
#1
Dec1912, 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 



#2
Dec1912, 12:14 PM

Mentor
P: 10,864

I would expect an error in your integration, as the first expression is not zero and the following one is.




#3
Dec1912, 12:44 PM

P: 61

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] 



#4
Dec1912, 02:24 PM

P: 21

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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