Physics 443, Solutions to PS 5
1. Angular Momentum
(a) Since l = 1 and m = ±1, 0, we have that



2 0 0
1


2
2
L = h̄  0 2 0  , and Lz = h̄  0
0 0 2
0
√



0
2 √0
0
 √


L+ = h̄  0 0
2  , and L− = h̄  2
0 0
0
0

0 0
0 0 
.
0 −1

0 0

√0 0  .
2 0
Then
L2 χm = h̄2 l(l + 1)χm = h̄2 2χm
where l = 1.
Lz χm = h̄mχm .
and
q
L± χm = h̄ l(l + 1) − m(m ± 1)χm±1
To calculate Ly , we use that
√



2
0
0
0
1
0
√ 
h̄ 
h̄
L+ − L−
h̄  √

Ly =
=  − 2
0
2  = √  −1 0 1  = J.
√
2i
2i
i
i 2
0 −1 0
0
− 2 0

(b) Using that Ry (θ) = exp(iLy θ/h̄) = exp(Jθ), and doing a Taylor
expansion we find that
1
1
1
= I + θJ + θ2 J 2 + θ3 J 3 + θ4 J 4 ...
2
3!
4!
1 2 2 1 3
1 4 2
= I + θJ + θ J − θ J − θ J ...
2
3!
4!
= I + J cos θ + J 2 (1 − cos θ)
√


