Reading through my quantum mechanics book I've stumbled on a question any help would be great.
Suppose we had a 3-D square well i.e. $$V(x,y,z)=\begin{Bmatrix} 0 \ \text{if}\ 0\leq x \leq a \ , \ 0\leq y \leq b \ , \ 0\leq z \leq c \ \\ \infty \ \text{otherwise} \end{Bmatrix}$$
For the first part of my question is finding the eigenvalues of energy. I firstly used the wave equation: $$\Psi (x,y,z)=A\sin\left(\frac{n_x \pi x}{a}\right)\sin\left(\frac{n_y \pi y}{b}\right)\sin\left(\frac{n_z \pi z}{c}\right)$$ and using $K_x=\frac{n_x \pi}{a}$ and by subbing this into the Schrodinger equation got my eigenvalues of energy to be: $$E=\frac{\pi^2 \hbar^2}{2m}\left ( \frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2} \right )$$
So my question leads to how would one compute the eigenstates of $\hat{H}$?
and also, say we had now our boundaries set to $a=b=c$, for a given value of energy $E$, is there only a single eigenstate?
I understand that at the 1st excited state there is degeneracy as there are 3 different wavefunctions with the same energy. And since there are many wavefunctions corresponding to different energies this would mean there are many eigenstates - so to that question the answer is no?