# Infinite potential well and spin $\frac{1}{2}$

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.

1. Do we have to add spin eigenstate factor to the answer?

2. Do we have to add the energy of the "new" particle in calculating first order correction?

3. $V$ is represented by matrix and the eigenstate is represented by real function. How can we solve the $(b)$