Solutions to Problems
in
Quantum Mechanics
P. Saltsidis, additions by B. Brinne
1995,1999
0
Most of the problems presented here are taken from the book
Sakurai, J. J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley,
1985.
Contents
I Problems 3
1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . 5
2 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Theory of Angular Momentum . . . . . . . . . . . . . . . . . . 14
4 Symmetry in Quantum Mechanics . . . . . . . . . . . . . . . . 17
5 Approximation Methods . . . . . . . . . . . . . . . . . . . . . 19
II Solutions 23
1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . 25
2 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 36
3 Theory of Angular Momentum . . . . . . . . . . . . . . . . . . 75
4 Symmetry in Quantum Mechanics . . . . . . . . . . . . . . . . 94
5 Approximation Methods . . . . . . . . . . . . . . . . . . . . . 103
1
2 CONTENTS
Part I
Problems
3
1. FUNDAMENTAL CONCEPTS 5
1 Fundamental Concepts
0
1.1 Consider a ket space spanned byaig of a Her-
the eigenkets fj
mitianA. There is no degeneracy.
operator
(a) Prove that
Y(Aa)
0
;
a0
is a null operator.
(b) What is the signi cance of
Y
(A;a00)
?
a0a00
a00 =a0 ;
6
1
(c)AsetSz of a spin system.
Illustrate (a) and (b) using equal to
2
1
~
1.2 A spin system is knownS nwith
to be in an eigenstate of ^
2
eigenvalueh=2,nisxz-plane that
where ^ a unit vector lying in the
makes withz-axis.
an angle the positive
(a)Sx is measured. What is the probability of getting
Suppose
+h=2?
(b)Sx, that is,
Evaluate the dispersion in
h(SxSx:
;h i)2 i
(For your own peace of mind check your answers for the special
cases =0, =2, .)
and
1.3 (a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
(h j j i j i) 0
+ h ) (j +
for in such a way that the
any complex number then choose
preceding inequality reduces to the Schwarz inequility.
6
(b) Show that the equility sign in the generalized uncertainty re-
lation holds if the state in question satis es
Aj i Bj i
=
with purely imaginary.
(c) Explicit calculations using the usual rules of wave mechanics
show that the wave function for a Gaussian wave packet given by
"ipix(xxi) #
2
0 0
h ;h
hx0j i d2);1=4 exp ;
=(2
h4d2
satis es the uncertainty relation
qh
q
h( x)2ip)2:
h( i =
2
Prove that the requirement
hx0jxj ix0jpj i
= (imaginary number)h
is indeed satis ed for such a Gaussian wave packet, in agreement
with (b).
1.4xandpx be the coordinate and linear momentum in
(a) Let
one dimension. Evaluate the classical Poisson bracket
[x F(px)]classical:
(b)xandpx be the corresponding quantum-mechanical opera-
Let
tors this time. Evaluate the commutator
ipxa
x exp:
h
(c) Using the result obtained in (b), prove that
ipxa
expx0i (xjx0ix0jx0i)
j =
h
2. QUANTUM DYNAMICS 7
is an eigenstatex. What is the corre-
of the coordinate operator
sponding eigenvalue?
1.5 (a) Prove the following:
@
(i)p0jxj iihhp0j i
h =
0
Z@p
@
(ii) jxj idp0 p0)ih p0)
h = ( (
@p0
where p0)p0 i p0)p0j i are momentum-space wave
( = h j and ( = h
functions.
(b) What is the physical signi cance of
ix
exp
h
wherexis 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=Sz!Sz
; =
mc
write the Heisenberg equations of motion for the time-dependent
operatorsSx(t),Sy(t),Sz(t).Sx y z as func-
and Solve them to obtain
tions of time.
2.2x(t) be the coordinate operator for a free particle in one
Let
dimension in the Heisenberg picture. Evaluate
[x(t) x(0)]:
8
2.3 Consider a particle in three dimensions whose Hamiltonian is
given by
p2
~
H=V(x):
+~
2m
Byx p H]
calculating~~obtain
[
* +
dp2
~
hx pix :
~~=~rVi
;h
dtm
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
;ipa
exp:
j 0i
h
You may use
2 3 0 1
! !
2 1=2
x0h
1
4 5 @ A
hx0j ;1=4x;1=2 x0:
0i = exp ;
0
2
x0m!
(b) Obtain a simple expression that the probability that the state
ist=0. Does this probability change
found in the ground state at
fort>0?
2.5 Consider a function, known as the correlation f
unction, de ned
by
C(t)x(t)x(0)i
= h
wherex(t) is the position operator in the Heisenberg picture. Eval-
uate the correlation function explicitly for the ground state of a
one-dimensional simple harmonic oscillator.
2. QUANTUM DYNAMICS 9
2.6 Consider again a one-dimensional simple harmonic oscillator.
Do the following algebraically, that is, without using wave func-
tions.
(a) Construct a linear combinationxi is as
of j 0i and j 1i such that h
large as possible.
(b) Suppose the oscillator is in thet=0.
state constructed in (a) at
Whatt>0 in the Schr
is the state vector for odinger picture?
Evaluatexit>0
the expectation value h as a function of time for
using (i) the Schr
odinger picture and (ii) the Heisenberg picture.
(c)x)2 i as a function of time using either picture.
Evaluate h(
2.7 A coherent state of a one-dimensional simple harmonic oscil-
lator is de ned to be an eigenstate of the (non-Hermitian) annihi-
lationa:
operator
aj i j i
=
where is, in general, a complex number.
(a) Prove that
2
j =2
j ie;je ay j 0i
=
is a normalized coherent state.
(b) Prove the minimum uncertainty relation for such a state.
(c) i as
Write j
1
X
j if(n)jni:
=
n=0
2
Showf(n)jnis of the
that the distribution of j with respect to
Poisson form. Findn,E.
the most probable value of hence of
(d) Show that a coherent state can also be obtained by applying
the translation ( e;ipl=hpis the
nite-displacement) operator (where
momentumlis the displacement distance) to the
operator, and
ground state.
10
(e) Show i remains coherent under time-
that the coherent state j
evolution and calculate (t)i. (Hint: di-
the time-evolved state j
rectly apply the time-evolution operator.)
2.8 The quntum mechanical propagator, for a particle with mass
m, moving in a potential is given by:
Z
1
X
sin(nrx)nry)
sin(
K(x y E)dteiEt=hK(x y t 0)Ah r2
= =
2
0
E;n2
n
2m
whereAis a constant.
(a) What is the potential?
(b)Ain terms of the parameters describing
Determine the constant
them,retc. ).
system (such as
2.9 Prove the relation
d (x)
= (x)
dx
where (x) is the (x) the Dirac delta
(unit) step function, and
function. (Hint: study the e ect on testfunctions.)
2.10 Derive the following expression
h i
m!
Sclx2x2!T)x0xT
= ( +2 ) cos( ;
0 T
2sin(!T)
for the classical action for a harmonic oscillator moving from the
pointx0t=0xTt=T.
at to the point at
2.11 The Lagrangian of the single harmonic oscillator is
1 1
2
Lmx_m!2x2
= ;
2 2
2. QUANTUM DYNAMICS 11
(a) Show that
iS
hxbtbjxataiexpclG(0 tb 0 t)
=
ha
whereScl is the actionxclxa ta) to
along the classical path from (
(xb tb)Gis
and
G(0 tb ta) =
0
8 9
( N +1)
Z
< =
X
2
miNm1
2
limdy1:::dyNexp(yj+1yj)2"m!2yj
; ;
:
N!1
2 ih"hj=0 2"2
tb ;ta
where"= .
(N+1)
(Hint:y(t)x(t)xcl(t) be the new integration variable,
Let = ;
xcl(t) being the solution of the Euler-Lagrange equation.)
(b)Gcan be written as
Show that
( N +1)
Z
2
m
G=dy1:::dyNexp(;nT n)
lim
N!1
2 ih"
2y3
1
6 7
.
6 7
.
wheren=nT is its transpose. Write the symmetric
and
.
4 5
yN
matrix .
(c) Show that
Z Z
N=2
T
;
dy1:::dyNexp(;nT n)dNyen n = p
det
[Hint: by an orhogonal matrix.]
Diagonalize
N
2ih" 0 0
(d)det det NpN.j jmatrices j that con-
Let De ne
m
0
sistjrowsjcolumns N and whose determinants
of the rst and of
0
arepj. j+1 in minors show the following recursion
By expanding
formulapj:
for the
pj+1"2!2)pjpjj=1 ::: N(2.1)
=(2 ; ;
;1
12
(e) (t)"pjt=taj"and show that (2.1) implies that in
Let = for +
the"! (t) satis es the equation
limit 0
d2 2
=! (t)
;
dt2
with (t=ta) d (t=ta ) =1.
initial conditions = 0
dt
(f) Show that
( )
sm!im!
hxbtbjxataiexp[(x2x2)!T)xaxb]
= + cos( ; 2
b a
2 ihsin(!T)hsin(!T)
2
whereT=tbta.
;
2.12 Show the composition property
Z
dx1Kfx2 t2 x1 t1)Kfx1 t1 x0 t0)Kfx2 t2 x0 t0)
( ( = (
whereKfx1 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
!
ihe
[ "ijkBk
] =
i j
c
~
x
~~
wheremdtp; and the relation
=~eA
c
" !#
d2~d d~d~
x~1x~~x
~
m2eE+B;B :
= =
dtdt2cdtdt
(b) Verify the continuity equation
@ ~~
0
+j=0
r
@t
2. QUANTUM DYNAMICS 13
~
withjgiven by
!
0
~h( )eAj 2
j=:
=~;~j
r
mmc
2.14 An electron moves in the presence of a uniform magnetic eld
~
inz-directionB=Bz).
the ( ^
(a) Evaluate
[ ]
x y
where
eAxeAy
px py:
; ;
x
cyc
(b) By comparing the Hamiltonian and the commutation relation
obtained in (a) with those of the one-dimensional oscillator problem
showhow we can immediately write the energy eigenvalues as
!
h2k2eBjh1
j
Ek nn+
= +
2mmc2
wherehkis thepznis a
continuous eigenvalue of the operator and
nonnegative integer including zero.
2.15mandqin an impenetrable
Consider a particle of mass charge
cylinderRanda. Along the axis of the cylin-
with radius height
der runs a thin, impenetrable solenoid carrying a magnetic ux .
Calculate the ground state energy and wavefunction.
2.16 A particlein one dimension (;1
constant force derivable from
V= x ( >0):
14
(a) Is the energy spectrum continuous or discrete? Write down an
approximate expression for the energy eigenfunction speci ed by
E.
(b) Discuss brie yVis replaced be
what changes are needed if
V= jxj:
3 Theory of Angular Momentum
3.1 Consider a sequence of Euler rotations represented by
!
;i 3 ;i 2 ;i 3
D(1=2)( ) = exp exp exp
2 2 2
!
; + )=2 ; ;
ei( ei( )=2 sin
cos ;
2 2
= :
; )=2
ei( ei( + )=2 cos
sin
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-momentumj m=mmaxji is rotated
eigenstate j =
by"abouty-axis. Without using the
an in nitesimal angle the
explicitd(j) function, obtain an expression for the
form of the
m0 m
probability for the new rotated state to be found in the original
state"2.
up to terms of order
3.3 The wave function of a patricle subjected to a spherically
symmetricalV(r) is given by
potential
(x)x+y+3z)f(r):
~=(
3. THEORY O F ANGULAR MOMENTUM 15
~
(a) anL?l-value? If
Is eigenfunction of If so, what is the
~
not, whatlweL2 is
are the possible values of may obtain when
measured?
(b) What are the probabilities for the particle to be found in various
ml states?
(c) Suppose (x) an energy eigenfunc-
it~is
is known somehow that
tionE.V(r).
with eigenvalue Indicate howwe may nd
3.4 Consider a particle with an intrinsic angular momentum (or
spin)h. (One example%-
of one unit of of such a particle is the
meson). Quantum-mechanically, such a particle is described by a
ketvector%i~representation wave function
jxa
or in
%i(x)x ij%i
~=~
h
wherex ii~withi:th di-
j~correspondxspin in the
to a particle at
rection.
(a) Show explicitly%i(x) obtained
that~are
in nitesimal rotations of
by acting with the operator
~
"~~
u"i L+ S) (3.1)
=1 ; (
~
h
~h~~
whereL=r r. Determine S !
^
i
~~
(b)Land S commute.
Show that
~
(c) Show that S is a vector operator.
~1~
(d)%(x)p)%wherepis momentum oper-
Show~~=~~~the
that r (S
h2
ator.
3.5 We arej1j2 = 1 to form
to add angular momenta = 1 and
j=2 1 and 0 states. Using the ladder operator method express all
16
(nine)j meigenketsj1j2 m1m2i. Write your answer as
in terms of j
1 1
p p
jj=1 m=1i +i ::: (3.2)
= j + 0i ; j 0
2 2
wherem1 2 0 respectively.
+ and 0 stand for =1
3.6 (a) Construct a spherical tensor of rank 1 out of two di erent
~~
vectorsU=Ux Uy Uz)V=(Vx Vy Vz).T(1) in
( and Explicitly write
1 0
termsUx y zVx y z.
of and
(b) Construct a spherical tensor of rank 2 out of two di erent
~~
vectorsUandV.T(2)Ux y z
Write down explicitly in terms of
2 1 0
andVx y z.
3.7 (a) Evaluate
j
Xd( )m
(j)
2
j j
mm0
m=;j
1
forj(integer or half-integer) j= .
any then check your answer for
2
(b)j,
Prove, for any
j
Xmd( )j(j+1) +m+ 1):
(j) 2
2 2 1 0 1
j j = sin (3 cos2 ;
m0 m
2 2
m=;j
[Hint: This can be proved in many ways. You may, for instance,
2
examineJz using the spherical (irre-
the rotational properties of
ducible) tensor language.]
3.8xy,xz,x2y2) as components of a spherical
(a) Write and ( ;
(irreducible) tensor of rank 2.
4. SYMMETRY IN QUANTUM MECHANICS 17
(b) The expectation value
Q eh j m=jjz2r2)j j m=ji
(3 ;
is known as the quadrupole moment. Evaluate
0
eh j mjx2y2)j j m=ji
( ;
0
(wherem=j j; j; :::)inQand appropriate Clebsch-
1 2 terms of
Gordan coe cients.
4 Symmetry in Quantum Mechanics
4.1 (a) Assuming that the Hamiltonian is invariant under time
reversal, prove that the wave function for a spinless nondegenerate
system at any given instant of time can always be chosen to be
real.
(b) The wave functiont= 0 is given by
for a plane-wave state at
p x=h
aei~ ~ . Why does this not violate time-reversal
complex function
invariance?
0
4.2 (p) the momentum-space i,
Let~be wave function for state j
0 0
that (p)pj i.Is momentum-space wave function for the
is,~=~the
h
0 0 0 0
time-reversed i (p, (;p), (p), (;p)?
state~~~or~
j given by
Justify your answer.
4.3 Read section 4.3 in Sakurai to refresh your knowledge of the
quantum mechanics of periodic potentials. You know that the en-
ergybands in solids are described by the so called Bloch functions
full lling,
n k
(x+a)eika n k(x)
=
n k
18
whereaisnlabels the band, and the lattice
the lattice constant,
momentumkis restricted =a =a].
to the Brillouin zone [;
Prove that any Bloch function can be written as,
X
i
(x) n(x;Ri)eikR
=
n k
Ri
where the sumRi. (In this simble one di-
is over all lattice vectors
mensionalRiia, but the construction generalizes easily
problem =
to three dimensions.).
The n are called Wannier functions, and are impor-
functions
tant in the tight-binding description of solids. Show that the Wan-
nier functions are corresponding to di erent sites and/or di erent
bandsi:e:prove
are orthogonal,
Z
dx ?x;Ri) n(x;Rj) ij mn
(
m
Hint: ns in Bloch functions and use their orthonor-
Expand the
mality properties.
4.4 Suppose a spinless particle is bound to a xed center by a
potentialV(x) assymetrical that no energy level is degenerate.
~so
Using the time-reversal invariance prove
~
hLi =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
XXF(r)Y( )
m
lm
l
m
l
what kind of phaseFlm(r)?
restrictions do we obtain on
4.5 The Hamiltonian for a spin 1 system is given by
2 2 2
H=ASzB(SxSy:
+ ; )
5. APPROXIMATION METHODS 19
Solve this problem exactly to nd the normalized energy eigen-
states and eigenvalues. (A spin-dependent Hamiltonian of this kind
actually appears in crystal physics.) Is this Hamiltonian invariant
under time reversal? How do the normalized eigenstates you ob-
tained transform under time reversal?
5 Approximation Methods
5.1 Consider an isotropic harmonic oscillator in two dimensions.
The Hamiltonian is given by
p2
p2m!2
y
x
H0x2y2)
= + + ( +
2m2m2
(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 en-
ergy to rst order [that is the unperturbed energy obtained in (a)
plus the rst-order energy shift] for each of the three lowest-lying
states.
(c)H0Vproblem exactly. Compare with the perturba-
Solve the +
tion results obtained in (b).
q
p
p
0
[Younjxjnih=2m!(n+1 nn n:]
may use h = 0 + 0 )
n+1 n;1
5.2 A system that has three unperturbed states can be represented
by the perturbed Hamiltonian
0Ematrix 1
1
BE0aC
0 A
@b
1
abE2
20
whereE2>E1.aandbare to be regarded as per-
The quantities
turbations that are of the same order and are small compared with
E2E1. 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
fort<0.t 0 it is subjected to a time-dependent but spatially
For
uniform force (not potential!) in the x-direction,
;
F(t)F0et=
=
(a) Using time-dependent perturbation theory to rst order, obtain
the probability of nding the oscillator in its rst excited state for
t>0).t!1 nite) limit of your expression is
Show that the (
independent of time. Is this reasonable or surprising?
(b) Can we nd higher excited states?
q
p
p
0
[Younjxjnih=2m!(n+1 nn n:]
may use h = 0 + 0 )
n+1 n;1
1
5.4 Consider a composite system made up of two spin objects.
2
fort<0, the Hamiltonian does not depend on spin and can be
taken to be zero by suitably adjustingt>0,
the energy scale. For
the Hamiltonian is given by
4
~~
H=S1S2:
h2
Supposet 0. Find, as a function of
the system is in j + ;i for
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-dependentHas a perturbation switched
perturbation theory with
ont=0. Under what condition does (b) give the correct results?
at
5.5 The groundn=1,l=0) is subjected
state of a hydrogen atom (
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 mo-
mentump.
~Show, in particular, howyou may compute the angular
distribution of the ejected and de ned
electron (in terms of
withz-axis). Discuss brie y the similarities and the
respect to the
di erences between this problem and the (more realistic) photo-
electric e ect. (note: For the initial wave function use
3
2
1Z
;
0
(x)eZr=a:
~= p
n=1 l=0
a0
If you have a normalization problem, the nal wave function may
be taken to be
1
p x=h
(x)ei~ ~
~=
f
3
2
L
withLvery large, but you should be able to show that the observ-
ableL.)
e ects are independent of
22
Part II
Solutions
23
1. FUNDAMENTAL CONCEPTS 25
1 Fundamental Concepts
0
1.1 Consider a ket spaceaig of a Her-
spanned by the eigenkets fj
mitianA. There is no degeneracy.
operator
(a) Prove that
Y(Aa)
0
;
a0
is a null operator.
(b) What is the signi cance of
Y
(A;a00)
?
a0a00
;
a00 =a0
6
1
(c)AsetSz of a spin system.
