Energy eigenvalues of spin Hamiltonian

[vbrasic]

Homework Statement

The Hamiltonian of the positronium atom in the $1S$ state in a magnetic field $B$ along the $z$-axis is to good approximation, $$H=AS_1\cdot S_2+\frac{eB}{mc}(S_{1z}-S_{2z}).$$ Using the coupled representation in which $S^2=(S_1+S_2)^2$, and $S_z=S_{1z}+S_{2z}$ are diagonal, obtain the energy eigenvalues and eigenvectors of the Hamiltonian and classify them according to quantum numbers associated with constants of motion.

Homework Equations

Not really sure.

The Attempt at a Solution

The coupled representation as far as I know is just the total angular momentum representation. We have that both the electron and positron are spin half particles, so the total angular momentum basis is, $$|1\,1\rangle;\,|1\,0\rangle;\,|1\,-1\rangle;\,|0\,0\rangle.$$ However, I have no idea where to go from here.

 

[kuruman]

What if you wrote the Hamiltonian as a 4×4 matrix using the basis that you posted and then diagonalized it?

 

[vbrasic]

What if you wrote the Hamiltonian as a 4×4 matrix using the basis that you posted and then diagonalized it?

Okay, so to do that I have to see how each operator affects the basis, right? I'm not sure how that would work with $S_{1z}$ for example. That is, how do I compute $S_{1z}|1\,0\rangle$ for example. I suppose one way would be to decompose $|1\,0\rangle$ as $$|1\,0\rangle=\frac{1}{\sqrt{2}}\bigg(|+\,-\rangle+|-\,+\rangle\bigg).$$ Then, we have that $$S_{1z}=|1\,0\rangle=\frac{\hbar}{2}\frac{1}{\sqrt{2}}\bigg(|+\,-\rangle-|-\,+\rangle=\bigg)=\frac{\hbar}{2}|0\,0\rangle.$$ Is that sounding okay?

 

[kuruman]

That sounds fine.