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Wavefunctions of fermions and bosons

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider two noninteracting particles p and q each with mass m in a cubical box od size a. Assume the energy of the particles is

    [tex] E = \frac{3 \hbar^2 \pi^2}{2ma^2} + \frac{6\hbar^2 pi^2}{2ma^2} [/tex]

    Using the eigenfunctions
    [tex] \psi_{n_{x},n_{y},n_{z}} (x_{p},y_{p},z_{p}) [/tex]
    [tex] \psi_{n_{x},n_{y},n_{z}} (x_{q},y_{q},z_{q}) [/tex]

    write down the two particle wave functions which could describe the system when the particles are
    a) distinguishable, spinless bosons
    b) identical, spinless bosons
    c) identical spin-half fermions in a symmetric spin state
    d) identical spin half fermions in an antisymmetric spin state

    2. Relevant equations
    For a cube the wavefunction is given by

    [tex] \psi_{n_{x},n_{y},n_{z}} = N \sin\left(\frac{n_{x}\pi x}{a}\right)\sin\left(\frac{n_{y}\pi y}{a}\right)\sin\left(\frac{n_{z}\pi z}{a}\right) [/tex]

    [tex] E = \frac{\hbar^2 \pi^2}{2ma^2} (n_{x}^2 + n_{y}^2 +n_{z}^2) [/tex]

    3. The attempt at a solution
    for the fermions the wavefunction mus be antisymmetric under exhange
    c) [tex] \Phi^{(A)}(p,q) = \psi^{(A)} (r_{p},r_{q}) \chi^{(S)}_{S,M_{s}}(p,q) [/tex]
    where chi is the spin state

    so since the energy is 3 E0 for the first particle there possible value nx,ny,nz are n=(1,1,1) and the second particle n'=(1,1,2).

    we could select
    [tex] \Psi^{(A)} (x_{p},x_{q},t) = \frac{1}{\sqrt{2}} (\psi_{n}(x_{p})\psi_{n'}(x_{q} - \psi_{n}(x_{q})\psi_{n'}(x_{p}) \exp[-\frac{i(E_{n} + E_{n'})t}{\hbar} [/tex]
    A means it is unsymmetric
    the spin state chi could be
    [tex] \chi^{(S)}_{1,1}(p,q) = \chi_{+}(p) \chi_{-}(q) [/tex]
    S means it is symmetric

    For d it is similar but switched around

    for a) and b) i have doubts though
    For a) the bosons must be distinguishable so we could have WF like this

    [tex] \Psi_{1} (r_{p},r_{q},t) = \psi_{n}(r_{p})\psi_{n'}(r_{q}) \exp[-\frac{i(E_{n} + E_{n'})t}{\hbar} [/tex]
    for one of the particles. Under exchange this would be symmetric.

    b) if the bosons are identical then we simply have to construct a wavefunction that is smmetric like we did in part c for the fermions.

    Is this right? Please help!

    thanks for any and all help!
    Last edited: Feb 21, 2007
  2. jcsd
  3. Feb 21, 2007 #2
  4. Feb 21, 2007 #3
    by the way [tex] \chi= \chi_{S,M_{S}} [/tex]

    [tex] \chi_{+} [/tex] when Ms = +1/2
    [tex] \chi_{-} [/tex] when Ms = -1/2
  5. Feb 23, 2007 #4


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    Staff Emeritus
    Science Advisor
    Gold Member

    Hi stunner5000pt,

    I'm just learning this stuff myself, but I'll try to help. Your answer to part c looks correct. The spatial part of the wf was chosen to be antisymmetric, since the spin state is given to be symmetric, so that the *overall* wf is antisymmetric.

    No it wouldn't be! But that's okay! :smile: Because the particles are meant to be distinguishable. So I think your answer is correct...an acceptable wavefunction for a two-particle system is the product of the individual one-particle wavefunctions, *if* you know that one is in the state n, and
    the other in the state n', because you are able to tell the difference between them.

    Yes, you do have to construct a wavefunction that is symmetric. But no, it's not like part c, because in part c, you constructed a wf that was antisymmetric. :tongue2: Not only that, but your bosons are spinless, so you'd only have a spatial part to your wavefunction, and it would have to be symmetric on its own.

    Hope this helps
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