The time-independent Schrodinger equation for a particle moving along the x-axis is: $$H\psi (x) + V(x) \psi (x) = E \psi(x)$$
I know that the physically acceptable solutions to this equation can only be found for a specific value of E (so there are $n$ solutions for each $E_1$, $E_2$, ... , $E_n$). Why in general (I am aware that this is the case for the infinite square well for example), mathematically, is that the case?
Let’s take your example of an infinite square well.
We have, $V(x) = 0 $ iff $0<x<a $ and $\infty$ otherwise.
This is what we call a bound state. The total kinetic energy s less than the potential energy.
So we have the TISE
$$\partial_x ^2\psi = E \psi$$
This is a standard second order PDE with solution of the form
$$Asin(\sqrt{E}x) + Bcos(\sqrt{E}x)$$
By imposing the wavefunction must be 0 on the boundary, we find
$$B=0$$ And $$\sqrt{E}a = n \pi$$ where $\in \mathbb{Z}$
Giving $$E_n =(\frac{n \pi }{a})^2$$
As you can see, the energy levels are discrete. This is a quite general principle for bound states. The boundary conditions impose the discretisation.
Hope this helped.
