1
Quantum Mechanics (II): Homework 2
Due: March 23, 2016
Ex.1 10 Let Mxy be the mirror imaging operator with respect to x-y plane. Show that {Mxy , Ji } = 0 for i = x, y,
and [Mxy , Jz ] = 0.
Ex.2 Consider a state with orbital angular momentum l = 1 with the following form
|Ψi = C0 |10i + C−1 |1 − 1i + C1 |11i
~ Express the coefficients C in terms
(a) 8 Find a direction n̂ such that this state is an eigenstate of the operator n̂ · L.
of the angles θ, φ that define the direction n̂
(b) 7 Write down expressions for the eigenvectors of Ly , |1, Ly = 0i, |1, Ly = h̄i and |1, Ly = −h̄i. Repeat the same
calculations for the eigenstates |1, Lx i.
Ex.3 10 Suppose that a particle is in a state ψ with some definite value of the z-component of orbital angular
momentum Lz . Find the average value of Lx and Ly for this state.
Ex.4
(a) 10 Find the values of hjm| Jx |jmi and hjm| Jy2 |jmi.
(b) 5 Let J~ be the general angular momentum operators. Let j be the quantum number of the total angular moment.
Find all eigenvalues to 4Jx + 3Jz .
Ex.5
∂
∂
∂
, L̂± = h̄i e±iφ ±i ∂θ
− cot θ ∂φ
,
(a) 20 Show that by coordinate transformation, in spherical coordinates (r, θ, φ), L̂z = h̄i ∂φ
and
1 ∂2
1 ∂
∂
2
2
L = −h̄
+
sin θ
.
sin θ ∂θ
∂θ
sin2 θ ∂φ2
(b) 5 Verify the commutation relation [J2 , Jk ] = 0 for k = 1, 2, and 3.
(c) 5 Let σ = (σx , σy , σz ) be the Pauli matrices. Show that {σi , σj } = 2δij .
(d) 5 Given any two vectors, a and b, show that
(σ · a) (σ · b) = a · b + iσ· (a × b) .
2
(e) 5 Show that L2 = (r × p)2 = r2 p2 − (r · p) + ih̄r · p.
Ex.6 Verify the following relations in the Schwinger’s model of the angular momentum:
(a) 5 [Jz , J± ] = ±h̄J± .
(c) 5 [J2 , J] = 0
2
(b) 5 [J+ , J− ] = 2h̄Jz .
(d) 5 J2 = h̄2 N̂ ( N̂2 + 1), where N̂ = a†+ a+ + a†− a− .
(e) 10 The Schwinger’s model has the constraint: a†+ a+ + a†− a− = 2j. Motivated by this, we can ”solve” a+ in terms
of a−
q
a†+ = a+ = 2j − a†− a− ,
√
√
Thus, J+ = h̄ 2j − n− a− , J+ = h̄a†− 2j − n− , Jz = h̄(j − n− ). Verify that they satisfy the commutation relations
for angular momentum. These are known as the Holstein-Primakoff representation of angular momentum.
Ex.7 Some of problems in Ex. 5(a) can be also done in a different way. Consider a wavefunction in spherical
coordinates, ψ(r, θ, φ). If we rotate this system about x-axis by x , i.e.,
 0 
 
x
1 0 0
x
 y 0  =  0 1 x   y  ,
0 −x 1
z
z0
(r, θ, φ) is changed into (r0 , θ0 , φ0 ), where r0 = r, θ0 ≡ θ + δθ, and φ0 ≡ φ + δφ.
(a) 5 Express δθ and δφ in terms of θ, φ, and x .
(b) 5 By expanding ψ(r0 , θ0 , φ0 ), find L̂x .
Ex.8 In the Schwinger model, angular momenta consist of two types of harmonic oscillators. This fact can be also
demonstrated in a two dimensional model for harmonic oscillators
Ĥ =
p̂2y
mω 2 (x̂2 + ŷ2 )
p̂2x
+
+
.
2m 2m
2
2
This problem is separable and can be solved by introducing two annihilation operators
r
mω
ip̂x
âx =
x̂ +
,
2h̄
mω
r
mω
ip̂y
ây =
ŷ +
.
