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Identical particles in 1D infinite well

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to find the mean square distance between the 2 particles. Before I can do that, I need the expectation of [tex]x_1^2[/tex] and [tex]x_2^2[/tex] , then [tex]x_1x_2[/tex]. I an on the first part and got stuck.

    2. Relevant equations

    [tex]<X_1^2>=\int_0^L \int_0^L x_1^2 |\psi_{n,m}(x_1,x_2)|^2 dx_1dx_2[/tex]
    This is used because the spin state is antisymmetric.
    [tex]\psi_n(x_1)=\sqrt{\frac{2}{L}}\sin (\frac{n \pi}{L}x_1)[/tex]
    [tex]\psi_m(x_1)=\sqrt{\frac{2}{L}}\sin (\frac{m \pi}{L}x_1)[/tex]
    [tex]\psi_n(x_2)=\sqrt{\frac{2}{L}}\sin (\frac{n \pi}{L}x_2)[/tex]
    [tex]\psi_m(x_2)=\sqrt{\frac{2}{L}}\sin (\frac{m \pi}{L}x_2)[/tex]

    3. The attempt at a solution
    I play around a bit and got this:

    [tex]\int_0^L \int_0^L x_1^2 |\psi_{n,m}(x_1,x_2)|^2 dx_1dx_2[/tex]
    [tex]=\frac{1}{2}[\int_0^L \psi_m(x_2)^2 \int_0^L x_1^2 \psi_n(x_1)^2dx_1dx_2+2\int_0^L \psi_n(x_2)\psi_m(x_2) \int_0^L x_1^2 \psi_n(x_1)[/tex][tex]\psi_m(x_1)dx_1dx_2+\int_0^L \psi_n(x_2)^2 \int_0^L x_1^2 \psi_m(x_1)^2dx_1dx_2][/tex]
    [tex]=\frac{1}{2}[L^2(\frac{1}{3}-\frac{1}{2n^2\pi^2})+\frac{8}{L^2}\int_0^L \sin (\frac{n \pi}{L}x_2)\sin (\frac{m \pi}{L}x_2)dx_2\int_0^L x_1^2[/tex][tex]\sin (\frac{n \pi}{L}x_1)\sin (\frac{m \pi}{L}x_1)dx_1+L^2(\frac{1}{3}-\frac{1}{2m^2\pi^2})][/tex]
    [tex]=\frac{1}{2}[L^2(\frac{2}{3}-\frac{1}{2n^2\pi^2}-\frac{1}{2m^2\pi^2})+\frac{8}{L^2}\int_0^L \sin (\frac{n \pi}{L}x_2)\sin (\frac{m \pi}{L}x_2)dx_2\int_0^L x_1^2[/tex][tex]\sin (\frac{n \pi}{L}x_1)\sin (\frac{m \pi}{L}x_1)dx_1][/tex]

    Am I correct so far? I am stuck from this point. Thanks for the help!
  2. jcsd
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