Joel Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 2
DO ALL WORK BY HAND – IN OTHER WORDS, DON’T USE MATHEMATICA OR ANY CALCULATORS.
You may need to use one or more of these:
r
3
1
0
0
Y1 =
cos θ
Y0 = √
4π
4π
Y11
=−
r
3 iφ
e sin θ .
8π
1. In the class notes it is shown that the top orbital angular momentum state is
given by
1/2
(2l + 1)!
1
Yll (θ, φ) = (−1)l
(sin θ)l eilφ := cl (sin θ)l eilφ .
4π
2l l!
In this problem you will fill in the details in the derivation of an arbitrary
Ylm (θ, φ) by applying the lowering operator
∂
∂
−iφ
L− = −~e
− i cot θ
∂θ
∂φ
m times to Yll .
(a) By applying L− to Yll , show that
√
Yll−1 (θ, φ) = −cl ei(l−1)φ 2l sinl−1 θ cos θ .
In principle, we can apply L− repeatedly to generate all of the Ylm , but
this gets very messy. The easy way is based on part (b) below.
(b) Prove the following result:
d
L− [eimφ f (θ)] = ~ei(m−1)φ sin1−m θ
sinm θ f (θ) .
d cos θ
(Hint: You will need to show
1 df
df
=−
d cos θ
sin θ dθ
and
d sin θ
= − cot θ .)
d cos θ
As a consequence, you have shown that
d
1
L− [eilφ sinl θ] = ~ei(l−1)φ sin1−l θ
sinl θ sinl θ = L− Yll .
d cos θ
cl
1
(c) Now show that
1
(L− )2 Yll = (L− )2 [eilφ sinl θ]
cl
2 i(l−2)φ
sin2−l θ
=~ e
d2
l
sin θ sinl θ
d(cos θ)2
√ √
and hence also that (using (L− )2 Yll = ~2 2l 4l − 2 Yll−2 )
s
d2
1
l−2
sin2l θ .
ei(l−2)φ sin2−l θ
Yl (θ, φ) = cl
2l(2l − 1)2
d(cos θ)2
The important point to realize here is that the sin1−l θ term in the middle
is canceled when the operator L− is applied twice. As you will see next,
this holds true for all higher powers of L− .
(d) To see what the general result is, apply L− a third time to show
1
(L− )3 Yll = (L− )3 [eilφ sinl θ]
cl
= ~3 ei(l−3)φ sin3−l
d3
l
sin
θ
sinl θ
d(cos θ)3
and hence also show
s
Yll−3 = cl
1
d3
sin2l θ
ei(l−3)φ sin3−l θ
2l(2l − 1)(2l − 2)3 · 2
d(cos θ)3
(e) Show that the result in part (d) can be generalized to
s
dl−m
(l + m)!
sin2l θ .
eimφ sin−m θ
Ylm (θ, φ) = cl
(2l)!(l − m)!
d(cos θ)l−m
This is just equation (1.24) in the notes on Angular Momentum.
2. In class we defined angular momentum in the classical sense as L = r × p,
and then interpreting r and p as quantum mechanical operators, we were able
to derive the commutation relations [Li , Lj ] = iεijk Lk . We then showed that
rotating a (scalar) wave function ψ(x) leads to a rotated state ψR (x) given by
ψR (x) = e−(i/~)θ·L ψ(x). In the case of an infinitesimal rotation θ ≪ 1, this
can be written as ψR (x) = [1 − (i/~)θ · L]ψ(x). This is why L is called the
generator of rotations.
A more general approach is to essentially reverse this logic. We have seen
that spin one-half particles are represented by a two-component spinor (or
vector), because the z component of spin can take one of two values (±1/2).
2
A spin-one particle has three z-components of spin (0, ±1) and would be
represented by a three-component vector. In other words, a particle with spin
is represented by a multi-component vector. Furthermore, the usual approach
in quantum mechanics is to start from a classical observable quantity and
then construct the corresponding operator that acts on the Hilbert space
of states of the system. In this case, we start from a spatial rotation in
three dimensions, and then construct the corresponding rotation operator. In
other words, corresponding to an infinitesimal rotation R(θ), we define the
generator of rotations J by
U (R(θ)) = 1 −
i
θ·J
~
where U (R(θ)) is the quantum mechanical operator that acts on the state ψ.
In class we saw that translations commute, i.e., that the translation group is
abelian. However, rotations in three dimensions do not commute in general,
and the rotation group is not abelian. In this problem you will derive the
commutation relations of the generators Ji using this fact.
In three dimensions, the matrices that rotate a vector about the x, y and z
axes are


