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Additional quantum states of the infinite square well

  1. Nov 27, 2015 #1
    The quantum states ##\psi(x)## of the infinite square well of width ##a## are given by

    ##\psi(x) = \sqrt{\frac{2}{a}}\sin\Big(\frac{n \pi x}{a}\Big),\ n= 1,2,3, \dots##

    Now, I understand ##n \neq 0##, as otherwise ##\psi(x)## is non-normalisable.

    But, can't we get additional states for ##n=-1,-2,-3,\dots##?

    Of course, they have the same position probability densities and energy eigenvalues as the corresponding positive $n$ states, but still, don't they *exist* at all?
     
  2. jcsd
  3. Nov 27, 2015 #2

    Orodruin

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    No, those are not additional states. In fact, they are the same states as they only differ by an arbitrary phase factor.
     
  4. Nov 27, 2015 #3
    Well, do we not rely on the negative sign on the wavefunction to distinguish between bosons and fermions?

    For bosons and fermions, we might as well call these the same particle as the wavefunction differs also by a phase factor ??? :frown:
     
  5. Nov 27, 2015 #4

    Orodruin

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    No we do not. We rely on the symmetry properties under exchange of particles in multi-particle states. Relative phase factors matter, overall phase factors do not.
     
  6. Nov 27, 2015 #5
    Thanks! I get it now!
     
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