PHYS851 Quantum Mechanics I, Fall 2009
HOMEWORK ASSIGNMENT 11
Topics Covered: Orbital angular momentum, center-of-mass coordinates
Some Key Concepts: angular degrees of freedom, spherical harmonics
1. [20 pts] In order to derive the properties of the spherical harmonics, we need to determine the action
d
of the angular momentum operator in spherical coordinates. Just as we have hx|Px |ψi = −i~ dx
hx|ψi,
~
~ =R
~ × P~ and our knowledge of momentum
we should find a similar expression for hrθφ|L|ψi.
From L
operators, it follows that
d
d
d
d
d
d
~
−z
−x
+ ~ey z
+ ~ez x − y
hrθφ|ψi.
hrθφ|L|ψi = −ı~ ~ex y
dz
dy
dx
dz
dy
dx
Cartesian coordinates are related to spherical coordinates via the transformations
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
and the inverse transformations
p
x2 + y 2 + z 2
p
x2 + y 2
θ = arctan(
)
z
y
φ = arctan( ).
x
Their derivatives can be related via expansions such as
r=
∂x =
∂θ
∂φ
∂r
∂r +
∂θ +
∂φ .
∂x
∂x
∂x
Using these relations, and similar expressions for ∂y and ∂z , find expressions for hrθφ|Lx |ψi, hrθφ|Ly |ψi,
and hrθφ|Lz |ψi, involving only spherical coordinates and their derivatives.
∂x r =
∂x θ =
x
r = sin θ cos φ
z2 √ x
= cos θrcos φ
r 2 z x2 +y 2
2
∂x φ = − x2x+y2 xy2 = − csc θrsin φ
So
d
dx
= sin θ cos φ ∂r +
cos θ cos φ
∂θ
r
∂y r =
∂y θ =
y
r = sin θ sin φ
z2 √ y
= cos θrsin φ
r 2 z x2 +y 2
∂y φ =
x2 1
x2 +y 2 x
So
d
dy
∂z r =
=
csc θ cos φ
r
sin θ sin φ∂r + cos θrsin φ ∂θ
z
r
= cos θ
−
csc θ sin φ
∂φ
r
=
+
csc θ cos φ
∂φ
r
1
√
2
x2 +y 2
= − sinr θ
∂z θ = − zr2 z 2
∂z φ = 0
d
So dz
= cos θ∂r − sinr θ ∂θ
Now
hrθφ|Lx |ψi = −i~(y
d
d
− z )hrθφ|ψi
dz
dy
So we can say
d
d
−z
Lx = i~ y
dz
dy
= −i~ r sin θ cos θ sin φ∂r − sin2 θ sin φ∂θ − r sin θ cos θ sin φ∂r − cos2 θ sin φ∂θ + cot θ cos φ∂φ
Which means
hrθφ|Lx |ψi = −i~ (− sin φ∂θ − cot θ cos φ∂φ ) hrθφ|ψi
Similarly we can say
d
d
−x )
dx
dz
= −i~ r sin θ cos θ cos φ∂r + cos2 θ cos φ∂θ − cot θ sin φ∂φ − r sin θ cos θ cos φ∂r + sin2 θ cos φ∂θ
Ly = −i~(z
so that
hrθφ|Ly |ψi = −i~ (cos φ∂θ − cot θ sin φ∂φ ) hrθφ|ψi
Lastly, we have
d
d
−y )
dy
dx
2
= −i~ r sin θ sin φ cos φ∂r + sin θ cos θ sin φ cos φ∂θ + cos2 φ∂φ
+r sin2 θ sin φ cos φ∂r − sin θ cos θ sin φ cos φ∂θ + sin2 φ∂φ
Lz = −i~(x
so that
hrθφ|Lz |ψi = −i~∂φ hrθφ|ψi
2
2. [15pts] From your previous answer and the definition L2 = L2x + L2y + L2z , prove that
hrθφ|L2 |ψi = −~2
∂2
1 ∂
∂
1
sin θ
+
2
sin θ ∂θ
∂θ sin θ ∂ 2 φ2
hrθφ|ψi.