Illustrate (a) and (b) using equal to
2
(a) i is an arbitrary state ket. Then
Assume that j
Y(Aa) iAa)aa icY(A;a)jai
Y( X X
0 0 00 00 0 00
; j = ; j i h j
a00
| {z }=
a0 a0 a00
ca00 a00 a0
X Y( ;
a00 2fa0 g
00 = 0
=caa00a0)ja00i:(1.1)
a00 a0
(b) i we will have
Again for an arbitrary state j
2 3 2 3X 1
00 00
Y Y
( ( ;
4A;a)5 iAa)5z }| {
4
ja000iha000 i
= j j
a0a00a0a00
; ;
a00 =a0 a00 =a0 a000
6 6
X Y
(a000a00)
;
=a000j ia000i =
h j
0 00
;
a000 a00 =a0
6
Xaa
=a000j i aa000ia0j ija0i )
h 000 j = h
a0
a000
2 3
00
Y
( ;
0 0
4Aa)5aa :(1.2)
= j ih j
a0
a0a00
;
a00 =a0
6
Soa0i.
it projects to the eigenket j
26
(c)Szh=2(j +ih+j ; j ;ih;j ). This operator has eigenkets j +i and j ;i
It is
withh=2h=2 respectively. So
eigenvalues and -
Y(Sa)Sa1)
Y(
0 0
; = ;
z z
a0 a0
"hh#
= (j +ih+j ; j ;ih;j ) ; (j +ih+j + j ;ih;j )
2
"h2 #
h
(j +ih+j ; j ;ih;j ) + (j +ih+j + j ;ih;j )
2 2
0
z }| {
=hjhjh2j (1.3)
[; ;ih;j ][ +ih+j ] = ; ;i h;j +ih+j =0
where we have used that j +ih+j + j ;ih;j =1.
Fora0h=2 we have
=
h
Y Y
(Sza00)Sza001) Sz + 1
; ( ;
2
= =
a0a00h=2a00h=2h=2
; ; +
a00 =a0 a00 =h=2
6 6
" #
1hh
= (j +ih+j ; j ;ih;j ) + (j +ih+j + j ;ih;j )
h2 2
1
=hj:(1.4)
+ih+j = j +ih+j
h
Similarlya0h=2 we have
for = ;
h
Y Y
(Sza00)Sza001) Sz ; 1
; ( ;
2
= =
a0a00h=2a00h=2h=2
; ; ; ; ;
a00 =a0 a00 =; h=2
6 6
" #
1hh
= ; (j +ih+j ; j ;ih;j ) ; (j +ih+j + j ;ih;j )
h2 2
1
=hj:(1.5)
; (; ;ih;j ) = j ;ih;j
h
1
~
1.2 A spin system is knownS nwith
to be in an eigenstate of ^
2
eigenvalueh=2,nisxz-plane that
where ^ a unit vector lying in the
makes withz-axis.
an angle the positive
(a)Sx is measured. What is the probability of getting
Suppose
+h=2?
1. FUNDAMENTAL CONCEPTS 27
(b)Sx, that is,
Evaluate the dispersion in
h(SxSxi)2i:
;h
(For your own peace of mind check your answers for the special
cases =0, =2, .)
and
Sincenmakes withz-axis and is
the unit vector ^ an angle the positive
lyingxz-plane, it can be written in the following way
in the
n=ez +^ (1.6)
^ex sin
^ cos
So
~
S n=Sz +Sx = [ (S-1.3.36),(S-1.4.18)]
^ sin
"hcos #h#
"
= + :
(j +ih+j ; j ;ih;j ) cos (j +ih;j + j ;ih+j ) sin (1.7)
2 2
~
Since theS nwithh=2 it has to
system is in an eigenstate of ^ eigenvalue
satisfay the following equation
~~~
S njS n h=2jS n :(1.8)
^ ^ +i = ^ +i
From (1.7) we have that
!
~hcos sin
S n=:(1.9)
^
sin ;
cos
2
The eigenvalues and eigenfuncions of this operator can be found if one solves
the secular equation
!
cos
~h=2 ; h=2sin =0 )
det(S n; I) =0 ) det
^
h=2sin ;h=2cos ;
h2h2h2h
; + 2 =0 2 =:(1.10)
cos2 ; sin2 ) ; =0 )
4 4 4 2
!
~ahave that
SinceS n +i we will
the system is in the eigenstate j ^
b
! ! ! ( )
hh
cos sin aaacos +bsin =a
= ) )
sin ; b2basin ;bcos =b
cos
2
1 2
; cos 2 sin
2
b=a=a atan:(1.11)
=
sin 2sin cos 2
2 2
28
~
ButS n +i to be normalized, that is
we want also the eigenstate j ^
a2b2a2a2a2a2 sin2 =cos2
+ =1 ) + tan2 =1 ) cos2
2 2 2
r 2
)a2a= (1.12)
= cos2 ) cos2 = cos
2 2 2
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
ofSz operator by
the
~ j j ;i
jS n :(1.13)
^ +i =cos +i +sin
2 2
(a) From (S-1.4.17) we know that
1 1
p p
jSx :(1.14)
+i = j +i + j ;i
2 2
Soh=2Sx is measured is given by
the propability of getting + when
!
2
1
~2 j j
p p
+j ^
hSx S n +i = 1 h+j + h;j cos +i + sin ;i
2 2
2 2
p p
= 1 1 2
cos + sin
2 2
2 2
1 1
= cos2 + sin2 +cos sin
2 2 2 2 2 2
1 1 1
= = ):(1.15)
+ sin (1 + sin
2 2 2
For = 0 whichSz +i eigenstate we have
means that the system is in the j
2 1 1
jSx Sz :(1.16)
h +j +ij = (1) =
2 2
For = =2 which meansSx +i eigenstate we
that the system is in the j
have
2
jSx Sx :(1.17)
h +j +ij =1
For = whichSz ;i eigenstate we have
means that the system is in the j
2 1 1
jSx Sz :(1.18)
h +j ;ij = (1) =
2 2
1. FUNDAMENTAL CONCEPTS 29
(b) We have that
2
h(SxSxi)2iSxiSxi)2:(1.19)
;h = h ; (h
As we know
h
Sx = (j +ih;j + j ;ih+j ) )
2
h2
2
Sx = (j +ih;j + j ;ih+j ) (j +ih;j + j ;ih+j ) )
4
h2h2
2
Sx:(1.20)
= (j +ih+j + j ;ih;j ) =
| {z }
4 4
1
So
h
hSxi = cos h+j + sin h;j (j +ih;j + j ;ih+j ) cos j +i + sin j ;i
2 2 2 2 2
h h h
= cos )
sin + sin cos = sin
2 2 2 2 2 2 2
h2
(hSxi)2 and
= sin2
4
h2
2
hSxi = cos h+j + sin h;j cos j +i +sin j ;i
2 2 4 2 2
h2 2 h2
=:(1.21)
[cos2 + sin ] =
4 2 2 4
So substituting in (1.19) we will have
h2h2
h(SxSxi)2i ) :(1.22)
;h = (1 ; sin2 = cos2
4 4
and nally
h2
h( Sx)2i (1.23)
=
=0 j Sz +i
4
h( Sx)2i (1.24)
= 0
= =2 j Sx +i
h2
h( Sx)2i :(1.25)
=
=0 j Sz ;i
4
30
1.3 (a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
(h j j i j i) 0
+ h ) (j +
for in such a way that the
any complex number then choose
preceding inequality reduces to the Schwarz inequility.
(b) Show that the equility sign in the generalized uncertainty re-
lation holds if the state in question satis es
Aj i Bj i
=
with purely imaginary.
(c) Explicit calculations using the usual rules of wave mechanics
show that the wave function for a Gaussian wave packet given by
"ipix(xxi) #
2
0 0
h ;h
hx0j i d2);1=4 exp ;
=(2
h4d2
satis es the uncertainty relation
qh
q
h( x)2ip)2i:
h( =
2
Prove that the requirement
hx0jxj ix0jpj i
= (imaginary number)h
is indeed satis ed for such a Gaussian wave packet, in agreement
with (b).
(a) Weci the following relation holds
know that for an arbitrary state j
hcjci:(1.26)
0
Thisci i j i is a complex number,
means that if we choose j = j + where
we will have
(h j j i j i) 0 ) (1.27)
+ h ) (j +
2
h j i h j i j i j j i:(1.28)
+ + h + j h 0
1. FUNDAMENTAL CONCEPTS 31
If = j i=h j i the previous relation will be
we now choose ;h
2
h j ih j i h j ih j i j h j ij
h j i ; ; + 0 )
h j i h j i h j i
2
h j ih j i j ij:(1.29)
j h
Notice that the equality sign in the last relation holds when
jci i j i i j i (1.30)
= j + =0 )j = ;
that i i are colinear.
is if j and j
(b) The uncertainty relation is
1
h( A)2ih( B)2iA B]ij2:(1.31)
j h[
4
To prove this relation we use the Schwarz inequality (1.29) for the vectors
j iAjai iBjai which gives
= and j =
2
h( A)2ih( B)2iA Bij:(1.32)
j h
The equality sign in this relation holds according to (1.30) when
Ajai Bjai:(1.33)
=
On the other hand the right-hand side of (1.32) is
1 1
2
jA BijA B]ij2A Bgij2 (1.34)
h = j h[ + j hf
4 4
which means that the equality sign in the uncertainty relation (1.31) holds if
1
jA Bgij2A Bgi =0
hf =0 )hf
4
(1:33)
)ajA B+ B Ajai ajB)2jai hajB)2jai =0
h =0 ) h ( + (
) + ajB)2jai: (1.35)
( )h ( =0
Thus the equality sign in the uncertainty relation holds when
Ajai Bjai (1.36)
=
with purely imaginary.
32
(c) We have
hx0jxj ix0jx;hxi)j ix0hx0j ixihx0j i
h ( = ;h
=x0xi)hx0j i:(1.37)
( ;h
On the other hand
hx0jpj ix0jp;hpi)j i
h (
@
=ihhx0j ipihx0j i (1.38)
; ;h
@x0
But
" #
@0@ihpix0x0xi)2
( ;h
hxj ix0j i ;
= h
@x0@x0h4d2
"ipi 1 #
h
=x0j ix0xi) (1.39)
h ; ( ;h
h2d2
So substituting in (1.38) we have
ih
hx0jpj ipihx0j ix0xi)x0j ipihx0j i
= h + ( ;h h ; h
2d2
ihih
=x0xi)x0j ix0jxj i )
( ;h h = h
2d2d2
2
;i2d2
hx0jxj ix0jpj i:(1.40)
= h
h
1.4xandpx be the coordinate and linear momentum in
(a) Let
one dimension. Evaluate the classical Poisson bracket
[x F(px)]classical:
(b)xandpx be the corresponding quantum-mechanical opera-
Let
tors this time. Evaluate the commutator
ipxa
x exp:
h
(c) Using the result obtained in (b), prove that
ipxa
expx0i (xjx0ix0jx0i)
j =
h
1. FUNDAMENTAL CONCEPTS 33
is an eigenstatex. What is the corre-
of the coordinate operator
sponding eigenvalue?
(a) We have
@x@F(px)@x@F(px)
[x F(px)]classical ;
@x@px@px@x
@F(px)
=:(1.41)
@px
(b)xandpx are treated as quantum-mechanical operators we have
When
" #
1
X X
ipxa1ia)npnia)n
( 1 (
x
x expx nx pn]
= = [
hhn!n!hn x
n=0 n=0
1
X X
1ia)n n;1
(
=pkx px]pn;k;1
[
x
n!hn x
n=0 k=0
1 n;1 1
X X X
1ia)nn(ia)n;1
(
=ih)pkpn;k;1pn;1a)
( = (;
x x
n!hnn! hn;1 x
n=1 k=0 n=1
1
X
1ian;1ipxa
=apxaexp:(1.42)
; = ;
(n;hh
1)!
n=1
(c) We have now
ipxaipxaipxa
(b)
xexpx0ixjx0iaexpx0i
j = exp ; j
hhh
ipxaipxa
=x0x0iaexpx0i
exp j ; j
hh
ipxa
=x0a)x0i:(1.43)
( ; exp j
h
ipxa
Sox0ixwithx0a.
exp j is an eigenstate of the operator eigenvalue ;
h
So we can write
ipxa
jx0aiCexpx0i (1.44)
; = j
h
whereCis a constant which due to normalization can be taken to be 1.
34
1.5 (a) Prove the following:
@
(i)p0jxj iihhp0j i
h =
0
Z@p
@
(ii) jxj idp0 p0)ih p0)
h = ( (
@p0
where p0)p0j i p0)p0j i are momentum-space wave
( = h and ( = h
functions.
(b) What is the physical signi cance of
ix
exp
h
wherexis the position operator and is some number with the
dimension of momentum? Justify your answer.
(a) We have
(i)
1
z }| {
Z Z
hp0jxj ip0jxdx0jx0ihx0j idx0hp0jxjx0ihx0j i
= h =
Z Z
ip0 x0
(S;1:7:32)
h
=dx0x0hp0jx0ihx0j idx0x0Ae;x0j i
= h
Z
Z@ @
ip0 x0 ip0 x0
h h
=Adx0e;ih)hx0j iihdx0Ae;x0j i
( = h
0 0
Z@p@p
@@
=ihdx0hp0jx0ihx0j iihhp0j i )
=
@p0@p0
@
hp0jxj iihhp0j i: (1.45)
=
@p0
(ii)
Z@
Z
h jxj idp0h jp0ihp0jxj idp0 p0)ih p0) (1.46)
= = ( (
@p0
where jp0i p0)p0j i p0).
we have used (1.45) and that h = ( and h = (
1. FUNDAMENTAL CONCEPTS 35
ix
(b) The operator exp gives translation in momentum space. This can
h
be justi ed by calculating the following operator
" #
1 n 1 n
X X
ix ix i
1 1
p expp =p xn]
= [
hn!hn!h
n=0 n=0
1 n n
X X
1i
=xn;kp x]xk;1
[
n!hk=1
n=1
1 n n 1 n
X X( X
1i i
1
=ih)xn;1n(;ih)xn;1
; =
n!hk=1n!h
n=1 n=1
1 n;1 1 n
X X
1i i ix
1
=xn;1ih) =
(;
(n;hhn!h
1)!
n=1 n=0
ix
=: (1.47)
exp
h
So when this commutatorp0i of the momentum oper-
acts on an eigenstate j
ator we will have
ix ix ix
p expp0ipexpp0ip0jp0i )
j = j ; exp
hhh
ix ix ix
exppexpp0ip0p0i )
= j ; exp j
hhh
ix ix
pexpp0ip0p0i:(1.48)
j = ( + ) exp j
hh
Thus we have that
ix
expp0iAjp0 (1.49)
j + i
h
whereAis 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=Sz!Sz
; =
mc
write the Heisenberg equations of motion for the time-dependent
operatorsSx(t),Sy(t),Sz(t).Sx y z as func-
and Solve them to obtain
tions of time.
Let us rst prove the following
[AS BS]CSAH BH]CH:(2.1)
= ) [ =
Indeed we have
h i
[AH BH]ASU UyBSASBSU;BSASU
= Uy U = Uy Uy
=AS BS]CSCH:(2.2)
Uy [ U = Uy U =
The Heisenberg equation of motion gives
dSx 1!
1
(S;1:4:20)
=S H]S !Sz]ihS)!Sy (2.3)
[ = [ = (; = ;
dtihxihxihy
dSy 1!
1
(S;1:4:20)
=S H]S !Sz]ihS)!Sx (2.4)
[ = [ = ( =
dtihyihyihx
dSz 1 1
(S;1:4:20)
=S H]S !Sz]Sz = constant:(2.5)
[ = [ = 0 )
dtihzihz
Di erentiating once more eqs. (2.3) and (2.4) we get
d2SxdSy
(2:4)
=!=!2S)Sx(t)Acos!t+Bsin!t)Sx(0)A
; ; = =
dt2dtx
d2SydSx
(2:3)
=!=!2S)Sy(t)Ccos!t+Dsin!t)Sy(0)C:
; = =
dt2dty
But on the other hand
dSx
=!S)
;
dty
;A!sin!t+B!cos!t=C!cos!t;D!sin!t)
;
A=DC=B:(2.6)
;
2. QUANTUM DYNAMICS 37
So, nally
Sx(t)Sx(0)!t;Sy(0)!t(2.7)
= cos sin
Sy(t)Sy(0)!t+Sx(0)!t(2.8)
= cos sin
Sz(t)Sz(0):(2.9)
=
2.2x(t) be the coordinate operator for a free particle in one
Let
dimension in the Heisenberg picture. Evaluate
[x(t) x(0)]:
The Hamiltonian for a free particle in one dimension is given by
p2
H=:(2.10)
2m
This means that the Heisenberg equationsxand
of motion for the operators
pwill be
" #
@p(t)p2(t)
1 1
=p(t) H(t)]p(t) =0 )
[ =
@tihih2m
p(t)p(0) (2.11)
=
" #
@x(t)p2(t)p(t)p(0)
1 1 1
(2:11)
=x H]x(t) =p(t)ih= = )
[ = 2
@tihih2m2mihmm
t
x(t)p(0)x(0): (2.12)
= +
m
Thus nally
ttiht
[x(t) x(0)]p(0)x(0) x(0)p(0) x(0)]:(2.13)
= + = [ = ;
mmm
2.3 Consider a particle in three dimensions whose Hamiltonian is
given by
~
p2
H=V(x):
+~
2m
38
Byx p H]
calculating~~obtain
[
* +
dp2
~
hx pix :
~~=~rVi
;h
dtm
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 usx p H]
rst~~
calculate the commutator [
2
"p#p3
3 3 2
X X
~2
4 5
[x p H]~ p +V(x)xipi jV(x)
~~=x~~=~
+
2mi=1m
2
j=1
#
X"~
X
p2
=xi jpixipi V(x)]:(2.14)
+ [
2mi
ij
The commutator in (2.14) will give
"prst
#
2
1 1 1
xi jxi p2]pj[xi pj]xi pj]pj)pjih ijih ijpj)
= [ = ( + [ = ( +
2m2mjm2m
2
1ih
=ih ijpj ijpj:(2.15)
2 =
2mm
The second commutator can be calculated if we Taylor expand the function
P
V(x)xiV(x)anxnan
~in terms of~= with
which means that we take
n i
independentxi. So
of
" #
1 n;1
X X X X
[pi V(x)]pi anxnanpi xn]anxkpi xi]xn;k;1
~= = [ = [
i i i i
n n
n=0 k=0
n;1
X X(; X X
@
=anih)xn;1ihannxn;1ihanxn
= ; = ;
i i
@xi n i
n n
k=0
@
=ihV(x): (2.16)
;~
@xi
The right-hand side of (2.14) now becomes
X X(;
ih@
[x p H] ijpjpiih)xiV(x)
~~=~
+
mi@xi
ij
ih~
=p2ih~ x):(2.17)
~;xrV(~
m
2. QUANTUM DYNAMICS 39
The Heisenberg equation of motion gives
d1p2
~
(2:17)
~
~ p=~ p H]~ x)
x~[x~=xrV(~)
;
dtihm
* +
dp2
~
hx pix (2.18)
~~=~rVi
;h
dtm
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
d1 1
hnjx pjninj~ p H]niEhnjx pjniEnhnjx pjni):
~~=x~j~~;~~=0
h [ = (
dtihihn
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:
j 0i
h
You may use
2 3 0 1
! !
2 1=2
x0h
1
4x5m!A
@
hx0j ;1=4x;1=2 x0:
0i = exp ;
0
2
0
(b) Obtain a simple expression that the probability that the state
ist=0. Does this probability change
found in the ground state at
fort>0?
(a) We have
;ipa
j t=0i = exp j 0i )
h
;ipa
(Pr:1:4:c)
hx0j t=0ix0x0aj 0i
= h exp j 0i = h ;
h
2x;a! 3
2
0
1
4x5
= ;1=4x;1=2:(2.19)
exp ;
0
2
0
40
(b) This probability is given by the expression
;ipa2
2
j t=0ij:(2.20)
h0j = j hexp j 0ij
h
It is
Z
;ipa;ipa
hexpdx0h0jx0ihx0j exp j 0i
j 0i =
hh
2
2
Zx! 3
0
1
4 5
=dx0 ;1=4x;1=2 ;1=4x;1=2
exp ;
0 0
2
0
2x;a!x
3
2
0
1
4x5
exp ;
2
0
" #
Z
1
=dx0 ;1=2x;1x02x02a2ax0
exp ; + + ; 2
0
2
2
0
" !#
Zx
1aa2a2
2
p exp ; ; 2
=dx0x02x0 + +
x0 2x2 2 4 4
0
! !
p
a2a2
1
= x0:(2.21)
exp ; p = exp ;
4x2 x0 4x2
0 0
So
!
a2
2
j t=0ij:(2.22)
h0j = exp ;
2x2
0
Fort>0
iHt2
2 2
j tijt)j t=0ij t=0ij
h0j = j h0j U( = j h0j exp ; j
h
2
t=h 2
0
=e;iE t=0i t=0ij:(2.23)
h0j = j h0j
2.5 Consider a function, known as the correlation function, de ned
by
C(t)x(t)x(0)i (2.24)
= h
wherex(t) is the position operator in the Heisenberg picture. Eval-
uate the correlation function explicitly for the ground state of a
one-dimensional simple harmonic oscillator.
