# QM: 1D Potential Well Spring - Energy Levels

1. Nov 27, 2017

### RJLiberator

1. The problem statement, all variables and given/known data
1D Potential V(x) = mw^2x^2/2, part of a harmonic oscillator.
Suppose that the spring can only be stretched, so that the potential becomes V=infinity for x<0. What are the energy levels of this system?

2. Relevant equations

3. The attempt at a solution

I argued my way though this problem by the following:
We know that V(x) = infinity
V(0) = 0
V(x) = 0 otherwise

From our typical energy levels we know E_n = ħw(n+1/2) for n=0,1,2,3,...

But there is a barrier at x =0. Therefore we need x=0 to have E=0.
Energy levels are thus:
E_n = ħw(n+1/2) with n=1,3,5,7,...

One can see this though the wave function graphs: https://i.stack.imgur.com/rb340.gif

Is that argued properly? Did I find the right solution?

2. Nov 27, 2017

### BvU

What are the boundary conditions at $x=0$ ?

3. Nov 27, 2017

### RJLiberator

I thought it was V(0) = 0. Did I interpret this incorrectly? Since the spring can only be stretched and not compressed?

4. Nov 27, 2017

### BvU

I mean the conditions imposed on the solution . You want to make a distinction between the independent variable $x$ and the solution $\ \psi(x) \$ that has to satisfy a second order equation ($\ {\mathcal H}\psi = E\psi\$ in your case) plus two boundary conditions.

5. Nov 27, 2017

### BvU

I don't want to confuse you, though: your line of thinking is correct; it's just the wording that is unfortunate:
The solution for x < 0 is $\ \Psi(x) = 0\$ and at $\ x=0 \$ the boundary condition is that the $\ \Psi \$ has to be continuous.

The first derivative does not have to be continuous at $\ x=0 \$ because the potential function isn't continuous either (I silently hope a theoretician will improve on this somewhat).

At the turning point (Where $\ V(x) = E\$) the situation is different and both $\ \Psi \$ and its first derivative have to be continuous.

6. Dec 4, 2017

### RJLiberator

@BvU sorry for not getting back to you earlier in this thread -- but your help was spot on. It was a much more basic question then I assumed and I got full credit for it, partially thanks to your guidance.
Cheers.