Solutions to Problems
in
Quantum Mechanics
P. Saltsidis, additions by B. Brinne
1995,1999
о
Most of the problems presented here are taken from the book
Sakurai, J. J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley,
1985.
Contents
I Problems	3
1	Fundamental Concepts........................... 5
2	Quantum Dynamics............................... 7
3	Theory of Angular Momentum.................... 14
4	Symmetry in Quantum Mechanics................. 17
5	Approximation Methods......................... 19
II Solutions	23
1	Fundamental Concepts...........................25
2	Quantum Dynamics...............................36
3	Theory of Angular Momentum.................... 75
4	Symmetry in Quantum Mechanics..................94
5	Approximation Methods.........................103
1
2
CONTENTS
Part I
Problems
3

1. FUNDAMENTAL CONCEPTS
5
1 Fundamental Concepts
1.1	Consider a ket space spanned by the eigenkets {|o/}} of a Her-
mitian operator A. There is no degeneracy.
(a)	Prove that
П(л -«')
fl'
is a null operator.
(b)	What is the significance of
(c)	Illustrate (a) and (b) using A set equal to Sz of a spin | system.
1.2	A spin | system is known to be in an eigenstate of S  h with
eigenvalue 7i/2, where h is a unit vector lying in the жг-plane that
makes an angle 7 with the positive г-axis.
(a)	Suppose Sx is measured. What is the probability of getting
+a/2?
(b)	Evaluate the dispersion in Sx, that is,

(For your own peace of mind check your answers for the special
cases 7 = 0, тг/2, and 7г.)
1.3	(a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
({а| + А*{/3|ИМ + А|Я)>0
for any complex number A; then choose A in such a way that the
preceding inequality reduces to the Schwarz inequility.
6
(b) Show that the equility sign in the generalized uncertainty re-
lation holds if the state in question satisfies
AA|a) = AAB|a)
with A purely imaginary.
(c) Explicit calculations using the usual rules of wave mechanics
show that the wave function for a Gaussian wave packet given by

satisfies the uncertainty relation
Prove that the requirement
(ж'|Дж|о!) = (imaginary number)(a/|Ap|a)
is indeed satisfied for such a Gaussian wave packet, in agreement
with (b).
1.4 (a) Let x and p..- be the coordinate and linear momentum in
one dimension. Evaluate the classical Poisson bracket
[ж, ^(Po;)]c;assjca; 
(b) Let x and p..- be the corresponding quantum-mechanical opera-
tors this time. Evaluate the commutator
(грха
x. exp ——
\ n
(c) Using the result obtained in (b), prove that

2. QUANTUM DYNAMICS
7
is an eigenstate of the coordinate operator x. What is the corre-
sponding eigenvalue?
1.5 (a) Prove the following:

where фа{р'} = (p'|a) and ф/з(р') = (p'|/3) are momentum-space wave
functions.
(b) What is the physical significance of
/ ix3\
exp hr) ’
where x is the position operator and E 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
и (eB\ Q c
D = - — 5г =
\mc/
write the Heisenberg equations of motion for the time-dependent
operators S^t), SyQ), and Sz(t). Solve them to obtain SXiyiZ as func-
tions of time.
2.2	Let x(t) be the coordinate operator for a free particle in one
dimension in the Heisenberg picture. Evaluate
[ж(£),ж(0)].
8
2.3	Consider a particle in three dimensions whose Hamiltonian is
given by
n2
H=^- + V(g).
By calculating [ж • p, H] obtain
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
You may use
(s'|0) = 7Г 1/<4ж01/<2ехр
тш
(b) Obtain a simple expression that the probability that the state
is found in the ground state at t = 0. Does this probability change
for t > 0?
2.5	Consider a function, known as the correlation function, defined
by
(7(£) = (ж(^)ж(О)),
where x(t) is the position operator in the Heisenberg picture. Eval-
uate the correlation function explicitly for the ground state of a
one-dimensional simple harmonic oscillator.
2. QUANTUM DYNAMICS
9
2.6	Consider again a one-dimensional simple harmonic oscillator.
Do the following algebraically, that is, without using wave func-
tions.
(a)	Construct a linear combination of |0) and |1) such that (ж) is as
large as possible.
(b)	Suppose the oscillator is in the state constructed in (a) at t = 0.
What is the state vector for t > 0 in the Schrodinger picture?
Evaluate the expectation value (ж) as a function of time for t > 0
using (i) the Schrodinger picture and (ii) the Heisenberg picture.
(c)	Evaluate ((Are)2) as a function of time using either picture.
2.7	A coherent state of a one-dimensional simple harmonic oscil-
lator is defined to be an eigenstate of the (non-Hermitian) annihi-
lation operator a:
«|A> = A|A>,
where A is, in general, a complex number.
(a)	Prove that
|A) = e-lAl2/2eAat|0)
is a normalized coherent state.
(b)	Prove the minimum uncertainty relation for such a state.
(c)	Write |A) as
oo
1Л) = 1Ж4
n=0
Show that the distribution of |/(n)|2 with respect to n is of the
Poisson form. Find the most probable value of n, hence of E.
(d)	Show that a coherent state can also be obtained by applying
the translation (finite-displacement) operator e~ipl^ (where p is the
momentum operator, and I is the displacement distance) to the
ground state.
10
(e)	Show that the coherent state |A) remains coherent under time-
evolution and calculate the time-evolved state |A(£)). (Hint: di-
rectly apply the time-evolution operator.)
2.8	The quntum mechanical propagator, for a particle with mass
m, moving in a potential is given by:
«	x . sin(nrs) sin(nry)
A(s,y;E) = / dte /nK(x,y,t,0) =	----z-----------
Jo	n E - ^n2
where A is a constant.
(a)	What is the potential?
(b)	Determine the constant A in terms of the parameters describing
the system (such as m, r etc. ).
2.9	Prove the relation
= 6(x)
ax
where 0(ж) is the (unit) step function, and <У(ж) the Dirac delta
function. (Hint: study the effect on testfunctions.)
2.10	Derive the following expression
muj г 2	2	1
Scl = n f mx (ж0 + 2st)cOs(ll?T) - X0XT\
2 Sin(Lc?I ) L	J
for the classical action for a harmonic oscillator moving from the
point x0 at t = 0 to the point xt at t = T.
2.11	The Lagrangian of the single harmonic oscillator is
C 1-2	22
L = -mx---mux x
2	2
2. QUANTUM DYNAMICS
11
(a)	Show that
(xbtb kX) = exp
lScl
Nh.
G(Q,tb-,0,Q
where Sci is the action along the classical path xci from (xa,ta) to
(xb, tb) and G is
G(0,tb-,0,Q =
(w+1)	( . N
where e =
(Hint: Let y(f) = x(T) — xci(T) be the new integration variable,
xci(T) being the solution of the Euler-Lagrange equation.)
(b)	Show that G can be written as
У1
where n =
and nT is its transpose. Write the symmetric
L yN
matrix a.
(c)	Show that
r	tvn!2
/ dyr... dyNexp(-nTan) = / dNye~nT<Tn =	---
J	J	Y deter
[Hint: Diagonalize a by an orhogonal matrix.]
(d)	Let ( deta = detcr'N = p^. Define j x j matrices <r'- that con-
sist of the first j rows and j columns of er'N and whose determinants
are pj. By expanding cr'-+1 in minors show the following recursion
formula for the pj:
P)+1 = (2 - eV)Pi - Pj-! j =
(2.1)
12
(e)	Let </>(£) = epj for t = ta + je and show that (2.1) implies that in
the limit e —> 0, ф(ф) satisfies the equation
with initial conditions ф(ф = ta) = 0,	= 1.
(f)	Show that
I muj	I 1ТПШ 2	2	I
(xbtb\xata) = J .	. [(xb + sJcos(^T) - f
у 2/kiTi sm(LL?i ) (27zsm(LL?i )	J
where T = tb — ta.
2.12	Show the composition property
У dx1Kf(x2,t2;x1,t1)Kf(x1,t1;x0,t0) = Kf(x2,t2,x0,t0)
where Aj(s15s0,^o) 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
[IIj, Пу] — I 1 SijkBk
\ C /
where П =	= p — — and the relation
at 1 c
ax flii	i / ax	ax \
-TV = ^ = e E + — — x В-В X—]
dt2 dt [ 2c у dt	dt J
(b) Verify the continuity equation
2. QUANTUM DYNAMICS
13
with j given by
—) Ar
тс/
2.14 An electron moves in the presence of a uniform magnetic field
in the г-direction (B = Bz).
(a)	Evaluate
where
(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
k’k.n —
h2k2
-----k
2m
1
n H—
2
where hk is the continuous eigenvalue of the pz operator and n is a
nonnegative integer including zero.
2.15 Consider a particle of mass m and charge q in an impenetrable
cylinder with radius R and height a. Along the axis of the cylin-
der runs a thin, impenetrable solenoid carrying a magnetic flux Ф.
Calculate the ground state energy and wavefunction.
2.16 A particle in one dimension (—oo < x < oo) is subjected to a
constant force derivable from
V = Xx, (A > 0).
14
(a)	Is the energy spectrum continuous or discrete? Write down an
approximate expression for the energy eigenfunction specified by
E.
(b)	Discuss briefly what changes are needed if V is replaced be
V = A|s|.
3 Theory of Angular Momentum
3.1	Consider a sequence of Euler rotations represented by
P(1/2)(a,/3,7)
(—	( — i(T2(3\ f—го'з'У
exp I—2—J exP I —2— / eXP \—2—
/ e-«(“+7)/2 cos f -e-’(“-7)/2 sin f \
\ e«(“-7)/2 sin |	ei(«+7)/2 cos | J 
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 ф. Find ф.
3.2	An angular-momentum eigenstate |j, m = rnmax = j) is rotated
by an infinitesimal angle e about the у-axis. Without using the
explicit form of the <-Qm function, obtain an expression for the
probability for the new rotated state to be found in the original
state up to terms of order £2.
3.3	The wave function of a patricle subjected to a spherically
symmetrical potential V(r) is given by
ф(х) = (ж + y + 3z)f(r).
3. THEORY OF ANGULAR MOMENTUM
15
(a)	Is ф an eigenfunction of L? If so, what is the Z-value? If
not, what are the possible values of I we may obtain when L2 is
measured?
(b)	What are the probabilities for the particle to be found in various
mi states?
(c)	Suppose it is known somehow that 'ф(х) is an energy eigenfunc-
tion with eigenvalue E. Indicate how we may find V(r).
3.4	Consider a particle with an intrinsic angular momentum (or
spin) of one unit of h. (One example of such a particle is the g-
meson). Quantum-mechanically, such a particle is described by a
ketvector |p) or in x representation a wave function
д\х) = (ж;г|е)
where \x,i) correspond to a particle at x with spin in the Z:th di-
rection.
(a)	Show explicitly that infinitesimal rotations of £>г(ж) are obtained
by acting with the operator
иг-= 1 -	• (I + 5)	(3.1)
where L = x V. Determine S !
г
(b)	Show that L and S commute.
(c)	Show that S is a vector operator.
(d)	Show that V x g(x) = ^(S • p)g where p is the momentum oper-
ator.
3.5	We are to add angular momenta Ji = 1 and j2 = 1 to form
j = 2,1, and 0 states. Using the ladder operator method express all
16
(nine) j. m eigenkets in terms of IJ1J2;	Write your answer as
(3.2)
where + and 0 stand for m12 = 1,0, respectively.
3.6	(a) Construct a spherical tensor of rank 1 out of two different
vectors U = (Ux,Uy,Uz) and V = (Vx,Vy,Vz). Explicitly write T^o in
terms of г and V^ty>z.
(b) Construct a spherical tensor of rank 2 out of two different
vectors U and V. Write down explicitly T±2±10 in terms of
and 14,у,z-
3.7 (a) Evaluate
£2 Ит1'(Я!2ш
for any j (integer or half-integer); then check your answer for J = |.
(b) Prove, for any j,
12 ш2|4'т(Я|2 = VQ + l)sin/? + m'2 + 1(3cos24- 1).
m=-j
[Hint: This can be proved in many ways. You may, for instance,
examine the rotational properties of using the spherical (irre-
ducible) tensor language.]
3.8 (a) Write xy, xz, and (x2 — y2) as components of a spherical
(irreducible) tensor of rank 2.
4. SYMMETRY IN Q UANTUM MECHANICS
17
(b) The expectation value
Q =	= j\(3z2 - r2)\a,j,m = j}
is known as the quadrupole moment. Evaluate
e(a, j, m'\(x2 - y2)|a, j,m = j},
(where m' = j, j — 1, j — 2,... )in terms of Q and appropriate Clebsch-
Gordan coefficients.
4 Symmetry in Quantum Mechanics
4.1 (a) Assuming that the Hamiltonian is invariant under time
reversal, prove that the wave function for a spinless nondegenerate
system at any given instant of time can always be chosen to be
real.
(b) The wave function for a plane-wave state at t = 0 is given by
a complex function ezp'xl%. Why does this not violate time-reversal
invariance?
4.2	Let (pfjf) be the momentum-space wave function for state |a),
that is, </>(j7) = (/7|a).Is the momentum-space wave function for the
time-reversed state 0|a) given by </>(j7, ф( — /7), </>*(j7), or ф*( — /7)?
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
фп,к fulffilling,
фп,к(х + a) = егкафП1к(х)
18
where a is the lattice constant, n labels the band, and the lattice
momentum к is restricted to the Brillouin zone [—тг/а, тг/а].
Prove that any Bloch function can be written as,
Ri
where the sum is over all lattice vectors Ri. (In this simble one di-
mensional problem Ri = ia, but the construction generalizes easily
to three dimensions.).
The functions фп are called Wannier functions, and are impor-
tant in the tight-binding description of solids. Show that the Wan-
nier functions are corresponding to different sites and/or different
bands are orthogonal, i.e. prove
Hint: Expand the фпз in Bloch functions and use their orthonor-
mality properties.
4.4	Suppose a spinless particle is bound to a fixed center by a
potential У(ж) so assymetrical that no energy level is degenerate.
Using the time-reversal invariance prove
<i) = o
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
I m
what kind of phase restrictions do we obtain on F/m(r)?
4.5	The Hamiltonian for a spin 1 system is given by
H = AS2z + B(S2-S2y).
5. APPROXIMATION METHODS
19
Solve this problem exactly to find 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
pf_ +
2m 2m
(*2 + У2)
9
тш
(a)	What are the energies of the three lowest-lying states? Is there
any degeneracy?
(b)	We now apply a perturbation
V = 8тш2ху
where 8 is a dimensionless real number much smaller than unity.
Find the zeroth-order energy eigenket and the corresponding en-
ergy to first order [that is the unperturbed energy obtained in (a)
plus the first-order energy shift] for each of the three lowest-lying
states.
(c)	Solve the Ho + V problem exactly. Compare with the perturba-
tion results obtained in (b).
[You may use (п'|ж|п) = y?i/2maj(Vra + 18n',n+i + y/n8n>in_i).]
5.2	A system that has three unperturbed states can be represented
by the perturbed Hamiltonian matrix
f Ei 0 a \
0 Ei 6 I
k a* b* E2 )
20
where E% > Ei. The quantities a and b are to be regarded as per-
turbations that are of the same order and are small compared with
Ez — Ei. Use the second-order nondegenerate perturbation theory
to calculate the perturbed eigenvalues. (Is this procedure correct?)
Then diagonalize the matrix to find the exact eigenvalues. Finally,
use the second-order degenerate perturbation theory. Compare
the three results obtained.
5.3	A one-dimensional harmonic oscillator is in its ground state
for t < 0. For t > 0 it is subjected to a time-dependent but spatially
uniform force (not potential!) in the x-direction,
F(t) = Foe-^
(a)	Using time-dependent perturbation theory to first order, obtain
the probability of finding the oscillator in its first excited state for
t > 0). Show that the t —> oo (r finite) limit of your expression is
independent of time. Is this reasonable or surprising?
(b)	Can we find higher excited states?
[You may use (п'|ж|п) =	+ l<5n>,n+1 + ^/m^-i).]
5.4	Consider a composite system made up of two spin | objects,
for t < 0, the Hamiltonian does not depend on spin and can be
taken to be zero by suitably adjusting the energy scale. For t > 0,
the Hamiltonian is given by
Suppose the system is in | -|---) for t < 0. Find, as a function of
time, the probability for being found in each of the following states
I + +)> I + “)> I _ +)> I-}'•
(a)	By solving the problem exactly.
5. APPROXIMATION METHODS
21
(b)	By solving the problem assuming the validity of first-order
time-dependent perturbation theory with H as a perturbation switched
on at t = 0. Under what condition does (b) give the correct results?
5.5	The ground state of a hydrogen atom (n = 1,Z = 0) is subjected
to a time-dependent potential as follows:
V(x,t) = VoCos(A;,z — art).
Using time-dependent perturbation theory, obtain an expression
for the transition rate at which the electron is emitted with mo-
mentum p. Show, in particular, how you may compute the angular
distribution of the ejected electron (in terms of 0 and ф defined
with respect to the г-axis). Discuss briefly the similarities and the
differences between this problem and the (more realistic) photo-
electric effect, (note: For the initial wave function use
If you have a normalization problem, the final wave function may
be taken to be
/
with L very large, but you should be able to show that the observ-
able effects are independent of /_.)
22
Part II
Solutions
23