2. QUANTUM DYNAMICS 41
The Hamiltonian for a one-dimensional harmonic oscillator is given by
p2(t)
1
H=m!2x2(t):(2.25)
+
2m2
So the Heisenberg equations of motion will give
" #
dx(t)p2(t)
1 1
1
=x(t) H]x(t) +m!2x2(t)
[ =
dtihih2m2
h i h i
1 1
1
=x(t) p2(t)m!2x(t) x2(t)
+
2mih2ih
2ihp(t)
=p(t) = (2.26)
2ihmm
" #
dp(t)p2(t)
1 1
1
=p(t) H]p(t) +m!2x2(t)
[ =
dtihih2m2
h i
m!2m!2
=p(t) x2(t)ihx(t)]m!2x(t):(2.27)
= [;2 = ;
2ih2ih
Di erentiating once more the equations (2.26) and (2.27) we get
d2x(t)dp(t)
1
(2:27)
=!2x(t)x(t)Acos!t+Bsin!t)x(0)A
= ; ) = =
dt2mdt
d2p(t)dx(t)
1
(2:26)
=!2p(t)p(t)Ccos!t+Dsin!t)p(0)C:
= ; ) = =
dt2mdt
But on the other hand from (2.26) we have
dx(t)p(t)
= )
dtm
p(0)D
;!x(0)!t+B!cos!t=!t+!t)
sin cos sin
mm
p(0)
B=D=m!x(0):(2.28)
;
m!
So
p(0)
x(t)x(0)!t+!t(2.29)
= cos sin
m!
and the correlation function will be
1
(2:29)
C(t)x(t)x(0)ix2(0)i!t+p(0)x(0)i!t:(2.30)
= h = h cos h sin
m!
42
Since we are interested in the ground state the expectation values appearing
in the last relation will be
hhh
hx2(0)ia+ay)(a+ay)jaayj 0i = (2.31)
= h0j ( 0i = h0j
s2m!sh2m!2m!
mh!
hp(0)x(0)iih0jaya)(a+ay)j 0i
= ( ;
2 2m!
hh
=ih0jaayji:(2.32)
; 0i = ;
2 2
Thus
hhh
C(t)!t;isin!t=e;i!t:(2.33)
= cos
2m!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 combinationxi is as
of j 0i and j 1i such that h
large as possible.
(b) Suppose the oscillator is in thet=0.
state constructed in (a) at
Whatt>0 in the Schr
is the state vector for odinger picture?
Evaluatexit>0
the expectation value h as a function of time for
using (i) the Schr
odinger picture and (ii) the Heisenberg picture.
(c)x)2i as a function of time using either picture.
Evaluate h(
(a) ic0jc1jxi is as large as
We want to nd a state j = 0i + 1i such that h
possible. i should be normalized. This means
The state j
q1
2 2
2
jc0jc1jc1jc0j:(2.34)
+ j =1 )j = ;j
Wec0c1 in the following form
can write the constands and
0
c0c0jei
= j
q
(2:34)
1 1 2
c1c1jei ei c0j:(2.35)
= j = 1 ;j
2. QUANTUM DYNAMICS 43
Thexi in a one-dimensional simple harmonic oscillator is given
average h
by
hxi jxj ic 0h0jc 1h1jx(c0jc1j 1i)
= h =( + ) 0i +
2 2
=c0jxjc 0c1h0jxjc 1c0h1jxjc1jxj 1i
j h0j + 1i +
sh0i sh0i + j h1j
2
=c0ja+ayjc 0c1a+ayj 1i
j h0j 0i + h0j
2 2
sm!sm!
h2h
+c 1c0a+ayjc1ja+ayj 1i
h1j 0i + j h1j
sh2m!sh2m!
=c 0c1c 1c0)c 0c1)
( + = 2 <(
s2m!2m!
q1 ;j
h2
= 1 0)jc0jc0j (2.36)
2 cos( ;
2m!
q
h
wherex=a+ay).
we have used that (
2m!
What wec0j 1 0 that make the
need is to nd the values of j and ;
averagexi as large as possible.
h
q1 2
@hxic0j j c0 j =1
6
2 2
2
q1j
=0c0jc0jc0j =0
) ;j ; ) 1 ;j ;j
@jc0j 2
;jc0j
1
)jc0j = p (2.37)
2
@hxi
=0 1 0) 1 0n n2:(2.38)
); sin( ; =0 ) = + Z
@ 1
Butxi maximum we want also
for h
@2hxi
<0n=2k k2:(2.39)
) Z
2
@ 1 =
1 1 max
So we can write that
1 1 1
+2k )
0 0 0
p p p
j iei ei( ei :(2.40)
= j 0i + j 1i = (j 0i + j 1i)
2 2 2
We 0 = 0. Thus
can always take
1
p
j i:(2.41)
= (j 0i + j 1i)
2
44
(b) t0i i. So
We have j = j
1 1
t=h
0 1
p p
j t0 tit t0 t0ie;iHt=hj ie;iEe;iE t=hj 1i
= U( =0)j = = j 0i +
2 2
1 1
p p
=e;i!t=2je;i!3t=2je;i!t=2e;i!tj:(2.42)
0i + 1i = j 0i + 1i
2 2
(i) In the Schr
odinger picture
hxiS t0 tjxSj t0 tiS
= h
" # " #
1 1
p p
=ei!t=2h0jei!3t=2h1jxe;i!t=2je;i!3t=2j 1i
+ 0i +
2 2
1 1
=ei(!t=2;!3t=2)h0jxjei(!3t=2;!t=2)h1jxj
2 2
sh1i +
shsh0i
1 1
=e;i!tei!t!t:(2.43)
+ = cos
2
2m!2m!2m!
2
(ii) In the Heisenberg picture we have from (2.29) that
p(0)
xH(t)x(0)!t+!t:
= cos sin
m!
So
hxiH jxHj i
= h
"p(0) !" #
#
1 1 1 1
p p p p
=x(0)!t+!tj 0i + j 1i
h0j + h1j cos sin
m!
2 2 2 2
1
1 1 1
=!th0jxj!th1jxj!th0jpj 1i
cos 1i + cos 0i + sin
2 2 2
m!
1
1
+!th1jpj 0i
sin
2
sm!sh1mh!
s
h1
1
=!t+!t+!t(;i)
cos cos sin
2
2m!2m!2m!2
2
s
1mh!
+!ti
sin
s2m!2
h
=!t: (2.44)
cos
2m!
(c) It is known that
h( x)2ix2ixi2 (2.45)
= h ;h
2. QUANTUM DYNAMICS 45
In the Sch
odinger picture we have
2h3h
s
2
2
4m!( 5
x2a+ay)a2ayaayaya) (2.46)
= = ( + + +
2m!
2
which means that
hxi2 t0 tjx2j t0 tiS
= h
S
" # " #
1 1
p p
=ei!t=2h0jei!3t=2h1jx2e;i!t=2je;i!3t=2j 1i
+ 0i +
2 2
h ih
1 1 1
=ei(!t=2;!t=2)h0jaayjei(!3t=2;!3t=2)h1jaayjayaj 1i
0i + 1i + h1j
2 2 2
2m!
hhh
i
1 1 1
=:(2.47)
+2 + =
2 2 2
2m!2m!
So
hh2h2
(2:43)
h( x)2iS!t=!t:(2.48)
= ; cos sin
2m!2m!2m!
In the Heisenberg picture
"p(0) #
2
x2t)x(0)!t+!t
( = cos sin
H
m!
p2(0)
=x2(0)!t+!t
cos2 sin2
m2!2
x(0)p(0)p(0)x(0)
+!tsin!t+!tsin!t
cos cos
m!m!
h2
2
=a2ayaayaya)!t
( + + + cos
2m!
mh!2
2
;a2ayaayaya)!t
( + ; ; sin
2m2!2
s
ihmh!sin!t
2
+a+ay)(aya)
( ;
m!4m!2
s
ihmh!sin!t
2
+aya)(a+ay)
( ;
m!4m!2
h2
2
=a2ayaayaya)!t
( + + + cos
2m!
46
h2ih
2 2
;a2ayaayaya)!t+aya2)!t
( + ; ; sin ( ; sin 2
2m!2m!
hhh
2
=aayaya)a2!t+ay!t
( + + cos 2 cos 2
2m!2m!2m!
ih
2
+aya2)!t (2.49)
( ; sin 2
2m!
which means that
hx2 jx2 iH
iH = h j
H H
" #
h1 1
p
= p h0j + h1j
2m!
2 2
h i
2 2
y y 2 y y 2
+ + cos 2!t+ cos 2 + ( ; ) sin 2
"aaaaaa!tiaa!t
#
1 1
p p
j 0i + j 1i
2 2
h i
h
=aayjaayjayaj 1i
h0j 0i + h1j 1i + h1j
4m!
hh
=:(2.50)
[1 +2 +1] =
4m!m!
So
(2:44) hh2h2
h( x)2iH!t=!t:(2.51)
= ; cos sin
2m!2m!2m!
2.7 A coherent state of a one-dimensional simple harmonic oscil-
lator is de ned to be an eigenstate of the (non-Hermitian) annihi-
lationa:
operator
aj i j i
=
where is, in general, a complex number.
(a) Prove that
2
j =2
j ie;je ay j 0i
=
is a normalized coherent state.
(b) Prove the minimum uncertainty relation for such a state.
2. QUANTUM DYNAMICS 47
(c) i as
Write j
1
X
j if(n)jni:
=
n=0
2
Showf(n)jnis of the
that the distribution of j with respect to
Poisson form. Findn,E.
the most probable value of hence of
(d) Show that a coherent state can also be obtained by applying
the translation ( e;ipl=hpis the
nite-displacement) operator (where
momentumlis the displacement distance) to the
operator, and
ground state.
(e) Show i remains coherent under time-
that the coherent state j
evolution and calculate (t)i. (Hint: di-
the time-evolved state j
rectly apply the time-evolution operator.)
(a) We have
h i
2 2
j =2 ay j =2 ay
aj ie;jae e;ja e (2.52)
= j 0i = j 0i
sinceaj 0i = 0. The commutator is
"
1 1
ha eia a# ha (a) i
X X
1 1
ay y n y n
= ( )n =
n!n!
n=0 n=0
1 n 1 n
h i(
X X( X X(
1 1
= nay)k;1a ayay)n;k nay)n;1
=
n!n!
n=1 k=1 n=1 k=1
1 1
X X
1 1
ay
= n(ay)n;1 ( ay)n e :(2.53)
= =
(n;n!
1)!
n=1 n=0
So from (2.52)
2
j =2 ay
aj ie;j e j i (2.54)
= j 0i =
which i is a coherent state. If it is normalized, it should satisfy
means that j
also j i = 1. Indeed
h
2 2
a j ay j a
h j ie e;je e;je e ay j 0i
= h0j j 0i = h0j
48
p
X
2 1
j
=e;j mh0janjay)mjay)mjm!jmi]
( )n ( 0i [( 0i =
n!m!
n m
p p
X X
2n!m! 2 1
j j 2
=e;j mhnjmie;j j )n
( )n = (j
n!m!n!
n m n
2 2
j j
=e;jej: (2.55)
=1
(b) According to problem (1.3) the state should satisfy the following relation
xj ic pj i (2.56)
=
wherex x; jxj i,p p; jpj icis a purely imaginary
h h and
number.
Since i is a coherent state we have
j
aj i j i jay j :(2.57)
= ) h = h
Using this relation we can write
shsh
xj ia+ay)j i +ay)j i (2.58)
= ( = (
2m!2m!
and
shsh
hxi jxj i ja+ay)j i jaj i jayj i)
= h = h ( = (h + h
sh2m!2m!
= + ) (2.59)
(
2m!
and so
sh
xj ix;hxi)j iay i:(2.60)
=( = ( ; )j
2m!
q
Similarlyp=imh!aya) we have
for the momentum ( ;
2
smh!smh!
p
pj ii(aya)j ii(ay )j i (2.61)
= ; = ;
2 2
2. QUANTUM DYNAMICS 49
and
smh!smh!
hpi jpj iih jaya)j ii(h jayj i jaj i)
= h = ( ; = ; h
2
smh!2
=i( ) (2.62)
;
2
and so
smh!
pj ip;hpi)j ii(ay i )
= ( = ; )j
2
s
2
(ay ii pj i:(2.63)
; )j = ;
mh!
So using the last relation in (2.60)
shsi
2
xj ii)pj ipj i (2.64)
= (; = ;
2m!mh!m!
| {z }
purely imaginary
and thus the minimum uncertainty condition is satis ed.
(c) The coherent state can be expressed as a superposition of energy eigen-
states
1 1
X X
j inihnj if(n)jni:(2.65)
= j =
n=0 n=0
forf(n) we have
the expansion coe cients
2 2
j =2 ay j =2
f(n)nj inje;je e;jnje ay j 0i
= h = h j 0i = h
1 1
X X
2 1 2 1
j =2 j =2
=e;jnj ay)mje;j mhnjay)mj 0i
h ( 0i = (
m!m!
m=0 m=0
1
p
X
2 1 2 1
j =2 ;j j =2
p
=e;j mm!hnjmie n ) (2.66)
=
m!
n!
m=0
2
(j j )n
2 2
jf(n)j j ) (2.67)
= exp(;j
n!
2
whichf(n)jnis of the Poisson
means that the distribution of j with respect to
2
typen= j .
about some mean value j
50
Thenis given by the maximum of the distribution
most probable value of
2
jf(n)j which can be found in the following way
2
(j j )n+1
2
2 2
exp(;j j )
jf(n+1)j j
j
(n+1)!
= = 1 (2.68)
2
2 (j j )n
2
jf(n)jn+1
exp(;j j )
n!
2
which meansnis j .
that the most probable value of j
(d) Weipl=h) j 0i is an eigenstate of the an-
should check if the state exp (;
nihilationa. We have
operator
h ij 0i
;ipl=h)
aexpipl=h)a e( (2.69)
(; j 0i =
sinceaj 0i =0. For the commutator in the last relation we have
n n
1 1 n
ha eiil!a p]il!p[a p]p
X X X
1 ; 1 ;
(;ipl=h) n k;1 n;k
= [ =
n!hn!hk=1
n=0 n=1
s
!
1 n
X X
1ilnmh!
;
=pn;1i
n=1 k=1
smn!h2
! !
n;1
1
X
h!;il1ilp
;
=i
2 1)!
n=1
rm!h(n;h
;ipl=h)
=le( (2.70)
2h
where we have used that
smh!smh!
y
[a p]i[a a;a]i:(2.71)
= =
2 2
So substituting (2.70) in (2.69) we get
rm!
a[expipl=h)l[expipl=h) j 0i] (2.72)
(; j 0i] = (;
2h
whichipl=h) j 0i is a coherent state with eigen-
means that the state exp (;
q
valuelm! .
2h
(e) Using the hint we have
1
X
(2:66) 2 1
; ; ;j j =2
p
j (t)it)j ieiHt=hj ieiHt=he njni
= U( = =
n!
n=0
2. QUANTUM DYNAMICS 51
1 1
X X 1
; it
2 1 2 1
(2:3:9)
; t=h ;j j =2 ) ;j j =2
n
h
2
p p
=eiEe njnieh!(n+e njni
=
n!n!
n=0 n=0
1
X
n
2 1
; ; ;j j =2
p
=ei!tei!t=2e njni
n!
n=0
n
1 ;
X
2 (
; ;j e; i!t ei!t) (2:66) ; ;
j =2
p
=ei!t=2e niei!t=2j ei!ti (2.73)
j =
n!
n=0
Thus
; ; ; ; ;
aj (t)iei!t=2aj ei!ti ei!tei!t=2j ei!ti
= =
;
= ei!tj (t)i:(2.74)
2.8 The quntum mechanical propagator, for a particle with mass
m, moving in a potential is given by:
Znrx)nry)
1
X
sin( sin(
K(x y E)dteiEt=hK(x y t 0)Ah r2
= =
2
0
E;n2
n
2m
whereAis a constant.
(a) What is the potential?
(b)Ain terms of the parameters describing
Determine the constant
them,retc. ).
system (such as
We have
Z Z
1 1
K(x y E)dteiEt=hK(x y t 0)dteiEt=hhx tjy 0i
0 0
Z
1
;
=dteiEt=hhxjeiHt=hjyi
0
Z
1
Xh
;
=dteiEt=hxjeiHt=hjnihnjyi
0
n
Z
1
X
; t=h
n
=dteiEt=heiExjnihnjyi
h
0
n
Z
1
X
;En )t=h
= n(x) n(y)ei(Edt
0
n
52
"ih#
1
X
;
;En
= n(x) n(y)ei(E +i")t=h
lim
"!0
E;Eni"0
+
n
X
ih
= n(x) n(y):(2.75)
E;En
n
So
X (x) (y)AX
ihsin(nrx)nry)
sin(
= )
n
n
h2r2 2
n
n n
2m
sAE;EE;n
h2r2
(x)nrx) Enn2:(2.76)
= sin( =
n
ih2m
For a one dimensional in niteLthe energy
square well potential with size
eigenvalueEn n(x) are given by
and eigenfunctions
sn x h
2
2
2
(x) Enn2:(2.77)
= sin =
n
LL2mL
Comparingr)L= and
with (2.76) we get =
L r
(0 for 0
r
V= (2.78)
1 otherwise
while
A2r2hr
=A=i:(2.79)
)
ih
2.9 Prove the relation
d (x)
= (x)
dx
where (x) is the (x) the Dirac delta
(unit) step function, and
function. (Hint: study the e ect on testfunctions.)
Forf(x) we have
an arbitrary test function
Zd (x)dZdf(x)
Z
+1 +1 +1
f(x)dx= (x)f(x)]dx; (x)dx
[
;1dx;1dx;1dx
2. QUANTUM DYNAMICS 53
Zdf(x)
+1
+1
= (x)f(x)dx
;
;1
0 dx
+1
=f(x)f(x)f(0)
lim ; =
x!+1
0
Z
+1
= (x)f(x)dx)
;1
d (x)
= (x): (2.80)
dx
2.10 Derive the following expression
h( +2 ) cos( ; i
m!
Sclx2x2!T)x0xT
=
0 T
2!T)
sin(
for the classical action for a harmonic oscillator moving from the
pointx0t=0xTt=T.
at to the point at
The Lagrangian for the one dimensional harmonic oscillator is given by
1 2 1
L(x x)mx_m!2x2:(2.81)
_ = ;
2 2
From the Lagrange equation we have
@Ld@Ld
(2:81)
;m!2x;mx) =0 )
=0 ) ; ( _
@xdt@x_dt
x+!2x=0: (2.82)

which is the equation of motion for the system. This can be solved to give
x(t)Acos!t+Bsin!t(2.83)
=
with boundary conditions
x(t=0)x0A (2.84)
= =
x(t=T)xTx0!T+Bsin!T)Bsin!T=xTx0!T)
= = cos ; cos
xTx0!T
; cos
B=:(2.85)
sin!T
54
So
xTx0!T
; cos
x(t)x0!t+!t
= cos sin
sin!T
x0!tsin!T+xT!t;x0!Tsin!t
cos sin cos
=
sin!T
xT!t+x0!(T;t)
sin sin
= ) (2.86)
sin!T
xT!cos!t;x0!cos!(T;t)
x(t) =:(2.87)
_
sin!T
With these at hand we have
Z Z
T T
1 2 1
S=dtL(x x)dt2mx_m!2x2
_ = ;
2
0 0
"
Zd#
T
1 1
=dt1m(xx)mxx;m!2x2
_ ; 
2 2 2
0
Zdt
T
mT
1
=mdtx[x+!2x]xx_
;  +
2
0 2
0
(2:82) m
=x(T)x(T)x(0)x(0)]
[ _ ; _
2
mxT!x0!
=xT!T;x0)xTx0!T)
( cos ; ( ; cos
2!Tsin!T
sin
h i
m!
=x2!T;x0xTx0xTx2!T
cos ; + cos
0
2sin!TT
h i
m!
