Spin orbit coupling and effect on the spatial distribution of states

How to understand the spin orbit coupling and the effect on the spatial distribution of states?

As we learned from the quantum mechanics textbook, the spin orbit coupling has the following conclusions:

1) We start with orbital states |l, lz> and |S, Sz>. Here the basis we can choose |l, lz, S, Sz>, as the common eigenstates of l2, lz, S2, Sz, which commutes with the Hamiltonian.

2) After spin orbit coupling, (e.g. by introducing the term S dot l into the Hamiltonian. Now the |l, lz, S, Sz> are not the eigenstates any more, because lz and Sz do not commute with Hamiltonian anymore.

3) The eigenstates of H are described using the common eigenstates of j2, jz, l2 and S2, where j=l+S, written as |j, l, S, jz>.

4) The unitary transformation between the two set of basis can be found using Cleberch-Gordon coeffcients (CG coefficient), which were already solved and can be looked up from text books.

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Next, we look at how the spatial distribution is for the new eigenstates, using the simplest case l=1 and S=1/2.

There are totally 6 states, basically 3 orbital X 2 spin in the old basis. In the coupled basis, the new basis are:

for l=1, or p orbital

|lz=1> = (x+iy)/sqrt(2)

|lz=0> = z

|lz=1> = (x-iy)/sqrt(2)

Therefore, in terms of x, y and z, the wavefunction for Sz=1/2 is

|j=3/2, jz=-3/2>=|-1,-1/2>

|j=3/2, jz=-1/2>=sqrt(1/3)|-1,1/2>+sqrt(2/3)|0,-1/2>

|j=3/2, jz=1/2>=sqrt(2/3)|0,1/2>+sqrt(1/3)|1,-1/2>

|j=3/2, jz=3/2>=|1,1/2>

|j=1/2, jz=-1/2>=-sqrt(2/3)|-1,1/2>+sqrt(1/3)|0,-1/2>

|j=1/2, jz=1/2>=-sqrt(1/3)|0,1/2>+sqrt(2/3)|1,-1/2>

Perturbation theory

We can work out the new states using perturbation theory.

The old Hamiltonian is

H=d2/dr+l2/r2+1/r+HmSz 

It is basically the Hamiltonian for hydrogen atom plus Zeeman energy. Here the Hm is the molecular field which represents the effect of the exchange interaction.

If we only consider the angular momentum part the Hamiltonian, the eigenstates are |lz,Sz>, with Sz=1/2 to be lower in energy.

Now we introduce the perturbation of ldotS.

This involves the degenerate perturbation, we now look at Sz=1/2 states. The old basis are:

|lz=-1,Sz=1/2> = sqrt(1/3)|j=3/2, jz=-1/2>-sqrt(2/3)|j=1/2, jz=-1/2>

|0,1/2> = sqrt(2/3)|j=3/2, jz=1/2>-sqrt(1/3)|j=1/2, jz=1/2>

|1,1/2>= |j=3/2, jz=3/2>

Using 2ldotS=(l+S)2-l2-S2=j2-l2-S2=j2-2-3/4=j2-11/4, here we will ignore the 11/4 for simplicity which will not affect the result.

The matrix elements are

7/4, 0, 0

0, 11/4, 0

0, 0, 15/4

It is clear that these states are not degenerate any more. The ground state is actually |lz=-1,Sz=1/2> for which the orbital electron cloud density are not spherical, but a doughnut-like shape in the x-y plane.

Therefore, the mystery is solved. It is unlike what I understood before: a spherical shaped wavefuncton becomes elongated toward a certain direction due to spin-orbit coupling. Instead, a spherical symmetric states is always a set of degenerate states which split when the spin-orbit coupling is introduced. In this case, the ground state is often not degenerate anymore and is not spherical either of course.

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15/4/3+3/4*2/3=5/4+1/2=7/4

⅔*15/4+1/3*3/4=5/2+1/4=11/4

15/4