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Excited states in the QHO

by copernicus1
Tags: states
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copernicus1
#1
Nov16-12, 10:23 AM
P: 83
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!
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Einj
#2
Nov16-12, 10:36 AM
P: 305
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.
copernicus1
#3
Nov16-12, 10:47 AM
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.

Einj
#4
Nov16-12, 10:58 AM
P: 305
Excited states in the QHO

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


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