1. FUNDAMENTAL CONCEPTS
25
1 Fundamental Concepts
1.1 Consider a ket space spanned by the eigenkets {|a')} of a Her-
mitian operator A. There is no degeneracy.
(a)	Prove that
П(А - a')
is a null operator.
(b)	What is the significance of
П
(Л-И?
a' — a"
(c)	Illustrate (a) and (b) using A set equal to Sz of a spin | system.
(a)	Assume that |a) is an arbitrary state ket. Then
П(А-а')|о!)
= П(л -a') 52 la//)	= 52 е»" П(л - а')Ю
а'	а"	а" а'
са"
=	(1-Ц
а" а'
(b)	Again for an arbitrary state |a) we will have
i

(A - a")
a1 — a11
(A - a")
a1 — a11
a"
III

a
a"±a'
a' — a"
a"'a‘
a

(A - a")
a1 — a11
(1-2)
So it projects to the eigenket |a').
26
(c)	It is Sz = 7i/2(|+)(+| — |—)(—I). This operator has eigenkets |+) and |—)
with eigenvalues h/2 and -h/2 respectively. So
a’	a'
= |(I+X+I - l-X-l) - f(l+X+l + l-X-l)
X |(I+X+I - I-X-D + |(I+X+I + l-X-l)
= [-»I-X-I№I+X+I] = -»2l-> Fi+X+I = 0, (1.3)
where we have used that |+)(+| + |—){—| = 1.
For a' = h/2 we have
„ {Sz ~ n"l) _ S2 + fl
a'W2 ^2 — a" + ^/2
i [|(I+X+I - l-x-l) + |(I+X+I + l-x-l)
= jhi+x+i = i+x+i-
Similarly for a' = —h/2 we have
JJ (5*2	«")
a"±a'
a' — a"
(1-4)
jj (5*2 aw)
a"±a'
a' — a"
„ (5, - <"!) = &-|l
.'J-V2 “ft/2 - °" -ft/2 - ft/2
[|(I+X+I - l-X-l) - |(I+X+I + l-X-l)
-|(-ft|-X-l) = l-X-l-
(1-5)
1.2	A spin I system is known to be in an eigenstate of S  h with
eigenvalue 7i/2, where h is a unit vector lying in the жг-plane that
makes an angle 7 with the positive г-axis.
(a)	Suppose Sx is measured. What is the probability of getting
+П/2?
1. FUNDAMENTAL CONCEPTS
27
(b)	Evaluate the dispersion in Sx, that is,
(For your own peace of mind check your answers for the special
cases 7 = 0, тг/2, and 7r.)
Since the unit vector n makes an angle 7 with the positive г-axis and is
lying in the жг-plane, it can be written in the following way
n = ez cos 7 + ex sin 7
(1.6)
So
h
S-n = Sz cos 7 + Sx sin 7 =	[(S-1.3.36),(S-1.4.18)]
= I (l+X+l - l-X-1)1 cos7 + [I (|+)(-| + |-)(+|)l siny.1.7)
2
Since the system is in an eigenstate of S  n with eigenvalue Д/2 it has to
satisfay the following equation
S  n\S  n- +) = h/2\S  n- +).	(1.8)
From (1.7) we have that
j л . h I cos 7 sin 7
S  n = —
2 \ sin 7 — cos 7
The eigenvalues and eigenfuncions of this operator can be found if one solves
the secular equation
(1-9)
! x/d a \ n 1 x I ^/2cos7 - A ^/2sin7 A n
det(S  n — XI) = 0 => det ' . .	. '	' .	= 0 =>
v	\	7i/2sm7	—n/2cos7 —A /
Д2 x h .
T = 0^A = ±-.	(1.10)
+ 2	+2
/4	Q	Q fl t П
-----cos 7 + A----------sin 7 =
4	4
Since the system is in the eigenstate |S  n; +) = I j we will have that
Ti ( cos 7 sin 7 \ ( a A _ h ( a
2 I sin 7 — cos 7 I I b I 2 \ b
7	1 — cos 7	2 sin2
b = a---;---- = a—;— -----
sin 7	2 sin cos
a cos 7 + b sin 7 = a
a sin 7 — b cos 7 = b
7
= a tan —.
2
(1-11)
28
But we want also the eigenstate |S  n; +) to be normalized, that is
a2 + b2 = 1
2 । 2 j 2 7 I . 2	272-27	27
a + a tan — = 1 => a cos —a sin — = cos —
2	2	2	2
2	2 7 ^	, /	2 7	7
a = cos — => a = ±i/cos — = cos —,
2	V 2	2’
(1-12)
where the real positive convention has been used in the last step. This means
that the state in which the system is in, is given in terms of the eigenstates
of the Sz operator by
|S-n;+) = cos j|+) + sin^|-).
(1-13)
(a)	From (S-l.4.17) we know that

(1-14)
So the propability of getting +7i/2 when Sx is measured is given by
|<5Ж; +|S  n- +>|
|	| sin7 = |(1 + sin7).	(1-15)
For 7 = 0 which means that the system is in the |S2; +) eigenstate we have
l(S>;+|S,;+)|! = 1(1) = j.	(1.16)
For 7 = 7Г/2 which means that the system is in the |5Ж;+) eigenstate we
have
+|5\r; +)|2 — 1-
(1.17)
For 7 = 7Г which means that the system is in the |S2; —) eigenstate we have
|(^;+|S2;-)|2 = |(1) = |.	(1-18)
1. FUNDAMENTAL CONCEPTS
29
(b)	We have that
{{Sx - (Sx))2) = (S2) - ({Sx)f.	(1.19)
As we know
S. = |(l+)(-l + l-)(+l)^
д2
si = T(l+)(-l + l-)(+l)(l+)(-l + l-)(+l)^
h2	h2
Si = T(l+)(+l + l-X-l) = y-	(1-20)
i
So
(Sx) =	cos|(+l +sin|(-| 7.7 L . - cos — sm —|— sm 2	2	2	2	№ 1 S* T		-l + l-x+l) tl . -8in7	cos	jl+) + sm jl-)
		7 — cos 2	I - 2			
(W)2 = {Si) =	h2 . 2	, — sin 7 and 4	' cos|(+l +sin|(-|	£ 4	cos		— )	
tl2 2 7	. 2 7 ti2 = — cos2 - + sm2 - = —. 4 L 2	2	4 So substituting in (1.19) we will have h2 {{Sx - (Sx))2) = T(1 -			sin2	2 7) = J- COS	7-	(1-21) (1-22)
and finally	((A5’®)2)7=0;|Sz;+) {(ASX) )y-v/2;\Sx;+} {(ASX) )7=0;|Sz;->			£ 4 ’ = 0, £ 4 		(1-23) (1-24) (1-25)
30
1.3 (a) The simplest way to derive the Schwarz inequality goes as
follows. First observe
(И + A*{/?!) (H + A|Z?>) > 0
for any complex number A; then choose A in such a way that the
preceding inequality reduces to the Schwarz inequility.
(b) Show that the equility sign in the generalized uncertainty re-
lation holds if the state in question satisfies
AA|a) = AAB|a)
with A purely imaginary.
(c) Explicit calculations using the usual rules of wave mechanics
show that the wave function for a Gaussian wave packet given by
(z'|a) = (27rd2)-1/4exp
satisfies the uncertainty relation
Prove that the requirement
(ж'|Дж|о!) = (imaginary number)(a/|Ap|a)
is indeed satisfied for such a Gaussian wave packet, in agreement
with (b).
(a)	We know that for an arbitrary state |c) the following relation holds
(c|c) > 0.	(1.26)
This means that if we choose |c) = |a) + A\/3) where A is a complex number,
we will have
«a| + A-(/?|M|a) + A|/3>) > 0
+ |A|2(/?|/?) > 0.
(1-27)
(1-28)
1. FUNDAMENTAL CONCEPTS
31
If we now choose A = — (/?|a)/{fl\/3) the previous relation will be
\ala/ m W) + m < u
(a|a)W) > |</3|«>|2.	(1.29)
Notice that the equality sign in the last relation holds when
|c) = |a) + Л|/?> = 0 H = -Л|/?>	(1-30)
that is if |a) and |/?) are colinear.
(b)	The uncertainty relation is
((ДА)2)((ДВ)2)>||([А,В]}|2.	(1.31)
To prove this relation we use the Schwarz inequality (1.29) for the vectors
|a) = AA|a) and \/3} = AB|u) which gives
((AA)2)((AB)2) > |(AAAB)|2.	(1.32)
The equality sign in this relation holds according to (1.30) when
AA|a) = AAB|u).	(1.33)
On the other hand the right-hand side of (1.32) is
|(ДАДВ)|! = 11([A, B]> |2 + | |({ДА, ДВ})|2	(1.34)
which means that the equality sign in the uncertainty relation (1-31) holds if
1 |({ДА, ДВ})|2 = 0 =4 ({ДА, ДВ}) = 0
=> (a\NANB + ABAA|u) = 0 (14-3) A*(u|(AB)2|u) + A(u|(AB)2|u) = 0
=> (A + A*)(u|(AB)2|u) = 0.	(1.35)
Thus the equality sign in the uncertainty relation holds when
AA|u) = AAB|u)	(1.36)
with A purely imaginary.
32
(c) We have
(1-37)
On the other hand
d
dx'
(1.38)
But
d
dx'
d i{p)x'
dx'
h
4d2
2 "I
(1.39)
h
1
' 2d?
So substituting in (1.38) we have
9 Й
(ж'|Др|а) = (р)(ж» + —
ih
—i2d?
ih
2d?
(l-«)
a
1.4 (a) Let x and px be the coordinate and linear momentum in
one dimension. Evaluate the classical Poisson bracket
Iх i F(px)]classical 
(b) Let x and px be the corresponding quantum-mechanical opera-
tors this time. Evaluate the commutator
/ ipxa
ж, exp ( ——
\ n
(c) Using the result obtained in (b), prove that
(sis7) =
1. FUNDAMENTAL CONCEPTS
33
is an eigenstate of the coordinate operator x. What is the corre-
sponding eigenvalue?
(a)	We have
classical
dx dF(px) _ dx dF(px)
dx dpx dpx dx
dF(px)
dpx
(1-41)
(b)	When x and px are treated as quantum-mechanical operators we have
fipxa
x, exp ——
\ n
hn n\
n=0
„ n! К
n=0
(c) We have now
fipxa
exp hr
Eid^^E/U^k
n=0	fc=O
^2,	1	/ ia \n 1	(ipxa\
Л?	=~“x[,m
(1-42)
(6) f грха\	/ zpxG,\
= exp I I x\x ) — a exp I J Iх /
I ('ЬРхО-У I n	('ЬРхО-У I м
= x exp I 1 \x / — a exp I 1 \x /
= (s'— a) exp |s').	(1-43)
So exp |s') is an eigenstate of the operator s with eigenvalue s' — a.
So we can write
I /	\	( ^Px^\ I n
|ж — a) = G exp I —— J |ж ),
(1-44)
where C is a constant which due to normalization can be taken to be 1.
x
34
1.5 (a) Prove the following:
(i)	(р'|ж|а) =
'ii) </?|ж|ог> = j dp' ф^р')Л^фа{р'),
where фа(р'} = (p'la) and фр{р'} = (p'|/3) are momentum-space wave
functions.
(b) What is the physical significance of
where x is the position operator and H is some number with the
dimension of momentum? Justify your answer.
(a) We have
(i)
= J dx'x'(p'\x'){x'\a) (S-=‘32)
= A J dx'~^ (е~^) ('Z^-)(J;/|
= ^~Q~i Уdx> {p'=
(р'|ж|а) = ih-f-(p'\a).
№|a) = j dp'{(3\p'){p'\x\a) = j dp'ф^р')Л—фа(р'),
ip' x'
(1-45)
(1-46)
where we have used (1.45) and that (/3|p') = фр(р') and (р'|ск) = фа(р')-
i
d
1. FUNDAMENTAL CONCEPTS
35
(b) The operator exp gives translation in momentum space. This can
be justified by calculating the following operator
P, exp
n
[p, xn]
00	1	A-V'-1	/z-X °° 1 ZirV
Hexp (“!“)	t1-47)
So when this commutator acts on an eigenstate |pz) of the momentum oper-
ator we will have
/ ix~A
P, exp I — I
„	/ ix\
^exp —
\ n, j
Г /ТхЕ\
v)|p>
p exp
Ip')
гх^\
-T) Ip>
h J
(p' + H) exp
p exp
p exp
(IX~A
exp
|p') - P exp
ixt
h
Ip') •
(1.48)
Thus we have that
Ip') = A|p' + =),
(1.49)
where A is a constant which due to normalization can be taken to be f.
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
л = - --- J А = оАг,
\mcj
write the Heisenberg equations of motion for the time-dependent
operators Sx(t), Sy(t\ and Sz(t). Solve them to obtain as func-
tions of time.
Let us first prove the following
[As, Bs] — Cs => [Ад, Bh] — C'h-	(2.1)
Indeed we have
[AH, Bh] = [U]ASU, U]BSU] = U]ASBSU - tfBsAsU
= W [As, Bs] U = WCSU = CH.	(2.2)
The Heisenberg equation of motion gives
= 4[5a,,^5J(S"='2°)^HS5,) = -^5„ (2.3)
at	irt	irt	irt
4[S„,1v5J(s’=”,^(aSI) = u.SI, (2.4)
at	irt	irt	irt
= — [5г, Я] = —[5г,^5г]	=4’2°) 0 => 5г = constant. (2.5)
at	in	in
Differentiating once more eqs. (2.3) and (2.4) we get
= —	— :^2SX => Sx(t) = Acosoot + В sincjt => 5ж(0) = A
=	(==) —oo2Sy => Sy(t) = C cosoot + D sincjt => A/(0) = C.
But on the other hand
dSx
dt
—Aco sin oot + Boo cos oot
OoSy —r*
—Coo cos oot — Doo sin oot =>
С = —B.	(2.6)
2. QUANTUM DYNAMICS
37
So, finally
Sx(t)	=	Sx(0) cos wt —	Sy (0) sin Cerf	(2.7)
Sy(t)	=	S'y(0) cos co’/, +	5'^(0) sin co’/,	(2.8)
Sz{t)	=	Sz(&).	(2.9)
2.2	Let s(t) be the coordinate operator for a free particle in one
dimension in the Heisenberg picture. Evaluate
|z(/),z(0)].
(2.Ю)
The Hamiltonian for a free particle in one dimension is given by
2m
This means that the Heisenberg equations of motion for the operators x and
p will be
dp(t)
dt
X*)
dx(t)
dt

