# Linear Harmonic Oscilator - QM

1. Jan 13, 2007

### Brewer

Just a quickie:

A particle is in the first excited Eigenstate of energy E corresponding to the one dimensional potential V(x) = $$\frac{Kx^2}{2}$$. Draw the wavefunction of this state, marking where the particles KE is negative.

Now my question.

The first excited state will be n=1 correct? The first excited state is not the ground state under a different name is it?

So if it is n=1, then the wavefunction will look like a sin wave? And the KE will be negative on the left hand side of the sin wave (i.e. where a graph of sin(x) will be negative.)?

Ta guys.

2. Jan 13, 2007

### Dick

I think you'll have to do more than just guess. But here's a hint. The first excited state is indeed not the ground state. It has one node in the wavefunction. The ground state has zero.

3. Jan 13, 2007

### Brewer

Well I've drawn the right form of the wavefunction at least. Whether or not I've correctly identified where the KE is negative I don't know. Back to the text book I go!

4. Jan 13, 2007

### Dick

Great idea! Once you have the wavefunctions remember KE is an operator on wavefunctions.