Physics 215A PS 4 Solutions

Physics 215A PS 4 Solutions
Anthony Karmis
February 15, 2010
Problem 1
If we have a particle interacting with an electromagnetic field our hamiltonian
will be:
H = 1 (¯h
∇
2m ı ⃗ − q ) 2
A
c ⃗ + qΦ
For an electron, q = −e. Given a magnetic field B ⃗ = Bẑ and electric field
⃗E = εŷ, we have Φ = −εy. There is more freedom in choosing A, ⃗ as long it
satisfies B ⃗ = ∇ ⃗ × A. ⃗ Using the Landau( gauge, we ) have A ⃗ = −Byˆx, which
additionally satisfies the Coulomb gauge ⃗∇ · A ⃗ = 0 . Using all these facts, we
have:
H =
1
(¯h
∇
2m ı ⃗ − q ) 2
A
c ⃗ + qΦ
= − ¯h2
2m ∇2 + ıq¯h A
mc ⃗ · ⃗∇ +
q2
2mc 2 A2 + qΦ
= − ¯h2
2m ∇2 + ıe¯h
mc By∂ x + e2 B 2
2mc 2y2 + eεy
= − ¯h2
2m ∇2 + ı¯hωy∂ x + 1 2 mω2 y 2 + eεy
with ω = eB
mc . Note that [H, p x] = [H, p z ] = 0, so the eigenstates of
H will be eigenstates of p x and p z , i.e. free waves. Assuming ψ(x, y, z) =
e ıpxx/¯h e ıpzz/¯h Y (y), we have:
[
]
¯h 2 (
Hψ = k
2
2m x + kz) 2 p 2 y +
2m − ω¯hk xy + 1 2 ω2 y 2 + eεy Y (y)
where k i = p i /¯h. We see that the problem has now been reduced to a
one-dimensional problem:
H y = p2 y
2m − ω¯hk xy + 1 2 ω2 y 2 + eεy
1

In order to find the eigenenergies, we first complete the square:
H y = p2 y
2m + 1 2 mω2 (y − y 0 ) 2 − 1 2 mω2 y 2 0
where we have introduced y 0 = (¯hωk x − eε)/mω 2 = ¯hkxc
eB
just a harmonic oscillator with energy:
(
E = n + 1 )
¯hω − 1 2 2 mω2 y0
2
Combining this with the above results, we have:
− εmc2
eB
. This is
E n (k x , k z ) =
=
(
n + 1 )
¯hω − 1 2 2 mω2 y0 2 + ¯h2 (
k
2
2m x + kz)
2
(
n + 1 )
¯hω + ¯hk x λ − 1 2
2 mλ2 + ¯h2 kz
2
2m
where λ = cε
B
. These correspond to the eigenstates:
(
ψ (x, y, z) = e ıkxx e ıkzz mω
) (√ )
1/4 1 mω
√
π¯h 2n n! H n
¯h (y − y 0) e − 1 mω
2 ¯h (y−y0)2
2

Problem 2
Part (a)
〉
∣ 1, 1 = ∣ 〉
d 1/2, 1/2 ∣
〉p 1/2, 1/2
n
( ∣∣1/2, 〉
〉
∣ 1, 0 =
〉d 1/2 ∣
p 1/2, -1/2
n + ∣ 〉
1/2, -1/2 ∣
〉p 1/2, 1/2 〉
∣ 1, -1 = ∣ 〉
d 1/2, -1/2 ∣
〉p 1/2, -1/2
n
Part (b)
n
)
/ √ 2
P ∣ ( ∣∣1/2, 〉
〉 )
1, 1 = P
〉d 1/2 ∣
p 1/2, 1/2
n
= ∣ 〉
1/2, 1/2 ∣
〉n 1/2, 1/2
p
= ∣ 1, 1
〉d
P ∣ ( ∣∣1/2, 〉
〉
1, 0 = P
〉d 1/2 ∣
p 1/2, -1/2
n + ∣ 〉 )
1/2, -1/2 ∣
〉p 1/2, 1/2 / √ 2
n
( ∣∣1/2, 〉
〉
= 1/2 ∣ 1/2, -1/2
p + ∣ 〉 )
1/2, -1/2 ∣
〉n 1/2, 1/2 / √ 2
p
n
P ∣ ∣1, -1 〉 d
= P ∣ ∣1, 0 〉 d
( ∣∣
= P 1/2, -1/2 〉 p∣
∣1/2, -1/2 〉 )
n
= ∣ 〉
1/2, -1/2 ∣
〉n 1/2, -1/2
p
= ∣ ∣1, -1 〉 d
Using this, as well as the fact that P 2 = 1, we have:
〈
1, m
′
d
∣ ∣
(
⃗Sp − ⃗ S n
) ∣∣
1, m d
〉
⇒ 〈 1, m ′ ∣ ( ⃗
d Sp − S ⃗ ) ∣∣1, 〉
n md
= 〈 ∣
1, m ′ d
∣P ( 2 Sp ⃗ − S ⃗ )
n P 2∣ 〉
∣1, m d
= 〈 1, m ′ ∣
d P
(PS ⃗ p P − PS ⃗ )
n P P ∣ 〉
1, md
= 〈 1, m ′ ∣ ( d PS ⃗ p P − PS ⃗ ) ∣∣1, 〉
n P md
= 〈 1, m ′ ∣ ( ⃗
d Sn − S ⃗ ) ∣∣ 〉
p 1, m d
= − 〈 1, m ′ ∣ ( ⃗
d Sp − S ⃗ ) ∣∣1, 〉
n md
= 0
Part (c)
H HF,d = − c 0
a 3 ⃗µ d · ⃗µ e
0
= 2c 0µ N µ d
¯h 2 a 3 0
= 2c 0µ N µ d
¯h 2 a 3 0
(g p
⃗ Sp + g n
⃗ Sn
)
· ⃗S e
(
gp + g n
2
3
⃗S d + g p − g n
2
(
⃗Sp − ⃗ S n
) ) · ⃗S e