2h̄
mω
In this scheme, we label the state by the corresponding numbers for each operators (nx , ny ). One can, however, use
the energy eigenvalue and the eigenvalue of Lz to label the state:
(a) 5 Show that L̂z = ih̄(âx â†y − â†x ây ).
(b) 10 If we define
1
â± = √ (âx ∓ iây ) ,
2
show that L̂z = h̄(â†+ â+ − â†− â− ) and Ĥ = h̄ω(â†+ â+ + â†− â− + 1). Verify that [â+ , â†+ ] = 1, [â− , â†− ] = 1, and [L̂z ,
Ĥ] = 0, so that we can diagonalize L̂z and Ĥ simultaneously. If the eigenvalue of L̂z is mh̄, show that the energy
eigenvalue can be written as E = h̄ω(2k + |m| + 1) with k = 0, 1, 2, 3, · · ·. (This result can be also derived in a more
traditional way, see Ex. 12.3.7.)
(c) 10 If we introduce complex numbers to represent (x, y) by z = x + iy, show that
r
mω
d
mω
h̄
exp −
|z|2
exp
|z|2 ,
â+ =
mω
2h̄
dz
2h̄
r
h̄
mω 2 d
mω 2 exp
â− =
exp −
|z|
|z| .
mω
2h̄
dz ∗
2h̄
From these expressions, find the ground state.
Ex. 9 10 Consider a system with angular momentum j = 1. The state of the system is
|ψi = α|1, −1i + β|1, 0i + γ|1, 1i,
where α, β, and γ are three complex numbers. Calculate hJi in terms of α, β, and γ.
Ex. 10
(a) 10 Consider the eigenstate of the orbital angular momentum |l, mi. We have shown that its functional form is
l
h r | l, mi = Cl (r) Ym
(r̂),
where Cl (r) is some arbitrary function satisfying the normalization condition
Z ∞
dr r2 |Cl (r)|2 = 1.
0
Find the functional form of |l, mi in the Fourier k space, i.e., find a similar functional form for h k |l, mi.
(b) 10 Consider a particle represented by the wavefunction
ψ = α(x + y + 2z)e−βr ,
p
where r = x2 + y 2 + z 2 , and α and β are positive real constants. Find the total angular momentum of the particle
and the probability of getting Lz = +h̄ if the z-component of angular momentum of the particle were measured.
Ex.11 10 The Schwinger model of angular momentum essentially does the 2nd quantization for the spins: Consider
J = h̄2 σ, where σ are the Pauli matrices. Show that the 2nd quantization of the J operator in terms of boson creation
and annihilation operators leads to exactly the same formulation as the Schwinger model. Note that one can also
do the 2nd quantization in terms of fermion operator, but the J operator thus obtained can only have the quantum
number 1/2.
Ex.12 10 Consider a particle of mass m in an attractive delta-shell potential of radius a, charaterized by the strength
of a postive parameter g
V (r) = −
h̄2 g
δ(r − a).
2m
Determine the condition on g and a such that there is a bound state for the particl with angular momentum l = 0.
3
Bonus problem 1 (+1) Uncertainty relation for Lz and φ
∂
so that similar to [x, p] = ih̄, we have [φ, Lz ] = ih̄. It is therefore tempting to conclude
It is known that Lz = h̄i ∂φ
∆φ∆Lz ≥
h̄
.
2
(1)
This equality, however, is not correct as it implies that when ∆Lz = 0, ∆φ = ∞, which is in conflict with the fact
that 0 ≤ φ ≤ 2π. The underlying reason for the failure of Eq.(1) is that Lz is not Hermitian for all the wavefunctions
considered in deriving Eq.(1): The Hermiticity of Lz requires the wavefunction being periodic in φ, and obviously,
if Ψ(φ) is periodic, φΨ(φ) encountered when [φ, Lz ] acts on Ψ is not periodic. Then what is the correct uncertain
relation for Lz and φ ? We shall derive the correct one in the following.
(a) Definition of hφi and hφ2 i
Even though φ and φ + 2π are the same for periodic
R 2π for integration of φ for performing the
R π functions, the domain
average is ambiguous. For example, the values of −π φ|Ψ(φ)|2 dφ and 0 φ|Ψ(φ)|2 dφ are different even if Ψ(φ) is
periodic in φ. To overcome this difficulty, we first note that even in the evaluation of the
R ∞average of x, one has an
alternative definition that can reproduce the same result. Consider the function f (a) = −∞ x2 |Ψ(x + a)|2 dx whose
R∞
minimu occurs at x0 . Show that x0 = hxi ≡ −∞ x|Ψ(x)|2 dx and f (x0 ) = (∆x)2 . Therefore, one can also define hxi
as the value of a for which f (a) is minimum and define (∆x)2 as f (x0 ).