=x2x2)!T;x0xT:(2.88)
( + cos 2
2sin!TT 0
2.11 The Lagrangian of the single harmonic oscillator is
1 1
2
Lmx_m!2x2
= ;
2 2
(a) Show that
iS
hxbtbjxataiexpclG(0 tb ta)
= 0
h
whereScl is the actionxclxa ta) to
along the classical path from (
(xb tb)Gis
and
G(0 tb ta) =
0
2. QUANTUM DYNAMICS 55
( N +1)
Zm8iXm9
N
< =
2
1
2
limdy1:::dyNexp(yj+1yj)2"m!2yj
; ;
:
N!1
2 ih"hj=0 2"2
tb ;ta
where"= .
(N+1)
[Hint:y(t)x(t)xcl(t) be the new integration variable,
Let = ;
xcl(t) being the solution of the Euler-Lagrange equation.]
(b)Gcan be written as
Show that
( N +1)
Z
2
m
G=dy1:::dyNexp(;nT n)
lim
N!1
2 ih"
2y3
1
6 7
.
6 7
.
wheren=nT is its transpose. Write the symmetric
and
.
4 5
yN
matrix .
(c) Show that
Z Z
N=2
T
;
p
dy1:::dyNexp(;nT n)dNnen n =
det
[Hint: by an orthogonal matrix.]
Diagonalize
N
2ih" 0 0
(d)det det NpN.j jmatrices j that con-
Let De ne
m
0
sistjrowsjcolumns N and whose determinants
of the rst and of
0
arepj. j+1 in minors show the following recursion
By expanding
formulapj:
for the
pj+1"2!2)pjpjj=1 ::: N(2.89)
=(2 ; ;
;1
(e) (t)"pjt=taj"and show that (2.89) implies that in
Let = for +
the"! (t) satis es the equation
limit 0
d2 2
=! (t)
;
dt2
with (t=ta) d (t=ta) =1.
initial conditions = 0
dt
56
(f) Show that
( )
sm!im!
hxbtbjxataiexp[(x2x2)!T)xaxb]
= + cos( ; 2
b a
2 ihsin(!T)hsin(!T)
2
whereT=tbta.
;
(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
hxbtbjxataidx1dx2:::dxNhxbtbjxNtNihxNtNjxNtN:::
= lim i
;1 ;1
N!1
hxi+1ti+1jxitii:::hx1t1jxatai:(2.91)
It is
; ;ti )=h ;
i+1
hxi+1ti+1jxitiixi+1eiH(txiixi+1jeiH"=hjxii
= h j j = h
1 1
"
; ( mp2 + m!2x2)
h
2 2
=xi+1eixii (since " is very small)
h j j
1
" p2 "
; ; m!2 x2
h 2m h
2
=xi+1eieixii
h j j
1
" " p2
; m!2 x2 ;
h i
2 h 2m
=eixi+1eixii:(2.92)
h j j
For the second term in this last equation we have
Z
" p2 " p2
; ;
h 2m h 2m
hxi+1eixiidpihxi+1jeipiihpijxii
j j = j
Z
p2
"
i
;
h 2m
=dpieixi+1jpiihpijxii
h
Z
pi 2
1 "
; ;xi
i
h 2m
=dpieieip (xi+1 )=h
2 h
h i
Z
" m2 m2
1 ;i p2 ;2pi m (xi+1 ;xi )+ (xi+1 ;xi )2 ; (xi+1 ;xi )2
i
2mh "
"2 "2
=dpie
2 h
Z
2
i" m2
1 "
;xi )2 ; ;pi m ;xi
2mh 2mh "
"2 i+1
=e(xdpiei [pi (xi+1 )]
2 h
2. QUANTUM DYNAMICS 57
s
i" m2
1 2hm
;xi
2mh
"2 i+1
=e(x )2
2
rhi"
mim
;xi )2
i+1
2h"
=e(x:(2.93)
2 hi"
Substituting this in (2.92) we get
1
i m 1
m(x )2 "m!2 xi]
2
;xi ;
i+1
h 2" 2
hxi+1ti+1jxitiie[ (2.94)
=
2 ih"
and this into (2.91):
Zi
hxbtbjxataiS[x] =
Dxexp
8
( N +1)
Zmh
N
=
X
2
1
limdx1:::dxNexp(xj+1xj)2"m!2x2:
; ;
j
:
N!1
2 ih"hj=0"2
2
Lety(t)x(t)xcl(t)x(t)y(t)xcl(t)x(t)y_t)x_t) with
= ; ) = + ) _ = ( + (
cl
boundaryy(ta)y(tb):For this new integration variable we
conditions = = 0
have Dx= Dyand
Z
tb
S[x]S[y+xcl]y+xcl y_x_dt
= = L( + )
cl
ta
2
Z@ @ @ @ 3
2 2
tb
L L L L
1 1 2
4
=xcl x_y+y_y2y_
L( ) + + + 5
cl
2 2
ta@x@x_@2x2@x_
xcl xcl xcl xcl
" !# Z h
tb tb
Z
i
tb
@Ld@L
L
1 2 1
=Scly @;y+my_m!2y2dt:
+ + ;
2 2
@x_@xdt@x_
ta ta xcl ta
So
ZiiZ
h i
tb
1 2 1
hxbtbjxataiSclmy_m!2y2dt
= Dyexp + ;
2 2
hhta
iS
=expclG(0 tb ta) (2.95)
0
h
with
G(0 tb 0 ta) =
( N +1)
Zm8iXm9
N
< =
2 1
2
limdy1:::dyNexp(yj+1yj)2"m!2yj:
; ;
:
N!1
2 ih"hj=0 2"2
58
(b) For the argument of the exponential in the last relation we have
Xyy"m!y
iNm2 1 (y0 =0)
2 2
( ; ) ; =
j
hj=0"j+1 j 2
2
XyyyyyyX"m!y y
iNm2iN 1 (yN +1 =0)
2 2
( + ; ; ) ; =
j+1 j j j+1 i ij j
j
hj=0"j+1hi j=1 2
2
X X
mN i"m!2 N
;yi ijyjyi i j+1yjyi i+1 jyj)yi ijyj:(2.96)
(2 ; ; ;
2"ihi j=1 2hi j=1
where the last step is written in will be
such a form so that the matrix
symmetric. Thus we have
( N +1)
Z
2
m
G=dy1:::dyNexp(;nT n) (2.97)
lim
N!1
2 ih"
with
2:::3:::0 0 3
2
2 ;1 0 0 0 1 0 0
6:::7:::0 0 7
6
6 ;1 2 ;1 0 0 7 6 0 1 0 7
6 7 6 7
6 7 6 7(2.98)
;1 2 0:::0 0
m0:::0i"m!2 0 0 1
6 7:
6 7
= 6 7 + 6 7
. . . . . . . . . .
6 . . 7 6 . . 7
2"ih.h. . .
2
. . . . . . . . . .
6 7 6 7
6:::2. . 7 6 7
4 0 0 0 ;1:::1 0 5
5 4 0 0 0
0:::;1:::0 1
0 0 2 0 0 0
(c) byU. is symmetric the
We can diagonalize a unitary matrix Since
following will hold
y y
=U DU) TUT D(U)TUT DU= )U=U:(2.99)
= =
So byR. So
we can diagonalize an orthogonal matrix
=RT DRandR=1 (2.100)
det
which
Zmeans that Z Z
T T T
Rn=
; n ; RT Rn ;
dNnendNnendN e
= =
Z
Z Z
2 2 2
; a1 ; a2 ;
1 2 N
=d 1e d 2e :::d Ne aN
s s s
N=2 N=2
qQ
p
=:::= =
N
a1a2aNai det
D
i=1
N=2
p
= (2.101)
det
2. QUANTUM DYNAMICS 59
whereai are the D.
diagonal elements of the matrix
(d) From (2.98) we have
!
N
2ih"
det =
m
8:::0:::0 0 9
2 3 2 3>
> 2 ;1 0 0 1 0 0
>
> >
> 6 7 6 >
>:::0:::0 0 7 >
>
> ;1 2 ;1 0 7 6 0 1 0 7 >
> >
> 6 7 6
> >
<6 7 6 7 =
6
0 ;1 2 0 0 0 1
6:::0:::0 0 7 >
7 6 7
det 6"2!2 6 7 =
7 ;
. . . . . . . . . .
> 6 . . . . . 7 6 . . . . . 7 >
> >
. . . . . . . .
> 6 7 6 7 >
>
> 6 7 6 7 >
>:::2. . >
> 0 0 0 ;1 5 4 0 0 0 >
4
> >
> >
::::;1:::1 0 5>

0 0 0 2:::0 1
0 0 0
0
det NpN: (2.102)
0
Wej jmatrices jjrowsjcolumns of
de ne that consist of the rst and
0
. So
N
2"!;1:::0 3
2 2
2 ; 0 0
6"2 2 7
;1 2 ; 0 0
6 7
6!:::0 7
6 7
0 ;1 0 0
6:::0 7
6 . . . . . 7:
0
. . . .
det j+1 =det6 7
. . . .
6"!;. 7
6:::2 ; . 7
2 2
0 0 1 0
6 7
6 7
4 5
0:::;1"2!2 ;1
0 2 ;
0:::0"2!2
0 ;1 2 ;
From the above it is obvious that
0 0 0
det j+1"2!2) j j )
= (2 ; det ; det
;1
pj+1"2!2)pjpjj=2 3 ::: N(2.103)
= (2 ; ; for
;1
withp0p1"2!2.
= 1 and =2 ;
(e) We have
(t) (taj")"pj
+
) (taj+1)")"pj+1"2!2)"pj"pj
+( = =(2 ; ;
;1
= (taj")"2!2 (taj") (taj;")
2 + ; + ; +( 1)
) (t+") (t)"2!2 (t) (t;"):(2.104)
= 2 ; ;
60
So
(t+") (t) (t) (t;")"2!2 (t) )
; = ; ;
(t+"); (t) (t); (t;")
;
" "
=!2 (t) )
;
"
0 0
(t) (t;")d2
;
lim!2 (t)!2 (t):(2.105)
= ; ) = ;
"!0
"dt2
From (c) we have also that
(ta)"p0 ! 0 (2.106)
=
and
d (ta") (ta)"(p1p0)
+ ; ;
(t) = =p;p0
=
dta""1
="2!2:(2.107)
2 ; ; 1 ! 1
The general solution to (2.105) is
(t)Asin(!t+ ) (2.108)
=
and from the boundary conditions (2.106) and (2.107) we have
(ta)Asin(!ta ) =!tan n2 Z (2.109)
=0 ) + = 0 ) ; +
which (t)Asin!(t;ta), while
gives that =
d
(2:107)
0
=A!cos(t;t) (ta)A!)
) =
dta
1
A!=1A= (2.110)
)
!
Thus
sin!(t;ta)
(t):(2.111)
=
!
(f) Gathering all the previous results together we get
"m #
1=2
N
(N+1)
p
G= lim
N!1
2 ih"det
2. QUANTUM DYNAMICS 61
2 3
! ;1=2
m1=2ih"N
2
4 5
="det
lim
N!1
2 ihm
;1=2
m1=2m1=2
(d) (e)
;1=2
="pN (tb)]
lim = [
N!1
s2 ih2 ih
m!
(2:111)
=: (2.112)
2 ihsin(!T)
So from (a)
iS
hxbtbjxataiexpclG(0 tb ta)
= 0
sh
h i
m!im!
(2:88)
=exp(x2x2)!T;xbxa:
+ cos 2
2 ihsin(!T)hsin!Tb a
2
2.12 Show the composition property
Z
dx1Kfx2 t2 x1 t1)Kfx1 t1 x0 t0)Kfx2 t2 x0 t0)
( ( = (
whereKfx1 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
dx1Kfx2 t2 x1 t1)Kfx1 t1 x0 t0)
( (
s
Zm"im(x;x) #
2
2 1
=dx1 exp
2 i2 1 2
2 1
"im(x;x) #
smh(t;t)h(t;t)
2
1 0
exp
2 ih(t1t0)h(t1t0)
; 2 ;
s
m1imx2imx2
2 0
= exp exp
2 2
2 1 1 0 2 1 2 1
"im2 #
Z ih(t;t)(t;t)h(t;t)h(t;t)
imimim
dx1x2x2x2x1x0
exp + ; 2x1 ; 2
1 1
2h(t2t1)h(t2t1)h(t2t1)h(t2t1)
; 2 ; 2 ; 2 ;
62
s ( " #)
m1imx2x2
2 0
= exp +
2 ih(t2t1)(t1t0)h(t2t1)t1t0)
; ; 2
( #; ( ;
Zim"im"xx#)
1 1
0
dx1x2x1 2 +
exp + ;
1
2 (t1) (
2 1 1 0 2 1 0
sh(t;t)t;"t)h(t;#)t;t)
(
m1imx2x2
2 0
= exp +
2 ih(t2t1)(t1t0)h(t2t1)t1t0)
; ; 2 ; ( ;
(
Zm"t;t#
;
2 0
dx1 exp
2 1 1 0
"h2ih(t;t)(t;t") ##)
2t2t0imx2(t1t0)x0(t2t1)
; ; + ;
x2x1
;
1
2 1 1 0 2 1 1 0
sim(t;t)(t;t)h(t;t)(t;t) #)
( "
m1imx2(t1t0)x2(t2t1)
; + ;
2 0
= exp
2
2 1 1 0 2 1 1 0
8 3
22 ih(t;t)(t;t)h(t;t)(t;t) 9
" #" #
2
Z
; +
2 0 2 1 0 0 2 1
4dxexp1 1
:h(t;t)(t;t)x; 5
2 (t2t0)
;
2 1 1 0
(im1x(t;t)x(t;t)] )
2
[ +
2 1 0 0 2 1
exp ;
2h(t2t1)(t1t0)t2t0)
; ; ( ;
v
s (
u
u
m1 2ih(t2t1)(t1t0)im1
; ;
t
= exp
2
2 1 1 0 2 0 2 1 1 0
"2 ih(t;t)(t;t)m(t;t)h(t;t)(t;t)
x2(t1t0)(t2t0)x2(t2t1)(t2t0)
; ; + ; ;
2 0
;
(t2t0)
;
#)
x2(t1t0)2x2(t2t1)2x02(t1t0)(t2t1)
; ; ; ; 2x2 ; ;
2 0
2 0
sm(t;t)
=
2 ih(t2t1)
(im";
x2(t1t0)(t2t0t1t0)x2(t2t1)(t2t0t2t1)
; ; ; + + ; ; ; +
2 0
exp ;
2h(t2t0)(t2t1)(t1t0)
; ; ;
#)
2x2x02(t1t0)(t2t1)
; ;
(
2 0 2 1 1 0
" #
st;t)(t;t)(t;t)
mim(x2x0)2
;
= exp
2 ih(t2t0)h(t2t0)
; 2 ;
=Kfx2 t2 x0 t0): (2.113)
(
2. QUANTUM DYNAMICS 63
2.13 (a) Verify the relation
!
ihe
[ "ijkBk
] =
i j
c
~
x
~~
wheremdtp; and the relation
=~eA
c
" !#
d2~d d~d~
x~1x~~x
~
m2eE+B;B :
= =
dtdt2cdtdt
(b) Verify the continuity equation
@ ~~
0
+j=0
r
@t
~
withjgiven by
!
0
~~2
~h=( r )eAj j:
j= ;
mmc
(a) We have
eAieAjee
[ pi pjpi Aj]Ai pj]
] = ; ; = ; [ ; [
i j
cccc
!
ihe@Ajihe@Aiihe@Aj@Ai
= ; = ;
c@xic@xjc@xi@xj
!
ihe
="ijkBk:(2.114)
c
We have also that
2 3 2 3
dxi~2~2
1 1 1
4 5 4 5
=x H]x +e =x
[ =
dtihiihimihim
2 2
1 1
=xi xi xi pj]xi pj]g
f[ ] + [ ]g = f[ + [
j j j
ih2mjih2mj j
2ih i
= ij = )
2ihmjm
64
2 3
" #
2
d2xidxi~
1 1
4 5
= H= +e
dt2ihdtihmim
2
1e
= ]
f[ ] + [ ]g + [
i j j j i j
ih2m2ihmi
" #
1iheiheeeAi
(2:114)
="ijkBk"ijkBkpi
+ + ;
j j
2m2ihccihmc
ee
="ikjBk "ijk Bkpi ]
(; + ) + [
j
2m2cjihm
exjxje@
=m"ijkBk"ikjBk ; )
;
2m2cdtdtm@xi
"x!~! #
d2xe~x
~~
m2ieEiB;B )
= +
dt2cdtidti
" !#
d2~1~~
x~x~~x
m2eE+B;B :(2.115)
=
dt2cdtdt
(b) The time-dependent Schr
odinger equation is
0 1
2
@ 1eA~
0 0 0
@ A
ihhxj t0 tixjHj t0 tixjp;e j t0 ti
=~+
h = h
@t2mc
2~3~3
2
0 0
1eA(x)eA(x)
0 0 0 0 0
4~~5~~5h~
4
=ihrihrxj t0 tie (x)hxj t0 ti
; ; ; ; +
2mcc
" #
1e~~e~~e2
0 0 0 0 0 0 0 0
~~
=h2rihrA(x)ihA(x)A2(x) (x t)
;~+~ ~~
r + r +
2mccc2
0 0
+e (x) (x t)
~~
1e~e~~
2
0 0 0 0 0 0 0 0
=h2r (x t)ihrA~ (x t)ihA(x) (x t)
;~+~+~ ~
r
2mcc
#
e~~e2
0 0 0 0 0 0 0
+ihA(x) (x t)A2(x) (x t)e (x) (x t)
~ ~+~~+~~
r
cc2
" #
1e~e~e2
2
0 0 0
=h2r +ihrA~ +2ihA~ +A2 +e :(2.116)
; r
2mccc2
Multiplying we get
the last equation by
@
ih =
@t
2. QUANTUM DYNAMICS 65
" #
1e~~2e~~e2
2
0 0 0 2 2
;h2 r +ihrAj jihA r +A2j je j j:
+2 +
2mccc2
The complex conjugate of this eqution is
@
;ih =
"@t #
1e~~2e~~e2
2
0 0 0 2 2
;h2 r ;ihrAj jihA r +A2j je j j:
; 2 +
2mccc2
Thus subtracting the last two equations we get
h i
h2
2 2
0 0
; r ; r
2m
e~2e~~
0 0 0
+ihrA~j jihA~ r + r )
+ (
mcmc
!
@@
=ih + )
@t@t
h i
h2e~2e~2
0 0 0 0 0
~~~
; r ; r +ihrA~j jihA~ j )
r + (r j
2mmcmc
@2
=ihj j )
@t
h i h i
@2h~~e~~2
0 0 0
j j r )Aj j )
= ; r =( + r
@tmmc
" #
@2~h~e~2
0 0
j j r )Aj j =0 )
+ r =( ;
@tmmc
@ ~~
0
+j=0 (2.117)
r
@t
0 e 2
~h~~2
withj= r )Aj j:and = j
=( ; j
m mc
2.14 An electron moves in the presence of a uniform magnetic eld
~
inz-directionB=Bz).
the ( ^
(a) Evaluate
[ ]
x y
where
eAxeAy
px py:
; ;
x
cyc
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
!
h2k2eBjh1
j
Ek nn+
= +
2mmc2
wherehkis thepznis a
continuous eigenvalue of the operator and
nonnegative integer including zero.
~ ~
TheB=Bzcan beA(x)
magentic eld ^ derived~
from a vector petential
of the form
ByBx
Ax Ay Az:(2.118)
= ; = =0
2 2
Thus we have
eAxeAyeByeBx
(2:118)
[ px pypx py ;
] = ; ; = +
x y
cc2c2c
eBeBiheBiheB
=px x]y py] = +
; [ + [
2c2c2c2c
eB
=ih: (2.119)
c
(b) The Hamiltonian for this system is given by
0~1
2
1eA1 1 1
2 2
@~A
H=p;p2H1H2 (2.120)
= + + = +
2mc2mxmymz
2 2
1 2 1 2 1
whereH1H2p2. Since
+ and
x y z
2m 2m 2m
" #
1eBy2eBx2
[H1 H2]px py p2 =0 (2.121)
= + + ;
4m2c2cz
2
there exists ak niH1 and
set of simultaneous eigenstates j of the operators
H2.hkis the continiouspzk ni its
So if eigegenvalue of the operator and j
eigenstate we will have
p2h2k2
z
H2jk nik ni:(2.122)
= j =
2m2m
2. QUANTUM DYNAMICS 67
OnH1 is similar to the Hamiltonian of the one-dimensional
the other hand
oscillator problem which is given by
1
1
H=p2m!2x2 (2.123)
+
2m2
withx p]ih. In order to use the eigenvalues of the harmonic oscillator
[ =
1
Enh!n+ we should have the same commutator between the squared
=
2
operators in the Hamiltonian. From (a) we have
eB c
x
[ ih) ih:(2.124)
] = [ ] =
x y
ceBy
SoH1 can be written in the following form
2
1c2eBj
1 1 1 j
x
2 2 2
H1 + = +
2mxmymymeBc2
2 2
!
