Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two-particle QM Infinite Well

  1. Feb 24, 2008 #1


    User Avatar
    Gold Member

    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?
    Last edited by a moderator: May 3, 2017
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted