Physics 505 Homework No. 5 Solutions S5-1grothserver.princeton.edu/~groth/phy505f12/sol05.pdf ·...

Physics 505 Homework No. 5 Solutions S5-1

1. Angular momentum uncertainty relations. A system is in the lm eigenstate of L2, Lz.

(a) Show that the expectation values of L± = Lx ± iLy, Lx, and Ly all vanish.

Solution

〈ψlm | L± | ψlm〉 = h̄√

l(l + 1) −m(m± 1) 〈ψlm | ψl,m±1〉 = 0, since ψlm ψl,m±1 areorthogonal. Note in the special cases m = ±l for L±, the square root kills the expectationvalue! Lx = (L+ + L−)/2 and Ly = −i(L+ − L−)/2, so these expectations must be zeroas well.

End Solution

(b) Determine ∆Li =√

〈L2i 〉 − 〈Li〉2. Verify the generalized uncertainty relation holds

for all pairs of angular momentum components. Comment on ∆Lx∆Ly for the casesm = 0 and m = l.

Solution

〈Lz〉 = mh̄,⟨

L2z

⟩

= m2h̄2, ∆Lz = 0. The state ψlm is symmetric in x and y, so⟨

L2x

⟩

=⟨

L2y

⟩

=(⟨

L2⟩

−⟨

L2z

⟩)

/2 = h̄2(l(l + 1) −m2)/2. ∆Lx = ∆Ly = h̄√

(l(l + 1) −m2)/2.

∆Lx∆Lz = 0 ≥ | 〈[Lx, Lz]〉 |/2 = | 〈−ih̄Ly〉 | = 0 .

This is not very exciting! Similarly, ∆Ly∆Lz = 0, also not exciting. However,

∆Lx∆Ly = h̄2(l(l + 1) −m2)/2 ≥ | 〈[Lx, Ly]〉 |/2 = | 〈ih̄Lz〉 | = h̄2m/2 .

In the case m = 0, there is no angular momentum about the z-axis. The particle is foundas close to the z-axis as possible. In this case the minimum uncertainty product for thex and y components is 0 and this is well satisfied since all the angular momentum is inthe x and y components (but with vanishing expectation value). The only case where theequality sign would be used in the uncertainty relation is the case l = 0 in which case allcomponents and their squares are zero.

The case m = l corresponds to the maximum angular momentum component alongthe z-axis. One might visualize the particle in the xy-plane rotating about the z-axis.Of course, it can’t be exactly in the xy-plane and its out of plane motion produces somecomponents of Lx and Ly which average to 0, but have some spread around the average.The uncertainty relation becomes

∆Lx∆Ly = h̄2(l(l + 1) − l2)/2 = h̄2l/2 = | 〈[Lx, Ly]〉 |/2 = | 〈ih̄Lz〉 | = h̄2l/2 .

In this case, the angular momentum components in the x and y directions have the mini-mum possible uncertainty product.

End Solution

Copyright c© 2012, Edward J. Groth


2. Fun with angular momentum commutators.

(a) Suppose the vector operators A and B commute with each other and L. Show that

[A · L,B · L] = ih̄(A × B) · L .

Solution

We use the summation convention and the completely anti-symmetric tensor and justgrind it out!

[A · L,B · L] = AiLiBjLj −BjLjAiLi

= AiBj(LiLj − LjLi)

= AiBj [Li, Lj]

= AiBj(ih̄ǫijkLk)

= −AiBj(ih̄ǫikjLk)

= +AiBj(ih̄ǫkijLk)

= ih̄(A × B)kLk

= ih̄(A × B) · L .

End Solution

(b) Suppose V is a vector operator which might be a function of x and p, so it doesn’tnecessarily commute with L. Show that

[L2,V ] = 2ih̄(V × L − ih̄V ) .

You might need the relation derived in lecture for a vector operator, [Li, Vj] = ih̄ǫijkVk.