L2 = −~2 [(sin φ∂θ + cot θ cos φ∂φ )(sin φ∂θ + cot θ cos φ∂φ )
+ (cos φ∂θ − cot θ sin φ∂φ )(cos φ∂θ − cot θ sin φ∂φ ) + ∂φ2
= −~2 sin2 φ∂θ2 + cot θ sin φ cos φ∂θ ∂φ − csc2 θ sin φ cos φ∂φ + cot θ sin φ cos φ∂φ ∂θ + cot θ cos2 φ∂θ
+ cot2 θ cos2 φ∂φ2 − cot2 θ cos φ sin φ∂φ + cos2 φ∂θ2 − cot θ cos φ sin φ∂θ ∂φ + csc2 θ sin φ cos φ∂φ
− cot θ sin φ cos φ∂φ ∂θ + cot θ sin2 φ∂θ + cot2 θ sin2 θ∂φ2 + cot2 θ sin φ cos φ∂φ + ∂φ2
= −~2 ∂θ2 + cot θ∂θ + (1 + cot2 θ)∂φ2
Noting that
1
∂θ sin θ∂θ = ∂θ2 + cot θ∂θ
sin θ
and
1 + cot2 θ =
the proof is complete.
3
1
sin2 θ
3. [10 pts] We can factorize the Hilbert space of a 3-D particle into radial and angular Hilbert spaces,
H(3) = H(r) ⊗ H(Ω) . Two alternate basis sets that both span H(Ω) are {|θφi} and {|ℓmi}. As the
angular momentum operator lives entirely in H(Ω) , we can use our results from problem 11.1 to derive
an expression for hθφ|Lz |ℓmi. Combine this with the formula Lz |ℓmi = ~m|ℓmi, to derive and then
solve a differential equation for the φ-dependence of hθφ|ℓmi. Your solution should give hθφ|ℓmi in
terms of the as of yet unspecified initial condition hθ|ℓmi ≡ hθ, φ|ℓmi
. What restrictions does this
φ=0
solution impose on the quantum number m, which describes the z-component of the orbital angular
momentum? Since mmax = ℓ, what restrictions are then placed on the total angular momentum
quantum number ℓ?
hθφ|Lz |ℓmi = −i~
∂
hθφ|ℓmi
∂φ
and
hθφ|Lz |ℓmi = ~mhθφ|ℓmi
Thus
−i~
∂
hθφ|ℓmi = ~mhθφ|ℓmi
∂φ
The solution to this simple first-order differential equation is
hθφ|ℓmi = hθ0|ℓmieimφ
Since the wave-function must be single valued, we require m to be a whole integer.
As mmax = ℓ, this implies that ℓ must be a whole integer also.
4
4. [10 pts] Using L± = Lx ± iLy we can use the relation L+ |ℓ, ℓi = 0 and the expressions from problem
11.1 to write a differential equation for hθφ|ℓℓi. Plug in your solution from 11.3 for the φ-dependence,
and show that the solution is hθφ|ℓℓi = cℓ eiℓφ sinℓ (θ). Determine the value of the normalization coefficient cℓ by performing the necessary integral.
We have hθφ|L+ |ℓℓi = 0
This implies hθφ|Lx |ℓℓi + ihθφ|Ly |ℓℓi = 0
Using the expressions from 11.1 gives (− sin θ∂θ − cot θ cos φ∂φ + i cos φ∂θ − i cot θ cos φ∂φ )hθφ|ℓℓi = 0
This simplifies to (i(cos φ + i sin φ)∂θ − cot θ(cos φ + i sin φ)∂φ )hθφ|ℓℓi = 0
Factoring out the eiφ gives (i∂θ − cot θ∂φ )hθφ|ℓℓi = 0
Plugging in the solution from 11.3 gives (i∂θ − iℓ cot θ)hθ0|ℓℓie−iℓφ = 0
which reduces to
∂θ hθ0|ℓℓi = ℓ cot θhθ0|ℓℓi
Now ∂θ sinℓ θ = ℓ sinℓ−1 θ cos θ = ℓ cot θ sinℓ θ
So the solution is
hθφ|ℓℓi = cℓ eiℓφ sinℓ θ
Rπ
R 2π
The normalization integral is 0 sin θdθ 0 dφ |hθφ|ℓℓi|2 = 1
Rπ
R 2π
2ℓ
With our solution this becomes |cℓ |2 0 sin
θdθ
0 dφ sin θ = 1
R
π
Performing the phi integral gives 2π|cℓ |2 0 sin θdθ sin2ℓ θ = 1
R1
u-substitution with u = cos θ gives 2π|cℓ |2 −1 du (1 − u2 )ℓ = 1
R1
Since the integrand is even, this reduces to 4π|cℓ |2 0 du (1 − u2 )ℓ = 1
From Mathematic we get 2π|cℓ |2 Γ[1/2]Γ[ℓ+1]
Γ[ℓ+3/2] = 1
q
Γ(ℓ+3/2)
which gives cℓ = 2πΓ[1/2]Γ[ℓ+1]
Thus we have
s
Γ(ℓ + 3/2)
sinℓ θeiℓφ
2πΓ[1/2]Γ[ℓ + 1]
q
1 3iφ
35
3
For the special case ℓ = 3 this gives hθφ|33i = 8 e
π sin θ, which agrees with the spherical
hθφ|ℓℓi =
harmonic Y33 (θ, φ) up to a non-physical phase factor.
