(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the expectation value: <(x1-x2)^2> for two non-interacting particles in the infinite square well. If one is in state [tex]\psi_1( n \neq l) [/tex] and the other in state [tex]\psi_n[/tex] find the expectation value for a) distinguishable particles, b) bosons, c) fermions

2. Relevant equations

[tex]\psi_n = \sqrt{\frac{2}{a}}sin(\frac{n \pi x}{a})[/tex]

[tex]\psi_1 = \sqrt{\frac{2}{a}}sin(\frac{ \pi x}{a})[/tex]

3. The attempt at a solution

verification for a):

[tex]<{\Delta x}^2> = <x^2>_1 + <x^2>_n - 2<x>_1<x>_n[/tex]

after integration I get:

[tex]<{\Delta x}^2> = 4a(\frac{1}{3} + \frac{1+\delta_{n,m=1}}{2\pi^2} - 4n\delta_{n,m=1})[/tex]

My Problem

While I don't know whether the above is correct, assuming it is, when I do this for identical particles, I get the extra term (in addition to the above):

[tex]\mp 2|<x>_1,n|^2 = \int \frac{4}{a^2} sin(\frac{\pi x}{a})sin(\frac{n \pi x}{a})dx[/tex]

which (using http://integrals.wolfram.com/index.jsp") comes to:

http://integrals.wolfram.com/Integrator/MSP?MSPStoreID=MSPStore188022599_0&MSPStoreType=image/gif [Broken]

evaluating this integral, the sins go to zero, but the cosines leave a 1 and a couple +/- 1's

Unfortunately, in the denominator, one of the terms has a (n-1) term. Does this mean that n cannot equal 1 even for Bosons? That is to say that it's meaningless to measure the spatial displacement between particles in the same state?

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# Homework Help: Two-particle QM Infinite Well

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