1 .
itr
p(0)
1.
itr
—p(0) + ж(0).
m
Thus finally
i(i)
ifi
^,H] = x^ =
ifi
[i(l),i(0)] = ^p(O) + i(0),i(0)
(2-11)
Ar2P(<)«=т (-1’	-
2min	m
(2.12)
t r ,	, чп iht
= — b(o), 4°) = —
m	m
(2.13)
1
1
m
2.3	Consider a particle in three dimensions whose Hamiltonian is
given by
P
2m
38
By calculating [x  p, H] obtain
To identify the preceding relation with the quantum-mechanical
analogue of the virial theorem it is essential that the left-hand side
vanish. Under what condition would this happen?
Let us first calculate the commutator [x • p, H]
[f • p, H]
3	3 p2
4=1	j = l
(2.14)
The first commutator in (2.14) will give
+ ki.Pjfe) = T (PjiMij + ihStjPj)
Zji / L	Zj! / L	Zj! / L
(2-16)
The second commutator can be calculated if we Taylor expand the function
V(ж) in terms of Xi which means that we take V(ж) = атаж” with an
independent of ж^. So
[Pn У(ж)]
OO
рь E a^xi
n=0
n-l	я
Еа™Е(-^)ж" 1 = -ih'^annxf 1 = -ih-x—^2 anx™
n &=0	n	n
—^(^)-	(2.16)
OXi
The right-hand side of (2.14) now becomes
[ж-p, Я] = ^2 —SijPjPi + Е(-^)ж«^7^(^)
ij	i	г
= -f-ihx-^V(x).	(2.17)
m
2. QUANTUM DYNAMICS
39
The Heisenberg equation of motion gives
J - -	1 r- - m (2-17> ?	-
—x • p = — Is • p, H\ =--------x-YV(x)^-
dt	in	m
d	/ »2 \
-(s-p) = L -(?WV),	(2.18)
dt	/
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
• p\n) = ^(n| [x  p, H] |n) = 4- (En(n\x  p\n) - En{n\x  p\n}) = 0.
dt	in	in
So to have the quantum-mechanical analogue of the virial theorem we can
take the expectation values with respect to a stationaru state.
2.4 (a) Write down the wave function (in coordinate space) for the
state
eXP \1TJ '°''
You may use
(a/|0) = 7Г 1/<4s01/<2 exp
(b) Obtain a simple expression that the probability that the state
is found in the ground state at t = 0. Does this probability change
for t > 0?
(a) We have
|a, t = 0)
(a/|a, t = 0)
f—ipa\
exp \~ir)
{x'exp	|0) (Pr=4’c) {x' - a|0)
/ ,	\ 2'
— 1/4 —1/2	1 I x a |
7Г ' x0 exp —x ----------
\ x0 )
(2.19)
40
(b) This probability is given by the expression
l<0|«,t = 0)|2 = |(exp |0>|2-
(2.20)
It is
тг V2x01 exp
./2 । „2
/dx exp
/ a2 \	1
1
-2^|
O /	2	2X1
4/,9	~ ,a a a \
— x12- 2x'~ + — + —
2x2\	2	4	4 /
/ a2 \
eXPl-4^T^I<> = eXp(-4^J'	(2'21)
So
/	a2\
|(0|a,t = 0)|2 = exp	•
(2.22)
For t > 0
l(0|a,i)|! = |(0|M(i)|a,i = 0)|! = |<0|exp (-^) |a,i = 0)|!
= |e-’w/a(0|a,t = 0)|2 = |(0|a,t = 0)|2.	(2.23)
2.5 Consider a function, known as the correlation function, defined
by
ОД = (s(t)s(O)),	(2.24)
where x(t) is the position operator in the Heisenberg picture. Eval-
uate the correlation function explicitly for the ground state of a
one-dimensional simple harmonic oscillator.
2. QUANTUM DYNAMICS
41
The Hamiltonian for a one-dimensional harmonic oscillator is given by
Я =	(2.25)
So the Heisenberg equations of motion will give
dx(U) dt		p2(t) x [*(<). 2m +	2-2G)	
	- = = 2inm	m	+ ^mcv2— [s(t), x		(2.26)
dp(l,') dt	= 4 [?(;),»] = 4 in,	in,	, . p2(t) л PG), 9 7 + 2mLd 2m 2	V(t)	
	9 - - H-2«]	О = 9-й [ 2гЯж(/)] Lilli	= — mci?2x(t). (2.27)	
Differentiating once more the equations (2.26) and (2.27) we get
d2x(t)
dt2
d2p(t)
dt2
PU)
m
p(0)	D .
----cos ut + — sin ut =>
m	m
D = — ттш?ж(0).	(2.28)
-----^dt^ —	x(t) = Acosut + Bsinut => ж(0) = A
--------------------------(2=6) —ш2р(Т) => p(t) = C cosut + D sinut => p(0) = C.
m at
But on the other hand from (2.26) we have
d,x(l)
dt
—Lc?s(0) sin Ltd + B(jJ cos Ltd =
b = ''""
mu
So
x(t) = ж(0) cosut + sinLcJt	(2.29)
тш
and the correlation function will be
C(t) = (s(t)s(O)) (2=9) (ж2(0)) costed + (р(О)ж(О))----sinLc?t. (2.30)
42
Since we are interested in the ground state the expectation values appearing
in the last relation will be
= <°i^(» + »,)(»+»,)i°) = ^z;(0|<,<,,|0> =
ZjIILuJ	lI'IiUJ
 I TTlfUjJ	/1/+ \Z	+\IV
(p(O)s(O)) = id—— J—— (0|(af -a)(a + at)|0)
у La у lIiILu
= -Ao|aat|O) = -4.	(2.32)
£	La
Thus
C(t) = ------cos a?/ — i----sinced = ----(2.33)
2mw	2mw	2mw
2.6 Consider a one-dimensional simple harmonic oscillator. Do the
following algebraically, that is, without using wave functions.
(a)	Construct a linear combination of |0) and |1) such that (s) is as
large as possible.
(b)	Suppose the oscillator is in the state constructed in (a) at t = 0.
What is the state vector for t > 0 in the Schrodinger picture?
Evaluate the expectation value (s) as a function of time for t > 0
using (i) the Schrodinger picture and (ii) the Heisenberg picture.
(c)	Evaluate ((As)2) as a function of time using either picture.
(a) We want to find a state |a) = co|0) + Ci|l) such that (s) is as large as
possible. The state |a) should be normalized. This means
|<=o|2 + |C1|2 = 1 =$ |C1| = p - Ы2.	(2.34)
We can write the constands c0 and in the following form
Co = |c0|e’5°
ci = He"1 (2= } etS^l - |c0|2.	(2.35)
2. QUANTUM DYNAMICS
43
The average (x) in a one-dimensional simple harmonic oscillator is given
by
{x} = (а|ж|а) = (cq(0| + c*(l|) ж (co|0) + cjjl))
= |co |2(0|ж |0) + с£с1(0|ж|1) + с*со(1|ж|0) + |сх|2(1|ж|1)
= \co\2\  -----(0|a + af|0) + с*осг\  --(0|a + af|l)
V Zmcv	V 2mcv
+c*c0\  ---(1 |a + af|0) + \ci\2\  ---(1 |a + af|l)
V 2mw	V 2тиз
=	(COC1 + ClCo) = 2W—^(CoC!)
V 2rnuj	V 2тиз
= 21/ —cos^ - <Уо)|со|а/1 - |c0|2,	(2.36)
where we have used that x = \ a + od).
What we need is to find the values of |co| and Si — So that make the
average (ж) as large as possible.
= 0	=> У1 - |e0|2 - |Co|i '“J?1 1 - |e0|2 - |co|2 = 0
ЭЫ	У1 - |c0|2
=> Ы =	(2-37)
d(x\
=0	=> — sin(Si — So) = 0 =>	= So + шт, n 6 Z. (2.38)
UOi
But for (ж) maximum we want also
< 0 => n = 2k, ktZ.	(2.39)
x _x
l di— d\max
So we can write that
|a) = e’5o^|0) + e’(5o+2^)	11) = eiS° -^(|0) + |1)).	(2.40)
д/2	д/2	д/ 2
We can always take <Уо = 0. Thus
w = -U°>+11»-	<2-41)
44
(b) We have |a, t0) = |«)- So
= U(t,t0 = 0)|a,to) = e
е~ш'
= -Le-iE°t/n\0} + -j=e~iElt/n\l}
1 e-^i/2 (|0) + e-^|l)} .(2.42)
(i) In the Schrodinger picture
(x)s = {a,t0;t\xs\a,t0;t}s
= X (e^i/2/0|+ e^
26
=	, l icvt	=	.
2 V 2ттш? 2 V 2ттш? V 2mcu
(ii) In the Heisenberg picture we have from (2.29) that
s#(t) = ж(0) cos cvt + ^(0) sin Ltd.
таз
So
е~ш'
(2.43)
(а|жн|а)
-/H0!+ -^(!| 40)cosurt + ^-smud |Q> + —^|1>
A/	\/	\	• * Мм	/ a/	a/ £
I COS Lc?t(0|s 11) + I cos wt(l Is|0) + sin Lc?t(o|p|l)
222 тш
+ 5^— sincui(l|p|0)
2 muj
. h	. h	1	z\/
-\ -----cos Ltd + -a/---cos Ltd + ---smcvt(-i)\ ——
2 у Imuj	2 у Zmuj	Zmw	у 2
1	. /mfujj
2mijj V 2
/ h
\ ----COS ujt.
V 2muj
(2.44)
(c) It is known that
((Да:)2) = (ж2) - (a:)2
(2-45)
1е^/2-^/2)(1|ж|0)
2. QUANTUM DYNAMICS
45
In the Schodinger picture we have
-----(a2 + od2 + aod + oda), (2.46)
2mw
which means that
(ex, to, t{x |cx? to, t)g
2

So
h
2rnw
। 1 ^«((X'3t/2—ал
h
2rnw
h
2rnw
(2.47)
<(A*)2)s (2^>
Д 2	• 2
----cos сЛ = ----sm сЛ.
2тпш	2тш
(2.48)
In the Heisenberg picture
^(t)
ж(0) cos иЛ +-- sin иЛ
тш
ж2 (0) cos2 wt +	(0) sin2 wt
т2ш2
ж(0)р(0)	.	р(0)ж(0)
+---------cos иЛ sm иЛ +-------cos иЛ sm иЛ
тш	тш
П / 2	+2	+	+\	2»
----(а +<г +aa'+a'a)cos сЛ
2тш
mllLiJ /9	+	+ \ . 2
— -—-—-(« +«' — aa' — a'a)sm cvt
2m2(jj2
i hmtuv x x 4sm2utf
---A/--------(a + a')(a1 — a) —  
тш V 4m<jj---2
i / hmtuv x	x sin 2cvt
+--V -----(a _ а)(а + a )—о—
mcv у 4mw	2
^/2 C +	+\	2»
----(a +«' +aa'+a'a)cos сЛ
2тш
46
/ *?	+2	+	+ \ • *?	X ^Z . +2 о. . л
------(а + а1 — аа' — oTo)sin wt-\	(ат —a )sm2Lczt
2т^	7	2тш-1
\	о о	42 л
------(аа|+а|а) + -------a cos 2ш/, + -----a' cos2u?/,
2mw	2mw	2muj
Z Zz .4.2 о. , _
+-----(aT -a)sin2wt,
2mcv''
(2.49)
which means that
(xh)h
X last + a^a + a2 cos 2 a?/, + «Д cos 2 a?/, + z(a^ — a2) sin 2 a?/,
----- [(0|aa^|0) + (1 |aa^|l) + (l|a^a|l)]
-	[1 + 2 + 1] —
4ттш?
h
mw
(2.50)
So
// л \2\	(2.44) h	h о	. о
((As)2)h =--------------------cos2 wt =--------sm2wt.
2rnw 2rnw	2mijj
(2.51)
2.7 A coherent state of a one-dimensional simple harmonic oscil-
lator is defined to be an eigenstate of the (non-Hermitian) annihi-
lation operator a:
a|A) = A|A>,
where A is, in general, a complex number.
(a)	Prove that
|A) = e-lAl2/2eAat|0)
is a normalized coherent state.
(b)	Prove the minimum uncertainty relation for such a state.
2. QUANTUM DYNAMICS
47
(c)	Write |A) as
oo
lA) = 52 f(n)\n}-
72=0
Show that the distribution of |/(n)|2 with respect to n is of the
Poisson form. Find the most probable value of n, hence of E.
(d)	Show that a coherent state can also be obtained by applying
the translation (finite-displacement) operator (where p is the
momentum operator, and I is the displacement distance) to the
ground state.
(e)	Show that the coherent state |A) remains coherent under time-
evolution and calculate the time-evolved state |A(t)). (Hint: di-
rectly apply the time-evolution operator.)
(a) We have
a|A) = e-|A|2/2«eAat |0> = e"|A|2/2 [a, eAat] |0>,	(2.52)
since a|0) = 0. The commutator is
So from (2.52)
a|A) = e-lAl2/2AeAat|0> = A|A),	(2.54)
which means that |A) is a coherent state. If it is normalized, it should satisfy
also (A |A) = 1. Indeed
<A|A> = (0|eA*ae-|A|2eAat|0) = e-|A|2(0|eA*aeAat|0)
48
= е"|А|2 Е -Г“f(A*)"Am(O|a"|(at)m|O> [(<?)т|0) = V^|m)]
п,т п-т-
(b) According to problem (1.3) the state should satisfy the following relation
Дж|А) = cAp|A),	(2.56)
where As = x — (A|s|A), Ap = p — (A|p|A) and c is a purely imaginary
number.
Since |A) is a coherent state we have a|A) = A|A> => (A|af = Using this relation we can write Ж|А) - \l (a + af)|A) - J V Zmcv	У and (P = (A|a?|A) =	+ «*)W — l/9 (A + A ) V Zmuj and so As|A) - (s (s))|A) - Similarly for the momentum p = p|A) - V~iJ (a1 a)|A) — г	= (A|A*.	(2.57) —(A + «f)| A)	(2.58) Zmuj = (АД«А|<.|А} + (А|<-’|А)) V Zmw (2.59) —(at_A*)|A).	(2.60) — a) we have /mhw , +	, s,,, </	(«’ A)|A)	(2.61)
2. QUANTUM DYNAMICS
49
and
, /rnhw	,
=	— (A -A)	(2.62)
and so
Ap|A) = (p - <p>)|A> =	- A*)|A) =>
(a+— A*)|A> = -n/-|-Ap|A).	(2.63)
V mruv
So using the last relation in (2.60)
I ti I 2	?
AslA) = \------(—i)\—z—AplA) =---------------AplA) (2.64)
1 1 V2m^v 7V mhw n 1 ^_rru^ n 1 V 1
purely imaginary
and thus the minimum uncertainty condition is satisfied.
(c) The coherent state can be expressed as a superposition of energy eigen-
states
oo	oo
M = J>)("|A) = £/(")!"}.	(2.65)
n=0	n=0
for the expansion coefficients f(n) we have
/(n) = (n|A) = (n|e-|A|2/2eAat|0) = e-|A|2/2(n|eAat|0)
OO 1	oo-l
= e-|A|2/2{„|	-(A«*)”|0) =	£ —A”1(n|(«t)"i|O)
0ТП •	r\ ТП •
m=0
= e—lAl2/2	-^-AmVnd(n|m) = e~^2~^=Xn =>	(2.66)
~ m!	v и,!
m=0	V l{'*
l»|2 = ^ехр(-|А|2)	(2.67)
which means that the distribution of |/(n) |2 with respect to n is of the Poisson
type about some mean value n = |A|2.
50
The most probable value of n is given by the maximum of the distribution
|/(n)|2 which can be found in the following way
№ + i)|2
№)l2
ДЖ exp(—|A|2)	n + 1-
(2.68)
which means that the most probable value of n is |A|2.
(d) We should check if the state exp (—ipl/K) |0) is an eigenstate of the an-
nihilation operator a. We have
a exp {—ipl/K) |0) = [a, el-4’'/*)] |p)	(2.69)
since a|0) = 0. For the commutator in the last relation we have
г	n 00 1 (.-il\n	°° 1 /_n
n=0 П- \ П /	п=1 П- \ П / fc=l
1 (-НУ
n!\ h
A „_! mKv
k=l v z
. mKjj i—il
V 2 \ h
1	(—Up
Л)!
where we have used that

(2.70)
/ mhuj	X
[a,p] = zU—[a,a1
So substituting (2.70) in (2.69) we get
mtiw
2
(2.71)
a [exp (—ipl/K) |0)] = I
(2.72)
which means that the state exp(—ipl/K) |0) is a coherent state with eigen-
(e) Using the hint we have
| A(Z)> = U{t) | A> = е-ж/а|А) (=6) e~iHt^ £ e-lAl2/2 -L Xn |ra>
n=0 Vn'.
2. QUANTUM DYNAMICS
51
Thus
1	z x CO
-^Me-|A^/2 i An|^ (^)
Vnl	n=0
e-i^n e-^/2e-|A|2/2_LA«|^
n=o '	vn!
e-iut/2 ^2	c-|Ae~,u-T/2 (Ле	)	(2Л>6)
n=0	Vn}-
n=0
~^M»+|)e-|A|72 1 дп|^
V?z!
е-г^/2|Ае-^
(2.73)
a\X(t)) = e~iwt^2a\Xe~iwt) = Xe~iwte-iwt^2\Xe~iwt)
= Ae"^|A(t)).	(2.74)
2.8 The quntum mechanical propagator, for a particle with mass
m, moving in a potential is given by:
roo
K(x, y;E) = dtetEt'*K(x, y,t,0) = Aj2
sin(nrs) sin(nry)
Yr
2m
2
where A is a constant.
(a)	What is the potential?
(b)	Determine the constant A in terms of the parameters describing
the system (such as m, r etc. ).
We have
K(x,y,E)
[°° dteiEt/nK(x, y-1, 0) = Г dteiEt/n{x,t\y, 0)
/0	Jo
[°° dteiEt/n(x\e-iHt/n\y)
/0
/»oo	, __
/ dteiEt^{x\e-iHt^\n}{n\y}
dteiEt^^e-iEnt/H{x\n}{n\y}
roo
^Фп(х)ф*п(у) /
n	J0
ei(E-En)t/ndt
52
So
n
^фп(х)ф*п(у)1ш1
п
iti
E — En
sin(nrs),
1,1 ci(E-En+ie)t/n
Е — Еп + is
n
En
sin(nrs) sin(nry)
E - ^n2
2m
h2r2 2
——n.
2m
(2-75)
(2.76)
For a one dimensional infinite square well potential with size L the energy
eigenvalue En and eigenfunctions фп(х) are given by
h2
2m
(2.77)
Comparing with (2.76) we get = r => L = у and
for 0 < x < -
r
otherwise
(2.78)
while
A 2r л 2hr
— = — => A = г------.
1П 7Г	7Г
(2.79)
2.9 Prove the relation
ф = <s(l)
dx
where 0(ж) is the (unit) step function, and <У(ж) the Dirac delta
function. (Hint: study the effect on testfunctions.)
For an arbitrary test function /(ж) we have
dx
df(x)
—-—dx
dx
2. QUANTUM DYNAMICS
53
(2.80)
2.10 Derive the following expression
тш
2 sin^T)
x2 + 2ж^) cos(cuT) — X0Xt
for the classical action for a harmonic oscillator moving from the
point x0 at t = 0 to the point xt at t = T.
The Lagrangian for the one dimensional harmonic oscillator is given by
£(ж,ж) =	— ^mw2x2.	(2.81)
From the Lagrange equation we have
d£ d d£ n (2.81)	9 d
—------= 0 => — тш x------------(mx) = 0 =>
ox dt ox	dt
x + (v2x = 0.	(2.82)
which is the equation of motion for the system. This can be solved to give
x(t) = A cos Ltd + В smart	(2.83)
with boundary conditions
x(t = T)
x0 = A	(2.84)
xt = x0 cos N + В sin aT => В sin aT = xt — x0 cos aT =>
Xt — Xq COS CjjT
sinwT
(2.85)
54
So
s(t)
i(t)
xt — xq coswT .
x0 cos иЛ H-----------—------sm иЛ
since?!
s0 cos (jjt sin (vT + xt sin art — x0 cos ojT sin art
sin ojT
xt sin art + x0 sin ш(Т — t)
sincejT
xtoj cos art — xooj cos oj(T — t)
sincejT
(2.86)
(2.87)
With these at hand we have
s
о
о
(ж,ж) = J dt \^mx2 — ^таз2x
, d
2mdt
^mxx — ')гпа2:г2