2 2
1eBjc2
j
2 1
=mx:(2.125)
+
2
2mymceB
j eBj
In this form!with to have
it is obvious that we can replace
mc
!
h2k2eBjh1
j
Hjk niH1jk niH2jk nik nin+k ni
= + = j + j
2
"hk eBjh! m#mc2
2
2
j 1
=n+k ni:(2.126)
+ j
2mmc2
2.15mandqin an impenetrable
Consider a particle of mass charge
cylinderRanda. Along the axis of the cylin-
with radius height
der runs a thin, impenetrable solenoid carrying a magnetic ux .
Calculate the ground state energy and wavefunction.
~
InB= 0 the Schr
the case where odinger equation of motion in the
cylindrical coordinates is
h2
; [r2 = 2 )
m
h ]E i
2 2 2
h2 @ 1 @ 1 @ @
;x)E x)
+~=~(2.127)
+ + ( 2 (
2 2 2
m @ @ @ @z2
68
2mE
If z) )R( )Z(z)k2 = we will have
we write ( = ( and
h2
d2R1dRR( )Z(z)d2
( )Z(z) )Z(z) +
+ (
d 2 d 2d 2
d2Z
+R( ) )k2R( ) )Z(z) =0 )
( + (
dz2
1d2R1dR1d2 d2Z2
1
+ +k= 0(2.128)
+ +
R( )d 2R( ) d 2 )d 2Z(z)dz2
(
with a z)R z) 0) a) =0.
initial conditions ( = ( = ( = (
So
1d2Z2d2Z2
;
=l)lZ(z)Z(z)A1eilzB1eilz (2.129)
; + =0 ) = +
Z(z)dz2dz2
with
;
Z(0)A1B1Z(z)A1eilzeilzCsinlz
= 0 ) + =0 ) = ; =
Z(a)Csinla=0la=n )l=lnnn= ::
= 0 ) ) = 1 2 :
a
So
Z(z)Csinlnz(2.130)
=
Now we will have
1d2R1dR1d2
+k2l2 =0
+ + ;
R( )d 2R( ) d 2 )d 2
(
2
d2R dR1d2
) + 2(k2l2) = 0
+ + ;
R( )d 2R( )d )d 2
(
1d2
)m2 )eim :(2.131)
= ; ) ( =
( )d 2
with
( +2 ) )m2:(2.132)
= ( ) Z
So the Schr
odinger equation is reduced to
2
d2R dR2
+m+ 2(k2l2) =0
; ;
R( )d 2R( )d
2. QUANTUM DYNAMICS 69
" #
d2R1dRm2
)k2l2)R( ) =0
+ + ( ; ;
d 2 d 2
" #
d2R1dRm2
p p p
) + +R( ) = 0
1 ;
(k2l2) 2
;
d(k2l2 )2k2l2 d(k2l2 )
; ; ;
p p
)R( )A3Jm(k2l2 )B3Nm(k2l2 ) (2.133)
= ; + ;
In aB3Nm !1
the case at hand in which ! 0 we should take = 0 since
when ! 0. From the other boundary condition we get
p p
R(R)A3Jm(Rk2l2)Rk2l2 m (2.134)
= 0 ) ; =0 ) ; =
where m -thm-thJm. This means
is the zero of the order Bessel function
that the energy eigenstates are given by the equation
2
p
2mE 2 2
m m
=Rk2l2k2l2n=
; ) ; = ) ;
m
2
2 2
"RhaR
#
h2 2 2
m
)E=n (2.135)
+
2mR2a
while the corresponding eigenfunctions are given by
z
m
(x)AcJm( )eim n(2.136)
~= sin
nm
Ra
withn= :::andm2 Z.
1 2
~
NowB=Bz. We can then write
suppose that ^
! !
B 2
a
^
A~= ^ :(2.137)
=
2 2
~
The Schr presence of the magnetic eld
odinger equation in theBcan be
written as follows
2~3~3
2
1eA(x)eA(x)
~~
4ihr;~5ihr;~5 (x)E (x)
4~~
; ; =
2mcc
" !#
h2@@1@ie
) ^z+ ^ ;
; +^
2m
"@@ @@z @ @ hc2
!#
1ie
^z+ ^ (x)E (x):(2.138)
+^~=~
;
@ @z @ hc2
70
@ ie
MakingD ; we get
now the transformation
@ hc 2
" # " #
h2@@1@@1
; ^z+ ^D ^z+ ^D (x)E (x)
+^ +^~=~
2m@ @z @ @z
" #
h2@2@1@2
1
)D 2x)E x) (2.139)
; +~=~
+ + ( (
2m@ 2 @ 2@z2
2
@ ie e
whereD 2A= we get
= ; . Leting
@ hc 2 hc 2
! !
@2@@2@
2ie
D 2A2iA;A2:(2.140)
= ; ; = ; 2
@ 2hc2 @ @ 2@
Following the same ( z)R( ) )Z(z))
procedure we used before (i.e. = (
we will get the same equations with the exception of
"@@#d d
2 2
;iA;A2m2 iA+(m2A2) :
2 = ; ) ; 2 ; =0
@ 2@ d 2d
The solutionel . So
to this equation is of the form
l2el iAlel m2A2)el l2iAl+(m2A2)
; 2 +( ; =0 ) ; 2 ;
q
2iA m2A2)
;4A2iA im
; 4( ;
2 2
)l= =i(A m)
=
2 2
which means that
m)
( )C2ei(A:(2.141)
=
But
0 0
( +2 ) )A m=mm2 Z
= ( )
0 0
)m=m;A)m2:(2.142)
( Z
This means that the energy eigenfunctions will be
0 z
m
(x)AcJm( )eimn(2.143)
~= sin
nm
Ra
butmis not an integer. As a result the energy of the ground state will
now
be
" #
h2 2 2
m
+
E=n(2.144)
2mR2a
2. QUANTUM DYNAMICS 71
0 0
wherem=m;Ais not zerom2 Z
now in general but it corresponds to
0
suchm;A<1. Notice also that if we require the ground state to
that 0
beB, we obtain ux quantization
unchanged in the presence of
0
e hc
2 m
0 0 0
m;A=0m)m2:(2.145)
) = = Z
hc2 e
2.16 A particlein one dimension (;1
constant force derivable from
V= x ( >0):
(a) Is the energy spectrum continuous or discrete? Write down an
approximate expression for the energy eigenfunction speci ed by
E.
(b) Discuss brie yVis replaced be
what changes are needed if
V= jxj:
(a) In the case under construction there is only a continuous spectrum and
the eigenfunctions are non degenerate.
From the discussion on WKBE>V(x)
approximation we had that for
Z
q
Ai
(x)m[E;V(x)]dx
= exp 2
I
[E;V(x)]1=4h
Z
q
Bi
+m[E;V(x)]dx
exp ; 2
[E;V(x)]1=4h
!
Z
q
x0
c1
=m[E;V(x)]dx;
sin 2
[E;V(x)]1=4hx 4
p
!
Z 0
=E=
c2m xE1=2
= sinxdx;
;
[E;V(x)]1=4h 4
x
2 3
s
c2E3=2m
2
4 5
=x;
sin ; ;
[E;V(x)]1=4 h4
3
2
c1
=q3=2 + (2.146)
sin
[q]1=4 3 4
72
h i
1=3
2m
whereq= Exand = .
;
h2
OnEthe other hand when
!
Z
q
x
c2 1
(x) =m( x;E)dx
exp ; 2
II
[ x;E]1=4h
x0 =E=
" #
Z
q
x
c2 1
= expm( x;E)d(2m x)
; 2
[ x;E]1=4h2m
x0 =E=
c3 2
=q)2=3:(2.147)
exp ; (;
[;q]1=4 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
Hj iEj ipjHj ipjEj i
= ) h = h
p2
)pj xj iEhpj i
h + =
2m
p2d
) (p)ih (p)E (p)
+ =
2mdp
!
d;ip2
) (p)E; (p)
=
dph 2m
!
d (p)ip2
;
)E;dp
=
(p)h 2m
!
;ip3
) (p)Ep;c1
ln = +
h 6m
"i p!#
3
) E(p)cexpEp:(2.148)
= ;
h 6m
We also have
Z Z
0 0 0
(E;E)EjEidphEjpihpjEi E(p) Ep)dp
= h = = 0 (
Zi
(2:148)
2 0 2 0
=cjdpexpE;E)pjcj h (E;E) )
j ( 2
h
1
p
c=: (2.149)
2 h
So
" !#
1ip3
p
(p)Ep:(2.150)
= exp ;
E
2 h h 6m
2. QUANTUM DYNAMICS 73
These are the Hamiltonian eigenstates in momentum space. For the eigen-
functions in coordinate space we have
Z Z
i p3
1 ipx ;Ep
(2:150)
h 6m
h
p
(x)dphxjpihpjEidpee
= =
2
"
Z h #
1p3iE
p
=dpexpi;xp:(2.151)
;
2 h h 6mh
Using now the substitution
pp3u3
u= ) = (2.152)
(h2m )1=3h 6m3
we have
" #
Z
+1
(h2m )1=3iu3iE
p
(x)duexpxu(h2m )1=3
= ; ;
;1
2 h 3h
"
Ziu#
3
+1
p ;
=duexpiuq (2.153)
;1
2 3
h i
1=3
2m
where =q= Ex. So
and ;
h2
Zu! Zu!
3 3
+1 +1
(x)ducosuq=uqdu
= p ; p cos ;
;1 0
2 3 3
R
+1
u3
sinceuqdu= 0. In terms of the Airy functions
sin ;
;1
3
!
Z
+1
1u3
Ai(q)uqdu(2.154)
= p cos ;
0 3
we will have
p
(x)Ai(;q):(2.155)
=
Forqj , leading terms in the asymptotic series are as follows
large j
1 2
Ai(q)q3=2 q>0 (2.156)
p exp ;
2 q1=4 3
2
1
Ai(q)q)3=2 q<0 (2.157)
p sin (; +
(;q)1=4 3 4
74
Using these approximations in (2.155) we get
1
2
(q)q3=2 forE>V(x)
p sin +
q1=4 3 4
1 2
p
(q)q)3=2 forE exp ; (;
3
2 (;q)1=4
as expected from the WKB approximation.
(b)V= jxj we have bound states and therefore the energy spec-
When
trum is discrete. So in this case the energy eigenstates heve to satisfy the
consistency relation
Z
q
x1
1
dx2m[E; jxjn+ h n=0 1 2 :::(2.159)
] =
2
x2
E E
Thex1x2 = . So
turning points are = ; and
Z Z
q q
E= E=
1
n+ h=dx2m[E; jxjm[E; x]dx
] = 2 2
2
;E= 0
ZE
1=2
p E=
=m ;xd(;x)
;2 2
0
E=
p p
2E3=2E3=2
2
=m ;x m )
;2 2 =2 2
3 0
3
1 1
2=3
3
E3=2n+ hE[3n+ h]
2 2
= p ) = )
42=3(2m )1=3
4
2n2m h 3
2=3
1
3 +
2
4 5
En:(2.160)
= p
4m
2
3. THEORY O F ANGULAR MOMENTUM 75
3 Theory of Angular Momentum
3.1 Consider a sequence of Euler rotations represented by
!
;i 3 ;i 2 ;i 3
D(1=2)( ) = exp exp exp
2 2 2
!
; + )=2 ; ;
ei( ei( )=2 sin
cos ;
2 2
= :
; )=2
ei( ei( + )=2 cos
sin
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
!
; + )=2 ; ;
ei( ei( )=2 sin
cos ;
2 2
D(1=2)( ) = (3.1)
; )=2
ei( ei( + )=2 cos
sin
2 2
while the same rotation will be represented by
0insin in;n) 1
cos ; (;
(S;3:2:45) z x y
2 2 2
@in+n)insin A:(3.2)
D(1=2)( n) =
^
(; sin cos + sin
x y z
2 2 2
Since these two operators must have the same e ect, each matrix element
should be the same. That is
! !
; + )=2
ei( inz sin
cos = cos ;
2 2
! + ) 2
(
) cos =cos cos
2 2 2
( + )
) =2 cos2 cos2 ; 1
cos
2
" 2 #
( + )
) =:(3.3)
arccos 2cos2 cos2 ; 1
2 2
3.2 An angular-momentumj m=mmaxji is rotated
eigenstate j =
by"abouty-axis. Without using the
an in nitesimal angle the
76
explicitd(j) function, obtain an expression for the
form of the
m0 m
probability for the new rotated state to be found in the original
state"2.
up to terms of order
The rotated state is given by
iJy"
jj jiRR(" y)jj jid(j)(")jj jij ji
= ^ = = exp ; j
"iJ"(;i)"#h
2 2
y
2
=Jyj ji (3.4)
1 ; + j
h2
2h
up"2.Jy in terms of the ladder operators
to terms of order We can write
)J;J
J+JxiJy
= +
+ ;
)Jy:(3.5)
=
J=JxiJy
;
; 2i
Subtitution of this in (3.4), gives
"""#
2
jj jiRJ+J)J+J)2j ji (3.6)
= 1 ; ( ; + ( ; j
; ;
2h2
8h
We know that for the ladder operators the following relations hold
q
J+jj mih(j;m)(j+m+1)jj m+1i (3.7)
=
q
Jjj mih(j+m)(j;m+1)jj m; 1i (3.8)
=
;
So
q
(J+J)jj jiJjj jih2jjj j; 1i (3.9)
; = ; = ;
; ;
q
(J+J)2jj jih2j(J+J)jj j; 1i
; = ; ;
; ;
q
=h2j(J+jj j;Jjj j; 1i)
; 1i ;
;
q q q
=h2j2jjj jij;j j; 2i
; ; 2(2 1)j
and from (3.6)
q q
""2"2
jj jiRj jijjj j;jjj jij(2j;j j; 2i
= j + 2 1i ; 2 + 2 1)j
2 8 8
!
q q
"2""2
=jjj jijjj j;j(2j;j j;:
1 ; + 2 1i + 1)j 2i
4 2 4
3. THEORY O F ANGULAR MOMENTUM 77
Thus the probability for the rotated state to be found in the original state
will be
2
2
2 1 ;"2 ! =1 ;
jj jjj jiRjj "j+O("4):(3.10)
h =
4 2
3.3 The wave function of a particle subjected to a spherically
symmetricalV(r) is given by
potential
(x)x+y+3z)f(r):
~=(
~
(a) anL?l-value? If
Is eigenfunction of If so, what is the
~
not, whatlweL2 is
are the possible values of may obtain when
measured?
(b) What are the probabilities for the particle to be found in various
ml states?
(c) Suppose (x) an energy eigenfunc-
it~is
is known somehow that
tionE.V(r).
with eigenvalue Indicate howwe may nd
(a) We have
(x)xj ix+y+3z)f(r):(3.11)
~ ~=(
h
So
" !#
1 1
(S;3:6:15)@2@@
hxjL2j ih2 (x):(3.12)
~~= ; +~
sin
sin2 @ 2 @ @
sin
If (x) terms of spherical coordinates (
we~inx=rsin cos y=
write
rsin sin z=rcos ) we will have
(x)rf(r) cos +sin sin + ):(3.13)
~= (sin 3 cos
Then
1@2rf(r) @rf(r)
sin
(x) ; ) + )
~= (cos sin = ; (cos sin (3.14)
sin2 @ 2 @ sin
sin2
78
and
!
h;3 sin i
1@@rf(r)@2
sin (x) +(cos +sin ) cos =
~= sin
sin @ @ sin @
h;6sin cos +(cos +sin )(cos ; )(3.15)
i
rf(r)
2
sin2:
sin
Substitution of (3.14) and (3.15) in (3.11) gives
1
hxjL2j ih2rf(r) +sin )(1 +sin2 )
~~= ; ; (cos ; cos2 ; 6 cos
sin
1
=h2rf(r) (cos +sin )
2sin2 +6 cos
sin
=h2rf(r) cos +sin sin +3 ]h2 (x)
2 [ sin cos~)
= 2
L2 (x)h2 (x)h2 (x)l(l+l)h2 (x)
~=~=~=~(3.16)
2 1(1 + 1)
~
which (x)L2l=1.
means~is en eigenfunction of with eigenvalue
that
(b)l= (x) terms of
Since we already know that~in
1 we can try to write
theY1m( ). We know that
spherical harmonics
szs4
s
3 3
Y10 =z=rY10
= cos )
4 4 r3
q 9 8 q
(x+iy)
3 2 ;1 +1
= <
Y1+1 = ;
1 1
8 r 3
qx=rqY;Y
)
(x;iy)
;1 3 ;1
:
Y1y=ir2 Y1Y1+1
= +
8 r 3
So we can write
s2 h p
i
;1 ;1
(x)rf(r)Y10Y1Y1+1iY1+1iY1
~= 3 2 + ; + +
s2 3
h i
p
;1
=rf(r)Y10i)Y1i;Y1+1:(3.17)
3 2 +(1 + +( 1)
3
But this means that the part of the statem
that depends on the values of
can be written in the following way
h i
p
j imN3l=1 m=0ii)jl=1 m=i)jl=1 m=1i
= 2j +(1 + ;1i +(1 ;
and if we want it normalized we will have
1
2
p
jNjN=:(3.18)
(18 + 2 + 2) = 1 )
22
3. THEORY O F ANGULAR MOMENTUM 79
So
9 2 9
2
P(m=0)l=1 m=0j ij (3.19)
= j h = =
22 11
2 1
2
P(m=l=1 m=+1j ij (3.20)
+1) = j h = =
22 11
2 1
2
P(m=l=1 m== ij:(3.21)
;1) = j h 1j = =
22 11
(c) E(x) an energy eigenfunction then it solves the Schr equation
If~is odinger
" #
;h2@2@L2
2
(x) (x) E(x)V(r) E(x)
~+~;~+~=
E
2m@r2r@rE h2r2
E E(x)
~
" #
;h2d2d2
2
)Ylmrf(r)]rf(r)]rf(r)]V(r)rf(r)Ylm =
[ + [ ; [ +
2mdr2rdrr2
Erf(r)Ylm )
" #
1h2d2 2
0 0
V(r)E+f(r)rf(r)]f(r)rf(r)]f(r) )
= [ + + [ + ;
rf(r)mdrrr
2
1h2
0 0 00 0
V(r)E+f(r)f(r)rf(r)f(r)]] )
= [ + + + 2
rf(r)m
2
00 0
h2rf(r)r)
+4f(
V(r)E+: (3.22)
=
2mrf(r)
3.4 Consider a particle with an intrinsic angular momentum (or
spin)h. (One example%-
of one unit of of such a particle is the
meson). Quantum-mechanically, such a particle is described by a
ketvector%i~representation wave function
jxa
or in
%i(x)x ij%i
~=~
h
wherex ii~withi:th di-
j~correspondxspin in the
to a particle at
rection.
(a) Show explicitly%i(x) obtained
that~are
in nitesimal rotations of
by acting with the operator
~
"~~
u"i L+ S) (3.23)
=1 ; (
~
h
80
~h~~
whereL=r r. Determine S !
^
i
~~
(b)Land S commute.
Show that
~
(c) Showthat S is a vector operator.
~1~
(d)%(x)p)%wherepis momentum oper-
Show~~=~~~the
that r (S
h2
ator.