Solution

Again, grinding it out works:(

[L2,V ])

i= LjLjVi − ViLjLj

= LjViLj + Lj[Lj , Vi] − LjViLj + [Lj , Vi]Lj

= ih̄ (LjǫjikVk + ǫjikVkLj)

= ih̄ (ǫjikLjVk + ǫjikVkLj)

= ih̄ (ǫjikVkLj + ǫjik[Lj, Vk] + ǫjikVkLj)

= ih̄ (2ǫjikVkLj + ih̄ǫjikǫjklVl)

= ih̄ (−2ǫijkVkLj − ih̄ǫijkǫjklVl)

= ih̄ (+2ǫikjVkLj + ih̄ǫijkǫjlkVl)

= ih̄ (+2ǫikjVkLj − ih̄ǫijkǫljkVl)

= ih̄ (+2ǫikjVkLj − 2ih̄δilVl)

= ih̄ (+2ǫikjVkLj − 2ih̄Vi)

= 2ih̄ (V × L − ih̄V )i .



Note that we used the fact that ǫijkǫljk = 2δil. This is easy to see. Pick an i, say 1. Thenthe only non-zero values of ǫijk are those with j, k = 2, 3 or j, k = 3, 2. In the secondtensor, the only non-zero values will occur for l = 1, the sign will be the same as the first,and there are two contributions.

End Solution

3. Classically, a particle moving in a spherically symmetric potential has the Hamiltonian

H =p2

r

2m+

L2

2mr2+ V (r) ,

where pr = r · p/r. For quantum mechanics, we must define

pr =1

2

(

1

r(r · p) + (p · r)

1

r

)

, (1)

with the Hermitian operator

pr =h̄

i

(

∂

∂r+

1

r

)

. (2)

Show the operator defined in equation (1) is the same as that in equation (2). Show thatpr in equation (2) is Hermitian (consider ψ(r, θ, φ) and ϕ(r, θ, φ)) and that when used inthe Hamilton, pr of equation (2) gives the correct Schroedinger equation. Also show thatthe operator (h̄/i)(∂/∂r) is not Hermitian!

Solution

r = rer. The momentum operator in spherical coordinates is

h̄

i∇ =

h̄

i

(

er∂

∂r+ eθ

1

r

∂

∂θ+ eφ

1

r sin θ

∂

∂φ

)

.

The first term in equation (1) becomes (h̄/2i)(∂/∂r). The second term is a bit more work.It’s most easily evaluated in a mix of Cartesian and spherical coordinates. Note that themomentum operator in the second term operates on r, 1/r, and whatever might be to theright. When it operates on whatever is to the right, we get a term that’s the same as thefirst term. So let’s just evaluate the second term when the p operates on r and 1/r.

1

2p ·(r

r

)

=h̄

2i

(

∂

∂x

x

r+

∂

∂y

y

r+

∂

∂x

z

r

)

=h̄

2i

(

1

r− x2

r3+

1

r− y2

r3+

1

r− z2

r3

)

=h̄

2i

(

3

r− x2 + y2 + z2

r3

)

=h̄

i

1

r,



and putting this together with the two other pieces, we see that indeed, the operator inequation (1) is the same as that in equation (2).

The adjoint of pr is

p†r =h̄

−i

(

∂

∂r+

1

r

)

.

We consider

〈prϕ | ψ〉 =

∫ ∞

0

r2 dr

∫ π

0

sin θ dθ

∫ 2π

0

dφ

(

− h̄i

(

∂

∂r+

1

r

)

ϕ∗(r, θ, φ)

)

ψ(r, θ, φ) .

The angular integrations can be done independently of the radial integration, so we justassume they have been done (or will be done) and drop the angular variables from thediscussion. We’ll use an integration by parts to move the derivative from the ϕ to ψ. Thederivative also acts on r2.

〈prϕ | ψ〉 =

∫ ∞

0

(

− h̄i

(

∂

∂r+

1

r

)

ϕ∗(r)

)

ψ(r) r2 dr

= − h̄i

(∫ ∞

0

∂ϕ∗

∂rψ r2 dr +

∫ ∞

0

1

rϕ∗ψ r2 dr

)

= − h̄i

(

ϕ ∗ ψ r2∣

∣

∣

∣

∞

0

−∫ ∞

0

ϕ∗ ∂ψ

∂rr2 dr −

∫ ∞

0

ϕ∗ψ (2r) dr +

∫ ∞

0

1

rϕ∗ψ r2 dr

)

=h̄

i

(

0 +

∫ ∞

0

ϕ∗ ∂ψ

∂rr2 dr +

∫ ∞

0

ϕ∗ 2

rψ r2 dr −

∫ ∞

0

ϕ∗ 1

rψ r2 dr

)

=

∫ ∞

0

ϕ∗

(

h̄

i

(

∂

∂r+

1

r

)

ψ

)

r2 dr

= 〈ϕ | prψ〉 ,

and pr is Hermitian. Note that the evaluation of the limits gives 0 since the integrandmust converge at both ends!

If we do the same calculation with pr bogus = (h̄/i)(∂/∂r) we find

〈pr bogusϕ | ψ〉 − 〈ϕ | pr bogusψ〉 =h̄

i

⟨

ϕ

∣

∣

∣

∣

2

rψ

⟩

,

which is not zero in general, so pr bogus is not Hermitian.

Finally, the Laplacian, in spherical coordinates, is

∇2 =1

r2∂

∂rr2

∂

∂r+

1

r21

sin θ

∂

∂θsin θ

∂

∂θ+

1

r2 sin2 θ

∂2

∂φ2.



If we multiply this by −h̄2, we get p2 for the Schroedinger equation expressed in sphericalcoordinates. We’ve already seen in lecture that the angular part is L2/r2. So it remainsto show that the first term times −h̄2 is the same as p2

r. The first term is

−h̄2 1

r2∂

∂rr2

∂

∂r= −h̄2

(

∂2

∂r2+

2

r

∂

∂r

)

.

Squaring pr,

p2r = −h̄2

(

∂

∂r+

1

r

)(

∂

∂r+

1

r

)

= −h̄2

(

∂2

∂r2− 1

r2+

1

r

∂

∂r+

1

r

∂

∂r+

1

r2

)

= −h̄2

(

∂2

∂r2+

2

r

∂

∂r

)

.

End Solution

4. What if we like x instead of z?

(a) Find the eigenfunction, ψ, of L2 and Lx with eigenvalues 2h̄2 and h̄, respectively.

Solution

Since the eigenvalue of L2 is 2h̄2, the eigenfunction has l = 1. The eigenfunctions ofLz with l = 1 are

Y1,+1 = −√

3

8πsin θ e+iφ = −

√

3

8π

1

r(x+ iy)

Y1,0 = +

√

3

4πcos θ = +

√

3

4π

1

rz

Y1,−1 = +

√

3

8πsin θ e−iφ = +

√

3

8π

1

r(x− iy) .

If we rotate the coordinate system about the y-axis so the new x-axis is along the oldz-axis then there will be one unit of angular momentum along the new x-axis. This meansreplacing z by x, y by y, and x by −z. With these replacements, the eigenfunction Y1,+1

will have one unit of angular momentum along the x-axis (and of course, it will still be aneigenfunction of L2 with eigenvalue 2h̄2. The desired eigenfunction is then

ψ = −√

3

8π

1

r(−z + iy) = −

√

3

8π(− cos θ + i sin θ sinφ) .



We can check this by operating with Lx in polar coordinates.

Lxψ =h̄

i

(

− sinφ∂

∂θ− cosφ cot θ

∂

∂φ

)

(

−√

3

8π(− cos θ + i sin θ sinφ)

)

= − h̄i

√

3

8π

(

− sinφ sin θ − i sin2 φ cos θ − i cos θ cos2 φ)

= h̄

(

−√

3

8π(− cos θ + i sin θ sinφ)

)

= h̄ψ .

End Solution

(b) Express the ψ just found as a linear combination of eigenfunctions of L2 and Lz.

Solution

Of course, if we want to write

ψ(θ, φ) =

∞∑

l=0

+l∑

m=−l

clmYlm(θ, φ) ,

we find the clm by projecting ψ onto the Ylm,

clm =

∫ π

0

sin θ dθ

∫ 2π

0

dφY ∗lm(θ, φ)ψ(θ, φ) ,

but in this case, it’s easier just to look at ψ and the Y1,+1, Y1,0, and Y1,−1 and decide whatwe need:

ψ =1

2Y1,+1 +

1√2Y1,0 +

1

2Y1,−1 .