2 1	171	•
аз x N----------xx
J 2
0
MT)i(T) - z(0)i(0)]
m
~2
XT^rr, (xt cos азТ - x0)
.зтаз!
таЗ Г 2	rri	I 2 m]
----— XT COS 031 — X0Xt — XqXt + XQ COS 031
2 sm (vT l	j
—-----— (x? + s§) cos wT - 2xoxT .
2smLc?l L	J
-----—\xT - X0 COS031 )
smccJi
(2.88)
2.11 The Lagrangian of the single harmonic oscillator is
Г 1 ’2
L = -mx
La
1	2 2
----таз x
2
(a) Show that
'i'ScX
where Sa is the action along the classical path xci from (sa,ta) to
(xb,tb) and G is
{xbtb\xata) = exp
(7(0, tb\ 0, ta)
(7(0, tb', 0, ta) —
2. QUANTUM DYNAMICS
55
(W+1)	( JV r	1
2 eIP^gg(»+i-W)2-2«™^
where £ = ^.
[Hint: Let y(t) = x(t) — xci(t) be the new integration variable,
xci(t) being the solution of the Euler-Lagrange equation.]
(b)	Show that G can be written as
.	(w+i)
( m \ 2 г	/ t \
G = lim -----—	/ dyi ...dyf/exp(-n an)
N-t-oo \27Tine/ J
У1
where n =
and nT is its transpose. Write the symmetric
L Ух
matrix a.
(c)	Show that
J dyi ... dyNexp(—nTcm) = j dNne n <7n =
[Hint: Diagonalize a by an orthogonal matrix.]
(d)	Let	deter = deta'N = p^. Define j x j matrices a'- that con-
sist of the first j rows and j columns of a'N and whose determinants
are p:t. By expanding in minors show the following recursion
formula for the p,:
Pj+i = (2 “ £2^2)Pj “ Pj-i J =	(2-89)
(e)	Let <^(t) = epj for t = ta je and show that (2.89) implies that in
the limit e —> 0, <^(t) satisfies the equation
Ф
№ = -ш
with initial conditions = ta) = 0,	= 1.
56
(f)	Show that
I muj f гтпш 2	2	1
(xbtb\x„t„) = . /———-—-—Tz—exn< ——-—-——— (xf + x„ ) cos(ll?T) — 2xnxb >
X b 61 “ a/ у 2Tvihsin(wT) ' [2ti sin(^T)И 6	’ a b]f
where T = tb — ta.
(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
У dx\xt}{xt\ = 1	(2.90)
So
(xbtb\xata) = lim / dxidx2 ... dxN{xbtb\xNtN'){xNtN\xN_ltN_^ ... X
{xi+1ti+1\xiti} . . . {Xit^Xata). It is (xi+Ji+^Xiti) = (xi+^e-^+'-^^Xi) = {Xi = (.7y+1|c-’ft(2W +2тШ X )|Жг-) . £	 e 1	2 2 = (жг+1|е	Л2	|жг) = e »2	’(жг-+1|е ft2’»|st). For the second term in this last equation we have (жг+1 \е~г^ |жг) = У dpi{xi+1\e~^^\pi}{pi\xi} = У dpie~ln2^ ^Xi+1\pi){pi\xi) = Д- [ dpie-^e^'*1-^ 2tvTi J _	1	[ J ~г2^л1р1~2р‘2?(х‘+1~х 27Th J Рг = 	егтЛ г2 (^+1 xd f dpie~'2m 2irfi	J P	(2-91) +1|е-‘я’"Чи) (since e is very small) (2.92) s)+^(^+i-^)2-<(^+i-^)2 [pi Pi "^“(a'»+1 a'«)]
2. QUANTUM DYNAMICS
57
1	is m? („	„ =		e2mh e2 H'+l X> 2Yh = л/	e2fteH»+l Я V 2тгЫв Substituting this in (2.92) we get i /	,	f m \2 \^«+l^«+l	\ 9 'A ) \ £j I ( (r I (/& / and this into (2.91): (s6t6|sata) = / Psexp] ^S[s] !> =	у Tr2hm V is (2.93) A[^(x‘+i~xi)2~2£ma,2x‘]	(2.94)
J	к гь lim / dxi ... dxN (	— N-i-ooJ	\27vihe, Let y(t) = s(t) — sc/(t) => a boundary conditions y(ta) = у have Dx = Dy and	(”+!)	(  N г	1 \	2	I 1 *	ТП ,	ч 9 _L	2 2 ) exP j > > |_^(ж1+1 -	~	xi •(t) = y(U) + xci(U) => x(U) = y(£) + xci(U) with [tQ = 0. For this new integration variable we
xci,y + Xcfidt
= / £(xci,xc Jta r, d£ tb = Sci + — у ox . ta So (sjtj |sata) = with G(0,i6;0,ia) = lim / dyi... dyN N-i-oo J	x d£	d£ . . d2£	2 ! d2£	' l) + ъ-	У + V7V	У + 2 Л2-	У + 2 л”-”2	У Ox	Ox	Ox	Oxz xci	xci	xci	ХС1 Сь [д£ d	г*ь п .2	1	2 21 7 + / л	Y	У+ Umy-lm^y\dt Jta Ox dt \ Ox /	Jta L	J ° L	\	/ J xci 1 Dy exp |^ScZ + ^jt [^my2 - |m^2y2] dt^ [iS ,1 ^P[~Y G^tb-^ta)	(2.95) ( m \ 2	( i	m	2	1	22 <2nihJ	»>	2e"“W-
58
(b)	For the argument of the exponential in the last relation we have
 TV -	1	-i
Z ТП ,	2 J- 2 2 (3/0=0)
jg[Sfe«-w)2-^42] =
- N	N i
z r-ч m 2 I 2	\ Ъ "L 2 c (^+1=°)
ft > > RM+i + Уз - Уз+гУз - УзУз+1) "7 L	=
ri з=о Z£	n i,j=i 2
N	2 N
m x	lemw x
“^7^ > > \2У^зУз ~ УЛз+гУз ~ У^+1,зУз)-, уг<Ш-(2-96)
«j=l	i,j=l
where the last step is written in such a form so that the matrix a will be
symmetric. Thus we have
	G =	lim N—>oo	/ m >	(W+1) 2	[ dyi	... dyxexp^—n		Tan)		(2.97)		
			\2тгЛе)									
with												
	2	-1	0 ...	0	° 1		Г 1	0	0 .	. 0	01	
	-1	2	-1 ...	0	0		0	1	0 .	. 0	0	
m	0	-1	2 ...	0	0	•	9 гетит	0	0	1 .	. 0	0	
2eih						1 2h						;2.98)
	0	0	0 ...	2	-1		0	0	0 .	. 1	0	
	0	0	0 ...	-1	2 _		0	0	0 .	. 0	1	
(c)	We can diagonalize a by a unitary matrix U. Since a is symmetric the
following will hold
a = U^aDU aT = UTaD(U^T = UTaDU* = a => U = U*.	(2.99)
So we can diagonalize a by an orthogonal matrix R. So
a = RTaoR and det R = 1	(2.100)
which means that
л/det cr
2. QUANTUM DYNAMICS
59
where аг- are the diagonal elements of the matrix ap.
(d) From (2.98) we have
We define j X j matrices <r'- that consist of the first j rows and j columns of
a'N. So
	- 2 - £V2	-1	0	0	0
	-1	2 -	...	0	0	0
	0	-1	0	0	0
det <t'+1 = det				
	0	0	2 - £V2	-1	0
	0	0	-1	2	=2, ,2 — £ U)	-1
	0	0	0	-1	2 — e2<jj
From the above it is obvious that				
det	<r'-+1 = (2 - £2LJ2) det <t'-	— det <7'_х =;		
Pj+i = (2 - AuY)p:i - for j = 2,3,... ,N (2.103)
with po = 1 and pi = 2 — e2uj2.
(e) We have
) = Ф(1а + je)
Ф(фа + (.7 + l)s)
=/• Ф(1 + s)
epj
epj+i = (2 - £2Lo’2)£p) - sp^.x
2<^(fa + je) ~ е2^2ф{1а + je) - ф(1а + (j - l)g)
2<£(t) - £V<^) - ф{1 - £).	(2.104)
60
So
ф(ф + e) — ф(ф) = ф(Г) — ф(ф — е) — е2а>2ф(ф) =>
ф(Н-е)-фр) _ фр)-фр-е)
----------------------- = -ш2ф(1') =>
е
hm------------------= — ш2ф(ф) =>	= —ш2ф(ф).	(2.105)
From (с) we have also that
ф(фа) = epo -> 0	(2.106)
and
(1ф	ф(1а + е) - ф(1а) e(pi - р0)
-Т-(^а) = ------------------ = ---------- = Р1 “ Ро
at	ее
= 2-е2ш2- I ^1.	(2.107)
The general solution to (2.105) is
ф(1) = Asin^t + 5)	(2.108)
and from the boundary conditions (2.106) and (2.107) we have
ф(фа) = 0 => Asin(c<;ta + 5) = 0 => 6 = —wta + та n Q Z (2.109)
which gives that <^>(t) = Asincj(t — ta), while
= Ajcos(7 - ta) => ф\1ф) = Ajj (2^>7)
Аш = 1 A = -	(2.110)
Thus
, 4 sincj(t — tA	z
ф(1) =-----------aA.	(2.111)
(f) Gathering all the previous results together we get
m \(N+V
2iritieJ
у/*
л/ deta
2. QUANTUM DYNAMICS
61
(d)
(2.П1)
So from (a)
(x^t^
(2.88)
m \V2
2irihJ
m \V2
2тгЛ/
I тш
2тггЛ. sin(a;T)
'iScf
exp
h
/erti \N r1/2
I 2ше\
lim £ ------ deta
v->oo \ m I
т	\-1/2 (e) / m \]
hm epjv I = I -—rr )
G(0,t6;0,ta)
тш
--—— exp
2tviIi sm(ojT)
гтгш
-1/2
(2.112)
^xl + x^cosiNT-2xbxa] .
L2nsma;i L	JJ
2.12 Show the composition property
j dx1Kf(x2,t2', X1,tl')Kf(x1,tr, XoAo) = Kf(x2,t2',xoAo)
where Kf(xr, tp, x0, t0) is the free propagator (Sakurai 2.5.16), by
explicitly performing the integral (i.e. do not use completeness).
We have
j dx^K}(x2,t2,xutQKf(x1,ti-,xo,to)
im(x2 — xQ2
2h(t2 - G)
im(x1 — ж0)2
.	- t0) .
/ m
1\ Z—ГТ7--------г exp
у 2Tvih(t2 — ti)
/ m
- / -—-------г exp
У 27rz7i(ti —10)
m /	1	imx^
2тгг7гУ (t2 — ti)(ti — t0) F27i(t2-ti)	2h(t2 —
Г	im 2 im „
i	।	-------------2xix2------------
27z(t2-ti) 1	27z(t2 — ti)	2h(t2-t
• 9
imxQ
2h(t2 -tQ
2ж1Ж0
62
т /	1	Г im х%	Xq
2Tvih\l (t2 - ti)(ti - t0) t .(^2 - ti) + (ti - to)
/(im 1	1
dxi exp < —— 7-- + 7--
P [2h [(t2 -tj (tx -t0)
„ im x2	x0
1 — j. #1	7--------7 + 7--------7
[(t2 — ti) (ti — t0)
m /	1	| im
---7T i	---------г exp < —-
2тгг1т,у (t2 — tx)(tx — t0)	( 2ti
t2 — to
(t2 — tx)(tx — t0)
X1
2h t2 —10 im
7--7-------Г7-------Xi
im (t2 - ti)(ti - t0) h
^(ti — to) + Xo(t2 — ti)
(t2 — tx) (tx — t0)
m /	1	J im x^(ti — t0) + •'-'o(t2 — tx)
2Tvih\l (t2 - tx)(tx - t0) eXp ( 2П _	(t2 - tx)(tx - t0)
У dxi exp
~ to
(t2 — tx)(tx — t0)
x2^ti — to) + ^o(t2 — tx)
(t2 — t0)
im___________1_________[s2(tx - t0) + x0(t2 - tx)]2
2h (t2 — tx)(tx — to)	(t2 — to)
m	1
2тгЛ\] (t2 - tx)(tx - t0) \
Tr2ih(t2 — tx)(tx —10) f im 1
m(t2-t0) eXp ( 2h (t2 - ti)(ti - t0)
^(ti - t0)(t2 - t0) + a:g(t2 - ti)(t2 - t0)
^2(tx - tp)2 - a:g(t2 - ti)2 - 2a:2a:o2(ti - t0)(t2 - tx)
(t2 — t0)
m
27vih(t2 — tx)
im
2П
^2^1 ~ ^o)(^2 - t0 - ti + t0) + a:g(t2 - ti)(t2 - t0 - t2 + tx)
(t2 — t0)(t2 — tx) (tx — t0)
2x2xo2(ti — to)(t2 — tx)
(t2 — t0)(t2 — tx) (tx — t0)
m
27vih(t2 — t0)
exp
im(x2 - s0)2
2/i(t2 -10)
Kf(,x2,t2’,xo,to)-
(2.113)
2. QUANTUM DYNAMICS
63
2.13 (a) Verify the relation
(zTze \
where П =	= n — — and the relation
at г c
d2x t/П	1 (dx dt2 dt	[ 2c \ dl, (b) Verify the continuity equation 8p ^1 "^7 + V • J = ot with j given by j = (-") 3=(v>*vV) - J (a) We have [ПЙП;] = |ps-	— ,pj	— ihe dA; ihe dA; 		J_  	c_   c dxi	c dxj (^tie\ — I I \ C / We have also that dxi	1 г	m 1	П2 ——	=	— \X; ,H = — X;,	+ ed dt	z/i	ih	2m —	{[жг, 1У П? + П? [хг, П I / lZj Illi 2ih	r Пг —	9-+ — => 2inm	m	dx\ x В - В x —	. </t) 0 (—) Л#|2- \mcj = -| [Pi,Aj] - | [A,Pj] ihe (ЭА;	8A;\ 	 1 	J_ 	 	1 с у dxi	dxj J (2.114) 1 _ i Г	n2 ih	2m 1 Ш = ifi2m {[^bP;] П; + IL, [ж,-,р,]}
64
d2Xi
"dt2
(2.114)
d2Xi
m
dt2
(Ex
mdt2
n2
2m
1 dxj
ih dt '
1
iW2m2
1
2m2ih
——(
2m2c '
e f Xj	Xj
---— m I 6iik~T Bk — &iki Bl- —
2m2c V 3 dt 3 dt
eEi ---
2c
1 I x
e E ----I —
2c \ dt
1
ihm
Лв D TT	TT D
^ijkB^A-j T	SijkBjBj- T
c	c
€
Пр ф]
гпт
e
ihm
dt
x
dt
trim
e дф
m dxi
dt I .
(b) The time-dependent Schrodinger equation is
f-\ 2
p--
C I
-Д- -ihv’-
2m
c
• -ihv'-
c
— -Ti2V' • V' + -?nv' • A(f') + ih-Atx1)  V' + ^A2
2m [	с	c	c2
+еф(х'')ф(х'А')
1
e
e
e -?
e2
-I
— -Д2 V'2-0 + -ih fv' • А) ф + 2ih-A  VV +
2m	c v 7	c
e2
^A2
c2
(2.115)
еф\а, to’, t)
- еф^х'^х^а, to’, t)
(£') ф(х'А')
I
+ ефф. (2.116)
Multiplying the last equation by ф* we get
•Й / * 9 /
w -&Ф =
2. QUANTUM DYNAMICS
65
1
2m
-h2^*V2^ + -ill (V' • a) IV’I2 + 2ih-A •	+ ~?A2IV’I2 + e^lV’l2.
c v 1	c	c2.
The complex conjugate of this eqution is
-Лф—ф* =
dt
1
2m
2
-П2ф^'2ф* - -ih (V' • A) |< - 2ih-A • V’VV* + “?A2 \ф I2
c '	'	c	c2
+ еф\ф\2-
Thus subtracting the last two equations we get
2.14 An electron moves in the presence of a uniform magnetic field
in the г-direction (B = Bz}.
(a) Evaluate
where
[Пж, Щ],
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
Ek,n
h2k2 (\eB\h\ f 1\
-----+ I ------- I I + — ) •>
2m \ тс I \	2 J
where hk is the continuous eigenvalue of the pz operator and n is a
nonnegative integer including zero.
The magentic field В = Bz
of the form
can be derived from a vector petential А(ж)
4, = --^, Ay = —, Az = 0.	(2.118)
Thus we have
[Щ, Щ]
eAx eAyl (2.11s) Г eBy еВх'
[ft-—,ft-—]	= [ft+
eB	eB	iheB iheB
eB
ih—.	(2.119)
(b) The Hamiltonian for this system is given by
f-.\ 2
eA\ 1 „	1 „	1 9
p----=-----П2 +--П2 +--p2
c j	2m 2m y 2тп
(2.120)
where Hr = ^П2 + ^П2 and H2 = ^p2z. Since
[Hi, H2] — - -
4m2
eBy\2 f eBx\2 2
Px + ^) +
= 0
(2.121)
there exists a set of simultaneous eigenstates \k, n) of the operators Hi and
H2. So if hk is the continious eigegenvalue of the operator pz and \k, n) its
eigenstate we will have
(2.122)
2. QUANTUM DYNAMICS
67
On the other hand is similar to the Hamiltonian of the one-dimensional
oscillator problem which is given by
H = —p2 + Anw2x2	(2.123)
2m
with [ж,р] = Hi. In order to use the eigenvalues of the harmonic oscillator
En = hjj (n + 0 we should have the same commutator between the squared
operators in the Hamiltonian. From (a) we have
eB
[Hj;, Щ] — lh
г/П^сЧ .
=> [ ( —- ), Щ] = ih.
\ et> /
(2.124)
So Hi can be written in the following form
1	1 ^9	1	1 /Щс\2 leBl2
1 = 2m x+2m y~2m. у+2^п\~Мз) —&~
= —+	(—V.
2m y 2 у me J \ eB J
In this form it is obvious that we can replace uj with to have
me
(2.125)
H\k, n)
2.15 Consider a particle of mass m and charge q in an impenetrable
cylinder with radius R and height a. Along the axis of the cylin-
der runs a thin, impenetrable solenoid carrying a magnetic flux Ф.
Calculate the ground state energy and wavefunction.
In the case where В = 0 the Schrodinger equation of motion in the
cylindrical coordinates is
-J [VV] = 2£0 =»
-S +	+	+	=	(2.127)
68
If we write Ф(р, ф, г) = Ф(^>)/?(р)Я(г) and к2 = we will have
R(p)Z(z)d2$
dp2	p dp p	d^Z
+К(р)Ъ(фф^ + ^д(/,)ф(йг(г) = о
1 cPR 1	dR	1 Л2Ф	1 d2Z ?2
D/ A , 2 +	D/ \	3	I—2Л/7/2	+	<7/ \ 7 2	+ к —	0(2.128)
R(p) dp2	R(p)p	dp	р2Ф(ф) dф2	Z(z) dz2
with initial conditions Ф (pa, ф, z) = У(И,Ф,2) = ^(р,Ф,0) = ^(р,ф,а) = 0.
So
1 d2Z
1^-dZ2 ~
with
d?Z
hz2
~ilz (2.129)
) = C sin Iz
n = ±1, ±2,...
Z(0) = 0 => Ax +	= 0 => Z(z) = Al (eilz - e~ilz
7f
Z(a) = 0 => C sin la = 0 => la = itk => I = ln = n—
a
So
Z(z) = C sin lnz
Now we will have
1 d2R 1 dR 1 Л2Ф 2	2 _ n
R{p) dp2 ”1” R(p)p dp ”1” р2Ф(^>) d(p +
p2 d2R p dR 1 б/2Ф 2/72	72ч
R(p) ~dp2+R(p) ~dp+ Ф(<^>) d<^>2 +f>	“ ) =
=>	= -m2 => Ф(ф) = е±гтф.
Ф(<£) d<j)2	v '
with
Ф(^> + 2тг) = Ф(^>) =t m £ 2.
(2.130)
(2.131)
(2.132)
So the Schrodinger equation is reduced to
p d R p dR 2 । 2/12	/2
RMd^+RM^~m P{
2. QUANTUM DYNAMICS
69
9'
) - R(p) = 0
1	dR
d2R 1 dR
dp2 + p dp
cFR
d(y/k2 - 12рУ Vk2 - l2p d(y/k2 - l2p) I/
=> R{p) = AsJm(Vk2 - l2p) + B3Nm(\/k2 - l2p)
2-Z2
9
m
(k2-l2)p2_
R(p) = о
(2.133)
In the case at hand in which pa —> 0 we should take B3 = 0 since Nm —> oo
when p —> 0. From the other boundary condition we get
7?(7?) = 0 => A3Jm(RVk2 - I2) = 0 => RVk2 - I2 = кт„ (2.134)
where K,mv is the i/-th zero of the m-th order Bessel function Jm. This means
that the energy eigenstates are given by the equation
7Г\2
n—
. a J
Kmv
R2
(2.135)
while the corresponding eigenfunctions are given by
Фпти(х) = АсТтС^р)егтф sin (n—'\	(2.136)
it	\ a /
with n = ±1, ±2,... and m E Z.
Now suppose that В = Bz. We can then write
-	( BP2\ A / Ф \ A
A= Ь- и	<2-137)
\ 2p J \2тгрJ
The Schrodinger equation in the presence of the magnetic field В can be
written as follows
1
2m
—гД V
еА(ж)
c
= Е'ф(х)
h2 г
2m dp ”1” dz
\d	^d	a1
P я + z я +
dp	dz	p
a1	/ d	ie	Ф
p	\dc[>	кс2тг
d	ie	Ф	\
d</>	he	2tf	J
(2.138)
70
Making now the transformation	we get
ti1 2
2m
h2
2m
. d
PA~
L dp
dp2
where D2 =	-
A д
Z~d~z
i_a
pdp
^)2. Leting A =
Ф~Оф
P
—D2
p2 ф
. д
L др
а21
. д 71
ZW~ + ^~
dz р
'5 + ^ Ф(х) = ЕФ(ж),
(2.139)
ie_
he 2-л\
9	f д2	2ie Ф д	42
ф	\дф2	Кс2ттдф
Following the same procedure we used before (i.e. ф(р, ф, z) = /?(р)Ф(^>)^(г))
we will get the same equations with the exception of
d2 л д ,2
— - 2iA— - A
оф2 оф
The solution to this equation is of the form е1ф. So
we §et
= (— 2iA—— A2
\дф2 Адф
(2.140)
9;t о?2Ф . t/Ф , 9	,9.,
ф = -Ш2Ф => —— - 2iA— + (m2 - А2)Ф = 0.
аф2 аф
12е1ф - 2гА1е1ф + (m2 - А2)е1ф = Q I2 - 2iAl + (m2 - A2)
2iA ± J-^A2 - 4(m2 - A2) 2iA±2im .. л .
=> Z =-------------------------------------------= г(А ± m)
which means that
ф(<^>) = C2ei(A±m)r/>.	(2.141)
But
Ф(ф + 2тг) = Ф(^>) A±m = m' m' £ Z
=> m = ±(m' — A) m' £ Z.	(2.142)
This means that the energy eigenfunctions will be
ФптЛх) = А^тС^р)ет'ф sin (n~\	(2.143)
it	\ a /
but now m is not an integer. As a result the energy of the ground state will
be
h2
2m
(2.144)
2. QUANTUM DYNAMICS
71
where now m = m' — A is not zero in general but it corresponds to m' £ Z
such that 0 < m' — A < 1. Notice also that if we require the ground state to
be unchanged in the presence of B, we obtain flux quantization
,	.	e Ф ,	Ткт Нс .	„	.
m' — A = 0 => ------= m' => Ф =----------- тп! E Z. (2.145)
nc 2tf	e
2.16 A particle in one dimension (—oo < x < oo) is subjected to a
constant force derivable from
V = Xx, (A > 0).
(a)	Is the energy spectrum continuous or discrete? Write down an
approximate expression for the energy eigenfunction specified by
E.
(b)	Discuss briefly what changes are needed if V is replaced be
V = A|s|.
(a) In the case under construction there is only a continuous spectrum and
the eigenfunctions are non degenerate.
From the discussion on WKB approximation we had that for E > У(ж)
72
where q = а [у — ж] and a = (ypr)
On the other hand when E < У(ж)
fLj. IL/, ^2m<-Xx ~ E^2mX^
[Аг - E]i/« exp
= [4^Ч4-^/3]'	(2Л47)
We can find an exact solution for this problem so we can compare with
the approximate solutions we got with the WKB method. We have
77|a) = £|a) => (p|77|a) = (p|E|a)
p2
(p|^— + Аж|а) = E{p\a)
2m
p2	d
—a(p) + ihX —a(p) = Ea(p)
2m	dp
d .	-2 x
dp '*
da(p)
a(p)
dp '* '
—a(p} = — (E — L— ) a(p)
hX\ 2m J
ln“W =tA\Ep-sL +C1
i ( P3 у
hx \6m p
oe(p) = cexp
(2.148)
We also have
(2Л48)
{E\E') = j dp{E\p){p\E') = j a*E(p)aE/(p)dp
|c|2 [ dp exp \-L(E - E’)p] \с\22ттШ(Е - E'}
J	L/iA	J
1
\LLLx
(2.149)
So
1	" i (	\"
“eW = ?srxp 	(2,150)
d
c
2. QUANTUM DYNAMICS
73
These are the Hamiltonian eigenstates in momentum space. For the eigen-
functions in coordinate space we have
1
2тг1ц/Х
(2.150)	1 f , i22L N
1 = 7-----= / dpe n enx
2Mvx/X J
p3 i (E \
tA-x~x)p
(2.151)
hX6m
Using now the substitution
we have
where a =
p3
p
и —------------
(712mA)1/3 7iA6m
(712mA)1/3
2тгК-\/Х
a r+c
2тгл/А J-oc
з
----iuq ,
2тгл/А
u3
У
(2.152)
(2.153)
( u3	A a r+°°	( u3 \
cos------uq ] = ----= I cos---------uq ] du
\3 M TvVxJo \3 M
since sin (y- — uq) du = 0. In terms of the Airy functions
1 r+oo
Ai(q) = —= / cos
д/7Г Jo
\
— uq j du
we will have
(2.154)
(2.155)
a H— x
з
E
a
^(^) = ~^=Ai(-q).
утгл
For large |g|, leading terms in the asymptotic series are as follows
Ai^ ~ 2vW/4 eXP (~tg3/2) ’ 9>°	(2.156)
Ai^ K ^F(^)./4 3in	+ ib 4 <0	<2'157>
74
Using these approximations in (2.155) we get
a 1
7Д Sm
t%)
V’(q)
-g3/2 + -
1зЧ 4]
for E > У(ж)
a 1
wx(UUexp
for E < У(ж) (2.158)
-|(-<7)3/!
о
as expected from the WKB approximation.
(b) When V = AI x I we have bound states and therefore the energy spec-
trum is discrete. So in this case the energy eigenstates heve to satisfy the
consistency relation
[ dxJ2m\E — A|ж|] = (^ + 0 тгЛ., n = 0,1,2,...	(2.159)
The turning points are x± = — у and x% = y. So
dx\j2m\E — A|.r |] = 2 J	\j2m\E — Xx\dx
3 (n + 0 tf/z /£?\	[3 (jz + 0 7t?z]2/3
4VW W = 42/3(2тА)х/3
г /	, \	, -| 2/3
3 \ n + 0 ttTzA
4л/2ш
(2.160)
3. THEORY OF ANGULAR MOMENTUM
75
3 Theory of Angular Momentum
3.1	Consider a sequence of Euler rotations represented by

