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(kx,ky,kz)=(0,0,0) solution for a free particle with PBC?

  1. Dec 23, 2015 #1
    When dealing with Dirichlet boundary conditions, that is asking for the wavefunction to be exactly zero at the boundaries, it can be clearly seen that (0,0,0) is not a physical situation as it is not normalizable. (Wavefunction becomes just 0 then)

    However when dealing with periodic boundary conditions, the basis is spanned by ##e^{i\vec{k}.\vec{r}}## where the only condition on ##\vec{k}## is that it has to correspond with ##L^{3}## cubic periodicity.

    The problem now is that ##\vec{k}=0## does seem to give a non trivial solution with zero energy ##\Psi=constant## which is periodic and noralizable.

    How do I interpret this werid 'constant' term in the general wavefunction part?


    Kittel eigth edition, p137

  2. jcsd
  3. Dec 24, 2015 #2


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    Why do you think it is weird?
  4. Dec 25, 2015 #3


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    To the contrary, it's not weird, but it's weird to think that with the Dirichlet boundary conditions you'd have a momentum operator. There is none, because there is no self-adjoint operator generating translations in this case. We have discussed this many times in this forum. Just search for it!
  5. Dec 26, 2015 #4
    Oh I was thinking it was weird because I have never encountered such an extra constant term in the wavefunction so I was doubting my reasoning to arrive at the extra constant term. Based on your reactions I see that there is indeed nothing special about it, thanks.
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