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Expected values in infinite square well

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Aikon
#1
Dec19-12, 11:17 AM
P: 21
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
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mfb
#2
Dec19-12, 12:14 PM
Mentor
P: 11,819
I would expect an error in your integration, as the first expression is not zero and the following one is.
tadchem
#3
Dec19-12, 12:44 PM
P: 70
Quote Quote by Aikon View Post
[itex]
<x>=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx
[/itex]

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]

Aikon
#4
Dec19-12, 02:24 PM
P: 21
Expected values in infinite square well

Quote Quote by tadchem View Post
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]
Yeah, i got it...it is easier to use table of integrals.
it gives the expected [itex]<x>=\frac{a}{2}[/itex] for any value of n.
thanks


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