/ — iasa\	(—i<T2/3\
eXp I—2— ) eXP \ —2— J eXP V—2—
/ e-«(a+7)/2 cog £	_e-«(«-7)/2 д|п £ \
e«(«-7)/2 g|n gs(a+7)/2 CQg £ J
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 ф. Find ф.
In the case of Euler angles we have
t>p«(QiA7) =
e-h“+7)/2 cog |
e«(«-7)/2 g|n £
—e 7)/2sin^
e®(a+7)/2 cos £
(3.1)
while the same rotation will be represented by

Since these two operators must have the same effect, each matrix element
should be the same. That is
e ®(a+7)/2 cog U _ cog
2
— in % sin
cos-----
2
cos —
2
, 9	2 /5	2 (« + ?)
cos ф = 2 cos — cos-----------------1
2	2
ф = arccos
о 2 P 2
2 cos — cos
2
(3.3)
3.2	An angular-momentum eigenstate |J, m = mmax = j} is rotated
by an infinitesimal angle e about the у-axis. Without using the
76
explicit form of the d$,m function, obtain an expression for the
probability for the new rotated state to be found in the original
state up to terms of order £2.
The rotated state is given by
\jJ}r =	=	exp	Ъ Jy& \ h J.	\j,j)
= '1 iJ*e ,	J | - A h	2h2	|J’-77			(3.4)
up to terms of order £2. We can write Jy in terms of the ladder operators
J+ = Jx + iJy I т ___ J+ J—	/о
J- = Л - iJy ]^Jy- 2i •
Subtitution of this in (3.4), gives
We know that for the ladder operators the following relations hold
= hyj(j -m)(j + m + 1)|j,m + 1)	(3.7)
= HyJ(j + m)(j -m + 1)\j, m - 1)	(3.8)
So
(J+ -	= -hy/2j\j,j -1)	(3.9)
(J+-
=	- 1) - J-\jJ - 1))
= -hfij \y/2j\j,j) - ф(23 - 1) |j, J - 2)
and from (3.6)
= \j,j) + f - 1) - j) + |-2^/j(2j -	- 2)
z *	о	О
(£2 \	£	_ £2 
1 “	+ 2 V ^l-7’-7 - XH - Ш-7 -
3. THEORY OF ANGULAR MOMENTUM
77
Thus the probability for the rotated state to be found in the original state
will be

