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Excited states in the QHOby copernicus1
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#1
Nov1612, 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! 


#2
Nov1612, 10:36 AM

P: 323

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
Nov1612, 10:47 AM

P: 83

Thanks, I think I understand it now. Normally the timedependent 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
Nov1612, 10:58 AM

P: 323

Excited states in the QHO
That's correct. The timedependent 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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