Spin Orbit Coupling + Magnetic Field

[t0pquark]

Homework Statement

For the ground state of a certain atom, the spin-orbit interaction and magnetic field interaction give the Hamiltonian $H = E (J^2-L^2-S^2) + \frac{\hbar e B}{2mc}$.
Find the four matrix elements $\langle J, J_z \vert H \vert J', J_z \rangle$ for the states $\vert J = \frac{3}{2}, J_z = \frac{1}{2} \rangle$ and $\vert J = \frac{5}{2}, J_z = \frac{1}{2} \rangle$ (and their four possible combinations).

Relevant Equations

"Your answer should make use of a table of Clebsch-Gordon coefficients and require minimal math."

I am pretty confused where to even start with this question, which is not a good thing less than a week before the final :(. One thing in particular that I don't get is that I thought we were using the Clebsch-Gordon coefficients for $\vert jm \rangle$ states, not for $\vert J, J_z \rangle$.
Where should I start?

 

[DrClaude]

Are you sure the Hamiltonian is correct? I would expect some operator in the term for the magnetic field.

The Hamiltonian includes the operators $L^2$ and $S^2$, so you will need to express the $| J, J_z \rangle$ states in terms of eigenstates of $L^2$ and $S^2$. This is where the CG coefficients will come in.

 

[t0pquark]

Are you sure the Hamiltonian is correct? I would expect some operator in the term for the magnetic field.

The Hamiltonian includes the operators $L^2$ and $S^2$, so you will need to express the $| J, J_z \rangle$ states in terms of eigenstates of $L^2$ and $S^2$. This is where the CG coefficients will come in.

Oooops that was definitely me copying incorrectly. The Hamiltonian should be $H = E (J^2-L^2-S^2) + \frac{\hbar e B}{2mc}(L_z+2S_z)$

Okay so starting with the Clebsch-Gordon table I want to use the section with 2 x 1/2??
I think I would get $\vert J = \frac{5}{2}, J_z=\frac{1}{2} \rangle = \sqrt{\frac{2}{5}} \vert 1, -1/2 \rangle + \sqrt{\frac{3}{5}} \vert 0, +1/2 \rangle$ and $\vert J = \frac{3}{2}, J_z=\frac{1}{2} \rangle = \sqrt{\frac{3}{5}} \vert 1, -1/2 \rangle - \sqrt{\frac{2}{5}} \vert 0, +1/2 \rangle$

Do I then apply each term of the Hamiltonian to this? I'm still kind of fumbling around for what to do here.

 

[DrClaude]

Okay so starting with the Clebsch-Gordon table I want to use the section with 2 x 1/2??
I think I would get $\vert J = \frac{5}{2}, J_z=\frac{1}{2} \rangle = \sqrt{\frac{2}{5}} \vert 1, -1/2 \rangle + \sqrt{\frac{3}{5}} \vert 0, +1/2 \rangle$ and $\vert J = \frac{3}{2}, J_z=\frac{1}{2} \rangle = \sqrt{\frac{3}{5}} \vert 1, -1/2 \rangle - \sqrt{\frac{2}{5}} \vert 0, +1/2 \rangle$

I don't think this is correct. Just to be clear: what is the notation here?

Do I then apply each term of the Hamiltonian to this? I'm still kind of fumbling around for what to do here.

You are on the right track. When you have transformed the $| J, J_z \rangle$ into kets that are simultaneous eigenkets of all the operators in the Hamiltonian, you apply $H$ to those eigenkets and then take the inner product with the bra.

 

[t0pquark]

I don't think this is correct. Just to be clear: what is the notation here?

I think this is where I'm confused. My logic was that the total angular momentum must be 2 and the total spin angular momentum must be 1/2, so I would want the 2 x 1/2 section on the Clebsch-Gordan table, then I find the results for $\vert J = \frac{5}{2}, J_z = \frac{1}{2}$ and $\vert J = \frac{3}{2}, J_z = \frac{1}{2}$ by reading down. I guess I don't really understand what I'm doing here.

I can apply each term of the Hamiltonian if I know $j, l, s$ (like $L_z \vert l, m \rangle = \hbar m \vert l, m \rangle$), but I'm not sure what I get out of the Clebsch-Gordan table.

 

[DrClaude]

I think this is where I'm confused. My logic was that the total angular momentum must be 2 and the total spin angular momentum must be 1/2, so I would want the 2 x 1/2 section on the Clebsch-Gordan table, then I find the results for $\vert J = \frac{5}{2}, J_z = \frac{1}{2}$ and $\vert J = \frac{3}{2}, J_z = \frac{1}{2}$ by reading down. I guess I don't really understand what I'm doing here.

You are correct that you must have $L=2$, $S=1/2$.

I can apply each term of the Hamiltonian if I know $j, l, s$ (like $L_z \vert l, m \rangle = \hbar m \vert l, m \rangle$), but I'm not sure what I get out of the Clebsch-Gordan table.

You want to go from $|J,J_z \rangle$ to $|L, M_L, S, M_S\rangle$, so you have
$$
|J,J_z \rangle = \sum_{L,M_L,S,M_S} |L, M_L, S, M_S\rangle \langle L, M_L, S, M_S |J,J_z \rangle
$$
where the $\langle L, M_L, S, M_S |J,J_z \rangle$ are the CG coefficients.

Once you have these decompositions, you can know the action of each of the angular momentum operators on the kets $|J,J_z \rangle$ and you can build the matrix for the Hamiltonian.