(3.10)
3.3	The wave function of a particle subjected to a spherically
symmetrical potential V(r) is given by
Ф(х) = (x + y + 3z)f(r).
(a)	Is an eigenfunction of L? If so, what is the Z-value? If
not, what are the possible values of I we may obtain when L2 is
measured?
(b)	What are the probabilities for the particle to be found in various
mi states?
(c)	Suppose it is known somehow that ф(х) is an energy eigenfunc-
tion with eigenvalue E. Indicate how we may find V(r).
(a) We have
V’(f) = {x \ф) = (x + y + 3z)/(r).	(3.11)
So
(£|L2\ф) (s"3=6-15) -h2
Id2 Id
sin2 0 дф2 ”1” sin 0 80
(3.12)
If we write ф(х) in terms of spherical coordinates (ж = r sin 0 costly =
r sin 0 sin ф, z = r cos 0) we will have
ф(х) = r/(r) (sin 0 cos ф + sin 0 sin ф + 3 cos 0).	(3.13)
Then
1	т f(r} sin 0 с)	т f(r}
= ——Na. (cos^ “ sin$ =----------~д (cos^>T sin $3.14)
sm20d^ ’ sm20 dф'	’ sinF	’
78
and
——- — | sin 0— | tb(x) = ——— — [—3 sin2 0 + (cos ф + sin d>) sin 0 cos б] =
sin0 50\ дОГ sin0 dO L	v v	J
—h2
——— —6 sin в cos в + (cos ф + sin ^>)(cos2 в — sin2 0) (3.
sm O'-	'
Substitution of (3.14) and (3.15) in (3.11) gives
1
•) —;—-(cos ф + sin ^>)(1 — cos2 0 + sin2 0) — 6 cos 0
L sm 0
 1
——-2 sin2 0(cos ф + sin ф} + 6 cos 0
. sm 0
= 27i2r/(r) [sin 0 cos ф + sin 0 sin ф + 3 cos 0] = 2Ъ2ф(х) =>
Ь2ф(х) = 2ti2 ф(х) = 1(1 + l)7i2 ф(х') = Z(Z + Z)/i2^(+)	(3.16)
which means that ф(х) is en eigenfunction of L2 with eigenvalue Z = 1.
n2
(b) Since we already know that Z = 1 we can try to write ф(х) in terms of
the spherical harmonics Y™(0, ф). We know that
So we can write
0(£) =
(3.17)
But this means that the part of the state that depends on the values of m
can be written in the following way
\Ф)т = N [3a/2|Z = 1, m = 0) + (1 + i)|Z = 1, m = — 1) + (1 — i)|Z = l,m = 1)]
and if we want it normalized we will have
|7V|2(18 + 2 + 2) = 1 => N = -^=.	(3.18)
3. THEORY OF ANGULAR MOMENTUM
79
So
P(m = 0)	= \(l =	1, m	= 0|< = ^- =	1Г	(3.19)
P(m = +1)	= \(l =	1, m	= +!|< = ^ =	1 11’	(3.20)
P(m = —1)	= \(l =	1, m	==H< = ^ =	1 П’	(3.21)
(c) If 'Фе(х) is an energy eigenfunction then it solves the Schrodinger equation
-h2 Г d2 ,	2d,	L2
-x—	+ -—tpE\x) ~
2m dr2	r dr	nr2
—Yt
2m
^\rf(r)] + - J-H(r)] - 4tr/(r)] + V(r)rf(r)Yi
dr£	r dr	r£
m
V(r)
V(r)
V(r)
^r/(r)Yzm =>
E + [zr[f(r) + r/'wl + I[/w + r/'W1
1 h2
E + UfGA^	+ rf"^ + 2f'^
I J I I j ZjII I
L2 rr(r) + 4f(r).
2m rf(r)
-rM
(3.22)
3.4 Consider a particle with an intrinsic angular momentum (or
spin) of one unit of h. (One example of such a particle is the g-
meson). Quantum-mechanically, such a particle is described by a
ketvector |p) or in x representation a wave function
g\x) = (ж;г|р)
where \x,i} correspond to a particle at x with spin in the i:th di-
rection.
(a)	Show explicitly that infinitesimal rotations of рг(ж) are obtained
by acting with the operator
u? = 1-г|.(£ + 5)	(3.23)
80
where L = ~r X V. Determine S !
г
(b)	Show that L and S commute.
(c)	Show that S is a vector operator.
(d)	Show that V X q(x) =	• p)Q where p is the momentum oper-
ator.
(a)	We have
3	3
\o)	I®;	= ^2[\x-, i)p\x)d3x.	(3.24)
« = 1 J	4=1 J
Under a rotation R we will have
3 Г
\q'} = U(R)\q} = £ / U(R) [|f> ® |0] рг(х)^х
г=1 J
= j l-йж) ® \i}D^(R)p\x)d3x det=-1 j \x-i)'D[i\R')Ql(R~1x)d3x
4=1 J	4=1 J
= У~) У \x- i}Qux)d?x =>
^'(x) = T^\r)q1(R~1x) => p'(x) = Rp(R-1x).	(3.25)
Under an infinitesimal rotation we will have
R(5<^,n")r = r + 5r = r + 8ф(п x U) = r-\- e x r.	(3.26)
So
^(s) = /?(50)^(/?-1s) = R(8<J))q(x — e x x)
= q(x — e x x) + e x p(x — s x x).	(3.27)
On the other hand
p(x — £ X x) = p(x) — (e X x) • Vp(x) = p(x) — £  (x X V)p(x)
= p(x) — —£ Lq(x)	(3.28)
3. THEORY OF ANGULAR MOMENTUM
81
where V^(®) =	|0. Using this in (3.27) we get
= ^(ж) — -£ • Lq(x) + e x q(x).
But
(3.29)
£ X q = \£yQ3 - £zqj ex + (ez^ - ехрА ёу + (exQ2 - еурА ez (3.30)
or in matrix form
Q2'
\ (-Л' /
/ 0
£T 0
0
0
1
/ 0
0
\0
/ 0
£z
\ ~£y
°
-1
0 )
0
0
ih
&Z
0
£y
0
0
0
-1
0
—ih
0 /
Q2
03 /
1 \
0	-
0 /
0
0
— ih 0
-1
0
0
Q2
,3
ih
0
0 /
0 —ih
ih 0
о 0
0 Y
0
0/.
Q2
\ <?3 /
I
h
у
У
0
0
0
0
0
z
0
1
0
z
0 \ 1 ( Q
0
0 ) J \ Q'
which means that
I
h
£ X Q =
with (Sy)ki = -ihe^ki.
Thus we will have that
^(f) = ВДж) = 1 -
-^£•(1 + 5). (3.31)
(b)	From their definition it is obvious that L and S commute since L acts
only on the |ж) basis and S only on |i).
(c)	S is a vector operator since
[5*n	SjSi]km —	' [(	lh^€jim ( ih^Ejkl^ ^)tj7m]
—	^h	h £jkl£iml^
—	h	$imhjk	hij^km T hjmhki)
= h2	(5jm8ki - 8im8jk)
=	= y^jh€j^(—ih€
kml) — ^ihcij^Si') km •	(3.32)
82
(d) It is
V x q(x)
X q(x) =
^S-p)Q.	(3.33)
3.5 We are to add angular momenta = 1 and jz = 1 to form
j = 2,1, and 0 states. Using the ladder operator method express all
(nine) j, m eigenkets in terms of IJ1J2;Write your answer as
\j = l,m = 1) = 4||+,0) - 7|l0’+)’•••’
(3.34)
where + and 0 stand for m^z = 1,0, respectively.
We want to add the angular momenta ji = 1 and jz = 1 to form j =
\ji — J2I, • • •	+ J2 = 0,1,2 states. Let us take first the state j = 2, m = 2.
This state is related to \jimi; j/2^2) through the following equation
=	{jij2-,m1mz\j1jz-,jm)\j1jz-,m1mz}	(3.35)
m=mi +7712
So setting j = 2, m = 2 in (3.35) we get
|j = 2,m = 2) = 0\j2; ++ Ы2; jm)| ++) n=m‘I ++)	(3.36)
If we apply the J_ operator on this statet we will get
J_|J = 2, m = 2) = (Ji- + «7г-)| + +)
=> T^/(,7 + m)(j -m + l)|j = 2, m = 1) =
^\/(ji + ^i)(Ji -	+ l)|0+) + ^\Z(j2 + mz)(jz - m2 + 1)| + 0)
=> V4\j = 2,m = l) = у/2|0+) + д/2| + 0)
=> \j = 2,m = 1) = -±=|0+) + -^| + 0).	(3.37)
3. THEORY OF ANGULAR MOMENTUM
83
In the same way we have
J-\j — 2,m — 1) — —+ Л-)|0+) + —^(Л- + Л-)| + 0) =>
Vb\j = 2, т = 0) = -^= [д/2| - +) + д/2|00>] +	[д/2|00> + д/2| + -)] =>
Vf)\j = 2,m = 0} = 2|00>+ |-+> +| + ->=J>
|J = 2,m = 0> = J||00) + ^=| + -) + ^|-+)	(3.38)
V 3 yb yb
J-\j = 2,m = 0)
a/6 I j = 2,m = — 1)
\j = 2,m = -1)
\j = 2,m = -1)
+-ty2|o-) + 2=V2| - o)
yb	yb
|V2|0-) + ?V2| - 0) + tv^|0-) + ^72| - 0)
b	b	b	b
4l0-) + 4||-0)	(3.39)
J-\j —	2,m	—	— 1)	—	-д(Л-	+	Л-)|0—) + —^(Л-	+	^2-)| — 0) =>
V4\j =	2,m	=	-2)	=	V^|----) + ^a/2|-------)
|j =	2,m	=	-2)	=	I--).	(3.40)
Now let us return to equation (3.35). If j = 1, m = 1 we will have
|j = l,m = l) = a| + 0) + 5|0+)	(3.41)
This state should be orthogonal to all |j, m) states and in particular to |j =
2, m = 1). So
(j = 2,m = l|j = l,m = 1) = 0 =>	= 0 =>
a + 6 = 0=>a = —b	.	(3.42)
84
In addition the state |j = l,m = 1) should be normalized so
(j = l,m = l|j = l,m = 1) = 1 => |a|2 + |Ь|2 = 1 (3Д-2) 2|a|2 = 1 => |a| = ~^=.
By convention we take a to be real and positive so a = and b = —
That is
|J = l,m = l) = -L| + 0)--L=|0+).	(3.43)
Using the same procedure we used before
J_|j = l,m = l) = —/|(Л- + Л-)| + 0) — —+ Л-)|0+) =>
V2|;' = l,ni = 0) = -к [л/2|00> + V2| + -)] -	[V2| - +) + v"2|00)] =>
|) = l,m = 0} = 4l + ->- 4!-+>	(3-44)
J_|j = l,m = 0)
= l,m = -1)
|j = l,m = -1)
-^(Л- + Л-)| + -) -	+ ^2-)l _
2=V^|0-> - ±V2\ - 0)
(3-45)
Returning back to (3.35) we see that the state \j = 0, m = 0) can be written
as
|J = 0,m = 0) = cjOO) + c2| + -) + c3| - +).	(3.46)
This state should be orthogonal to all states |j, m) and in particulat to |j =
2, m = 0) and to j = 1, m = 0). So
(j = 2, m = 0|J = 0, m = 0) = 0 =>
2ci + c2 + c3 — 0	(3.47)
{j = 1, m = 0|j = 0, m = 0) = 0 => -^c2 - ~^=c3
c2 = c3.	(3.48)
3. THEORY OF ANGULAR MOMENTUM
85
Using the last relation in (3.47), we get
2cx T 2c2 — 0 =>• Ci T c2 — 0 =>• Ci — —c2.	(3.49)
The state |j = 0, m = 0) should be normalized so
(j = 0,m = 0|j = 0,m = 0) = 1	|cx|2 + |c2|2 + |c3|2
, , 1
=> Ы =
V V
1 ^3|c2|2 = 1
(3.50)
By convention we take c2 to be real and positive so c2 = c3 = and
cx = — Thus
(3.51)
So gathering all the previous results together
|j = 2,m = 2)
\j = 2,m = 1)
\j = 2,m = 0)
\j = 2,m = -1)
\j = 2,m = -2)
\j = l,m = 1)
\j = l,m = 0)
\j = l,m = -1)
\j = Q,m = 0)
(3.52)
3.6 (a) Construct a spherical tensor of rank 1 out of two different
vectors U = (Ux,Uy,Uz) and V = (Vx,Vy,Vz). Explicitly write T±V,o *п
terms of Ux,y,z and
(b) Construct a spherical tensor of rank 2 out of two different
vectors U and V. Write down explicitly T^±x о *п terms of Ux,y,z
and К>у>г.
86
(a)	Since U and V are vector operators they will satisfy the following com-
mutation relations
[tfj, Jj\ — iflSijkUk [14', J•
(3.53)
From the components of a vector operator we can construct a spherical tensor
of rank 1 in the following way. The defining properties of a spherical tensor
of rank 1 are the following
= S9C<‘>,	[J±,C<1>] = ^(lT<rt(2±'0C,(i)1. (3.54)
It is
[Jz,uz] (3=3) ож(и4)
Uz = Uo	(3.55)
[J+,UO] (3=4) V2hu+1 = {j+,uz] = {jx + ijy,uz]
(3=3) —ihUy + i(ih)Ux = —h(Ux + iUy) =>
U+1 = --^(Ux + iUy)	(3.56)
[J_,u0] (3=4) V2hu_1 = [J_,uz] = [Jx-iJy,uz]
(3=3) -ihUy - i(iK)Ux = h(Ux - iUy) =>
U^ = ^=(Ux-iUy)	(3.57)
So from the vector operators U and V we can construct spherical tensors
with components
u0 = uz
U+1 = -^Ux + iUy)
U-i = ^Ux-iUy)
v0 = vz
V+1 = --^V. + IVy)
V-i = ^(yx-iVy)
(3.58)
It is known (S-3.10.27) that if X^1) and are irreducible spherical tensors
of rank ki and k? respectively then we can construct a spherical tensor of
rank к
9192
(3.59)
3. THEORY OF ANGULAR MOMENTUM
87
In this case we have
tV) 1+i	= (11; +10| 11; 11)77+! Vo + (И; 0 + 1111; ll)V0V+i (3=2) 2=c+1% - = -1(с1 + .едк + |сдк + >к)	(з.бо)
tV) Jo	= (И; 00|11; 10)7Ш + (И; -1 + 1 |H; 10)V_i V+1 +(11;+1 - 1|11; 10)t/+iV_i (3.52)	1 TT	1 TT -	/-V-1V+1 + У	у Zj —	f-\UxVxiUxVy iUyVx + UyVy UxVx + iUxVy iUyVx Ц/И/] 2y 2 = +(СЛ - C,K)	(3.61)
	= (11; -10| 11; ll>Z7_i Vo + (11; 0 - 1111; ll)V0V_i (3.52)	1 TT	1 ггтг —	MJ-iVo + ^VoV_i л/2	д/2 = -^их-гиу)К + ^(Ух-гУу).	(3.62)
(b)	In the same manner we will have
T^ = (11; +1 + 1111; 2 + 2)U+1V+1 (=2) U+1V+1 = \(UX + iUy)(Vx + iVy)
= -^UxVx-UyVy + iUxVy + iUyVx)	(3.63)
T$ = <ll;O + 1|11;2 + 1>Z7OV+1 + <ll;+10|ll;2 + 1>Z7+1H)
(3.52)	1 _ _ _ _	1 TT
=	—^U0V+i + -/=I7+iVo
= ~l(UzVx + UXVZ + iUzVy + iUyVz)	(3.64)
To(2)	=	(ll;00|ll;20)t/oVo + (ll;-l + l|ll;20)t/_1V+1
+<11; +1 - l|ll;20>Z7+1
88
(3.52)
^+1K
уЗ^2^2 у62^ж	+ у g у/22^х ”1” Wy}
У	1 [2СШ - ±UXVX - l-UxVy + l-UyVx
-	\uyvy - \uxvx + l-uxvy - l-uyvx - \uyvy
y|(2C4K - Uxvx - UyVy)
(3.65)
(3.52)
(И; 0 - 1111; 2 - l)t/0V-i + (И; -Ю|11; 2 + l)^ Уо
—	-^UqV-i + —-^U-iVo
\{UZVX + UXVZ - iUzVy - iUyVz)
(3.66)
= (11; -1 - 1111; 2 - 2)C7_1y_1 (=2) U-yV-y = l(Ux- iUy)(Vx - iVy)
= \(UxVx-UyVy-iUxVy-iUyVx).
(3.67)
3.7 (a) Evaluate
t Н±.Св)|2т
m=-j
for any j (integer or half-integer); then check your answer for J = |.
(b) Prove, for any j,
Ц m2\dm'm(P)\2 = Уи + ^зт^ + т'2 + l(3cos2(3-l).
m=-j
[Hint: This can be proved in many ways. You may, for instance,
examine the rotational properties of using the spherical (irre-
ducible) tensor language.]
3. THEORY OF ANGULAR MOMENTUM
89
(a) We have
j
2m
m=-j
3
2
*
m=-j
m=-j
ni = -j
m=-j
i/- о •
= |e
= | O'm' I eiJyPIK Jz e~iJyPln \jm'}
=	ey)JzT>((3- ey)\jm'}.
But the momentum J is a vector operator so from (S-3.10.3) we
that
,iJypinjz ^2	e
_m=-j
(3.68)
will have
P*(/?; ey)JzT>(J3-, ey) = £ Rz3((3- ey)J3.
i
On the other hand we know (S-3.1.5b) that
cos /3 0
0	1
— sin/? 0
W; ?y) =
sin/? \
0
cos (3 )
So
2m
m=-j
(3.69)
(3.70)
| [- sin (3{jm'\Jx\jm') + cos 0{jm'\Jz\jm'}]
| - sin (3{jm'\
m' cos /?.
------\jiri') + hm' cos /?
(3.71)
90
For j = 1/2 we know from (S-3.2.44) that
\ b111 2	— sin| \ COS 2	/	(3.72)
So for m' = 1/2		
1/2 E \dml/2Wm = m=—\/2	1 • 2	, 1	2 ft ~2sm 2+2COS 2	
—	| cos (3 = m' cos (3	(3.73)
while for m' = —1/2		
1/2 E \dml/2Wm = m=—1/2	1 2 ft , 1 • 2 ft ~2COS 2+2sm 2	
—	— | cos (3 = m' cos (3.	(3.74)
(b) We have
£ ™2Н±(/з)12
=	12 rn2\{jm'\e-^l3/t\jm)\2
m=-j
m=-j
т=-з
— У2 O’m,le-?Jy^m2IJmXJmle?Jy^IJm^
m=-j
= ^2^m'	e-iJvP/hj2	3 E \jm)Um\ _m=-j	
= ^2^m'	e-iJvf3lTij2e		