(b) Following (a), we define (∆φ)2 as
Z π
min
φ2 |Ψ(φ + α)|2 dφ .
−π
The uncertainty for Lz is defined as
(∆Lz )2 =
Z
π
Ψ∗ (φ)(Lz − hLz i)2 Ψ(φ)dφ.
−π
Prove the following uncertainty relation
(∆Lz )2 · (∆φ)2
2
[1 − (3/π 2 )(∆φ)2 ]
≥
h̄2
.
4
Find (∆φ)2 when ∆Lz = 0.
Bonus problems (+2): Supersymmetric Quantum Mechanics
We have learned the usage of ladder operators both in the theory of harmonic oscillators and the theory for the
angular momentum. In the following, I will introduce you another usage.
(A) (a) Suppose Ψ0 (x) is the ground state (with energy E0 ) to the Hamiltonian
Ĥ− = −
h̄2 d2
+ V− (x)
2m dx2
where V− (x) ≡ V(x) − E0 and we have shifted E0 to be the new energy zero point so that Ĥ− Ψ0 (x) = 0. Show that
we can factorize Ĥ− as
h̄ d
h̄ Ψ00 (x)
Ĥ− = †  with  = √
+ W(x), W(x) ≡ − √
.
2m dx
2m Ψ0 (x)
W(x) is usually called the super-potential of this problem.
√
†
(b) Verify that [Â, Â ] = 2h̄ W0 (x) / 2m.
†
(c) Consider another Hamiltonian defined by Ĥ+ = ÂÂ . Show that Ĥ+ can be rewritten as
Ĥ+ = −
h̄2 d2
+ V+ (x),
2m dx2
where V+ (x) = −V− (x) + 2W2 (x). V+ (x) is called the supersymmetric partner potential.
−
(d) Let the energy spectrum of Ĥ+ and Ĥ− labeled by E+
n and En , and the corresponding eigenfunctions are labeled
4
−
by Ψ+
n and Ψn . Prove the following relations
−
E+
n = En+1 ,
1
ÂΨ−
Ψ+
n = q
n+1 ,
−
En+1
1
† +
Ψ−
n+1 = p + Â Ψn .
En
In other words, Ĥ+ and Ĥ− almost have the same spectrum except that E−
0 has no correspondence.
(B) Application The results of (d) have important applications: suppose that we know the solutions of some
Hamiltonian Ĥ− , then we can solve another supersymmetric partner Hamiltonian Ĥ+ ! We now apply this approach to
a constant potential when we know the√solution. For this purpose,
it is convenient to write everything in dimensionless
√
form. Verify that by re-defining x as mx/h̄ and W as W/ 2, we can write
σz
d2
1
b ≡ Ĥ+
− 2 + W2 (x) + W0 (x).
(2)
H
=
2
dx
2
Ĥ−
2
0
In other word, we have V± = (W
√ ± W )/2.
V+ = V0 > 0. Show that for |W| < 2V0 ,
Consider the supersymmetric partner to the constant potential
V− (x) = V0 1 −
2
√
.
cos h2 2V0 x
V− (x) is one of the famous reflectionless potential. From the above consideration, show that there is one bound
state to the Hamiltonian
h̄2 d2
h̄2 a2
b
h=−
−
sec h2 (ax),
2m dx2
m
and find the ground state and the ground state energy. Show that the excited states are the form
ψk =
−ik + a tanh(ax)
√
exp(ikx),
a2 + k 2
q
where k = 2mE/h̄2 and verify that these states have no reflection by taking the limit x → −∞.
(C) By considering W(x) = n tanh(x) in eq.(2) with n being some positive integer and successively applying the
above approach from n = 1 to n, show that there are n bound states to the Hamiltonian
1 d2
n(n + 1)
b
h=−
−
sec h2 (x),
2 dx2
2
with bound state energies being −1/2, −22 /2, −32 /2, · · ·, −n2 /2.