(a) We have
Z Z
3 3
X X
j%ix iihx ij%ix ii%i(x)d3x:(3.24)
=~~=~~
j j
i=1 i=1
UnderRwe will have
a rotation
Z
3
X
0
j%iU(R)j%iU(R)xiii]%i(x)d3x
=~ ~
= [j j
i=1
Z
3 3
XZxi~X~(x)
(1) R=1 (1)
;
=R~ iiDilR)%l(x)d3xdet=x iiDilR)%l(R1~d3x
j j ( j
i=1 i=1
3
XZ~
0
=x ii%i~d3x)
jx)
i=1
(1)
0 ; 0 ;
%ix)R)%l(R1~)%(x)R~(R1~:(3.25)
(~=x)~~=%x)
Dil (
Under an in nitesimal rotation we will have
R( n)~=~+ ~=~+ (^~)~+~ ~(3.26)
^rrrrn r=r"r:
So
0 ;
%(x)R( )%(R1~=R( )%(x;~ ~
~~=~x)~~"x)
=%(x;" ~+~ %(x;" ~:(3.27)
~~~x)"~~~x)
On the other hand
~
~~~x)~~;~x)~=~~;~(~r)~~
%(x;" ~=%(x)" ~ %(x)%(x)" x %(x)
( r~~
i
=%(x)~ L~(x)
~~;"~%~(3.28)
h
3. THEORY O F ANGULAR MOMENTUM 81
hr%( i
~~
where%(x)x)ii. Using this in (3.27) we get
r~~ ~j
i
ii
0
%(x)%(x)~ L~(x)~ %(x)~ L~(x)
~~=~~;"~%~+"~~;"~%~
hh
i
=%(x)~ L~(x)~ %(x):(3.29)
~~;"~%~+"~~
h
But
~~; ^ + ; ^ + ; ^
" %="y%3"z%2ex"z%1"x%1ey"x%2"y%1ez (3.30)
or in
0matrix form 1 0 1
1 0
10 1
;
z y
B%C"0""C%C
B B
0
= 0 ; @ A
@%2"x%2 =
A @ z
A
30 3
; 0
y x
2%0"1"0%1%1
0 130
1
0 0 0 0 0 1 0 ;1 0
6"B"B"B%C
C C C
2
4 @ 0 0 ;1 + @ 0 0 0 + @ 1 0 0 @ A =
A A A7
5B
x y z
0 1 0 ;1%3
0 0 0
2 0 1 0 10 00 130 1
1
0 0 0 0 ;
i0ih0ih0%
6 B C B C B C
;"xih+"y"zih0%2
4 @ 0 0 ; A @ 0 0 0 + @ 0 @ A
A A7 C
5B
h0ih0ih0%3
; 0 0 0 0
which means that
i~
~~;~S~~
" %=" %(x)
h
withih kl.
(S )kl = ;
Thus we will have that
i~~i~~
0
%(x)U"%(x)~ L+%(x)U"" L+:(3.31)
~~=~~="(~~)~( S)
1 ; S) =1 ;
~
h~h
~~~
(b) From theirLandLacts
de nition it is obvious that S commute since
~
onlyxiii.
on~basis and S only on j
the j
~
(c) S is a vector operator since
X
[Si Sj]kmSiSjSjSi]kmih) ikl(;ih) jlmih) jkl(;ih) ilm]
= [ ; = [(; ; (;
h i
X
=h2 ikl jmlh2 jkl iml
;
X
=h2 ij km im jk ij km jm ki)
( ; ; +
X
=h2 jm ki im jk)
( ;
X X X
=h2 ijl kmlih ijl(;ih kml)ih ijl(Sl)km:(3.32)
= =
82
(d) It is
~i~~~i~1 ( )l
r %(x)p %(x) i p %l(x)jiiS p %mjii
~~= = =
hhl h2 m
1
~
=S p)%: (3.33)
(~~
h2
3.5 We arej1j2 = 1 to form
to add angular momenta = 1 and
j=2 1 and 0 states. Using the ladder operator method express all
(nine)j meigenketsj1j2 m1m2i. Write your answer as
in terms of j
1 1
p p
jj=1 m=1i +i ::: (3.34)
= j + 0i ; j 0
2 2
wherem1 2 0 respectively.
+ and 0 stand for =1
We wantj1j2j=
to add the angular momenta = 1 and = 1 to form
jj1j2j ::: j1j2 1 2j=2,m=2.
; + =0 states. Let us take rst the state
Thisj1m1 j2m2i through the following equation
state is related to j
X
jj mij1j2 m1m2jj1j2 jmijj1j2 m1m2i (3.35)
= h
m=m1 +m2
Soj=2,m= 2 in (3.35) we get
setting
norm.
jj=2 m=2ij1j2 j1j2 jmij ++i = j ++i (3.36)
= h + + j
IfJoperator on this statet we will get
we apply the
;
Jjj=2 m=2iJ1J2 )j ++i
=( +
; ; ;
q
)h(j+m)(j;m+1)jj=2 m=1i =
q q
h(j1m1)(j1m1h(j2m2)(j2m2 +1)j +0i
+ ; +1)j 0+i + + ;
p p p
)j=2 m=1i = 2j 0+i + 2j +0i
4j
1 1
p p
)j=2 m=1i:(3.37)
j = j 0+i + j +0i
2 2
3. THEORY O F ANGULAR MOMENTUM 83
In the same way we have
1 1
p p
Jjj=2 m=1iJ1J2J1J2 )j +0i )
= ( + )j 0+i + ( +
; ; ; ; ;
2
h i2 h i
p p p p p
1 1
p p
6jj=2 m=0i = 2j ; +i + 2j 00i + 2j 00i + 2j + ;i )
2 2
p
6jj=2 m=0i = 2j + j ; +i + j + ;i )
s00i
2 1 1
p p
jj=2 m=0i = j 00i + j + ;i + j ; +i (3.38)
3
6 6
s2
Jjj=2 m=0iJ1J2 )j 00i
= ( +
; ; ;
3
1 1
p p
+J1J2J1J2 )j ; +i )
( + )j + ;i + ( +
; ; ; ;
6
s2 6
h i
p p p p p
1 1
p p
6jj=2 m= ;1i = 2j ; 0i + 2j 0;i + 2j 0;i + 2j ; 0i )
3
6 6
p p p p
2 2 1 1
jj=2 m= ;1i = 2j 0;i + 2j ; 0i + 2j 0;i + 2j ; 0i )
6 6 6 6
1 1
p p
jj=2 m= ;1i = j 0;i + j ; 0i (3.39)
2 2
1 1
p p
Jjj=2 m=J1J2J1J2 )j ; 0i )
;1i = ( + )j 0;i + ( +
; ; ; ; ;
2 2
p p p
1 1
p p
4jj=2 m= ;2i = 2j ; ;i + 2j ;;i )
2 2
jj=2 m=: (3.40)
;2i = j ;;i
Now letj=1,m= 1 we will have
us return to equation (3.35). If
jj=1 m=1iajbj 0+i (3.41)
= +0i +
This statej mij=
should be orthogonal to all j states and in particular to j
2 m=1i. So
1 1
p p
hj=2 m=1jj=1 m=1ia+b=0 )
=0 )
2 2
a+b=0a=b:(3.42)
) ;
84
Inj=1 m=1i should be normalized so
addition the state j
(3:42) 1
2 2 2
hj=1 m=1jj=1 m=1iajbjajaj:
=1 )j + j =1 ) 2j =1 )j = p
2
1 1
p p
Byatoa=b= ; .
convention we take be real and positive so and
2 2
That is
1 1
p p
jj=1 m=1i:(3.43)
= j +0i ; j 0+i
2 2
Using the same procedure we used before
1 1
p p
Jjj=1 m=1iJ1J2J1J2 )j 0+i )
= ( + )j +0i ; ( +
; ; ; ; ;
2 2
h i h i
p p p p p
1 1
2jj=1 m=0i = p 2j 00i + 2j + ;i ; p 2j ; +i + 2j 00i )
2 2
1 1
p p
jj=1 m=0i = j + ;i ; j ; +i (3.44)
2 2
1 1
p p
Jjj=1 m=0iJ1J2J1J2 )j ; +i )
= ( + )j + ;i ; ( +
; ; ; ; ;
2 2
p p p
1 1
p p
2jj=1 m= ;1i = 2j 0;i ; 2j ; 0i )
2 2
1 1
p p
jj=1 m=:(3.45)
;1i = j 0;i ; j ; 0i
2 2
Returning backj=0 m=0i can be written
to (3.35) we see that the state j
as
jj=0 m=0ic1jc2jc3j:(3.46)
= 00i + + ;i + ; +i
This state shouldj mij=
be orthogonal to all states j and in particulat to j
2 m=0ij=1 m=0i. So
and to
s2 1
1
p p
hj=2 m=0jj=0 m=0ic1c2c3
=0 ) + +
3
6 6
)c1c2c3 =0 (3.47)
2 + +
1 1
p p
hj=1 m=0jj=0 m=0ic2c3
=0 ) ;
2 2
)c2c3: (3.48)
=
3. THEORY O F ANGULAR MOMENTUM 85
Using the last relation in (3.47), we get
2c1c1c2c1c2:(3.49)
+2c2 =0 ) + =0 ) = ;
Thej=0 m=0i should be normalized so
state j
2 2 2 2
hj=0 m=0jj=0 m=0ic1jc2jc3jc2j =1
=1 )j + j + j =1 ) 3j
1
p
)c2j: (3.50)
j =
3
1
p
Byc2c2c3 = and
convention we take to be real and positive so =
3
1
p
c1 = ; . Thus
3
1 1 1
p j + ;i + j ; +i ; j 00i
p p
jj=0 m=0i =:(3.51)
3 3 3
So gathering all the previous results together
jj=2 m=2i = j ++i
1 1
p p
jj=2 m=1i = j 0+i + j +0i
2 2
q
2 1 1
p p
jj=2 m=0i = j 00i + j + ;i + j ; +i
3
6 6
1 1
p p
jj=2 m= ;1i = j 0;i + j ; 0i
2 2
jj=2 m= ;2i = j ; ;i (3.52)
1 1
p p
jj=1 m=1i = j +0i ; j 0+i
2 2
1 1
p p
jj=1 m=0i = j + ;i ; j ; +i
2 2
1 1
p p
jj=1 m= ;1i = j 0;i ; j ; 0i
2 2
1 1 1
p p p
jj=0 m=0i =:
j + ;i + j ; +i ; j 00i
3 3 3
3.6 (a) Construct a spherical tensor of rank 1 out of two di erent
~~
vectorsU=Ux Uy Uz)V=(Vx Vy Vz).T(1) in
( and Explicitly write
1 0
termsUx y zVx y z .
of and
(b) Construct a spherical tensor of rank 2 out of two di erent
~~
vectorsUandV.T(2)Ux y z
Write down explicitly in terms of
2 1 0
andVx y z.
86
~~
(a)UandVare vector operators they will satisfy the following com-
Since
mutation relations
[Ui Jj]ih"ijkUkVi Jj]ih"ijkVk:(3.53)
= [ =
From the components of a vector operator we can construct a spherical tensor
of rank 1 in the following way. The de ning properties of a spherical tensor
of rank 1 are the following
q
(1)
(1) (1) (1)
[Jz UqhqUq [J Uqh(1q)(2q)Uq:(3.54)
] = ] =
1
It is
(3:53) (3:54)
[Jz Uz]hUz )
= 0
UzU0 (3.55)
=
p
(3:54)
[J+ U0]hU+1J+ Uz]JxiJy Uz]
= 2 =[ = [ +
(3:53)
=ihUyi(ih)Uxh(UxiUy) )
; + = ; +
1
p
U+1UxiUy) (3.56)
= ; ( +
2
p
(3:54)
[J U0]hU1J Uz]JxiJy Uz]
= 2 =[ = [ ;
; ; ;
(3:53)
=ihUyi(ih)Uxh(UxiUy) )
; ; = ;
1
p
U1UxiUy) (3.57)
= ( ;
;
2
~~
SoUandVwe can construct spherical tensors
from the vector operators
with components
U0UzV0Vz
= =
1 1
p p
U+1UxiUy)V+1VxIVy)
= ; ( + = ; ( +
(3.58)
2 2
1 1
p p
U1UxiUy)V1VxiVy)
= ( ; = ( ;
; ;
2 2
(k1 ) (k2)
ItXqZq are irreducible spherical tensors
is known (S-3.10.27) that if and
1 2
ofk1k2 respectively then we can construct a spherical tensor of
rank and
rankk
Xh
(k1 ) (k2
Tq(k)k1k2 q1q2jk1k2 kqiXqZq ) (3.59)
=
1 2
q1 q2
3. THEORY O F ANGULAR MOMENTUM 87
In this case we have
(1)
T+1U+1V0U0V+1
= h11 +10j 11 11i + h11 0 + 1j 11 11i
1 1
(3:52)
p p
=U+1V0U0V+1
;
2 2
1 1
=UxiUy)VzUz(VxiVy) (3.60)
; ( + + +
2 2
(1)
T0U0V0U1V+1
= h11 00j 11 10i + h11 ;1 + 1j 11 10i
;
+h11 U+1V1
+1 ; 1j 11 10i
;
1 1
(3:52)
p p
=U1V+1U+1V1
; +
; ;
2 2
1 1
1 1
p p
=UxiUy)(VxiVy)UxiUy)(VxiVy)
( ; + ; ( + ;
2 2
2 2
1
p [ + ; + ; + ; ;
=UxVxiUxVyiUyVxUyVyUxVxiUxVyiUyVxUyVy]
2 2
i
p ( ;
=UxVyUyVx) (3.61)
2
T(1)U1V0U0V1
= h11 ;10j 11 11i + h11 0 ; 1j 11 11i
; ;
;1
(3:52) 1 1
=U1V0U0V1
; p + p
; ;
2 2
1 1
=UxiUy)VzUz(VxiVy):(3.62)
; ( ; + ;
2 2
(b) In the same manner we will have
(3:52)
(2)
1
T+2U+1V+1U+1V+1UxiUy)(VxiVy)
= h11 +1 + 1j 11 2 + 2i = = ( + +
2
1
=UxVxUyVyiUxVyiUyVx) (3.63)
; ( ; + +
2
(2)
T+1U0V+1U+1V0
= h11 0 + 1j 11 2 + 1i + h11 +10j 11 2 + 1i
(3:52) 1 1
p + p
=U0V+1U+1V0
2 2
1
=UzVxUxVziUzVyiUyVz) (3.64)
; ( + + +
2
(2)
T0U0V0U1V+1
= h11 00j 11 20i + h11 ;1 + 1j 11 20i
;
+h11 U+1V1
+1 ; 1j 11 20i
;
88
s2 s1 s1
(3:52)
=U0V0U1V+1U+1V1
+ +
; ;
6
s3 s6 s1 1
2 1
1 1
p
=UzVzUxiUy)(VxiVy)UxiUy)(VxiVy)
; ( ; + ; ( + ;
2 2
6 6
2
s3
1ii
1
=UzVzUxVxUxVyUyVx
2 ; ; +
2
6 2 2
ii1
1 1
;UyVyUxVxUxVyUyVxUyVy
; + ; ;
2 2 2
2 2
s1
=UzVzUxVxUyVy) (3.65)
(2 ; ;
6
T(2)U0V1U1V0
= h11 0 ; 1j 11 2 ; 1i + h11 ;10j 11 2 + 1i
; ;
;1
1 1
(3:52)
p p
=U0V1U1V0
+
; ;
2 2
1
=UzVxUxVziUzVyiUyVz) (3.66)
( + ; ;
2
(3:52)
1
T(2)U1V1U1V1UxiUy)(VxiVy)
= h11 ;1 ; 1j 11 2 ; 2i = = ( ; ;
; ; ; ;
;2
2
1
=UxVxUyVyiUxVyiUyVx):(3.67)
( ; ; ;
2
3.7 (a) Evaluate
j
Xd( )jm
(j)
2
j
mm0
m=;j
1
forj(integer or half-integer) j= .
any then check your answer for
2
(b)j,
Prove, for any
j
Xmjd( )jj(j+1) +m+ ; 1)
(j) 2
2 2 1 0 1
= sin:
(3 cos2
m0 m
2 2
m=;j
[Hint: This can be proved in many ways. You may, for instance,
2
examineJz using the spherical (irre-
the rotational properties of
ducible) tensor language.]
3. THEORY O F ANGULAR MOMENTUM 89
(a) We have
j
Xd( )jm
(j)
2
j
mm0
m=;j
j
X
; =h 0 2
y
=mjjmjeiJjmij
h j
m=;j
j
X
; =h 0 ; =h 0
y y
=mhjmjeiJjmijmjeiJjmi
j h j
m=;j
j
X
; =h 0 0 =h
y y
=mhjmjeiJjmihjmjeiJjmi
j j
m=;j
j
X
0 =h ; =h 0
y y
=jmjeiJmjjmihjmjeiJjmi
h j
m=;j
2 3
j
X
1
0 =h ; =h 0
y y
4 5
=jmjeiJJzjmihjmjeiJjmi
h j j
hj
m=;
1
0 =h ; =h 0
y y
=jmjeiJJzeiJjmi
h j
h
1
0 0
=jmj ey)JzD( ey)jjmi:(3.68)
h D ( ^ ^
h
ButJ~is a vector operator so from (S-3.10.3) we will have
the momentum
that
X
D ey)JzD( ey)Rzj( ey)Jj:(3.69)
( ^ ^ = ^
j
On the other hand we know (S-3.1.5b) that
0 0 1
cos sin
B C
R( ey):(3.70)
^ = @ 0 1 0 A
; 0
sin cos
So
j
Xd( )jm= hjmjJjjmi hjmjJjjmi]
1
(j)
2 0 0 0 0
j [; sin + cos
x z
mm0
h
m=;j
1J+J
+
;
0 0 0
= hjmjjmihmcos
; sin j +
h2
0
=mcos :(3.71)
90
Forj=1=2 we know from (S-3.2.44) that
!
cos ; sin
2 2
d(1=2)( ):(3.72)
=
mm0
sin cos
2 2
0
Som=1=2
for
1=2
Xd( )jm= ; sin2 + cos2
(j)
2 1 1
j
m1=2
2 2
2 2
m=;1=2
1 0
= =mcos (3.73)
cos
2
0
whilem==2
for ;1
1=2
Xd( )jm= ; cos2 + sin2
(j)
2 1 1
j
m1=2
2 2
2 2
m=;1=2
1 0
= =mcos :(3.74)
; cos
2
(b) We have
j
Xmjd( )j
(j)
2 2
m0 m
m=;j
j
X
0 ; =h 2
y
=m2jjmjeiJjmij
h j
m=;j
j
X
0 ; =h 0 ; =h
y y
=m2hjmjeiJjmijmjeiJjmi
j h j
m=;j
j
X
0 ; =h =h 0
y y
=m2hjmjeiJjmihjmjeiJjmi
j j
m=;j
j
X
0 ; =h =h 0
y y
=jmjeiJm2jjmihjmjeiJjmi
h j
m=;j
2 3
j
X
1
0 ; =h 2 =h 0
y y
4 5
=jmjeiJJzjmihjmjeiJjmi
h j j
h2
m=;j
1
0 ; =h 2 =h 0
y y
=jmjeiJJzeiJjmi
h j
h2
1
0 2 y 0
=jmj ey)Jz ey)jjmi:(3.75)
h D( ^ D ( ^
h
3. THEORY O F ANGULAR MOMENTUM 91
From (3.65) we know that
s1
(2)
2
T0JzJ2) (3.76)
= (3 ;
6
(2)
whereT0 is the 0-component of a second rank tensor. So
p
6 1
(2)
2
JzT0J2 (3.77)
= +
3 3
y y
andR)J2DR)J2D(R)DR)J2 we will have
since D( ( = ( =
Pmjd( )j =
j (j)
2 2
m=;j mm0
s2 1
1 1
0 0 0 2 y 0
hjmjJ2jjmijmj ey)Jz ey)jjm(3.78)
+ h:
D( ^ D ( ^ i
3 3
h3h2
We know that for a spherical tensor (S-3.10.22b)
k
X
(
y
D(R)Tq(k)DR)R)Tq(k) (3.79)
( = Dqk)(
0 0
q
q0 =;k
which means in our case that
2
X
(2)
0 2 y 0 0 0
hjmj ey)Jz ey)jjmijmjTq(2)Dq ey)jjmi
D( ^ D ( ^ = h ( ^
0 0
0
q0 =;2
2
X
(2)
0 0
= ey)hjmjTq(2)jjmi:(3.80)
Dq 0( ^
0 0
q0 =;2
0 0
But we know fromjmjTq(2)jmi =0. So
the Wigner-Eckart theorem that h j
0
6
=0
j
Xmjd( )j
(j)
2 2
mm0
m=;j
s2
1 1
(2) (2)
0 0
=h2j(j+1) ey)hjmjT0jmi
+ D00 ( ^ j
3h2h2 3
1 1 1
0 2 0
=j(j+1)d(2)( )hjmjJzJ2jjmi
+ ;
00
3 2 3
1 1 1
0
=j(j+1)d(2)( )m2j(j+1)
+ ;
00
3 2 3
92
1 1 1
0
=j(j+1) ;m2j(j+1)
+ (3 cos2 1) ;
3 2 3
0
1 1m2
1
=j(j+ +j(j+1)j(j+1) ; 1)
; 1) cos2 + + (3 cos2
2 6 3 2
1 0 1
=j(j+ +m2 ; 1) (3.81)
1) sin2 (3 cos2
2 2
1
whered(2)( )P2(cos ) ; 1).
we have used = = (3 cos2
00
2
3.8xy,xz,x2y2) as components of a spherical
(a) Write and ( ;
(irreducible) tensor of rank 2.
