# Spin states in a finite potential well

i have a question concerning an electron in an attractive potential well. Let's suppose the potential function is defined as

$$V = \left\{ \begin{array}{cl}0, & \mbox{for } z < 0\\ -|V_0|, & \mbox{for } 0 < z < a\\ 0 & \mbox{for } z > 0 \end{array}\right.$$

Describing the electron with the Dirac equation I have the dispersion relations:

$$E = \left\{\begin{array}{lr} \sqrt{p_1^2 + m^2}, \qquad\qquad \qquad\qquad \mbox{ for } z < 0 \mbox{ and } z > a\\ \sqrt{p_2^2 + m^2} - e|V_0|, \qquad \qquad \,\,\,\,\mbox{ for } 0 < a < a\end{array}\right.$$

where I chose the positive energy since I'm considering an electron. Then let us suppose I want to determine the bound states energy eigenvalues. In this situation $\ E^2 < m^2$ such that $\ p_{1}$ is purely imaginary. To get the eigenvalues quantization i have to impose the continuity conditions at $z=0, \, z=a$. My question concerns the spin part of the wave functions: can I assume it to be the same in all three zones, i.e, always spin up or always spin down? Since the energy is spin independent and I'm only interested in obtaining the possible eigenvalues is should not matter, is it correct?

Thank you.

Your square well does not have any term that interacts with a spin. So there is no reason to have different spins at different places, you just can change a wavefunction from $\psi$ to $\psi\chi$, where $\chi$ describes the spin part and remains the same everywhere.
$V(r)=-V_R f(r) - i W_v f(r) + ... + V_{SO}\frac{1}{r} \frac{d}{dr}f(r) \sigma \cdot l$
The last term with a coefficient $V_{SO}$ includes a spin-orbit ($\sigma \cdot l$) interaction and from this you can have different energies for different spins. If $V_{SO}=0$, yes, you have different $l$'s (orbital momenta), but the energy does not depend on the spin of the particle - the levels are degenerate for spins. This is where the Shell Model got so successful.
$f(r)$ is used to be Woods-Saxon, imaginary $i W_V$ speaks about reaction channels.