problem A particle with spin $\frac{1}{2}$, mass $m$ and electric charge $q$ is bounded in the 2-dimensional infinite potential well.
(a) Find the eigenvalue and eigenstate of this particle.
(b) Put a particle with spin $1/2$ in the well. The interaction of two particles is represented by $V = -\alpha \vec{S} \cdot \vec{B}$ and $\vec{B}$ has only $\hat z $ direction. Find the first and second order correction of the energy by perturbation theory.
I have three questions about the above problem. It is easy to solve (a).
$$\psi = \sqrt{\frac{2}{a}} \sqrt{\frac{2}{b}} \sin {\frac{n\pi x}{a}}\sin \frac{l\pi y}{b}$$ $$E = E_n + E_l = \frac{n^2 \pi^2 \hbar^2}{2ma^2} +\frac{l^2 \pi^2 \hbar^2}{2ma^2}$$
Here, I'm confused in writing answer of (a). The problem gives us the spin $1/2$. Do we have to add the spin eigenstate factor to the above one? Like below.
$$\psi = \sqrt{\frac{2}{a}} \sqrt{\frac{2}{b}} \sin {\frac{n\pi x}{a}}\sin \frac{l\pi y}{b} {( \alpha |+> + \beta |->)} $$
The $|+>, |->$ is spin up and down. If we have to add spin factor to the eigenstate in writing answer, how can we find $\alpha$ and $\beta$? It is the first question of this problem.
Now, in (b), do we have to add the energy of "new" particle to the first order correction? It means that
$$E' = E_{n'} + E_{l'} = \frac{n'^2 \pi^2 \hbar^2}{2m'a^2} +\frac{l'^2 \pi^2 \hbar^2}{2m'a^2}$$
should be added to the first correction. I think it is false, isn't it? I think we have to find the first and second order correction of the $V$. Am I right?
Moreover, $V$ is represented by $2 \times 2$ matrix. I don't know how to solve it.
Here I rearrange my questions.
Do we have to add spin eigenstate factor to the answer?
Do we have to add the energy of the "new" particle in calculating first order correction?
$V$ is represented by matrix and the eigenstate is represented by real function. How can we solve the $(b)$
