Note that this was just a classical calculation which we will apply to quantum states later. It is
correct for the EM forces, but, the electron is actually in a rotating system which gives an additional
~L·S~term (not from the B field!). This term is 1/2 the size and of opposite sign. We have already
included this factor of 2 in the answer given above.
Recall that
H 2 ∝L~·~S=1
2
[
J^2 −L^2 −S^2
]
and we will therefore want to work with states of definitej,ℓ, ands.
23.4.3 Perturbation Calculation for Relativistic Energy Shift
RewritingH 1 =−^18 p
(^4) e
m^3 c^2 asH^1 =−
1
2 mc^2
(
p^2
2 m) 2
we calculate the energy shift for a stateψnjmjℓs.While there is no spin involved here, we will need to use these states for the spin-orbit interaction
〈
ψnjmjℓs|H 1 |ψnjmjℓs〉
= −
1
2 mc^2〈
ψnjmjℓs∣
∣
∣
∣
∣
(
p^2
2 m) 2 ∣∣
∣
∣
∣
ψnjmjℓs〉
= −
1
2 mc^2〈
ψnjmjℓs∣
∣
∣
∣
∣
(
H 0 +
e^2
r) 2 ∣∣
∣
∣
∣
ψnjmjℓs〉
= −
1
2 mc^2〈
ψnjmjℓs∣
∣
∣
∣H
2
0 +e^2
rH 0 +H 0
e^2
r+
e^4
r^2∣
∣
∣
∣ψnjmjℓs〉
= −
1
2 mc^2[
E^2 n+〈
ψnjmjℓs∣
∣
∣
∣
e^2
r∣
∣
∣
∣H^0 ψnjmjℓs〉
+
〈
H 0 ψnjmjℓs∣
∣
∣
∣
e^2
r∣
∣
∣
∣ψnjmjℓs〉
+
〈
ψnjmjℓs∣
∣
∣
∣
e^4
r^2∣
∣
∣
∣ψnjmjℓs〉]
= −
1
2 mc^2[
E^2 n+ 2Ene^2〈
1
r〉
n+e^4〈
1
r^2〉
nl]
where we can use some of our previous results.
En = −1
2
α^2 mc^2 /n^2 =−e^2
2 a 0 n^2
〈
1
r〉
n=
(
1
a 0 n^2)
〈
1
r^2〉
=
(
1
a^20 n^3 (ℓ+^12 ))
〈
ψnjmjℓs|H 1 |ψnjmjℓs〉
= −
1
2 mc^2[(
−^12 α^2 mc^2
n^2) 2
+ 2
(
−^12 α^2 mc^2
n^2)
e^2
a 0 n^2+
e^4
a^20 n^3 (ℓ+^12 )]
= −
1
2 mc^2En(0)2[
1 −4 +
4 n
ℓ+^12]
= +
En(0)22 mc^2[
3 −
4 n
ℓ+^12