H HF,d = 2c (
0µ N µ d gp + g n
¯h 2 a 3 2
0
(
gp + g n
= 2c 0µ N µ d
¯h 2 a 3 0
2
⃗S d · ⃗S e + g p − g n
2
J 2 − S 2 e − S2 d
2
(
⃗Sp − ⃗ S n
)
· ⃗S e
)
+ g p − g n
2
(
⃗Sp − ⃗ S n
)
· ⃗S e
)
Acting this on our total spin state ∣ ∣j, m j
〉
=
∣ ∣ S d , m d
〉∣ ∣ S e , m e
〉
in order to get
the energy eigenvalues for the hyperfine hamiltonian, and defining ḡ = (g p +
g n )/2:
E HF,d = 〈 H HF,d
〉
= 2c 0µ N µ d
¯h 2 a 3 0
(
ḡ
〈
J 2 − S 2 e − S 2 d〉
2
+ g p − g n 〈 ( Sp ⃗ − S
2
⃗ )
n · ⃗S
〉 )
e
= (j (j + 1) − S e (S e + 1) − S d (S d + 1))ḡ c 0µ N µ d
a 3 0
(
= j (j + 1) − 1 ( ) )
1
2 2 + 1 − 1 (1 + 1) ḡ c 0µ N µ d
a 3 0
(
= j (j + 1) − 11 )
ḡ c 0µ N µ d
4 a 3 0
For S d = 1 and S e = 1 2 , we could have J = 1 2 or J = 3 2
. We are also given
g p = 5.56 and g n = −3.82, so ḡ = 0.87. Therefore, we have (in units of c0µNµ d
):
a 3 0
E HF,d
= −2ḡ, ḡ
= −1.74, 0.87
for the j = 1 2 and j = 3 2
states, respectively.
Part (d)
As λ ∼ 1/f ∼ 1/∆E,
λ d
λ p
= ∆E HF,p
∆E HF,d
⇒ λ d = ∆E HF,p
∆E HF,d
λ p
From above, we see that ∆E HF,d = 3ḡ. We are also given λ p = 21.4 cm.
To find ∆E HF,d , we proceed similarly to the previous calculation:
H HF,p = − c 0
a 3 ⃗µ p · ⃗µ e
0
= 2c 0µ N µ d
¯h 2 g
a 3 pSp ⃗ · ⃗S e
0
= 2c 0µ N µ d J 2 − Se 2 − Sp
2
¯h 2 g
a 3 p
2
0
4

Therefore, we have (in units of c0µNµ d
):
a 3 0
(
E HF,p = g p j (j + 1) − 3 )
2
= − 3 2 g p, 1 2 g p
for j = 0 and j = 1, respectively. This means ∆E HF,p = 2g p . Therefore, we
have:
λ d = ∆E HF,p
∆E HF,d
λ p
= 2g p
3ḡ λ p
g p
= 4 λ p
3 g p + g n
= 4 ( )
5.56
21.4 cm
3 5.56 − 3.82
= 91.2 cm
5

Problem 3
H = 1 (¯h
∇
2m ı ⃗ + e ) 2
A
c ⃗ − eΦ
The oft-neglected quadratic term we wish to investigate is:
H (1) =
e2 ∣ ∣ ⃗A 2
2mc 2
For a uniform magnetic fieldB, ⃗ we have A ⃗ = 1 2 (⃗ B × ⃗r):
H (1) =
=
=
e 2 ∣ 1
2mc 2 2 (⃗ B × ⃗r) ∣ 2
e 2 ( )
⃗B × ⃗r
8mc 2 ·
e 2 (
8mc 2 r 2 B 2 −
(
⃗B × ⃗r
)
(
⃗r · ⃗B
) ) 2
have:
Choosing our coordinates such that ⃗ B = Bẑ, we have:
H (1) = e2 B 2
8mc 2 (
x 2 + y 2)
Finding the correction to the first energy level (ψ 100 = √ 1 e −r/a0 ), we
πa 3
0
E (1) = 〈 100 ∣ ∣ 〉
H
(1) 100
∫ ∣∣ψ100 ∣
=
2
H (1) d 3 ⃗r
= 4π e2 B 2 ∫
1
8mc 2 πa 3 e ( −2∗r/a0 x 2 + y 2) r 2 dr
0
= e2 B 2 ∫
1
2mc 2 a 3 e ( −2∗r/a0 x 2 + y 2) r 2 dr
0
Noting that 〈 x 2〉 = 〈 y 2〉 = 1 3〈
r
2 〉 , we have:
E (1) = e2 B 2 ∫
1
3mc 2 a 3 0
= e2 B 2
4mc 2 a2 0
= 1 e 2 a 2 0
2 2mc 2 B2
e −2∗r/a0 r 4 dr
= − 1 2 χB2
with,
χ = − e2 a 2 0
2mc 2
6