2 sin θ 1√− cos θ
1√
+ cos θ
1

=
2√cos θ
2 sin θ  .
 − 2 sin θ
2
1 − cos θ − 2 sin θ 1 + cos θ
where we have used the fact that J 3 = −J and J 4 = −J 2 .
2. Griffiths 4.19.
1
(a)
[Lz , x] = [xpy − ypx , x] = [xpy , x] − [ypx , x] = −y[px , x] = ih̄y
[Lz , y] = [xpy − ypx , y] = [xpy , y] − [ypx , y] = x[py , y] = −ih̄x
[Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0
[Lz , px ] = [xpy − ypx , px ] = [xpy , px ] − [ypx , px ] = py [x, px ] = ih̄py
[Lz , py ] = [xpy − ypx , py ] = [xpy , py ] − [ypx , py ] = −px [y, py ] = −ih̄px
[Lz , pz ] = [xpy − ypx , pz ] = [xpy , pz ] − [ypx , pz ] = 0
(b)
[Lz , Lx ] =
=
=
=
[Lz , ypz − zpy ]
y[Lz , pz ] + [Lz , y]pz − z[Lz , py ] − [Lz , z]py
0 − ih̄xpz + ih̄zpx + 0
ih̄(zpx − xpz ) = ih̄Ly
(c)
[Lz , r2 ] =
=
=
=
[Lz , x2 ] + [Lz , y 2 ] + [Lz , z 2 ]
x[Lz , x] + [Lz , x]x + y[Lz , y] + [Lz , y]y + z[Lz , z] + [Lz , z]z
i2h̄xy − i2h̄yx + 0
0
[Lz , p2 ] =
=
=
=
[Lz , p2x ] + [Lz , p2y ] + [Lz , p2z ]
px [Lz , px ] + [Lz , px ]px + py [Lz , py ] + [Lz , py ]py + pz [Lz , pz ] + [Lz , pz ]pz
i2h̄px py − i2h̄py px + 0
0
(d)
p2
, L] + [V (r), L]
2m
1
[p2 , Lx ]x̂ + [p2 , Ly ]ŷ + [p2 , Lz ]ẑ
=
2m
+ [V (r), Lx ]x̂ + [V (r), Ly ]ŷ + [V (r), Lz ]ẑ
= 0
[H, L] = [
In the last step we take advantage of the fact that if [Lz , p2 ] =
0, then the same must be true for Lx , and Ly . The x,y and z
2
components of the angular momentum operator can be written as
differential operators that are functions only of θ and φ. Since the
operator does not have an r dependence it will commute with a
function V (r) that depends only on r.
3. Griffiths 4.20. From the Equation of Motion, we have that
i
d
hLi = h[H, L]i.
dt
h̄
We can calculate this commutator as follows
p2
[H, L] = [
, L] + [V, r x p].
(1)
2m
We showed in problem in problem 2 (Griffiths 4.19), that [Lz , p2 ] = 0.
The same is true for Lx and Ly so the first term vanishes. . Then
[V, r × p] = −r × [V, p] − [V (r), r] × p
The second commutator is zero since V (r) is a function of r. For the
second term, you can write p = −ih̄∇. Then
r × [V, p] = −ih̄r × [V, ∇] = ih̄r × ∇V
It follows that
d
hLi = r × (−∇V (r)).
dt
For a potential that depends only on the magnitude of r, we see that
the gradient of V (|r|) is in the r̂ direction, and r × r̂ = 0 giving us that
the angular momentum is conserved.
4. Griffiths 4.22. Being a state of maximum Lz , we get that L+ Yll = 0.
To get the functional form,we write L+ as a differential operator,
!
iφ
L+ = h̄e
∂
∂
+ i cot θ
.
∂θ
∂φ
Then
0 = L+ Yll
!
∂
∂
= h̄eiφ
+ i cot θ
Y l (θ, φ)
∂θ
∂φ l
!
1 ∂
∂
→0 =
+i
Y l (θ, φ)
cot θ ∂θ
∂φ l
3
We try separating variables and write Yll (θ, φ) = g(θ)h(φ) and then
!
∂
1 ∂
+i
g(θ)h(φ)
0 =
cot θ ∂θ
∂φ
1
∂
i ∂
→
g(θ) = −
h(φ) = k
cot θg(θ) ∂θ
h(φ) ∂φ
As usual since all of the θ dependence is on the left and all of the φ
dependence on the right, then both are equal to a constant, k. Then
dh(φ)
= ikdφ → h = (constant)eikφ
h
Also
Z
dg
= k cot θg → ln(g) = k cot θdθ
g
= k ln sin θ + constant
⇒ g = c sink θ
And so Yll (θ, φ) = c sink θeikφ . We use the fact that Yll is an eigenstate
of Lz with eigenvalue h̄l to determine k.
∂
c sink θeikφ
∂φ
h̄l = h̄kc sink θeikφ
⇒k = l
Lz Yll = h̄
We fix the normalization constant by integrating.
2
1 = |c|
Z 2π Z π
0
2
1 = 2π|c|
We use that
π
2
sin2l θ sin θdθdφ
0
Z
sin2l+1 (θ) dθ.
Γ(p)Γ(q)
2Γ(p + q)
0
In our case, q = 1/2, and p = l + 1. Putting it all together, we have
that
√
2 Γ(l + 1) π
1 = 4π|c|
2Γ(l + 3/2)
Z
sin2p−1 cos2q−1 =
4
In the particluar case that l = 3, we have
c=
v
u
u
t
Γ(l + 3/2)
√ =
Γ(l + 1)2π π
s
7!!
=
3!π25
s
35
64π
5. Griffiths, 4.27. An electron is in the spin state
3i
χ=A
.
4
(a) Determine the normalization constant A.
[χ† χ = 1 = |A|2 ( −3i 4 )
⇒A =
3i
4
= |A|2 (25)
1
]
5
(b) Find the expectation values of Sx , Sy , and Sz .
h̄
[hSx i = = χ Sx χ = |A| ( −3i 4 )
2
h̄ 4
1
( −3i 4 )
=
25
2 3i
= 0
0 1
1 0
h̄ 0 −i
hSy i = |A| ( −3i 4 )
2 i 0
1
h̄ −4i
=
( −3i 4 )
25
2 −3
24 h̄
= −
25 2
3i
4
h̄ 1 0
2 0 −1
1
h̄ 3i
=
( −3i 4 )
25
2 −4
7 h̄
= −
]
25 2
3i
4
†
2
2
hSz i = |A|2 ( −3i 4 )
5
3i
4
(c) Find the ”uncertainties” σSx , σSy , and σSz . (N ote : These sigmas
are standard deviations, not Pauli matrices!) [Remember that
σSi2 = hSi2 i − hSi i2 , and also that Si2 = 31 S 2 . Then
h̄2
h̄2
(1 − 0) =
4
4
2
2
24
h̄
= hSy2 i − hSy i2 = (1 −
)=
4
25
2
7
h̄2
2
2
= hSz i − hSz i = (1 −
)=
4
25
σS2 x = hSx2 i − hSx i2 =
σS2 y
σS2 z
h̄2 7 2
4 25
h̄2 24 2
]
4 25
(d) Confirm that your results are consistent with all three uncertainty
principles, namely
h̄
σSx σSy ≥ |hLz i|
2
and its cyclic permutations.
!2 [σSx σSy =
h̄
2
!2
σSy σSz =
h̄
2
7 24
≥
25 25
!2
σSz σSx =
h̄
2
24
≥
25
!
h̄
hSz i =
2
7
≥
25
h̄
2
!2 7
25
!
h̄
hSx i = 0
2
!
h̄
hSy i =
2
h̄
2
!2
24
]
25
6. Griffiths 4.28. For the most general normalized spinor χ where
χ=
with
a
b
= aχ+ + bχ− ,
1
, and χ− =
0
χ+ =
0
,
1
compute hSx i, hSy i, hSz i, hSx2 i, hSy2 i, and hSz2 i. Check that hSx2 i+hSy2 i+
hSz2 i = hS 2 i
h̄ ∗
(a
[hSx i =
2
0 1
b )
1 0
∗
6
a
b
=
h̄ ∗
(a b + b∗ a)
2
h̄ ∗
(a
hSy i =
2
h̄ ∗
hSz i =
(a
2
−ih̄ ∗
0 −i
a
b )
=
(a b − b∗ a)
i 0
b
2
h̄
1 0
a
∗
b )
= (|a|2 − |b|2 )
0 −1
b
2
∗
Since Sx2 = Sy2 = Sz2 = 31 S 2 , it follows that
1
1
hSx2 i = hSy2 i = hSz i2 = hS 2 i = h̄2
3
4
7