(b) The expectation value
Q eh j m=jjz2r2)j j m=ji
(3 ;
is known as the quadrupole moment. Evaluate
0
eh j mjx2y2)j j m=ji
( ;
0
(wherem=j j; j; :::)inQand appropriate Clebsch-
1 2 terms of
Gordan coe cients.
~
(a) Using the relations (3.63-3.67) weU=
can nd that in the case where
~
V=~the
xcomponents of a spherical tensor of rank 2 will be
(2)
1 1
T+2x2y2)ixyT(2)x2y2)ixy
= ( ; + = ( ; ;
;2
2 2
(2)
T+1xz+izy)T(2)xz;izy(3.82)
= ;( =
;1
q q
(2)
1 1
T0z2x2y2)z2r2)
= (2 ; ; = (3 ;
6 6
So from the above we have
(2) (2)
T+2T(2)T(2)T+1
; ;
(2)
;2 ;1
x2y2T+2T(2) xy= xz=:(3.83)
; = +
;2
2i2
(b) We have
Q=eh j m=jjz2r2)j j m=ji
(3 ;
3. THEORY O F ANGULAR MOMENTUM 93
p p
h jkT(2)k ji
(3:82) (W:;E:)
(2)
=eh j m=jjT0 j m=jij2 j0jj2 jjie
6 j = h p 6
2j+1
p
Q2j+1
p
) jkT(2)k ji:(3.84)
h =
hj2 j0jj2 jji
6e
So
0
eh j mjx2y2)j j m=ji
( ;
(3:83)
(2)
0 0
=eh j mjT+2 j m=jieh j mjT(2)j j m=ji
j +
;2
0
z jkT(2)
}| ji jkT(2)
{h
k h ji
k
0
p + 0 h 2j
=ehj2 j2jj2 jmie mj2 j;j2 jj; 2i p
j;2
2j+1 2j+1
Qhj2 j ;2jj2 j j; 2i
(3:84)
p 0
= m:(3.85)
j;2
hj2 j 0jj2 j ji
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 functiont= 0 is given by
for a plane-wave state at
p x=h
aei~ ~ . Why does this not violate time-reversal
complex function
invariance?
(a)ni in a nondegenerate energy eigenstate. Then
Suppose that j
H niHjniEnjniniei ni
j = = ) j = j
; ; =h
n
)n t0tieitH=hjnieitEni =
j =0 = = j
E t 2E t
n n
=h + +
n ) )j
h h
eitEniei(niei(n t0ti
j = j =
Z Z=0
2E t
n
+
)
h
)d3xjxihxjn t0tiei(d3xjxihxjn t0ti
~~j~~j =0
=0 =
Z Z
2E t
n
+
)h
h
)d3xhxjn t0tixid3xei(xjn t0tijxi
~=0 ~=~=0 ~
j
2E t
n
+
)
h
) n(x t)ei( n(x t): (4.1)
~=~
2Ent
So if = ; the wave function will be
we choose at any instant of time
h
real.
(b) In the case of a free particle the Schr
odinger equation is
p2h2
~
jniEjnix)E n(x)
= ) ; r n( =
2m2m
p x=h ; p x=h
~
) n(x)Aei~Bei~ ~ (4.2)
= +
; p x=h 0 p x=h
~
The n(x)ei~ n(x)ei~ ~ correspond to the
wave functions = and =
p2
sameE= and so there is degeneracy since these correspond
eigenvalue
2m
topipi. we cannot apply the previous result.
di ~and~So
erent state kets j j ;
0
4.2 (p) the momentum-space i,
Let~be wave function for state j
0 0
that (p)pj i.Is momentum-space wave function for the
is,~=~the
h
4. SYMMETRY IN QUANTUM MECHANICS 95
0 0 0 0
time-reversed i (p), (;p), (p), (;p)?
state~~~or~
j given by
Justify your answer.
In the momentum space we have
Z Z
0 0 0 0 0 0
j id3phpj ijpi id3p (p)jpi
=~~)j~~
=
Z Z
0 0 0 0 0 0
) id3p pj ijpi]d3phpj ipi:(4.3)
j~~=~ ~
= [h j
For the momentum it is natural to require
h jpj i jpj ~ )
~=~i
;h ~
; ;
h ~p ip p(4.4)
j~1j~1~
~ ) = ;
So
(4:4)
0 0 0 0
pjpip pipipi
~~=~j~)~=~(4.5)
; j j ;
up to a phase factor. So nally
Z Z
0 0 0 0 0 0
j id3phpj ipid3ph;pj ipi
=~j~=~j~
;
0 0 0 0
~
)pj i (p)pj i (;p):(4.6)
h~ ~=~=~
j = h;
4.3 Read section 4.3 in Sakurai to refresh your knowledge of the
quantum mechanics of periodic potentials. You know that the en-
ergybands in solids are described by the so called Bloch functions
full lling,
n k
(x+a)eika n k(x)
=
n k
whereaisnlabels the band, and the lattice
the lattice constant,
momentumkis restricted =a =a].
to the Brillouin zone [;
Prove that any Bloch function can be written as,
X
i
(x) n(x;Ri)eikR
=
n k
Ri
96
where the sumRi. (In this simble one di-
is over all lattice vectors
mensionalRiia, but the construction generalizes easily
problem =
to three dimensions.).
The n are called Wannier functions, and are impor-
functions
tant in the tight-binding description of solids. Show that the Wan-
nier functions are corresponding to di erent sites and/or di erent
bandsi:e:prove
are orthogonal,
Z
dx ?x;Ri) n(x;Rj) ij mn
(
m
Hint: ns in Bloch functions and use their orthonor-
Expand the
mality properties.
The de ning n k(x) is
property of a Bloch function
(x+a)eika n k(x):(4.7)
=
n k
P
i
We n(x;Ri)eikR satisfy the same relation
can show that the functions
Ri
X (x+a;R)e= [x;R;a)]ee
X
ikRi ik(Ri;a) ika
(
n i n i
Ri Ri
X
Ri ;a=Rj
j
=eika n(x;Rj)eikR (4.8)
Rj
which means that it is a Bloch function
X
i
(x) n(x;Ri)eikR:(4.9)
=
n k
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
;
j
multiplyeikRk.
this relation by and integrate over
X
i
(x) n(x;Ri)eikR
=
n k
Ri
Z Z
=a =a
X
; ;Rj)
j i
)dkeikR n k(x) n(x;Ri)eik(Rdk(4.10)
=
; =a ; =a
Ri
4. SYMMETRY IN QUANTUM MECHANICS 97
But
=a
Ze =a(R;R)]
ik(Ri ;Rj)
=a
2sin [
i j
;Rj)
i
eik(Rdk= =
; i(RiRj)RiRj
=a ; ;
; =a
2
= ij (4.11)
a
whereRiRjna,n2 Z. So
in the last step we used that ; = with
Z
=a
X
2
;
j
dkeikR n kx) n(x;Ri) ij
( =
; =a a
Ri
Z
a =a
;
i
) n(x;Ri)eikR n k(x)dk(4.12)
=
2
; =a
So using the orthonormality properties of the Bloch functions
Z
dx m(x;Ri) n(x;Rj)
ZZZa
2
0
; Rj 0
i
0 (
=eikR m k(x)eik n kx)dkdkdx
2
ZZa(2 ) Z
2
;ik0 Rj 0
i
0 (
=eikR m k(x) n kx)dxdkdk
2
(2
ZZa )
2
;ik0 Rj 0 0
i
=eikR mn (k;k)dkdk
(2 )2
Z
=a
a2a
;Rj)
i
= mneik(Rdk= mn ij:(4.13)
(2 )2
; =a 2
4.4 Suppose a spinless particle is bound to a xed center by a
potentialV(x) assymetrical that no energy level is degenerate.
~so
Using the time-reversal invariance prove
~
hLi =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
XXF(r)Y( )
m
lm
l
m
l
98
what kind of phaseFlm(r)?
restrictions do we obtain on
Since the Hamiltonian is invariant under time reversal
H H:(4.14)
=
Soni is an energyEn we will have
if j eigenstate with eigenvalue
H niHjniEn ni:(4.15)
j = = j
Ifnini can di er at most by a phase factor.
there is no degeneracy j and j
Hence
jniniei ni:(4.16)
~ j = j
For the angular-momentum operator we have from (S-4.4.53)
~~(4:16)~
hnjLjninjLjninjLjni )
= ;h ~ ~ = ;h
~
hnjLjni:(4.17)
=0
We have
Z Z
jnid3xjxihxjnid3xhxjnixi
=~~=~ ~
j
Z
(4:16)
=d3xhxjnixiei ni )
~j~= j
0 0 0
~~~
hxjnixjniei xjni:(4.18)
j = h = h
P P
SoxjniFlm(r)Ylm( )
if~=
we use h
l m
XF(r)Y( )eXF(r)Y( )
m i m
=
lm
lm l l
ml ml
X X
(S;4:4:57)
;m
)Flm(r)(;1)mYl )ei Flm(r)Ylm( )
( =
ml ml
Z Z
X X
0 0
;m
)YlmFlm(r)(;1)mYl )d ei YlmFlm(r)Ylm( )d
( =
0 0
ml ml
X X
0 0 = 0 0
)Flm(r)(;1)m m lei Flm(r) m l l
m
;m l
ml ml
0
;m0
0
)Flr)(;1)ei Flr)Flr)Flr)ei :(4.19)
( = 0 ( ) ( =(;1)m (
0 0
m0 m0
;m0 ;m0
4. SYMMETRY IN QUANTUM MECHANICS 99
4.5 The Hamiltonian for a spin 1 system is given by
2 2 2
H=ASzB(SxSy:
+ ; )
Solve this problem exactly to nd the normalized energy eigen-
states and eigenvalues. (A spin-dependent Hamiltonian of this kind
actually appears in crystal physics.) Is this Hamiltonian invariant
under time reversal? How do the normalized eigenstates you ob-
tained transform under time reversal?
Forl=m= 0 +1.Sz we
a spin 1 system 1 and ;1 For the operator
have
Szjl mihmjl milnjSzjl mihmhnjmiSz)nmhm nm (4.20)
= ) h = ) ( =
So
0 1 0 1
1 0 0 1 0 0
B C B C
2
Szh@Szh2 @ 0 0 0 A
= 0 0 0 ) =
A
0 0 ;1 0 0 1
ForSx we have
the operator
S+S1
+
;
1
Sxjl mi miS+j miSj mi !
= j 1 = 1 + 1
;
2 2
2
1 1
h1 njSxj mi njS+ mi njSj mi
1 = h1 j 1 + h1 1
;
2 2
q q
(S;3:5:39)
1 1
=h(1m)(2m) n m+1h(1m)(2m) n m;1:
; + + + ;
2 2
So
p
0 1 0 1
0 0 0
2 0
p
h0hp
B C B C
Sx = @ 0 0 2 A @ 2 0 0 A
+
p
2 2
0 0 0 0 2 0
p
0 1
0 2 0
p
hp
B C
= @ 2 0 2 A )
p
2
0 2
0 10 0 1
1 1
0
2 2
h2 2 0 2
B C B C
2
Sxh2:(4.21)
= @ 0 4 0 = @ 0 1 0 A
A
4
1 1
2 0 2 0
2 2
100
S+ ;S;
In theSy = we nd
same manner for the operator
2i
p
0 1
0 2 0
p
hp
B C
Sx = @ ; 2 0 2 A )
2ip
0 ; 2 0
0 1 0 1
1 1
0 ;
2 2
h2 ;2 0 2
B C B C
2
Sxh2:(4.22)
= ; @ 0 ;4 0 = @ 0 1 0 A
A
4
1 1
2 0 ;2 ; 0
2 2
Thus the Hamiltonian can be represented by the matrix
0A0B1
B C
H=h2:(4.23)
@ 0 0 0
A
B0A
To nd the energy eigenvalues we have to solve the secular equation
0Ah; 0Bh1
2 2
B C
det(H; I) 0 =0
= 0 ) det@ 0 ; A
Bh2Ah2
0 ;
h i
)Ah2 )2(; )Bh2)2 =0 (Ah2 )2Bh2)2 =0
( ; +( ) ; ; (
) (Ah2 ;Bh2)(Ah2 +Bh2) =0
; ;
) 1 2h2(A+B) 3h2(A;B):(4.24)
=0 = =
Ton i that c we have to
nd the eigenstate j corresponds to the eigenvalue
c
solve the following equation
0A0B1a1a1
0 0
B C B C B C
h2b= cb:(4.25)
@ 0 0 0 @ A @ A
A
B0Acc
For 1 =0
0A0B1a1
0
(
B C B C
0 0 0 @ A
h2b=0 )
@aA+cB= 0
AaB+cA= 0
B0Ac
(a=c(a=0
B
;
A
) 2 ) (4.26)
c=0
;cBcA=0
+
A
4. SYMMETRY IN QUANTUM MECHANICS 101
So
0 1 0 1
0 0
B C B C
jn0ibnorm: @ 1 )
= @ A = A
0 0
jn0i:(4.27)
= j 10i
In =h2(A+B)
the same way for
0A0B1a1a1aA+cB=a(A+B)
0 0 8
>
<
BbCA+B)bCb(A+B)
C B B
@ 0 0 0 @ A =( @ A ) 0 =
A
>
:
B0AccaB+cA=c(A+B)
(a=c
) (4.28)
b=0
So
0c1 0 1
1
1
B C norm: B C
p
jnA+Bi = @ 0 = @ 0 )
A A
c2 1
1 1
p p
jnA+Bi +1i ;1i:(4.29)
= j 1 + j 1
2 2
For =h2(A;B) we have
0A0B1a1a1aA+cB=a(A;B)
0 0 8
>
<
BbCA;B)bCb(A;B)
C B B
@ 0 0 0 @ A =( @ A ) 0 =
A
>
:
B0AccaB+cA=c(A;B)
(a=c
;
) (4.30)
b=0
So
0c1 0 1
1
1
B C norm: B C
p
jnA+Bi = @ 0 = @ 0 )
A A
;c2 ;1
1 1
p p
jnA;Bi +1i ;1i:(4.31)
= j 1 ; j 1
2 2
102
Now we are going to check if the Hamiltonian is invariant under time reversal
;1 2 ;1 2 ;1 2 ;1
H A SzB( Sx Sy )
= + ;
;1 ;1 ;1 ;1 ;1 ;1
=A Sz Sz B( Sx Sx Sy Sy )
+ ;
2 2 2
=ASzB(SxSyH:(4.32)
+ ; ) =
To nd the transformation of the eigenstates under time reversal we use the
relation (S-4.4.58)
jl mil ;mi:(4.33)
=(;1)mj
So
(4:33)
jn0i = j 10i = j 10i
=n0i (4.34)
j
(4.35)
1 1
p p
jnA+Bi +1i ;1i
= j 1 + j 1
2 2
1 1
(4:33)
p p
= ;1i +1i
; j 1 ; j 1
2 2
=nA+Bi (4.36)
;j
(4.37)
1 1
p p
jnA;Bi +1i ;1i
= j 1 ; j 1
2 2
1 1
(4:33)
p p
= ;1i +1i
; j 1 + j 1
2 2
=nA;Bi:(4.38)
j
5. APPROXIMATION METHODS 103
5 Approximation Methods
5.1 Consider an isotropic harmonic oscillator in two dimensions.
The Hamiltonian is given by
p2
p2m!2
y
x
H0x2y2)
= + + ( +
2m2m2
(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 en-
ergy to rst order [that is the unperturbed energy obtained in (a)
plus the rst-order energy shift] for each of the three lowest-lying
states.
(c)H0Vproblem exactly. Compare with the perturba-
Solve the +
tion results obtained in (b).
q
p
p
0
[Younjxjnih=2m!(n+1 nn n:]
may use h = 0 + 0 )
n+1 n;1
De ne step operators:
rm!ip
x
axx+
( )
2h
rm!m!
ipx
y
axx;
( )
2h
rm!m!
ipy
ayy+
( )
2h
rm!m!
ipy
y
ayy;:(5.1)
( )
2hm!
From the fundamental commutation relations we can see that
y y
[ax ax]ay ay]:
= [ = 1
104
De ning the number operators
y y
Nxaxax Nyayay
we nd
H0
N NxNy = ; 1 )
+
h!
H0h!(N+1):(5.2)
=
I.e. energyN:
eigenkets are also eigenkets of
Nxm nimjm ni
j =
Nym ninjm ni )
j =
Njm nim+n)m ni (5.3)
= ( j
so that
H0m niEm nm nih!(m+n+1)m ni:
j = j = j
(a) The lowest lying states are
state degeneracy
E0 0h!1
=
E1 0E0 1h!2
= =2
E2 0E0 2E1 1h!3
= = =3
(b)V= m!2xy.
Apply the perturbation
FullH0V)liEjli
problem: ( + j =
UnperturbedH0l0E0l0 i
problem: j i = j
Expand the energy levels and the eigenkets as
1 2
E=E0:::
+ + +
jlil0l1:::(5.4)
= j i + j i +
so that the full problem becomes
h i h i
1 2
(E0H0)l0l1:::=(V;:::)l0l1::::
; j i + j i + ; j i + j i +
5. APPROXIMATION METHODS 105
To 1'st order:
1
(E0H0)l1V;l0:(5.5)
; j i =( ) j i
Multiplyl0 j to nd
with h
1
hl0E0H0l1l0V;l0 i )
j ; j i = 0 = h j j
1 1
hl0l0l0Vjl0 i (5.6)
j i = = h j
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
1
hm0E0H0l1m0V;l0 i )
j ; j i = 0 = h j j
1
hm0l0m0Vjl0:(5.7)
j i = h j i
Now,m0l0 kl since only states corresponding to di er-
h j i is not necessarily
ent eigenvalues have to be orthogonal!
Insert a 1:
XhmjVjkihkjlimjli:
0 0 0 0 1 0 0
= h
k2D
This is the eigenvalue equation which gives the correct zeroth order eigen-
vectors!
Let us use all this:
1. The ground state is non-degenerate )
1 y y
= 0Vj 0 m!2h 0xyj 0 0ax+ax)(ay+ay) 0 i =0
h 0 j 0 i = 0 j 0 i h 0 j ( j 0
00
2. First 0 1 i. We need the matrix
excited state is degenerate j 1 i, j 0
elements 0Vj 0 0Vj 1 1Vj 0 1Vj 1 i.
h 1 j 1 i, h 1 j 0 i, h 0 j 1 i, h 0 j 0
h h!
y y y y y y
V= m!2xy= m!2ax+ax)(ay+ay)axay+axay+axay+axay)
( = (
2m!2
and
p
p
y
axm nimjm; niaxm nim+1m+1 nietc:
j = 1 j = j
106
Together this gives
V10 10V01 01
= =0
h! h!
y
V10 01 0axay 1
= h 1 j j 0 i =
2 2
h! h!
y
V01 10 0axay 1:
= h 1 j j 0 i =
2 2
(5.8)
TheV-matrix becomes
!
h!
0 1
1 0
2
1
and so the eigenvalues (= ) are
h!
1
=:
2
To get the eigenvectors we solve
! ! !
