Quantum Mechanics I, Correction Sheet 8, Spring 2013

Responsible for this sheet: J. Guillod (julien.guillod@unige.ch), office 212, Sciences I

May 7, 2013 (Ecole de Physique, Auditoire Stückelberg)

Prof. D. van der Marel (dirk.vandermarel@unige.ch)

Tutorials: J. Guillod (julien.guillod@unige.ch), O. E. Peil (oleg.peil@unige.ch)

I. RADIAL POTENTIALS

Consider a particle in three dimensions in a radial potential V (r),

p̂2 ~2 1 ∂ 2 L̂2
Ĥ = + V (r) = − 2
r+ + V (r) ,
2µ 2µ r ∂r 2µr2

where p̂ = −i~∇ is the momentum operator and L̂ = x ∧ p̂ is the angular momentum operator.

A. Radial equation

1. Since the potential is radial the Hamiltonian commutes with the angular momentum, and
therefore we can take a basis of eigenvectors of Ĥ, L̂2 and Lz ,

ψ`,m (x) = R` (r)Y`,m (θ, ϕ) ,

where the spherical harmonics Y`,m are the eigenfunctions of the angular momentum

L̂2 Y`,m = ~2 ` (` + 1) Y`,m , Lz Y`,m = ~mY`,m .

By acting with the Hamiltonian on this ansatz, we find

 2
~ 1 ∂2 ~2 ` (` + 1)

Ĥψ`,m = − r+ + V (r) R` Y`,m ,
2µ r ∂r2 2µr2

and therefore the Schrödinger equation Ĥψ`,m = Eψ`,m implies that the radial wave function
satisfies
 2
~ 1 ∂2 ~2 ` (` + 1)

− r+ + V (r) R` = E R` .
2µ r ∂r2 2µr2

2. The change of variable u` (r) = rR` (r), transforms the previous equation into
 2 2
~2 ` (` + 1)

~ ∂
− + + V (r) u` = E u` .
2µ ∂r2 2µr2

3. (*) By plugging R` (r) ≈ rs in the radial equation we find

 2
~ s (s + 1) ~2 ` (` + 1)

− + + V (r) − E rs ≈ 0 .
2µr2 2µr2

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 Therefore by assuming that the potential is bounded by 1/r near r ≈ 0, the dominant terms
are the first two, so we obtain the relation

s (s + 1) = ` (` + 1) ,

which has two solutions

s = `, s = −` − 1 .

Consequently the second order differential equation for R` admits two independent solutions:
R` (r) ≈ r` and R` (r) ≈ r−`−1 . For ` ≥ 1, the second solution in not square-integrable and
therefore, not admissible. For ` = 0, the second solution is like 1/r, but since ∆(1/r) =
−4πδ(r), ψ0,0 is not an eigenvector of the Hamiltonian. Therefore, the only physical solution
is u` (r) ≈ r`+1 which can be distinguished from the second one by adding to the radial
differential equation the boundary condition u` (0) = 0.

B. Hydrogen atom and harmonic oscillator

1. The equation for the hydrogen atom is

 2
` (` + 1) 2µZe2 1 2µE

∂
− + + 2 u` = 0 ,
∂r2 r2 ~2 r ~
and the one for the harmonic oscillator is
 2
` (` + 1) µ2 ω 2 2 2µE

∂
− − 2 r + 2 u` = 0 .
∂r2 r2 ~ ~

2. By using the change of variable u` (r) = sp v` (s), with s = rq , we have

 
∂ ∂s ∂ p p−1/q ∂
u` (r) = (s v` (s)) = qs s v` (s) + pv` (s) ,
∂r ∂r ∂s ∂s
and
∂2 2
 
p−2/q 2 ∂ ∂
u` (r) = qs qs v ` (s) + q (2pq + q − 1) s v ` (s) + p (pq − 1) v ` (s) .
∂r2 ∂s2 ∂s
Since no first derivatives are contained in both equations, we have to set 2pq + q − 1 = 0,
i.e. 1/q = 1 + 2p. That way the equation for the hydrogen atom becomes
2 2µZe2 −1/q 2µE
 
2 −4p ∂ −2/q −2/q
q s + pq (pq − 1) s − ` (` + 1) s + s + 2 v` = 0 ,
∂s2 ~2 ~
or equivalently
!
∂2 ` (` + 1) (1 + 2p)2 + p (p + 1) 2µZe2 2p−1 2µE
− + 2 2 s + 2 2 (1 + 2p)2 s4p v` = 0 .
∂s2 s2 ~ q ~ q

By choosing p = 1/2, i.e. q = 1/2, this equation becomes

 2
4` (` + 1) + 3/4 8µE 2 8µZe2

∂
− + 2 s + v` = 0 ,
∂s2 s2 ~ ~2
which is similar to the equation for the harmonic oscillator.

