Download we find

Spin-Orbit Interactions
⃗⃗
𝐻 ′ = −𝜇⃗ ⋅ 𝐵
The perturbation Hamiltonian is:
The magnetic dipole moment of the electron is related to its spin
angular momentum:
𝜇𝐵
𝜇⃗ = −𝑔𝑠
𝑆⃗
ℏ
𝑒ℏ
𝜇𝐵 = 2 𝑚
where 𝜇𝐵 is the Bohr magneton:
𝑒
and the magnetic field at the site of the electron is:
⃗⃗𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 = (
𝐵
𝑒
) 𝐿⃗⃗
4 𝜋𝜖𝑜 𝑚𝑒 𝑐 2 𝑟 3
The perturbation Hamiltonian becomes:
′
𝐻 =
𝑒2
⃗⃗
𝑆⃗⋅𝐿
4 𝜋𝜖𝑜 2 𝑚𝑒2 𝑐 2 𝑟 3
The factor of “2” comes from the Thomas precession.
𝐻′ = 𝛼
ℏ
⃗⃗
𝑆⃗⋅𝐿
2 𝑚𝑒2 𝑐 𝑟 3
The spin-orbit interaction undermines the usefulness of the |𝑛 𝑙 𝑚𝑙 𝑚𝑠 ⟩
states. The |𝑛 𝑙 𝑚𝑙 𝑚𝑠 ⟩ states are shown on the next page.
We can use 𝑚𝑙 and 𝑚𝑠 as “good quantum
numbers” to determine the stationary
states as long as we are able to specify
eigenvalues independently for the
observables 𝐿𝑧 and 𝑆𝑧 . These two
quantities are separately conserved
whenever there exist states of definite
energy in which 𝐿𝑧 and 𝑆𝑧 also have
definite values.
Recall that our perturbation Hamiltonian is:
𝐻′ = 𝛼
ℏ
⃗⃗
𝑆⃗⋅𝐿
2 𝑚𝑒2 𝑐 𝑟 3
We need eigenstates described by quantum numbers that are
eigenstates of the Hamiltonian H’. Why? Because we need to calculate
the first-order correction to the stationary states in the H-atom due to
⃗⃗ interaction (a.k.a. the 𝑆⃗ ⋅ 𝐿⃗⃗ interaction).
the – 𝜇⃗ ⋅ 𝐵
𝐸1 = ⟨𝜓|𝐻′|𝜓⟩ ~ ⟨𝜓|
𝑆⃗ ⋅ 𝐿⃗⃗
| 𝜓⟩
𝑟3
Since our H’ implies a dependence of the energy on the relative
orientation of 𝐿⃗⃗ and 𝑆⃗, the two vectors must be coupled together as a
result of this new dynamical variation of the energy. We can see the
coupling in the figure if we fix the energy by fixing the angle between 𝐿⃗⃗
and 𝑆⃗ while maintaining the z components of the two vectors.
Lz and Sz cannot be assigned definite values; however a state of definite
energy can still have a definite value of Jz. The total angular
momentum is conserved as long as the atom is isolated.
Let’s look at the following figure to see how we can construct states of
1
1
𝑗 = ℓ + and 𝑗 = ℓ − .
2
2
How do we go from the |𝑛 ℓ 𝑚ℓ 𝑚𝑠 ⟩ states to the |𝑛 ℓ 𝑗 𝑚𝑗 ⟩ states?
All “fine and good,” but how are these states eigenstates of the 𝑆⃗ ⋅ 𝐿⃗⃗
operator in our perturbation Hamiltonian, H’ ?
First of all, the total angular momentum of the atom is 𝐽⃗ = 𝐿⃗⃗ + 𝑆⃗, and
𝐽⃗ ⋅ 𝐽⃗ = (𝐿⃗⃗ + 𝑆⃗) ⋅ (𝐿⃗⃗ + 𝑆⃗)
Solving this for 𝑆⃗ ⋅ 𝐿⃗⃗ we find:
𝐽2 − 𝐿2 − 𝑆 2
𝑆⃗ ⋅ 𝐿⃗⃗ =
2
We can now find the expectation value for 𝑆⃗ ⋅ 𝐿⃗⃗ by using our new
eigenstates |𝑛 ℓ 𝑗 𝑚𝑗 ⟩:
For example:
〈𝐽2 〉 = ⟨𝑛 ℓ 𝑗 𝑚𝑗 |𝐽2 |𝑛 ℓ 𝑗 𝑚𝑗 ⟩ = 𝑗(𝑗 + 1)ℏ2
Continuing on with the other expectation values we find the following:
〈𝐽2 〉 − 〈𝐿2 〉 − 〈𝑆 2 〉
〈𝑆⃗ ⋅ 𝐿⃗⃗ 〉 =
2
𝑗(𝑗 + 1)ℏ2 − ℓ(ℓ + 1)ℏ2 − 𝑠(𝑠 + 1)ℏ2
=
2
We still have to calculate the expectation value of
1
〈 3〉
𝑟
1
𝑟3
.
𝛼𝑚𝑒 𝑐 3
2
= (
)
𝑛ℏ
ℓ(ℓ + 1)(2ℓ + 1)
Collecting our calculations, we find the following:
〈𝐻′〉 =
〈𝐻′〉𝑆⃗⋅𝐿⃗⃗
where
𝐸𝑛0 =
𝛼ℏ
2 𝑚𝑒2 𝑐
〈𝑛 ℓ 𝑗 𝑚𝑗 |
⃗⃗
𝑆⃗⋅𝐿
|𝑛
3
𝑟
ℓ 𝑗 𝑚𝑗 〉
𝛼 2 0 𝑗(𝑗 + 1) − ℓ(ℓ + 1) − 34
=
𝐸
𝑛 𝑛
ℓ(ℓ + 1)(2ℓ + 1)
2
1
𝑚 𝑐 2 𝛼2
2 𝑒
𝑛
.
Let’s combine our two contributions to the fine structure splitting:
1) The relativistic kinetic energy, and
2) The spin-orbit coupling
〈𝐻 ′ 〉𝑓𝑠 = 〈𝐻 ′ 〉𝑟𝑒𝑙 + 〈𝐻′〉𝑆⃗⋅𝐿⃗⃗
〈𝐻 ′ 〉𝑓𝑠
𝐸𝑛0 𝛼 2 4𝑛
𝛼 2 0 𝑗(𝑗 + 1) − ℓ(ℓ + 1) − 34
= −
[
− 3] +
𝐸
4𝑛2 ℓ + 12
𝑛 𝑛
ℓ(ℓ + 1)(2ℓ + 1)
〈𝐻
′〉
𝑓𝑠
𝛼2 0
1
3
=
𝐸𝑛 (
− )
1 4𝑛
𝑛
𝑗+
2
Now we can calculate the total energy of each 𝑛 𝑗 state.
𝐸𝑛𝑗 = 𝐸𝑛0 + 〈𝐻 ′ 〉𝑓𝑠
𝐸𝑛𝑗 =
−𝐸𝑛0 [1
𝛼2 𝑛
3
+ 2(
−
)]
𝑛 𝑗 + 12 4
The corresponding energy level diagram is shown in the following
figure:
These are the energy levels for the |𝑛 ℓ 𝑗 𝑚𝑗 ⟩ eigenstates for a oneelectron atom.