# Why is the set of physically acceptable solutions to the time-independent Schrodinger equation discrete? [duplicate]

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?

• This is not always the case. Only for "bound states" the energy levels are discreet. For unbound states, a spectrum of $E$ is possible (for example for the free particle in 1d $E = \hbar^2 k^2/2m$ for any $k$. By the way - $H$ already includes the potential $V$. The correct form of the equation is $H\psi(x) = E\psi(x)$.
– user245141
Oct 31 '19 at 14:33
• Possible duplicate: physics.stackexchange.com/q/65636/2451 Oct 31 '19 at 16:55

Let’s take your example of an infinite square well.

We have, $$V(x) = 0$$ iff $$0 and $$\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.