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 3. The correspondence between the parameters of the two problems is given by
1
`harm. = 2`coul. + , Eharm. = 4Ze2 , µω 2 = −8Ecoul. ,
2
and therefore the roles of the coupling constants and of the energy eigenvalues are inter-
changed. It is important to notice that this correspondence is not totally exact from a
physical point of view since both physical problems have integers `, but the relation between
`harm. and `coul. does not implies that. In fact this peculiar relation between the Coulomb
potential and the harmonic potential comes from the fact that there are the only two radial
potentials which have an “hidden symmetry” called a dynamical symmetry. In addition to
angular momentum, a second independent quantity is conserved for these two potentials:
the Lenz vector. Intuitively this symmetry is related to the fact that in classical mechanics
orbits are closed with the two potentials in consideration.

II. SYMMETRIES AND CONSERVED QUANTITIES

In quantum mechanics a
symmetry is represented by a one-parameter unitary group, i.e. a family

of unitary operator Ûα α∈R such that

Ûα Ûβ = Ûα+β .

The one-parameter unitary group act on states as

|ψi 7→ Ûα |ψi .

Since only brackets have physical meaning, we have

hφ|Â|ψi = hÛα φ|Â|Ûα ψi = hφ|Ûα† ÂÛα |ψi ,

and therefore, the action of the group can also be viewed as the transformation of observables,

 7→ Ûα† ÂÛα .


A one-parameter unitary group Ûα α∈R
is called as symmetry of the system if the Hamiltonian is
invariant, i.e.

Ĥ = Ûα† Ĥ Ûα .

1. The action of the symmetry on the scalar product is given by

hφ|ψi 7→ hÛα φ|Ûα ψi = hφ|Ûα† Ûα |ψi ,

which shows that the requirement of Ûα to be unitary correspond to the conservation of the
scalar product.

2. By multiplying the definition of a symmetry by Ûα we obtain

Ûα Ĥ = Ûα Ûα† Ĥ Ûα = Ĥ Ûα ,

which proves that Ûα is a symmetry if and only if

 
Ûα , Ĥ = 0 .

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 3. The action of the one-parameter group on the Schrödinger equation is given by
     
∂ ∂ ∂
i~ − Ĥ |ψi 7→ i~ − Ĥ Ûα |ψi = Ûα i~ − Ûα† Ĥ Ûα |ψi ,
∂t ∂t ∂t
and so the Schrödinger equation is invariant if and only if Ûα† Ĥ Ûα = Ĥ, i.e. if Ûα is a
symmetry.

4. By the Stone’s theorem, every one-parameter unitary group can be written as

Ûα = e−iαQ̂/~ ,
where Q̂ is an hermitian operator, which is called the generator of the symmetry. By taking
the derivative with respect to α, and evaluating at α = 0, we directly find that
d
i~
Ûα = Q̂Ûα , Û0 = Iˆ .
dα
In particular the generator of the symmetry is given by
d
Q̂ = i~ Ûα .
dα α=0

5. If Ûα is a symmetry, then

   d  d  
Q̂, Ĥ = i~ Ûα , Ĥ = i~ Ûα , Ĥ = 0,
dα α=0 dα α=0
 
and reciprocally if Q̂, Ĥ = 0, then

Ûα , Ĥ = e−iαQ̂/~ , Ĥ = 0 .
   

In particular there exists a basis of the Hilbert space formed from eigenvectors common to
Ĥ and Q̂.

6. By using Ehrenfest theorem, we obtain that Q̂ is a conserved quantity,

d 1 D E
Q̂ = Q̂, Ĥ = 0 .
dt i~
This result is the quantum analog of the Noether’s theorem which relates symmetries to
conserved quantities.