The sum of the squares of the coefficients is 1, as it should be.

End Solution

5. Algebra all the way. We used algebraic techniques in lecture to deduce that L2 and Lz

could be simultaneously diagonalized (that is, eigenfunctions could be eigenfunctions of L2

and Lz at the same time) and that the eigenvalues are l(l+1)h̄2 and mh̄, respectively, withl and m integers or half integers and with l ≥ 0 and m = −l,−l + 1, . . . l − 1, l. We thenabandoned the algebraic technique and solved a differential equation to find the orbitalangular momentum eigenfunctions. Here we will outline the use of algebraic techniques todeduce the orbital angular momentum eigenfunctions. We start by introducing x± = x±iy.



(a) Show that the following commutation relations hold (you may use relations alreadyderived in lecture):

[Lz, x±] = ±h̄x±[L±, x±] = 0

[L±, x∓] = ±2h̄z

[L2, x+] = 2h̄x+Lz + 2h̄2x+ − 2h̄zL+ .

Solution

[Lz, x] = ih̄y and [Lz, iy] = (−i)(i)h̄x = h̄x, so [Lz, x± iy] = h̄(±x+ iy) = ±h̄x±.

[L±, x±] = [Lx ± iLy, x± iy] = ±i[Lx, y] ± i[Ly, x] = −(±h̄z) + (±h̄z) = 0.

[L±, x∓] = [Lx ± iLy, x∓ iy] = ∓i[Lx, y] ± i[Ly, x] = +(±h̄z) + (±h̄z) = ±2h̄z.

For the final commutator, we use the following,

[AB,C] = ABC −BAC = ACB +A[B,C] − [C,A]B − ACB = A[B,C] + [A,C]B .

Also, from lectureL2 = L−L+ + h̄Lz + L2

z .

So[L−L+, x+] = L−[L+, x+] + [L−, x+]L+ = 0 − 2h̄zL+ ,

h̄[Lz, x+] = h̄2x+ ,

[L2z, x+] = Lz[Lz, x+] + [Lz, x+]Lz = h̄Lzx+ + h̄x+Lz = h̄x+Lz + h̄2x+ + h̄x+Lz ,

and adding it all up, we have

[L2, x+] = 2h̄x+Lz + 2h̄2x+ − 2h̄zL+ .

End Solution

(b) Show thatLzx+ |l, l〉 = x+Lz |l, l〉 + h̄x+ |l, l〉 = h̄(l + 1)x+ |l, l〉 ,

and

L2x+ |l, l〉 = h̄2l(l + 1)x+ |l, l〉+ 2h̄2(l + 1)x+ |l, l〉 = h̄2(l + 1)(l + 2)x+ |l, l〉 .

This means that x+ is the ladder or raising operator for states in which m = l.



One can look up the integral or integrate by parts l times.

∫ +1

−1

(

1 − x2)ldx =

(

1 − x2)lx

∣

∣

∣

∣

+1

−1

+ 2l

∫ +1

−1

(

1 − x2)l−1

x2 dx

=2l

3

(

1 − x2)l−1

x3

∣

∣

∣

∣

+1

−1

+22l(l − 1)

3

∫ +1

−1

(

1 − x2)l−2

x4 dx

=22l(l − 1)

3 · 5(

1 − x2)l−2

x5

∣

∣

∣

∣

+1

−1

+23l(l − 1)(l − 2)

3 · 5

∫ +1

−1

(

1 − x2)l−3

x6 dx

· · ·

=2ll!

3 · 5 · · · (2l − 1)

∫ +1

−1

x2l dx

= 22ll!

3 · 5 · · · (2l + 1)

= 22ll!

(2l + 1)!!.

We deduce that, up to a phase,

C =1

rl

√

(2l + 1)!!

4π 2ll!,

and

|l, l〉 =

√

(2l + 1)!!

4π 2ll!sinl θ eilφ .

Note that this is almost Yll(θ, φ). It’s missing (−1)l, but this phase is determined byconvention!

End Solution

From here one could go on to use L− to determine (up to a phase) all the angular mo-mentum eigenfunctions (for integer l). The normalization constants were given in lecture.However, it’s unlikely that this will lead to new insights, so this problem ends here!



