# [Physics] Finite potential well, parity of solutions

potentialquantum mechanicsschroedinger equationsymmetrywavefunction

I'm working through some problems for a QM exam and I've realised I don't really understand the concept of parity of solutions. I'm looking at a simple finite potential well problem: $$V(x)=0, \quad |x|>d$$
$$V(x) = V_0 \quad |x|<d$$
where $V_0$ is positive. I'm looking at bound states, so that $0<E<V_0$. I divide the x-axis into three regions, from left to right, in the standard way, so that my wavefunction is as follows:

$$\psi(x) = \begin{cases} Ce^{ax}, & x\in I\\ A \sin kx + B\cos kx, & x\in II \\ De^{-ax}, &x \in III \end{cases}$$

Where $a^2\equiv \frac{2m(V_0-E)}{\hbar^2}$ and $k^2\equiv \frac{2mE}{\hbar^2}$. Now this is the general wavefunction, but to make any progress I know I need to consider even and odd solutions (because the potential is symmetric). These I can find by imposing symmetry/antisymmetry on the wavefunction, so I get

$$\psi_S(x) = \begin{cases} Ce^{ax}, & x\in I\\ B\cos kx, & x\in II \\ Ce^{-ax}, &x \in III \end{cases}$$

$$\psi_{AS}(x) = \begin{cases} Ce^{ax}, & x\in I\\ A\sin kx, & x\in II \\ -Ce^{-ax}, &x \in III \end{cases}$$

But what ARE those functions? Am I right in thinking that "half" the eigenvalues of energy correspond to eigenfunctions which are symmetric (and the other half – antisymmetric)? So that the entire energy spectrum looks like this: $$\{ E^S_n , E^{AS}_n \}_{n \in \mathbb{N}}$$

Is my understanding correct?

It would be better to specify that $V_0 < 0$ as otherwise your problem is that of a potential barrier (no bound states) and not that of a finite potential well.