1
0
0


Rx (θ) =  0 cos θ − sin θ 
0
sin θ


Ry (θ) = 
cos θ
cos θ
0
0
1
− sin θ
sin θ
− sin θ

Rz (θ) =  sin θ
cos θ
0

0 
0 cos θ
cos θ


0
0


0 .
1
(a) Let θ = ε ≪ 1 and expand these to first order in ε to write each of them
in the form
Ri (ε) = I + εGi
i = (1, 2, 3) = (x, y, z)
where Gi is a 3 × 3 matrix.
(b) Show that

0 −1 0

[Rx (ε), Ry (ε)] = ε2  1
0
3


0 0 .
0 0
(c) Show that to order ε2 we can also write
Rz (ε2 )Ry (ε)Rx (ε) = Rx (ε)Ry (ε) .
(d) Carrying the result of part (c) over to quantum mechanics, we replace
each rotation matrix by it’s corresponding quantum mechanical operator
U (R(ε)) = 1 − (i/~)ε·J acting on an arbitrary state, and we can therefore
write
i
i
i
i
i
1 − εJx 1 − εJy − 1 − ε2 Jz 1 − εJy 1 − εJx |ψi
~
~
~
~
~
= 0.
Expand this to order ε2 and show that
[Jx , Jy ] = i~Jz .
By cyclically permuting the indices x → y → z → x in part (c), it is not
hard to show that in general we have
[Ji , Jj ] = i~εijk Jk
and thus the generators of rotation are in fact the angular momentum
operators.
3. Find the matrix element hl m|L2x |l mi.
4. Let the state function of an electron be
"r
#
r
1 0
2 1
ψ = R(r)
Y (θ, φ)χ+ +
Y (θ, φ)χ−
3 1
3 1
where R(r) is a radial wave function and χ± are the eigenfunctions of Sz .
(a) Show directly that the z-component of the electron’s total angular momentum is 1/2, and that the electron has orbital angular momentum
unity.
(b) What is the probability density for finding the electron with spin up at
(r, θ, φ)? What about spin down?
5. Consider an angular momentum 1 system represented by the state vector


1
1 

ψ = √  4 .
26
−3
What is the probability that a measurement of Lx yields the value 0? [Hint:
There are at least two ways to do this problem, and neither of them requires
a lot of algebra if you’re a little bit clever.]
4
6. Suppose the wave function of a particle in a spherically symmetric potential
is
2
2
2
2
ψ(x) = c(x + y + z)e−(x +y +z )/α
where c is a normalization constant. What is the probability that in a measurement one will find L2 = 2~2 and Lz = ~?
7. A spin 1/2 particle is in an eigenstate of Sx with eigenvalue +~/2 at time
t = 0. At that time it is placed in a magnetic field B = (0, 0, B) in which it
is allowed to precess for a time T . At that instant the magnetic field is very
rapidly rotated in the y-direction, so that its components are (0, B, 0). After
another time T a measurement of Sx is carried out. What is the probability
that the value ~/2 will be found?
8. Suppose two electrons are in the singlet state
1
|0 0i = √ (| + −i − | − +i) .
2
(1)
where | + −i is shorthand notation for 12 − 12 etc. Let Sa be the component
(2)
of spin for particle 1 in the direction â, and let Sb be the component of
spin for particle 2 in the direction b̂. Show that the expectation value of the
(1) (2)
operator Sa Sb is given by
(2)
hSa(1) Sb i = −
~2
cos θ
4
where â · b̂ = cos θ. (Hint: Since two vectors define a plane, you might as well
let â = ẑ and let b̂ lie in the xz-plane.)
9. Construct all possible total spin states that result from combining two spin 1
particles.
10. An electron in a hydrogen atom is in the combined spatial and spin state
"r
#
r
1 0
2 1
ψ = R21 (r)
Y (θ, φ)χ+ +
Y (θ, φ)χ−
3 1
3 1
where the radial function is given by
1 −3/2 r −r/2a0
R21 (r) = √ a0
e
a0
24
and
a0 =
~2
= 0.529 Å
me e 2
is the Bohr radius.
5
(a) If you measured L2 , what values could you get and with what probabilities?
(b) What if you measured Lz ?
(c) What if you measured S 2 ?
(d) What if you measured Sz ?
(e) What if you measured J 2 ?
(f) What if you measured Jz ?
(g) What is the probability density of finding the electron at the position
(r, θ, φ)?
(h) Since the operator Sz and the operator r commute, both can be simultaneously measured. What is the probability density of finding the particle
with spin up at the radius r?
6