(3.75)
3. THEORY OF ANGULAR MOMENTUM
91
From (3.65) we know that
h2)= Д(3Л2-^)	(3.76)
where T^ is the О-component of a second rank tensor. So
Л2 = ^d2) + U2	(3.77)
о	о
and since 'D(R) J2P^(7?) = J2P(/?)P^(7?) = J2 we will have
eL-, ™2l<£b« =
no	у on
We know that for a spherical tensor (S-3.10.22b)
р(Д)Т«р1(д) = £ V$(R)T^	(3.79)
qf=—k
which means in our case that
(jm'\-D(/3;et)J^(/3;et)\jm') = (jm'\ £ T^V^(/3; ?Mm')
g'=-2
= £ »^(/3;8,)0™'|T,(.2,|j">').	(3.80)
g'=-2
But we know from the Wigner-Eckart theorem that (Jm'|Tg/2^0|Jm') = 0. So
E \dmm' (£) Г
=	+ i) +	^){>K(2W)
= (j +1) + |4o (/?)<jm'IЛ2 - |J2 \im'}
= |j(J +x) + |4o(/?) m2-|j(J +x)
O	Z	L	О
92
=	+ i) +1(3 cos2 P - i) m'2 ~	+ i)
o	Z	L о
1	11
= “ 2J0 + COs2 +	+	+ 3J(J + X) + “2“(3 cos2 “ X)
= |j(j' + l)sm2/? + m'2|(3cos2/3 - 1)	(3.81)
where we have used </^(/?) = /^(cos /?) = |(3cos2/? — 1).
3.8 (a) Write xy, xz, and (ж2 — у2) as components of a spherical
(irreducible) tensor of rank 2.
(b) The expectation value
Q = e(a, j,m = j|(3z2 - r2)\a, j,m = j)
is known as the quadrupole moment. Evaluate
e(«, j, т'|(ж2 - y2)|a, j,m = j),
(where m' = j, j — 1, j — 2,... )in terms of Q and appropriate Clebsch-
Gordan coefficients.
(a)	Using the relations (3.63-3.67) we can find that in the case where U =
V = x the components of a spherical tensor of rank 2 will be
T+2 = | (x2 - y2) + ixy	| (ж2 - y2) - ixy
= — (xz + izy)	tI2^ = xz — izy (3.82)
Го(2) = 4(2г2 - x2 - у2) = ^(Зг2 - г2)
So from the above we have
(b)	We have
Q = e{aj,m = j\(3z2 -r2)\a,j,m = j}
3. THEORY OF ANGULAR MOMENTUM
93
(3.82)
\/6e(a,j,m = j|T0(2)|a, j, m = j) (ж=&) (J2; jO|j'2; jj)	д/бе
(aj ||T<2)\\aj) = ® y. л/2/, +	(3.84)
So
(3.83)
IT^Ioi, j,m = j
о
V 2j + 1
(3.84) Q {j2; j, -2|j2; j, j - 2)
V6 U2:.^\j2;.hJ}
+ e8m/J_2{j2- j - 2\j2-jj - 2)
(3.85)
2
\T^\a,j,m = j}
94
4 Symmetry in Quantum Mechanics
4.1 (a) Assuming that the Hamiltonian is invariant under time
reversal, prove that the wave function for a spinless nondegenerate
system at any given instant of time can always be chosen to be
real.
(b) The wave function for a plane-wave state at t = 0 is given by
a complex function егР'ж/\ Why does this not violate time-reversal
invariance?
(a)	Suppose that |n) in a nondegenerate energy eigenstate. Then
H0\n) = 0H\n) = En\n) => 0|n) = ей|п)
=> 0|Mo = 0;t) = Qe~itH^\n} = Qe~itE"^\n} =
eitE"/n0ln) = e
=? j a x\x\n, l0 = w, L? \x? = j a xe	bo = v,
=> Ф*п&^ = el(^”+5)^n(s,t).	(4.1)
So if we choose at any instant of time 8 = — the wave function will be
real.
(b)	In the case of a free particle the Schrodinger equation is
2	д2
= E\n) =>	= ЕФп(х)
=> фп(х) =	+ Ве~^1п	(4.2)
The wave functions = е~г^'х^ and <^(ж) = correspond to the
same eigenvalue E = and so there is degeneracy since these correspond
to different state kets |p) and | — p). So we cannot apply the previous result.
4.2	Let <^>(/7) be the momentum-space wave function for state |a),
that is, <^>(j7) = (p,|a).Is the momentum-space wave function for the
4. SYMMETRY IN Q UANTUM MECHANICS
95
time-reversed state 0|a) given by <^>(/7), ф(— j7), <^>*(/7), or ф*(—р'У?
Justify your answer.
In the momentum space we have
|a) =	=* I«) = J «^WOPI?)
0|a) = J <fpWI<W)] = I <Pp'maye\?\ (4.3)
For the momentum it is natural to require
(a|p|a) = — (a|p|a) =>
(а|0р0-1|а) => 0p0-1 = — p	(4.4)
So
0^/) (4=4) -p0|/) => 0|/) = | - /)	(4.5)
up to a phase factor. So finally
0|a) = j </3p'(/|a)*| -/) = j <Z3p'(-/|a)*|/)
=> {/|0|«) = <^(/) = <-/!«>* = <^*(-/)-	(4-6)
4.3	Read section 4.3 in Sakurai to refresh your knowledge of the
quantum mechanics of periodic potentials. You know that the en-
ergybands in solids are described by the so called Bloch functions
фп,к fullfilling,
фп,к(х + a) =
where a is the lattice constant, n labels the band, and the lattice
momentum к is restricted to the Brillouin zone [—тг/а, тг/а].
Prove that any Bloch function can be written as,
Фп,к(х) = ^Фп(х - Ri)etkRi
Ri
96
where the sum is over all lattice vectors Ri. (In this simble one di-
mensional problem Ri = ia, but the construction generalizes easily
to three dimensions.).
The functions фп are called Wannier functions, and are impor-
tant in the tight-binding description of solids. Show that the Wan-
nier functions are corresponding to different sites and/or different
bands are orthogonal, i.e. prove
j" dx(j>m[x Ri)(j)n(x Rj) ~
Hint: Expand the фпз in Bloch functions and use their orthonor-
mality properties.
The defining property of a Bloch function фп,к(х) is
фп,к(х + а) = гкафп,к(х).	(4.7)
We can show that the functions фп(х — Ri)etkRi satisfy the same relation
+ a - Ri)eikR‘ = ^фп[х - (Ri - a)]eik^~^eika
Ri	Ri
^ка^фп(х - Rj)eikRi	(4.8)
Rj
which means that it is a Bloch function
Фп,к^ = ЦФп(х-Щук^.
Ri
(4.9)
The last relation gives the Bloch functions in terms of Wannier functions.
To find the expansion of a Wannier function in terms of Bloch functions we
multiply this relation by e~tkR:> and integrate over k.
Фп,к(х) = ^фп(х- Ri)eikR‘
Ri
=> Г/а dke-ikRyn,k(x) = ^фп(х - Ri) Г/а eik^-R^dk (4.10)
4. SYMMETRY IN Q UANTUM MECHANICS
97
But
r/a eik(R.-R3)dk = eik(R'-R^ v/a = 2 sin [7г/а(Д-- Я;)]
2-тг/а	i(Ri — Rj) _T[a	Ri ~ Rj
r 2?T	.	.
= Sij—	(4.11)
where in the last step we used that Ri — Rj = na. with n G Z. So
j	dke~tkR}Cn,k(x') =	фп(х - Rijbij^
T^/a,	ft,	d
=> фп(х - R{) = — [ e~zkR^n,k(x)dk	(4-12)
27Г J-ir/a
So using the orthonormality properties of the Bloch functions
У dxф*m(x - Нф)фп(х — Rj)
= У J f ^^егк^ф*т к(х)е-гк'^фщкфх)акак'ах
= j J ±e^Ri-^k'R3 j ^*m k^x)^n k,^x)dxdkdk'
= JJ -^^'^'^^k-k'ykdk'
= 7^Smn Г/а eik^-R^dk = ^8mn8ij.	(4.13)
J—-п/а	27Г
4.4	Suppose a spinless particle is bound to a fixed center by a
potential V(s) so assymetrical that no energy level is degenerate.
Using the time-reversal invariance prove
(L) = 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
I ТП
98
what kind of phase restrictions do we obtain on F/m(r)?
Since the Hamiltonian is invariant under time reversal
HQ = QH.	(4.14)
So if |n) is an energy eigenstate with eigenvalue En we will have
HQ\n) = QH\n) = EnQ\n).	(4.15)
If there is no degeneracy |n) and 0|n) can differ at most by a phase factor.
Hence
|n) = 0|n) = e’5|n).	(4-16)
For the angular-momentum operator we have from (S-4.4.53)
{n\L|n) = —(n|Z|n) (4=6) —(n|Z|n) =>
(n|L|n) = 0	.	(4.17)
We have
0|n) = 0 У </3ж|ж)(ж|п) = У </3ж(ж|п)*0|ж)
= У d3x(x|п)*|ж) (4=6) ег5|п) =>
(ж'|0|п) = (х'\п)* = eiS{x'\n).	(4.18)
So if we use (x\n) =
E WYrtO, Ф) = ei5 £ Fim(r) Ф)
= e,>	Ф)
= e’1/
=>	£ r;„(r)(-l)”4„..-„5,., = <" £ F,„(r)5„..„5w
= ei5F,.m.(r) => г;,_„,(г) = (-1)“’F,.„.(r)e". (4.19)
4. SYMMETRY IN Q UANTUM MECHANICS
99
4.5 The Hamiltonian for a spin 1 system is given by
H = AS2 + B(S2 - S2).
Solve this problem exactly to find 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?
For a spin 1 system I = 1 and m = —1,0,+1. For the operator Sz we
have
S2|Z,m) = 7im|Z,m) => (Zn|S2|Z, m) = hm(n\m} => (S2)nm = hm8nm (4.20)
So
/ 1 0
sz = h oo
\ о 0
0
0
-1
/ 1
0
\0
о 0 \
о 0
о 1/
For the operator Sx we have
Sx\l,m)	=
(1,п|5ж|1,т)	=
(S—3.5.39)
S+ 2 S~ I1’"1) =	+ 2S- I1 ’
|(l,n|S+|l,m) + |(l,n|S_|l,m)
U _ m)(2 +	+ m)(2 -
So
Sx
S2X
(4-21)
\2 0 2/ Ц
° И
1 0
0 I/
100
In the same manner for the operator Sy = S+2iS we find
Thus the Hamiltonian can be represented by the matrix
h2
(4.23)
/ A 0 В \
0 0 0
\ В 0 A /
To find the energy eigenvalues we have to solve the secular equation
det(7? — Al) = 0 => det
( Ah2-X 0
0 -A
\ Bh2 о
Bh2 \
о = о
Ah2 — x /
=> (Ah2 - A)2(—A) + (Bh2)2X = 0 => A [(Ah2 - A)2 - (B7i2)2] = 0
=> X(Ah2 - A - Bh2)(Ah2 - A + Bh2) = 0
=> A1 = o, X2 = h2(A + B), X3 = h2(A-B).	(4.24)
To find the eigenstate |пдс) that corresponds to the eigenvalue Ac we have to
solve the following equation
For Ai = 0
h2
h2
/ A 0
0 0
\ в 0
в \
0
A /
(4.25)
/ A 0
0 0
\ в 0
0
0
a = — cM
o2	4	\
—c^- + cA = 0
a = 0
c — 0
(4.26)
4. SYMMETRY IN Q UANTUM MECHANICS
101
So
Ы
Ы
/ 0 \	/ 0 A
norm. i
к о /	\ " !
|10).
(4.27)
In the same way for A = h2(A + B)
/ A 0 В \
0 0 0
\ В 0 A /
f a = c
H b = 0
aA + cB
0
aB + cA
a(A + B)
b(A + B)
c(A + B)
(4.28)
So
|^a+b)
|^a+b)
(4.29)
For A = h2(A — B) we have
So
/ A 0
0 0
\ в 0
a = —c
b = Q
в \
0
A /
(A-B)
a A + cB
0
aB + cA
a(A-B)
b(A - B)
c(A - B)
(4.30)
|«A+b)
|«a-b)
( 1 \
T2|1’+lbi|1’-1>'
(4.31)
102
Now we are going to check if the Hamiltonian is invariant under time reversal
0Я0"1 = A0Sf0-x + B(0S|0-x - 0SJ0-1)
= А05г0-Х05г0-Х + B(0S3;0-x0S3;0-x -
= AS2Z + B(S| - Sj) = H.	(4.32)
To find the transformation of the eigenstates under time reversal we use the
relation (S-4.4.58)
0|Z,m) = ( —l)m|Z, —m).
(4.33)
So
0|n„)	=	0|1O) '	|10)
= l"o)	(4.34)
(4.35)
0|пл+в)	=	4=0|1,+1) + -^0|1,-1)
=	-|«А+в)	(4.36)
(4.37)
0|»л-в)	=	4=0|1,+1)-4=0|1,-1)
-^|1’-1> + T2|1’+1>
=	|пл-в).	(4.38)
5. APPROXIMATION METHODS
103
5 Approximation Methods
5.1 Consider an isotropic harmonic oscillator in two dimensions.
The Hamiltonian is given by
9
rnw
2m 2m
(*2 + y2)
(a)	What are the energies of the three lowest-lying states? Is there
any degeneracy?
(b)	We now apply a perturbation
V = SmuAxy
where 5 is a dimensionless real number much smaller than unity.
Find the zeroth-order energy eigenket and the corresponding en-
ergy to first order [that is the unperturbed energy obtained in (a)
plus the first-order energy shift] for each of the three lowest-lying
states.
(c)	Solve the HQ + V problem exactly. Compare with the perturba-
tion results obtained in (b).
[You may use (п'|ж|п) = Jh/2mw(y/n + lAn',n+1 + y/n8n>,n-i).]
Define step operators:
=	/ТПШ	ipx
al =	[mw	ipx
ay -	Imw .	ipv .
ai =	=	(5.1) V 2h v mw	'
From the fundamental commutation relations we can see that
[аж,4] = [«у, aj] = 1.
104
Defining the number operators
TVjj; = CL^CL^^ Ny — Q'yQ'y
we find
N = Nx + Ny = ^- \
Ъш
HQ = ^(7V + 1).	(5.2)
I.e. energy eigenkets are also eigenkets of N:
Nx | m, n) Ny | m, n) N | m, n)	= m | m, n = n | m, n) => = (m + n) \m,n)	(5-3)
so that
Ho | m, n) = Em,n | m, n) = huj(m + n + 1) | m, n
(a)	The lowest lying states are
state	degeneracy
E0>0 = huj Eito = Eoti = E-2,0 = EOt2 = Eiti = ЗЬш	1 2 3
(b)	Apply the perturbation V = 8mw2xy.
Full problem:	(Ho + V) 11) = E 11)
Unperturbed problem: Ho 11° ) = E° 11°)
Expand the energy levels and the eigenkets as
E = E° + A1+A2 + ...
\l) = |Z°) + 1П + ...	(5.4)
so that the full problem becomes
(Е°-Яо)[|/°)+ | Z1) + ...] = (V — A1 — A2...) [ | Z°) + | Z1) + ...].
5. APPROXIMATION METHODS
105
To Г st order:
(Я°-Я0)Ц1) = (У-А1)Ц°).	(5.5)
Multiply with (1° | to find
(l°\E° - Ho\ I1) = 0 = (1° | V - A111°) =>
A^Z^Z0) = A1 = (Z°|y|Z°)	(5.6)
In the degenerate case this does not work since we’re not using the right basis
kets. Wind back to (5.5) and multiply it with another degenerate basis ket
(m° | E° - Яо 111) = 0 = (m° | V - A1 | Z°) =>
A1(m°|Z°) = (m°|y|Z°).	(5.7)
Now, (m° | Z° ) is not necessarily 8ki since only states corresponding to differ-
ent eigenvalues have to be orthogonal!
Insert a 1:
^2 (m° | У | A:0 )(A:0 | Z° ) = Ax( m° | Z° ).
keD
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 =>
Aoo = <0,0 | V 10,0> = <W(0,0|.ry |0,0) ~ (0,0 | (^+4)^+4) | 0,0 ) = 0
2. First excited state is degenerate 11,0), | 0,1). We need the matrix
elements (1,0 | V 11,0), (1,0 | V | 0,1), (0,11 V 11,0), (0,11 V | 0,1).
A
= SmuPxy = 8maP----(«a.+«t)(«J,-|-«t) = —-(a^+aj.a^+a^a^+a^aj)
Zj! / bUJ	Zj
and
ax | m, n) = y/m | m — 1, n)
a* | m, n) = y/m+l | m + 1, n) etc.
106
Together this gives
Иодо	— Hl,01	= 0,		
		1	X		8tlW
Ио,01	= —<	1,0 \axa'y	|0,1 >	~ ~2~
		I	X		8hw
Индо	=	1,0 \a'xay	|0,1 >	~ ~T~
The У-matrix becomes
/ 0 1 \
"~2\ 1 0 J
(5.8)
and so the eigenvalues (= A1) are
To get the eigenvectors we solve
Д‘ = ±^
and get
|...)+ = -L(|o,i>+
|...)_ = -^( | o, i) — | i,o)),
E+ — tiw(2 + -),
E_ = hu(2 - |). (5.9)
3. The second excited state is also degenerate | 2,0), 11,1), | 0,2), so
we need the corresponding 9 matrix elements. However the only non-
vanishing ones are:
И1,20 — Ho,ll
-V - V
— '11,02 — '02,11 —
(5.10)
(where the y2 came from going from level 1 to 2 in either of the oscil-
lators) and thus to get the eigenvalues we evaluate
5. APPROXIMATION METHODS
107
which means that the eigenvalues are {0,	By the same method
as above we get the eigenvectors
|( 12,0) + л/2 11, » + |o,2>),
_L(_ |2,0) + 10,2»,
|( |2,0> - V2 11,1> + |0,2»,
E+ = ЩЗ + S),
Eo = ЗКш,
E_ = Ъш(3-8).
(c) To solve the problem exactly we will make a variable change. The poten-
tial is
тш2 [|(ж2 + у2) + 6xy\ =
= тш2
+ у)2 + (x - у)2) +	+ I/)2 - (i - y)2)
(5.11)
Now it is natural to introduce
x> =	+ У),
у = Jx'x~y^
1
p'x = 72^ +p'y^
p'v = ~/=(Px ~ p'v)-
1 у /2	1 у '
(5.12)
Note: [x',px] = [y',p'y] = Hi, so that (x', p'x) and (y', p'y) are canonically
conjugate.
In these new variables the problem takes the form
1	2
Я =	+p'y2) + ^[(1 + 6)x'2 + (1 - S)y'2].
So we get one oscillator with шх = ivy/1 + 6 and another with Шу = Шу/l — 8.
The energy levels are:
Eoto — Еш.
Elf0 = Нш + Пшх = Ьш(1 + -i/l + <^) =
= ?1ш(1 + 1 +	+ ...) = ?1ш(2 + |<У) + 0(<У2),
108
£0,1	—	bw	+	b.~Ey —    — £<^(2 — |<У)	+	0(<У2),
Е2,о	=	bw	+	27u<4 = ... = /z^(3	+ <У)	+	0(<J2),
Elfl	=	bw	+	ь^'.. + hu}'y = ... =	зЬш	+	O(£2),
£0,2	=	bw	+	2ЬиУу = ... = /uj(3	— <У)	+	0(<У2).
(5.13)
So first order perturbation theory worked!
5.2 A system that has three unperturbed states can be represented
by the perturbed Hamiltonian matrix
Ei 0 a
0 Ei b
k a* b* E2 )
where E2 > Ei. The quantities a and b are to be regarded as per-
turbations that are of the same order and are small compared with
E2 — Ei. Use the second-order nondegenerate perturbation theory
to calculate the perturbed eigenvalues. (Is this procedure correct?)
Then diagonalize the matrix to find the exact eigenvalues. Finally,
use the second-order degenerate perturbation theory. Compare
the three results obtained.
(a)	First, find the exact result by diagonalizing the Hamiltonian:
0
Ei — E 0 a
0 Ei — E b
a* b* E2- E
= (Ei - E) [(Ei - E)(E2 - E) - |5|2] + a [0 - a*(Ei - £?)] =
= (Ei - E)2(E2 - E) - (Ei - £)(|5|2 + |a|2).	(5.14)
So,	E = Ei or (Ei — E)(E2 — E) — (|5|2 + |a|2) = 0 i.e.
E2 - (Ei + E2)E + EiE2-(\a\2 + \b\2) = 0^
E = E+E? ±	_ EiE2 + |a|2 + |5|2 =
5. APPROXIMATION METHODS
109
Ei + E2
2
±\/-^ + l“l2+lbl2.
(5.15)
Since |a|2 + |6|2 is small we can expand the square root and write the three
energy levels as:
E =	+	^(1 + 1(|а|2+|Ь|2)(^-^)2 + ...) =
Z	Z \	— £?2	/
Ei - E2 ''
_ Ei + E2 Ei — E2 . . _	|a|2 + |6|2
“ ~2	Ei_E2 
(5.16)
(b)	Non degenerate perturbation theory to 2’nd order. The basis we use is
/ 1 \
|1)=	0
W
/ ° \
0
\ 1 /
|2) =
1
W
0 a
0 b
b* 0 ;
The matrix elements of the perturbation V =
0
0
a*
are
(1 I V |3) = a, (2 | V |3) = 6,	(1 | У | 2) = (A: | У | A:) = 0.
Since = (к | У | к) = 0 1’st order gives nothing. But the 2’nd order
shifts are
aS2>
д<2’
v |K1|2 = l(3| V 11)|2 = |a|2
Z J 770	770	77	77	77	77 ’
k^-l &1 —	£/l	—	£/2	—	-^2
v IW	=	1(31 V |2)|2	=	|6|2
•Z-7 770	770	77	77	77	77 ’
~	£/l	—	£/2	—	^2
v |yfc3|2 = |q|2	|Ь|2 = И + 1&12
/	77O	770	77	77 I 77	77	77	77
^3 -C>3 —	^2 —	bi — £>2
(5.17)
по
The unperturbed problem has two (degenerate) states 11) and | 2) with
energy Ei, and one (non-degenerate) state |3) with energy E2. Using non-
degenerate perturbation theory we expect only the correction to E2 (i.e. A^)
to give the correct result, and indeed this turns out to be the case.
(c)	To find the correct energy shifts for the two degenerate states we have
to use degenerate perturbation theory. The V-matrix for the degenerate
subspace is	, so 1’st order pert.thy. will again give nothing. We have
to go to 2’nd order. The problem we want to solve is (Ho + V) 11) = E 11)
using the expansion
11)	= 11°) + 111) + ... E = E° +	+ A<2) + ...	(5.18)
where Ho \l°) = E° \l°). Note that the superscript index in a bra or ket de-
notes which order it has in the perturbation expansion. Different solutions to
the full problem are denoted by different Z’s. Since the (sub-) problem we are
now solving is 2-dimensional we expect to find two solutions corresponding
to Z = 1,2. Inserting the expansions in (5.18) leaves us with
(e°-hq)[\i°)+ п + ...] =
(V - AW - A<2)...) [ I Z°) + IZ1) + ...] .	(5.19)
At first order in the perturbation this says:
(t;0-^)|z1) = (v-aW)|z°),
where of course = 0 as noted above. Multiply this from the left with a
bra (k° I from outside the deg. subspace
(k° I E° - HQ 111) = (k° I V I Z°)
v \k°){k°\V\l°)
- Ek
(5.20)
This expression for | Z1) we will use in the 2’nd order equation from (5.19)
(£?°-#o)|Z2) = V|ZX) - A<2)|Zo).
To get rid of the left hand side, multiply with a degenerate bra (m° |
(HQ\mQ) = EQ\m°))
{m°\E° - HQ\l2) = 0 = (m°|V |ZX) - A(2)(m°|Z°).
5. APPROXIMATION METHODS
111
Inserting the expression (5.20) for 111) we get
s
k£D
{m° I V I k° )(k° I V 11° }
E° - Ek
= A(2)(m°|Z°).
To make this look like an eigenvalue equation we have to insert a 1:
EE
n^D k^D
E° — Ek
Maybe it looks more familiar in matrix form
EMmnxn = N^xm
n^D
where
Mmn
E°-Ek
0/
are expressed in the basis defined by
subspace basis D = { 11), | 2 )}
Evaluate
M in the degenerate
M	_	V13V31	_	|q|2
£/l — £/3	£/1 — £/2
_	V23V31	_	a*b
p	pO	p	p 1
£j~\ —	—	£J2
M12 —
ab*
— £/3	— 1>2
IW	H2
V13V32
M->-> =--------------—---------------
22 TP ТРО TP TP
£?X — £/3	£/1 — £/2