A. Translation invariance

The translation operator is defined as

T̂α |ψ(x)i = |ψ(x − α)i .


1. The set T̂α α∈R
is a one-parameter group because

T̂α T̂β |ψ(x)i = T̂α |ψ(x − β)i = |ψ(x − β − α)i = T̂α+β |ψ(x)i .
Since
hφ(x)|ψ(x)i = hφ(x − α)|ψ(x − α)i = hT̂α φ(x)|T̂α ψ(x)i = hφ(x)|T̂α† T̂α |ψ(x)i ,
we obtain that T̂α† T̂α = I.
ˆ Since T̂ −1 = T̂−α this prove that T̂α is surjective, so T̂α T̂α† = Iˆ
α
and therefore T̂α is unitary.

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 2. By writing the exponential as a series, and using the Taylor expansion, we obtain
∞
X
e−iαP̂ /~ |ψ(x)i = e−α∂x |ψ(x)i = (−α∂x )n |ψ(x)i = |ψ(x − α)i = T̂α |ψ(x)i ,
n=0

where P̂ = −i~∂x is the momentum operator.

3. By Ehrenfest theorem, if the Hamiltonian is invariant under translation, then the momentum
P̂ is conserved.

4. By considering a particle in a periodic potential V (x+a) = V (x) of period a, the Hamiltonian

is invariant under T̂a . Since T̂a is unitary, the eigenvalue λ corresponding to an eigenvector
|ψ(x)i of T̂a is normalized, so we choose λ = e−ika , with k ∈ R. Therefore,

|ψ(x − a)i = T̂a |ψ(x)i = λ|ψ(x)i = e−ika |ψ(x)i ,

and the wavefunction |ψ(x)i is not a-periodic, but can be written as

|ψ(x)i = eikx u(x) ,

where u is periodic of period a. This is the Bloch theorem. Then the Schrödinger equation
becomes

~2 2 ~2
   
ikx 2
− ∂ + V (x) − E |ψ(x)i = e − (∂x + ik) + V (x) − E u(x) ,
2m x 2m

and can be solved with a Fourier series, because u(x) is periodic.

B. Time invariance

The time-translation operator or evolution operator is given by

Ûα |ψ(t)i = |ψ(t + α)i .

1. By using the Schrödinger equation, we have

∞
X
e−iαĤ/~ α∂t
|ψ(t)i = e |ψ(t)i = (α∂t )n |ψ(t)i = |ψ(t + α)i .
n=0

2. By defining the Hamiltonain as the generator of the time-invariance unitary group, by Stone’s
theorem we have that Ĥ is an hermitian operator and that

Ûα = e−iαĤ/~ .

We already prove that

d
i~ Ûα = Ĥ Ûα , Û0 = Iˆ ,
dα
which is equivalent to the Schrödinger equation.

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 C. (*) Rotation invariance

In two dimensions x = (x, y), the operator associated to a rotation of angle α is given by

R̂α |ψ(x)i = |ψ(Rα x)i ,

where Rα is the following rotation matrix

 
cos α − sin α
Rα = .
sin α cos α

1. The generator of the rotation symmetry is defined by

d
L̂z = i~ R̂α .
dα α=0

First of all we define the infinitesimal rotation generator,

   
d d cos α sin α 0 1
Az = Rα = = .
dα α=0 dα − sin α cos α α=0 −1 0

So by acting on a state and using the chain rule we obtain

d d
L̂z |ψ(x)i = i~ R̂α |ψ(x)i = i~ |ψ(Rα x)i
dα α=0 dα α=0
d
= i~ Rα x · ∇|ψ(x)i = i~ (Az x) · ∇|ψ(x)i
dα α=0
= − (Az x) · p̂|ψ(x)i = (xp̂y − y p̂x ) |ψ(x)i .

This proves that the generator of the rotation symmetry in the angular momentum

L̂z = xp̂y − y p̂x .

2. In three dimensions the rotation of angle α around the axis n, is generated by the angular
momentum L̂ along n, i.e. n · L̂,

R̂n,α = e−iαn·L̂/~ .

Therefore, an Hamiltonian which is invariant under the rotation along n induces the con-
served quantity n · L̂ .

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