# Excited states in the QHO

1. Nov 16, 2012

### copernicus1

Maybe the answer to this should be obvious, but if the quantum harmonic oscillator has a natural angular frequency \omega_0, why do the excited states vibrate with higher and higher angular frequencies? How do we obtain these frequencies?

Thanks!

2. Nov 16, 2012

### Einj

Maybe I didn't understand your question but this is quite the definition of "higher state". In order to increase an harmonic oscillator energy its frequency must grow.

3. Nov 16, 2012

### copernicus1

Thanks, I think I understand it now. Normally the time-dependent part would look like $$e^{-i\omega t},$$ but I suppose in this case it essentially looks like $$e^{-i(n+1/2)\omega t}.$$ So as the n value increases the frequency will increase. Does this look correct?

Thanks.

4. Nov 16, 2012

### Einj

That's correct. The time-dependent part, in fact, generally is $exp(-iHt)$ so, in your case $H=(n+1/2)\hbar\omega$ and you get exactly what you wrote.