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expected values in infinite square well |
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| Dec19-12, 11:17 AM | #1 |
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expected values in infinite square well
Ok...this must sound stupid, because i didn't found answer on the web and on my books...but i am having trouble with the infinite square well.
I want to calculate <x>. V(x)=0 for 0<=x<=a [itex] <x>=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx [/itex] Doing integration by parts i got to: [itex]\frac{2}{a}\left[\frac{a^2}{2n}-\int^a_0\frac{a}{2n}dx\right]=0[/itex] What i am doing wrong? Thank you |
| Dec19-12, 12:14 PM | #2 |
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I would expect an error in your integration, as the first expression is not zero and the following one is.
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| Dec19-12, 12:44 PM | #3 |
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set [itex] u=\frac{n\pi}{a}x [/itex] look up [itex] \int u \sin^2(u)du [/itex] in a Table of Integrals to find [itex] \frac{u^2}{4}-\frac{u\sin(2u)}{4}-\frac{cos(2u)}{8} [/itex] |
| Dec19-12, 02:24 PM | #4 |
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expected values in infinite square wellit gives the expected [itex]<x>=\frac{a}{2}[/itex] for any value of n. thanks |
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| infinite square well, quantam mechanics |
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