5
p
5. [10 pts] Using L− |ℓmi = ~ ℓ(ℓ + 1) − m(m − 1)|ℓ, m − 1i together with your previous answers to
derive an expression for hθφ|ℓ, m − 1i in terms of hθφ|ℓmi. Explain how in principle you can now
recursively calculate the value of the spherical harmonic Yℓm (θφ) ≡ hθφ|ℓmi for any θ and φ and for
any ℓ and m. Follow your procedure to derive properly normalized expressions for spherical harmonics for the case ℓ = 1, m = −1, 0, 1.
To construct the other m states for the same ℓ, we can begin from the expression
p
hθφ|L− |ℓmi = ~ (ℓ + m)(ℓ − m + 1)hθφ|ℓ, m − 1i
Now
hθφ|L− |ℓmi = hθφ|Lx |ℓmi − ihθφ|Ly |ℓmi
= −i~(− sin φ∂θ − cot θ cos φ∂φ )hθφ|ℓmi − ~(cos φ∂θ − cot θ sin φ∂φ )hθφ|ℓmi
= ~e−iφ (−∂θ + i cot θ∂φ )hθφ|ℓmi
Putting the pieces together gives
e−iφ (−∂θ + i cot θ∂φ )
hθφ|ℓmi
hθφ|ℓ, m − 1i = p
(ℓ + m)(ℓ − m + 1)
Starting from our expression for hθφ|ℓℓi, we can find hθφ|ℓ, ℓ − 1i by applying the above differential
formula. Successive iterations will then generate all the remaining hθφ|ℓmi states.
6
6. [10 pts] A particle of mass M is constrained to move on a spherical surface of radius a.
Does the system live in H(3) , H(r) , or H(Ω) ? What is the Hamiltonian? What are the energy levels
and degeneracies? What are the wavefunctions of the energy eigenstates?
Because the radial motion is constrained to a fixed value, it is only necessary to consider the dynamics
in H(Ω) .
The Hamiltonian is then
H=
L2
2M a2
Choosing simultaneous eigenstates of L2 and Lz , we have
H|ℓ, mi =
~2 ℓ(ℓ + 1)
|ℓ, mi
2M a2
so that
Eℓ =
~2 ℓ(ℓ + 1)
2M a2
and
dℓ = 2ℓ + 1
The wavefunctions are the spherical harmonics
hθφ|ℓ, mi = Ymℓ (θ, φ)
7
7. [10 pts] Two particles of mass M1 and M2 are attached to a massless rigid rod of length d. The rod
is attached to an axle at its center-of-mass, and is free to rotate without friction in the x-y plane.
Describe the Hilbert space of the system and then write the Hamiltonian. What are the energy levels
and degeneracies? What are the wavefunctions of the energy eigenstates?
Only a single angle, φ is required to specify the state of the system, where φ is the azimuthal
angle, thus the Hilbert space is H(φ) .
The Hamiltonian is then
H=
where
L2z
2I
2
M d2
d
=
I = 2M
2
2
is the moment of inertia. This gives
H=
L2z
M d2
Em =
~2 m2
M d2
The energy levels are then
where m = 0, ±1, ±2, ±3, . . .
The energy levels all have a degeneracy of 2, except for E0 , which is not degenerate.
The wavefunctions are given by
eimφ
hφ|mi = √
2π
8
8. [10 pts] For a two-particle system, the transformation to relative and center-of-mass coordinates is
defined by
~ =R
~1 − R
~2
R
~
~
~ CM = m1 R1 + m2 R2
R
m1 + m2
The corresponding momenta are defined by
d ~
P~ = µ R
dt
d ~
P~CM = M R
CM
dt
where µ = m1 m2 /M is the reduced mass, and M = m1 + m2 is the total mass. Invert these expres~ 1, R
~ 2 , P~1 , and P~2 in terms of R,
~ R
~ CM , P~ , and P~CM .
sions to write R
The solutions are
~1 = R
~ CM + m2 R
~
R
M
~2 = R
~ CM − m1 R
~
R
M
Writing P~ and P~CM in terms of P~1 and P~2 gives
m2 P~1 − m1 P~2
P~ =
m1 + m2
P~CM = P~1 + P~2
Inverting this gives
m1 ~
PCM + P~
P~1 =
M
m2 ~
P~2 =
PCM − P~
M
9