0xx
1
=
1yy
0
and get
1
p
j 1 0 E+h!(2
i+ = ( j 0 i + j 1 i) = + )
2
2
1
p
j 1 0 E=h!(2:(5.9)
i = ( j 0 i ; j 1 i) ; )
; ;
2
2
3. The second excited 0 1 2 i, so
state is also degenerate j 2 i, j 1 i, j 0
we need the corresponding 9 matrix elements. However the only non-
vanishing ones are:
h!
p
V11 20V20 11V11 02V02 11 = (5.10)
= = =
2
p
(where the 2 came from going from level 1 to 2 in either of the oscil-
lators) and thus to get the eigenvalues we evaluate
0 1 0 1
;
B C
0det@ 1 ( 2 = (2 2)
= 1 ; A = ; ; 1) + ;
0
1 ;
5. APPROXIMATION METHODS 107
which h!g. By the same method
means that the eigenvalues are f0
as above we get the eigenvectors
p
1
j 0 1 2 E+h!(3 )
i+ = ( j 2 i + 2 j 1 i + j 0 i) = +
2
1
p
j 0 2 E0h!
i0 = (;j 2 i + j 0 i) =3
2
p
1
j 0 1 2 E=h!(3 ):
i = ( j 2 i ; 2 j 1 i + j 0 i) ;
; ;
2
(c) To solve the problem exactly we will make a variable change. The poten-
tial is
h i
1
m!2x2y2) xy=
( + +
2
" #
1
=m!2x+y)2x;y)2)x+y)2x;y)2):(5.11)
(( +( + ( ; (
4 4
Now it is natural to introduce
1 1
0 0 0 0
p ( p ( +
x x+y) pxpxpy)
2 2
1 1
0 0 0 0
p p
y x;y) pypxpy):(5.12)
( ( ;
2 2
0 0 0 0 0 0 0 0
Note:x px]y py]ih,x,px)y,py ) are canonically
[ = [ = so that ( and (
conjugate.
In these new variables the problem takes the form
1m!2
02 0 0 0
H=pxpy2) )x2 )y2]:
( + + [(1 + +(1 ;
2m2
p p
0 0
So!x!1 and!y!1 .
we get one oscillator with = + another with = ;
The energy levels are:
E0 0h!
=
p
0
E1 0h!+h!xh!(1 ) =
= = + 1 +
1 1
=h!(1 +:::)h!(2 )O( 2)
+1 + = + +
2 2
108
0 1
E0 1h!+h!y:::=h!(2 )O( 2)
= = ; +
2
0
E2 0h!+2h!x:::=h!(3 )O( 2)
= = + +
0 0
E1 1h!+h!xh!y:::=3h!+O( 2)
= + =
0
E0 2h!+2h!y:::=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
0E0a1
1
BEbC
@ 0 A
1
abE2
whereE2>E1.aandbare to be regarded as per-
The quantities
turbations that are of the same order and are small compared with
E2E1. 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:
E1E0a
;
0E1Eb =
= 0 ;
abE2E
;
h i
2
=E1E)E1E)(E2E)bja[0a(E1E)] =
( ; ( ; ; ;j + ; ;
2 2
=E1E)2(E2E)E1E)(jbjaj:(5.14)
( ; ; ; ( ; + j )
2 2
So,E=E1E1E)(E2E)bjaj ) =0 i.e.
or ( ; ; ; (j + j
2 2
E2E1E2)E+E1E2ajbj ) =0 )
; ( + ; (j + j
s
E1E2E1E2 2
+ +
2 2
E=E1E2ajbj =
; + j + j
2 2
5. APPROXIMATION METHODS 109
s
E1E2E1E2 2
+ ;
2 2
=ajbj:(5.15)
+ j + j
2 2
2 2
Sinceajbj is small we can expand the square root and write the three
j + j
energy levels as:
E=E1
E1E2E1E2 1 2 2 2
+ ;
E=ajbj:::=
+ 1 + (j + j )( )2 +
2
2E1E2
2 ;
2 2
jajbj
+ j
=E1
+
E1E2
;
2 2
E1E2E1E2ajbj
+ ; j + j
E=:::)E2:
; ( = ;
2E1E2
2 ;
(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
B C B C B C
j j j:
1 i = @ 0 A 2 i = @ 1 A 3 i = @ 0 A
0 0 1
0a1
0 0
B C
The matrixV=bare
elements of the perturbation @ 0 0 A
ab0
hVja hVjb hVjkjVjki:
1 j 3 i = 2 j 3 i = 1 j 2 i = h =0
(1)
SincekjVjki = 0 1'st order gives nothing. But the 2'nd order
= h
k
shifts are
2 2 2
X
jVk1Vjaj
j j h 3 j 1 ij j
(2)
= =
=
1
0 0
E1EkE1E2E1E2
; ; ;
k=1
6
2 2 2
X
jVk2jVjbj
j h 3 j 2 ij j
(2)
= =
=
2
0 0
E2EkE1E2E1E2
; ; ;
k=2
6
2 2 2 2 2
X
jVk3jajbjajbj
j j j + j
(2)
= = +:
= ;
3
0 0
E3EkE2E1E2E1E1E2
; ; ; ;
k=3
6
(5.17)
110
The unperturbed problem has two (degenerate) states j 1 i and j 2 i with
energyE1, and one (non-degenerate)E2. Using non-
state j 3 i with energy
(2)
degenerate perturbation theoryE2 (i.e. )
we expect only the correction to
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 degenerateV-matrix for the degenerate
perturbation theory. The
!
0 0
subspace is , so 1'st order pert.thy. will again give nothing. We have
0 0
to go to 2'nd order. TheH0V)liEjli
problem we want to solve is ( + j =
using the expansion
(1) (2)
jlil0l1:::E=E0:::(5.18)
= j i + j i + + + +
whereH0l0E0l0 i. Note that the superscript index in a bra or ket de-
j i = j
notes which order it has in the perturbation expansion. Di erent solutions to
the fulll's. Since the (sub-) problem we are
problem are denoted by di erent
now solving is 2-dimensional we expect to nd two solutions corresponding
tol=1 2. Inserting the expansions in (5.18) leaves us with
h i
(E0H0)l0l1:::=
; j i + j i +
h i
(1) (2)
(V;:::)l0l1::::(5.19)
; j i + j i +
At rst order in the perturbation this says:
(1)
(E0H0)l1V;l0
; j i =( ) j i
(1)
where of course = 0 as noted above. Multiply this from the left with a
brak0 j from outside the deg. subspace
h
hk0E0H0l1k0Vjl0 i
j ; j i = h j
X
jk0k0Vjl0 i
ih j
)l1:(5.20)
j i =
E0Ek
;
k=D
6
Thisl1 i we will use in the 2'nd order equation from (5.19)
expression for j
(2)
(E0H0)l2Vjl1l0:
; j i = i ; j i
To get rid of the left hand side, multiplym0 j
with a degenerate bra h
(H0m0E0m0 i)
j i = j
(2)
hm0E0H0l2m0Vjl1m0l0:
j ; j i =0 = h j i ; h j i
5. APPROXIMATION METHODS 111
Insertingl1 i we get
the expression (5.20) for j
Xmjli:
hm0Vjk0k0Vjl0 i
j ih j
(2) 0 0
= h
E0Ek
;
k=D
6
To make this look like an eigenvalue equation we have to insert a 1:
Xnjlimjli:
X
hm0Vjk0k0Vjn0 i
j ih j
0 0 (2) 0 0
h = h
E0Ek
;
n2D k=D
6
Maybe it looks more familiar in matrix form
XMx= x
(2)
mn n m
n2D
where
X
hm0Vjk0k0Vjn0 i
j ih j
Mmn
=
E0Ek
;
k=D
6
0
xmmjl0 i
= h
are expressedl0Min the degenerate
in the basis de ned by j i. Evaluate
subspaceD= j 2 ig
basis f j 1 i
2
V13V31ajV13V32ab
j
M11 M12
= = = =
0 0
E1E3E1E2E1E3E1E2
; ; ; ;
2 2
V23V31abjV23jbj
j
M21 M22:
= = = =
0 0
E1E3E1E2E1E3E1E2
; ; ; ;
WithM, solve the eigenvalue equation (de ne
this explicit expression for
(2) 1
= E1E2), and take out a common factor )
( ;
E1 ;E2
!
2
jaj ab
;
0det2 =
=
abjbj
;
2 2 2 2
=aj )(jbj )ajbj =
(j ; ; ;j j
2 2
= 2ajbj
; (j + j )
2 2
) =0 jajbj
+ j
2 2
jajbj
+ j
(2) (2)
):(5.21)
=0 =
1 2
E1E2
;
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
fort<0.t 0 it is subjected to a time-dependent but spatially
For
uniform force (not potential!) in the x-direction,
;
F(t)F0et=
=
(a) Using time-dependent perturbation theory to rst order, obtain
the probability of nding the oscillator in its rst excited state for
t>0.t!1 nite) limit of your expression is
Show that the (
independent of time. Is this reasonable or surprising?
(b) Can we nd higher excited states?
q
p
p
0
[Younjxjnih=2m!(n+1 nn n:]
may use h = 0 + 0 )
n+1 n;1
(a) The problem is de ned by
p2m!2x2@V
;
H0V(t)F0xet= F= ; )
= + = ; (
2m2@x
Att= 0 0 i = j 0 i. We want to calculate
the system is in its ground state j
X
; t=h
n
j ticn(t)eEni
= j
n
0 1
Enh!(n+ )
=
2
wherecn(t) from its di . eqn. (S. 5.5.15):
we get
X
@t
nm
ihcn(t)Vnmei!cm(t)
=
@tm
VnmnjVjmi
= h
EnEm
;
!nm!(n;m) (5.22)
= =
h
WeVnm
need the matrix elements
; ;
VnmnjF0xet= miF0et= njxjmi =
= h ; h
shj = ;
p
p
;
=F0et= m n m;1m+1 n m+1):
; ( +
2m!
5. APPROXIMATION METHODS 113
Put it back into (5.22)
s
p
@hp
; ;
ihcn(t)F0et= n+1ei!tcn+1t)nei!tcn;1(t):
= ; ( +
@t2m!
Perturbationcn(t)c(0)c(1):::, and to zeroth
theory means expanding = + +
n n
order this is
@(0)
c(t)c(0) n0
= 0 ) =
n
@tn
To rst order we get
Z
t
X
1
0 0 t0
nm
c(1)(t)dtVnm(t)ei!c(0) =
=
n m
ihm
0
s
Z
p
p
F0ht 0 0
0 ; ;
=dtet= n+1ei!tc(0)t)nei!tc(0)t)
; ( + (
n+1 n;1
ih2m!0
Wen= 1, i.e. at rst order in perturbation
get one non-vanishing term for
theory witht= 0 there is just one non-zero
the H.O. in the ground state at
expansion coe cient
s
Z
p
F0ht 0
0 ;t0 =
c(1)(t)dtei!t 1 =
= ; 1
;
1
ih2m!1 0
0
s
" #
t
F0h1 1
;
=e(i! )t0
;
1
ih2m!i!;
0
s
F0h1 1
=e(i!; )t
1 ;
1
ih2m!i!;
and
X
; iE t
; iE t
n 1
h h
j tic(1)(t)ejnic(1)(t)ej:
= = 1 i
n 1
n
The probability of nding the H.O. in j 1 i is
2 2
j tijc(1)(t)j:
h 1 j = j
1
Ast!1
s
F0h1
c(1)const:
! =
1
1
ih2m!i!;
This is of course reasonable since applying a static force means that the
system asymptotically nds a new equilibrium.
114
(b) As remarked earliercn's at rst order,
there are no other non-vanishing
so no higher excited states can be found. However, going to higher order in
perturbation theory such states will be excited.
1
5.4 Consider a composite system made up of two spin objects.
2
fort<0, the Hamiltonian does not depend on spin and can be
taken to be zero by suitably adjustingt>0,
the energy scale. For
the Hamiltonian is given by
4
~~
H=S1S2:
h2
Supposet 0. Find, as a function of
the system is in j + ;i for
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-dependentHas a perturbation switched
perturbation theory with
ont=0. Under what condition does (b) give the correct results?
at
(a) TheS1z S2z i. Expand the interaction
basis we are using is of course j
potential in this basis:
~~
S1S2S1xS2xS1yS2yS1zS2z = fin this basisg
= + +
"
h2
= ( j + ih ;j + j ;ih + j )1 ( j + ih ; j + j ;ih + j )2+
4
+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 =
"
h2
= j ++ ih ; ; j + j + ;ih ; + j +
4
+ j ; + ih + ; j + j ; ; ih ++ j +
+i2( 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)
0 1
1 0 0 0
B C
0 ;1 2 0
B C
H= :(5.23)
B C
@ 0 2 ;1 0
A
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
;
0det=(;1 )2 2 ; 3
= ; ; 4 = +2
2
;1 ;
) =1 ;3
=1 :
! ! !
;1xx
2
=
2yy
;1
1
);x+2y=x)x=y= p
2
= ;3 :
! ! !
;1xx
2
= ;3
2yy
;1
1
p
);x+2y=x)x=y=
;3 ;
2
So, the complete spectrum is:
8
1
> p
j ++ i ; ; i
>
< j ( j + ;i + j ; + i with energy
2
>
>
1
:
p
( j + ;i ; j ; + i with energy ; 3
2
116
This was a cumbersome but straightforward way to calculate the spectrum.
~~~
A smarterS=S1S2 to nd
way would have been to use +
~~2~2~~~~1~~2~2
S2S2S1S2S1S2S1S2S2S1S2
= = + +2 ) = ; ;
2
3h2
~2~2
WeS1S2h2 1 1 +1 = so
know that = =
2 2 4
!
2
3
~~1h
S1S2S2 ;
=
2
2
1
Also, we know that two spin systems add up to one triplet (spin 1) and one
2
singlet (spin 0), i.e.
3h2 1
~~1
S=S1S2h21(1h2
1 (3 states) ) = ( + 1) ; ) =
2 2 4
:(5.24)
~~1 3h2 3
S=S1S2h2
0 (1 state) ) = (; ) = ;
2 2 4
4
~~
SinceH=S1S2 we get
h2
1
4 h2
E(spin=1)
= =
h2 4
4 h2
;3
E(spin=0):
= = ;3
h2 4
(5.25)
n
From Clebsch-Gordan decomposition j
we know that j ++ i ; ; i
o
1 1
p p
( j + ;i + j ; + i) are spin 1, and ( j + ;i ; j ; + i) is spin 0!
2 2
Let's get back on track and ndHis diagonal
the dynamics. In the new basis
and time-independent, so we can use the simple form of tthe time-evolution
operator:
i
U(t t0)exp;H(t;t0):
=
h
The initial state was j + ;i. In the new basis
n j j
1
p
j 1 i = j ++ i 2 i = j ; ; i 3 i = ( j + ;i + j ; + i)
2
o
1
p
j 4 i = ( j + ;i ; j ; + i)
2
5. APPROXIMATION METHODS 117
the initial state is
1
p
j:
+ ;i = ( j 3 i + j 4 i)
2
Actingt 0) on that we get
with U(
1i
j tiexp;Ht( j 3 i + j 4 i) =
= p
h
2
1i3i
p 3 i +
=exp;tjexp tj 4 i =
2
"i t hh
; 1
p
=exp( j + ;i + j ; + i)+
h
2
#
3i t1
p
+exp( j + ;i ; j ; + i) =
h(h2 i
1 ; ;
=ei!te3i!t)ei!te3i!t) j ; + i
+ j + ;i +( +
2
where
! : (5.26)
h
2
The probability i j tij
to nd the system in the state j is as usual j h
8 ti ti =0
>
h ++ j = h ; ; j
>
>
>
>
>
<
2 1 4i!t ;4i!t 1 2
j h + ; j = (2 + = (1 + ' 1 ; 4(
4 2
> tije+e)cos4!t)!t):::
>
>
>
>
> tije;e)cos4!t)!t):::
: 2 1 4i!t ;4i!t 1 2
j h ; + j = (2 ; = (1 ; ' 4(
4 2
(b) First order perturbation theory (use S. 5.6.17):
c(0) ni
=
n
Z
;it
0 t0 0
ni
c(1)(t)dtei!Vni(t):(5.27)
=
n
ht0
Here weH0Vgiven by (5.23)
have (using the original basis) =0,
jii
= j + ;i
118
jfi
= j ; + i
EnEi
;
!niEn
= = f =0g =0
h
Vfi
= 2
Vni n=f:
= 0 6
Inserting this into (5.27) yields
c(0)c(0)
= =1
i
j +;i
Z
it
c(1)c(1)dt2 i!t:(5.28)
= = ; = ;2
f j ;+ i
h
0
as the only non-vanishing coe cients 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
P(c(1)(t)c(2)(t):::ji!tj!t)2
j ; + i) = j + + = j 2 =4(
f f
to rst order, in correspondence with the exact result.
The approximation!t1 is no longer valid, so for a
breaks down when
given t:
h
!t1:
)
t
5.5 The groundn=1,l=0) is subjected
state of a hydrogen atom (
to a time-dependent potential as follows:
V(x t)V0kz;!t):
~= cos(
Using time-dependent perturbation theory, obtain an expression
for the transition rate at which the electron is emitted with mo-
mentump.
~Show, in particular, howyou may compute the angular
distribution of the ejected and de ned
electron (in terms of
withz-axis). Discuss brie y the similarities and the
respect to the
di erences between this problem and the (more realistic) photo-
electric e ect. (note: For the initial wave function use
3
2
1Z
;
0
(x)eZr=a:
~= p
n=1 l=0
a0
5. APPROXIMATION METHODS 119
If you have a normalization problem, the nal wave function may
be taken to be
1
p x=h
(x)ei~ ~
~=
f
3
2
L
withLvery large, but you should be able to show that the observ-
ableL.)
e ects are independent of
Ton=1 l=t= 0 the perturbation
begin with the atom is in the 0 state. At
V=V0cos(kz;!t)
is turned on. We want to nd the transition rate at which the electron is
emittedpf The initial wave-function is
with~.
momentum
3=2
1 1
;
0
(x)er=a
~= p
i
a0
and the nal wave-function is
1
p x=h
~
(x)ei~:
~=
f
L3=2
The perturbation is
h i
;!t) ; ;!t)
V=V0ei(kzei(kz
+
y ;
=ei!tei!t:(5.29)
V + V
Time-dependent perturbation theory (S.5.6.44) gives us the transition rate
2
2
y
wi (EnEih!))
= V ; ( +
!n
hni
becauseh!. The matrix element is
the atom absorbs a photon
2
2 2
y =
0
Vni V eikz
ni
4
and
Z
eikz~jeikzn=1 l=0d3xhkfeikzxihxjn=1 l=0 i =
=kf~j j
h j i =
ni
Ze1 1
;i~ x 3=2
kf ~
;
3 0
p
=d3x3=2eikxer=a =
L a0
Z
1
~
; x;kx3 );r=a0
~
f
=d3xei(k:(5.30)
p
L3=2 a3=2
0
120
Soeikz is the 3D Fourier transform of the initial wave-function (and some
ni
constant)~=kfk~. That can be extracted from (Sakurai problem
withq~;ez
5.39)
64 2
eikz =
h~1 i
ni
1
L3a5kfk~) 4
0
+(ez 2
;
a2
0
The transition rate is understood to be integrated over the density of states.
Wepfh~. As in (S.5.7.31), the volume
need~=kf
to get that as a function of
element is
dn
n2dnd n2d dpf:
=
dpf
Using
p2
n2(2 )2
f
2
kf = =
h2 L2
we get
dn1L2pfhL2pfL
2 2
= = =
dpfn(2 h)2Lpf h)2 h
2 (2 2
which leaves
2
L3kfL3p2
f
n2dnd d dpfd dpf
= =
(2 )3h(2 h)3
and this is the sought density.
Finally,
3 2
2 V02 64 2
f
wi pf = d dpf:
hk1Lp
i
!~
1
h4L3a5~4 h)3
0
+(k~)2 (2
;ez
a2 f
0
NoteL's cancel. The angular dependence is in the denominator:
that the
~2 [(j j
kfk~=kfcos ;k)~+kfsin (cos'~+sin'~)]2 =
;ezezexey
j j
2 2
=kfcos2 +k2kjkfcos +kfsin2 =
j j ; 2 j j j
2
=kfk2kjkfcos :(5.31)
+ ; 2 j
In a comparison between this problem and the photoelectric e ect as dis-
cussed in (S. 5.7) we note that since there is no polarization vector involved,
w has no dependence . On the other hand we did
on the azimuthal angle
not make any dipole approximationx-integral exactly.
but performed the