# Excited states in the QHO

by copernicus1
Tags: states
 P: 83 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.
 P: 324 Excited states in the QHO 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.