PROBLEM SET 6, FYS3110, 2016
SOLUTION
Problem 6.1a)
The probability of measuring spin up along the z-axis equals the sum of the probabilities
of finding the electron in a state with spin up:
X
P↑z =
| (hn, l, m| ⊗ h↑z |) |ψi |2
n,l,m
=
X
r
| hn, l, m| ⊗ h↑z |
n,l,m
2
|3, 2, 1i ⊗ |↓z i +
5
r
!
3
|2, 1, 1i ⊗ |↑z i |2
5
2
r
X 3
3
=
δn,2 δl,1 δm,1 =
5
5
n,l,m
P↓z = 1 − P↑z = 1 −
3
2
= ,
5
5
as could of course have been shown by direct calculation.
6.1b)
For l one can measure the values 2 and 1, corresponding to L2 = ~2 l(l + 1) = 6~2 and
2~2 , respectively. The calculations of the probabilities are completely analogous to those
in a):
r
2
X 2
X
2
2
| (hn, 2, m| ⊗ hms |) |ψi | =
Pl=2 =
δms ,↓ δn,3 δm,1 =
5
5
n,m,m
n,m,m
s
s
ms = {↑z , ↓z } denotes the possible spin states. Similarly,
3
Pl=1 = .
5
|ψi is a pure m = 1 state, and an Lz -measurement will give m~ = ~ with probability Pm=1 = 1. |ψi is similarly a pure s = 21 -state, and an S 2 -measurement will give
1 1
+ 1 ~2 = 43 ~2 with a probability Ps= 1 = 1.
2 2
2
6.1c)
First, note that
|↑z i ≡ s =
|↓z i ≡ s =
1
1
, ms =
2
2
1
1
, ms = −
2
2
1
PROBLEM SET 6, FYS3110, 2016
SOLUTION
Using the Clebsch-Gordan table for combining a l=2 states with a s=1/2 state, p. 188
in Griffiths, the 2x1/2 table, upper right corner, we re-write the first part of |ψi in terms
of |n, j, mj i-states:
r
1 1
2
|3, 2, 1i , −
|ψ1 i =
5
2 2
r
r r !
2
2 5 1
3 3 1
=
+
3, ,
3, ,
5
5
2 2
5 2 2
Similarly, from the table for combining l = 1 and s = 1/2, we find
r
1 1
3
|2, 1, 1i ,
|ψ2 i =
5
2 2
r 3 3 3
=
2, ,
5 2 2
The possible values of j are 52 and 32 , and a J 2 -measurement can thus give 35
~2 or
4
respectively. The probabilities are (dropping some steps of the calculations)
r r
X
2 2
4
Pj= 5 =
|
δn,3 δmj , 1 |2 =
2
2
5 5
25
n,m
15 2
~
4
j
Pj= 3 =
2
6
3
21
+ =
25 5
25
6.1d)
Jz can give 12 ~ and 32 ~ with the probabilities
Pm j = 1
2
2 √ 2
2
2
6
= + =
5 5
5
Pm j = 3 =
2
3
5
6.1e)
The radial probability density for a spin-up particle P↑z (r) is defined such that
1. Therefore
Z 2π Z π
P↑z (r) =
dφ
dθ sin θ r2 |ψ↑z (r, θ, φ)|2 .
0
0
The wavefunction is
r
ψ↑z (r, θ, φ) = (hr, θ, φ| ⊗ h↑z |) |ψi =
2
3
R21 (r)Y11 (θ, φ)
5
R∞
0
drP↑z (r) =
PROBLEM SET 6, FYS3110, 2016
SOLUTION
thus
Z
2π
Z
π
dθ r2 sin θ |ψ↑z (r, θ, φ)|2
0
0
Z 2π Z π
2
3 2
2
dφ
dθ sin θ Y11 (θ, φ)
= r |R21 (r)|
5
0
|0
{z
}
dφ
P↑z (r) =
=1
1 r4 − r
e a
=
40 a5
Griffiths 4.49 a) (Optional)
Normalize:
|χ|2 = |A|2 (1 + 4 + 4) = 9|A|2 .
(1)
So, choosing A to be real, A = 1/3.
The discussion in the remainder of this problem leans heavily on section 4.4.1 in the
book.
Griffiths 4.49 b) (Optional)
As usual, one can measure the eigenvalues of the operator Ŝz , i.e. ±~/2, often referred
to as ”spin up (↑)” and ”spin down (↓)”, respectively. Here we have
1
1
χ = (1 − 2i) χ+ + · 2 χ−
3
3
(2)
so the probability to measure spin up is
1
5
P↑ = |1 − 2i|2 = .
9
9
(3)
It immediately follows that P↓ = 1 − P↑ = 4/9, as can of course be computed explicitly
as well. The expectation value is thus
~
−~
~
hSz i = P↑ +
P↓ = .
(4)
2
2
18
Griffiths 4.49 c) (Optional)
This is exactly the situation discussed between Eq.[4.150] and [4.152] in the book. The
values one can measure are the eigenvalues of Ŝx , i.e. ±~/2. To find the probabilities,
we must change bases and express our state χ in the eigenspinors of Ŝx . The result is
given in [4.152] and gives, in our case,
1 1 − 2i
1
1
(x)
(x)
χ=
(5)
= √ (3 − 2i) χ+ + √ (−1 − 2i) χ−
2
3
3 2
3 2
3
PROBLEM SET 6, FYS3110, 2016
where
(x)
χ+
1
=√
2
1
1
SOLUTION
;
(x)
χ−
1
=√
2
1
−1
.
(6)
Thus,
~
1
13
P Sx = +
= |3 − 2i|2 =
2
18
18
(7)
~
5
P Sx = −
= .
2
18
(8)
and
This gives the expectation value
~
~
−~
~
2~
hSx i = · P Sx = +
+
· P Sx = −
=
.
2
2
2
2
9
(9)
Griffiths 4.49 d) (Optional)
Same calculation as in the previous subproblem, but with
~ 0 −i
Ŝy '
.
i 0
2
(10)
A simple calculation shows that this matrix, too, has eigenvalues ±~/2 which are thus
the values one can measure. Next, we need to find the two eigenspinors, i.e. the numbers
α and β obeying
~ α
~ 0 −i
α
=±
.
(11)
i 0
β
2
2 β
One finds
1
1
1
1
(y)
=√
; χ− = √
.
2 i
2 −i
A generic spinor can be expressed in terms of these eigenspinors,
a
(y)
(y)
= c1 χ + + c2 χ − .
b
(y)
χ+
Solving for c1 and c2 one finds
a − ib (y) a + ib (y)
a
= √ χ+ + √ χ− .
b
2
2
which is easily checked by insertion. In
1 1 − 2i
χ=
=
2
3
(12)
(13)
(14)
our case,
1
1
(y)
√ (1 − 4i) χ(y)
+ + √ χ− .
3 2
3 2
4
(15)
PROBLEM SET 6, FYS3110, 2016
SOLUTION
Thus,
and
17
18
(16)
~
1
P Sx = −
= .
2
18
(17)
P
~
Sy = +
2
=
This gives the expectation value
~
−~
~
4~
~
+
· P Sy = −
=
.
hSy i = · P Sy = +
2
2
2
2
9
5
(18)