Appendix. Since we were somewhat rushed with the coverage of Plms and Ylms, I includea few items here. Much more can be found in any reference on mathematical functionssuch as Abramowitz and Stegun or any quantum text.

Associated Legendre equation. After separation of variables θ and φ (so the solutionsfor φ are exp(±imφ)) the equation for θ becomes

(

1

sin θ

∂

∂θ

(

sin θ∂

∂θ

)

+ l(l + 1) − m2

sin2 θ

)

f(θ) = 0 .

This is the associated Legendre equation and it’s customary to change the variable toµ = cos θ.

(

(1 − µ2)d2

dµ2− 2µ

d

dµ+ l((l + 1) − m2

1 − µ2

)

Plm(µ) = 0 ,

where the non-singular (at µ = ±1) solution has been written as Plm(µ) which is knownas an associated Legendre function. When m = 0, the equation is known as the Legendreequation with regular solutions Pl0(µ) = Pl(µ) which are Legendre polynomials.

Pl(µ) =(−1)l

2ll!

(

d

dµ

)l

(1 − µ2)l .

Pl(µ) is an lth order polynomial in µ and is either even or odd depending on whether l iseven or odd. The normalization (be careful when consulting references, not everyone usesthe same) is Pl(1) = 1. The associated Legendre functions are given by (m ≥ 0),

Plm(µ) = (1 − µ2)m/2

(

d

dµ

)m

Pl(µ) =(−1)l

2ll!(1 − µ2)m/2

(

d

dµ

)l+m

(1 − µ2)l .

For m = l, the associated Legendre function is particularly simple. The farthest rightfactor is a polynomial in l in which the highest power is (−µ2)l Since the polynomial isdifferentiated 2l times, only this term survives. The (−1)l cancels the (−1)l in front. Thel derivatives produce a factor 2l! which combined with the other factors in front produces(2l−1) ·(2l−3) ·(2l−5) · · ·3 ·1 which is often abbreviated (2l−1)!! where !! is read “doublefactorial.” So Pll(µ) = (2l − 1)!!(1 − µ2)l/2 = (2l − 1)!! sinl θ. The associated Legendrepolynomials with different l, but the same m ≥ 0, are orthogonal on the interval −1 to+1,

∫ +1

−1

dµPlm(µ)Pl′m(µ) =2

2l + 1

(l +m)!

(l −m)!δll′ .

The normalized, orthogonal, complete eigenfunctions of L2 and Lz (for orbital angularmomentum) are the spherical harmonics which are defined in terms of the associatedLegendre functions for the polar angle and the azimuthal wave for the azimuthal angle.For m ≥ 0, these are

Ylm(θ, φ) = (−1)m

√

2l + 1

4π

(l −m)!

(l +m)!Plm(cos θ)eimφ ,



and for negative m, we use

Yl,−m(θ, φ) = (−1)mY ∗lm(θ, φ) .

The ortho-normality relation is

∫ π

0

sin θ dθ

∫ 2π

0

dφYlm(θ, φ)Y ∗l′m′(θ, φ) = δll′δmm′ .

The completeness relation is

∞∑

l=0

+l∑

m=−l

Ylm(θ, φ)Y ∗lm(θ′, φ′) =

1

sin θδ(θ − θ′) δ(φ− φ′) .

Some of the low order spherical harmonics are,

Y00 = +

√

1

4π

Y10 = +

√

3

4πcos θ

Y11 = −√

3

8πsin θeiφ

Y20 = +

√

5

16π

(

3 cos2 θ − 1)

Y21 = −√

15

8πsin θ cos θeiφ

Y22 = +

√

15

32πsin2 θe2iφ

Y30 = +

√

7

16π

(

5 cos3 θ − 3 cos θ)

Y31 = −√

21

64πsin θ

(

5 cos2 θ − 1)

eiφ

Y32 = +

√

105

32πsin2 θ cos θe2iφ

Y33 = −√

35

64πsin3 θe3iφ

· · ·


Physics 505 Homework No. 5 Solutions S5-1grothserver.princeton.edu/~groth/phy505f12/sol05.pdf ·...

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