;4i!t) = 21 (1 + cos4!t) ' 1 ; 4(!t)2 : : :
>
>
>
: jh ; + j ; t ij2 = 1 (2 ; e4i!t ; e;4i!t) = 1 (1 ; cos4!t) ' 4(!t)2 : : :
4
2
(b) First order perturbation theory (use S. 5.6.17):
c(0)
n = ni ; Z
;i t dt0ei!nit0 V (t0):
c(1)
(
t
)
=
ni
n
h t0
Here we have (using the original basis) H0 = 0, V given by (5.23)
j i i = j + ; i;
(5.27)
118
j f i = j ; + i;
Ei = fE = 0g = 0;
!ni = En ;
n
h
Vfi = 2;
Vni = 0; n 6= f:
Inserting this into (5.27) yields
(0)
c(0)
i = c j +; i = 1;
Zt
i
(1)
(1)
cf = c j ;+ i = ; h dt2 = ;2i!t:
(5.28)
0
as the only non-vanishing coecients up to rst order. The probability of
nding the system in j ; ; i or j + + i is thus obviously zero, whereas for
the other two states
P ( j + ; i) = 1
(2)
2
2
2
P ( j ; + i) = jc(1)
f (t) + cf (t) + : : : j = j2i!tj = 4(!t)
to rst order, in correspondence with the exact result.
The approximation breaks down when !t 1 is no longer valid, so for a
given t:
!t 1 ) ht :
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) = V0 cos(kz ; !t):
Using time-dependent perturbation theory, obtain an expression
for the transition rate at which the electron is emitted with momentum ~p. Show, in particular, how you may compute the angular
distribution of the ejected electron (in terms of and dened
with respect to the z-axis). Discuss briey the similarities and the
dierences between this problem and the (more realistic) photoelectric eect. (note: For the initial wave function use
Z 32
1
n=1;l=0(~x) = p a e;Zr=a0 :
0
5. APPROXIMATION METHODS
119
If you have a normalization problem, the nal wave function may
be taken to be
1
f (~x) =
L 32
ei~p~x=h
with L very large, but you should be able to show that the observable eects are independent of L.)
To begin with the atom is in the n = 1; l = 0 state. At t = 0 the perturbation
V = V0 cos(kz ; !t)
is turned on. We want to nd the transition rate at which the electron is
emitted with momentum ~pf . The initial wave-function is
1 3=2
1
i(~x) = p a e;r=a0
0
and the nal wave-function is
1 f (~x) = L3=2 ei~p~x=h:
The perturbation is
h
i
V = V0 ei(kz;!t) + e;i(kz;!t)
= V ei!t + V ye;i!t:
(5.29)
Time-dependent perturbation theory (S.5.6.44) gives us the transition rate
2
wi!n = 2h Vniy (En ; (Ei + h!))
because the atom absorbs a photon h!. The matrix element is
y 2 V02 ikz 2
Vni = 4 e ni and
Z
ikz ikz
~
e ni = h kf j e j n = 1; l = 0 i = d3xh ~kf j eikz j x ih x j n = 1; l = 0 i =
1 3=2
Z e;i~kf ~x
1
3
ikx
3
= d x L3=2 e p a e;r=a0 =
0
Z
= 3=2p1 3=2 d3xe;i(~kf ~x;kx3 );r=a0 :
(5.30)
L a0
120
So eikz ni is the 3D Fourier transform of the initial wave-function (and some
constant) with ~q = ~kf ; k~ez . That can be extracted from (Sakurai problem
5.39)
ikz 642
1
e ni = L3a5 h 1
i4
0 a2 + (~kf ; k~ez )2
0
The transition rate is understood to be integrated over the density of states.
We need to get that as a function of ~pf = h~kf . As in (S.5.7.31), the volume
element is
dn dp :
n2dnd
= n2d
dp
f
f
Using
2 )2
p2
kf2 = f2 = n (2
L2
h
we get
dn = 1 2L2pf = 2h L2pf = L
dp 2n (2h)2 Lp (2h)2 2h
which leaves
f
f
L3kf2
L3p2f
= (2)3h d
dpf = (2h)3 d
dpf
and this is the sought density.
Finally,
2
2
L3p2f
1
d
dpf :
wi!~pf = 2h V40 L643a5 h 1
i
4
0 a2 + (~kf ; k~ez )2 (2 h)3
0
n2dnd
Note that the L's cancel. The angular dependence is in the denominator:
~
2
kf ; k~ez = [(jkf jcos ; k) ~ez + jkf jsin (cos'~ex + sin'~ey )]2 =
= jkf j2cos2 + k2 ; 2kjkf jcos + jkf j2sin2 =
= kf2 + k2 ; 2kjkf jcos:
(5.31)
In a comparison between this problem and the photoelectric eect as discussed in (S. 5.7) we note that since there is no polarization vector involved,
w has no dependence on the azimuthal angle . On the other hand we did
not make any dipole approximation but performed the x-integral exactly.