With this explicit expression for M, solve the eigenvalue equation (define
A = Д(2)(Е1 — E2), and take out a common factor E ^_E )
0
/ |a|2 — A ab* \ _
у a*b |5|2 — A J
(|«|2-A)(|6|2-A)-|«W =
(5.21)
112
From before we knew the non-degenerate energy shift, and now we see that
degenerate perturbation theory leads to the correct shifts for the other two
levels. Everything is as we would have expected.
5.3 A one-dimensional harmonic oscillator is in its ground state
for t < 0. For t > 0 it is subjected to a time-dependent but spatially
uniform force (not potential!) in the x-direction,
F(t) = Foe~^
(a)	Using time-dependent perturbation theory to first order, obtain
the probability of finding the oscillator in its first excited state for
t > 0. Show that the t —> oo (r finite) limit of your expression is
independent of time. Is this reasonable or surprising?
(b)	Can we find higher excited states?
[You may use (п'|ж|п) = Jh/2mw(y/n + 16ni<n+1 + y/n8n,>n_i).]
(a)	The problem is defined by
=	+	V(t) = -Faxe-^ (F =
At t = 0 the system is in its ground state | a, 0) = | 0). We want to calculate
|a,t) = ^cn(i)e~Ent/n \n)
E°n = Щп + l)
where we get cn(t) from its diff. eqn. (S. 5.5.15):
V*	m
Vnm = {П I V I m)
Шпт = En Em = w(n - m)	(5.22)
We need the matrix elements Vnm
Vnm = {n | - Foxe~t/T | m) = -FQe~tlr^n | x | m) =
— Foe / (уТпА^ m_ i H- y/m T 1<УП m_|_i).
V 2тш
5. APPROXIMATION METHODS
113
Put it back into (5.22)
d , 4	/ h
rh—Cnit) = ~F°e \ о-----
dt	у 2muj
(yTz + le “^cn+1(t) +	.
Perturbation theory means expanding cn(t) = c^0) +	+ ..., and to zeroth
order this is
|U0)(*) = ° => 40) = ^no
\J l
To first order we get

4	=
m Jo m
ih у 2mw Jo	4	7
We get one non-vanishing term for n = 1, i.e. at first order in perturbation
theory with the H.O. in the ground state at t = 0 there is just one non-zero
expansion coefficient

To / h
ih v 2mco

ih у 2mco iw —	Q
^./ h 1 fi _ e(^-i)t
ih у 2mco iw —	'
and
I О = £	I n) = cP(f)e^ 11).
n
The probability of finding the H.O. in 11) is
|{1 |a,t>|2 = kfW-
As t —> oo
c(i) F°. П 1
ih V 2mw iw — -
= const.
This is of course reasonable since applying a static force means that the
system asymptotically finds a new equilibrium.
114
(b)	As remarked earlier there are no other non-vanishing cn’s at first order,
so no higher excited states can be found. However, going to higher order in
perturbation theory such states will be excited.
5.4 Consider a composite system made up of two spin | objects,
for t < 0, the Hamiltonian does not depend on spin and can be
taken to be zero by suitably adjusting the energy scale. For t > 0,
the Hamiltonian is given by
/4A\ - -
H = ТГ )sl-s2.
Suppose the system is in | -|--) for t < 0. Find, as a function of
time, the probability for being found in each of the following states
I + +X I + I - +), I----}'•
(a)	By solving the problem exactly.
(b)	By solving the problem assuming the validity of first-order
time-dependent perturbation theory with H as a perturbation switched
on at t = 0. Under what condition does (b) give the correct results?
(a) The basis we are using is of course \Siz^S2z\ Expand the interaction
potential in this basis:
Si • S2 = SixS2x + SiyS2y + SizS2z = {in this basis}
h2 Г
= T[(l + x-l + l->< + l)i( l + ><-1 + I-X +1)2+
+ *2(-| + ){-| + |->< + 1Ы- l + ><-I + l-X + IW
+ (l + X + l - l-><-1)1(l + >< +1 - l-X-1)2 =
h2 г
= y l+ + X—l + l+_X_+l +
+ 1 -+)(+-1 +1 --){++1 +
+ *2(l ++){--1 -1 +-)(-+1 +
-I -+)(+-1 +1 --){++1) +
5. APPROXIMATION METHODS
115
+ I + + )(++ I - I +-X+- I +
-I -+){-+ I + I I
In matrix form this is (using 11} = | + + } | 2} = | H— )
|3)= I -+) |4)= | --))
/ 1 0	0 0 \
(5.23)
\ 0 0	0 1 /
This basis is nice to use, since even though the problem is 4-dimensional we
get a 2-dimensional matrix to diagonalize. Lucky us! (Of course this luck is
due to the rotational invariance of the problem.)
Now diagonalize the 2x2 matrix to find the eigenvalues and eigenkets
0 = det i -1JA . ) = (—1 — A)2 — 4 = A2 + 2A — 3
\ Z	— 1 — Л j
=> A = 1, —3
A = 1 :
( — 1 2\/ж\_/ж\
(2 -1 J \ у J ~ \ у J
1
=> — x -\-2y = x=?x = y = -/=
A = -3 :
/ —1 2 \ / x \ _	/ x \
(2	-1Д г/ / = A у )
i
=> — x + 2y = —Зж => x = — у = -/=
So, the complete spectrum is:
f I+ + M-----I + - H I - +) with energy A
with energy — ЗД
116
This was a cumbersome but straightforward way to calculate the spectrum.
A smarter way would have been to use S = Si + S2 to find
s2 = S2 = S2 + S2+2S1 • S2 => Si- S2 = |(S2 - S[- S$)
We know that S2 = S% = Ti2| Q + 1) = so
Si • S2 = 1 (s2 -
\	2 J
Also, we know that two spinj systems add up to one triplet (spin 1) and one
singlet (spin 0), i.e.
S = 1 (3 states) => Si-S2 = j(ft2l(l + 1) - ^) = \h2
.	(5.24)
S = 0 (1 state) => Si • S2 = |(-¥) =
Since H = ^-Si - S2 we get
.	.	4A Ih2
E(spm=l) = p- — = A,
T,/ • X	4A-3&2
E(spm=0) =	= -3A-
(5.25)
From Clebsch-Gordan decomposition we know that 11 + + ), |---),
Tjd + _) + I - +)) } are spin 1, and ^( | + -) - | - +)) is spin 0!
Let’s get back on track and find the dynamics. In the new basis H is diagonal
and time-independent, so we can use the simple form of tthe time-evolution
operator:
W(t,t0) = exp —	— to)} •
The initial state was | -|— ). In the new basis
{ |1) = I ++), |2) = I -	|3) =-)=(| + -} + I -+)),
I4) = A( I + “) “ I “+)) }
5. APPROXIMATION METHODS
117
the initial state is
I +-> = 4(|3>+ |4)).
Acting with 7/(t, 0) on that we get
where
ш =	(5.26)
The probability to find the system in the state | /3) is as usual |( /3 | a, t )|2
(+ + | a, t) = (-------------| a, t) = 0
< |( + - | a, t )|2 = |(2 + e4itJjt + e~4itJjt) = | (1 + cosXA) ~ 1 - 4(^t)2 ...
|( —h | a, t)|2 = | (2 — e4llJjt — е~4шГ) = | (1 — созЗиЛ) ~ 4(^t)2 ...
(b) First order perturbation theory (use S. 5.6.17):
40) =
41}(t) = J' dt'e^V^.
(5.27)
Here we have (using the original basis) Ho = 0, V given by (5.23)
118
I/)	=	i-n
K	~ £i	= {E„ = 0} = 0,
vf.	=	2Д
Vni	=	0,	n±	f.
Inserting this into (5.27) yields
„(0)	_	(0)	_ 1
ci — c|-|—) —
c{p =	= -1- Гdt2A = -2iwt.	(5.28)
as the only non-vanishing coefficients up to first order. The probability of
finding the system in |-----) or | + +) is thus obviously zero, whereas for
the other two states
Л1 +-)) = 1
p(I - + » = kJ1!') + *(') +    I2 = |2M2 = 4(wf)2
to first order, in correspondence with the exact result.
The approximation breaks down when urt C 1 is no longer valid, so for a
given t:
л Ъ
wt < 1 => A < -.
5.5 The ground state of a hydrogen atom (n = 1,Z = 0) is subjected
to a time-dependent potential as follows:
V(x, t) = Vo cos(kz — wt).
Using time-dependent perturbation theory, obtain an expression
for the transition rate at which the electron is emitted with mo-
mentum p. Show, in particular, how you may compute the angular
distribution of the ejected electron (in terms of в and ф defined
with respect to the г-axis). Discuss briefly the similarities and the
differences between this problem and the (more realistic) photo-
electric effect, (note: For the initial wave function use
ф /=0(^)=	(_)
л/7Г \ttoJ
5. APPROXIMATION METHODS
119
If you have a normalization problem, the final wave function may
be taken to be
\L2 /
with L very large, but you should be able to show that the observ-
able effects are independent of L.)
To begin with the atom is in the n = 1,1 = 0 state. At t = 0 the perturbation
V = Vo cos(kz — wt)
is turned on. We want to find the transition rate at which the electron is
emitted with momentum p f. The initial wave-function is
1 Zl\3/2 ,
Фг(ж) = — — e~r/a°
д/7Г \а0/
and the final wave-function is
The perturbation is
у = y0	_|_ e-i(A:z-wt)]
= Veiwt + V]e~iwt.	(5.29)
Time-dependent perturbation theory (S.5.6.44) gives us the transition rate
Wj_>n =	|V^| 8(En - [Ei + Tiw))
because the atom absorbs a photon hw. The matrix element is
and
eikz]
/ ni
(kf\eikz\n = 1,1 = 0) = j d3x{kf\eikz\x){x\n = 1,1 = 0) =
-ikf-x	1	/1\3/2
АЛ________ ikx3___I___\ p-r/a0 _
L3/2 6 y/EVao)
У d3xe~^f '^-кхА-г/ао
(5.30)
120
So (elkz) . is the 3D Fourier transform of the initial wave-function (and some
constant) with q = kf — kez. That can be extracted from (Sakurai problem
5.39)
(«Л _ 64яг	1
” LV° [4 + й - M4
The transition rate is understood to be integrated over the density of states.
We need to get that as a function of pf = hkf. As in (S.5.7.31), the volume
element is
n2dnd£l = n2<7Q-—dpf.
dpf
Using
P2 _ п2(2тг)2
f ~ h2 ~ I2
we get
dn 1 2L2pf	2irh L2pf	L
dpf	2n (2тг/г)2	Lpf (2тг/г)2	2ivti
which leaves
2	L3k2f	L3p2f
n dnd£l = 'dSldpf = ,n 'dSldpt
[2тг)лп	(27ГП)-5
and this is the sought density.
Finally,
_ 2яУог64^ 1______________________Щ
’’ 4 L’“5 [й + d, - M4 (w P/'
Note that the L’s cancel. The angular dependence is in the denominator:
(kf — kezj = [(|kf \cos0 — fc) ez +	(cos(pex + simpey)] =
= \kf\2cos20 + k2 — 2k\kf\cos0 + \kf\2sin20 =
= k2 + k2 — 2k\kf\cos0.	(5.31)
In a comparison between this problem and the photoelectric effect as dis-
cussed in (S. 5.7) we note that since there is no polarization vector involved,
w has no dependence on the azimuthal angle ф. On the other hand we did
not make any dipole approximation but performed the ж-integral exactly.