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Old Oct4-05, 09:29 PM                  #1
Tuneman

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Quantum Physics question

I am having trouble with 2 problems about the harmonic oscilator.
I realize this isnt the homework section, however I feel that in my situation where time is limited, perhaps someone would be able to give me a certain amount of leeway.


First of all the question tells me to use the wave function for when n=1

So I have:
Wavefunction = A[1]*r^1/2 * e^-((r^2)/2)


where r= (mw/(hbar))^1/2 and A[1] is A sub 1.

I am wondering when I am multiplying this by the complex conjugate, is the complex conjugate going to have a e^+((r^2)/2). I dont think I am but for some reason, in a solution finding <x> which equals ](integral) psi* (times) x (times) psi] those exponential functions did not appear. So I was wondering how they cancelled out when you calculate <x>

also it asks me to calculate <p>, which I know equals m*d<x>/dt. My question is, if my <x> does not depend on t, because the time dependent part of the equation was cancelled out when solving for <x>, how can I find d<x>/dt?

Any help would be appreciated, I'm sure my questions or equations aren't too clear, so if you have any questions, I will try to clarify. Thank you.
 
Old Oct4-05, 11:06 PM                  #2
vanesch

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Originally Posted by Tuneman
I am having trouble with 2 problems about the harmonic oscilator.
I realize this isnt the homework section, however I feel that in my situation where time is limited, perhaps someone would be able to give me a certain amount of leeway.


First of all the question tells me to use the wave function for when n=1

So I have:
Wavefunction = A[1]*r^1/2 * e^-((r^2)/2)


where r= (mw/(hbar))^1/2 and A[1] is A sub 1.
This doesn't make sense: I'm sure you are missing a space variable in "r".
That explains then too why you don't find your exponential anymore once you've integrated over it...

cheers,
Patrick.

PS: and this really belongs in the homework section...
 
Old Oct4-05, 11:13 PM                  #3
Tom Mattson
 
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Tuneman, please don't double post. One thread is enough for one question.
 
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