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Two distinguishable particles space-spin wavefunctions

  1. Dec 30, 2004 #1

    Two (distinguishable) non-interacting electrons are in an infinite square well with hard walls at x=0 and x=a, so that the one particle states are
    [tex]\phi_n(x)=\sqrt{\frac{2}{a}} sin(\frac{n\pi}{a}x), E_n=n^2K[/tex] where [tex]K=\pi^2\hbar^2/(2ma^2)[/tex]

    My question is what are the spins and space-spin wavefunctions for the state with the lowest two energy eigenvalues?

    The answer for the ground system is:

    [tex] S=0[/tex]

    [tex]\phi_1(x_1,x_2)=\frac{2}{L} sin(\frac{\pi x_1}{L}) sin(\frac{\pi x_2}{L}) * \frac{1}{\sqrt{2}} (\chi_{up}(1)\chi_{down}(2) - \chi_{down}(1) \chi_{up}(2)) [/tex]

    (the first part denotes the space wave function, the second part denotes the spin wavefunction)

    [tex] E = 2 K[/tex] where
    [tex] K = \frac{\pi^2\hbar^2}{2mL^2}[/tex]

    Is this becuase since the wavefunction is symmetric, the spin wavefunction must be antinsymmetric (singlet). Therefore, the electrons are in opposite spins and S=0. However, I'm confused about how to find the spin-state of the non-ground system. I know that S=1 and it must be a symmetric configuration but there are three of them. How do I know which one it is in?
  2. jcsd
  3. Jan 14, 2005 #2
    If I well understood, you have a state with a given energy but three different wave functions (a degenerate state). Then, the system reaches all these states with a certain probability. The total wave function will be a linear combination of those three functions:


    The system can be found in one of its degenerate state with a PROBABILITY given by coefficients c1, c2 and c3.

    Last edited: Jan 14, 2005
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