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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?

Qmechanic
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    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 $H\psi(x) = E\psi(x)$.

     –  Oct 31 '19 at 14:33
  • Possible duplicate: https://physics.stackexchange.com/q/65636/2451 – Qmechanic Oct 31 '19 at 16:55

1 Answers1

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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.

Jake
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