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B Ψ=f(r)exp[-i(Et-pz+Φ/2)] a Q.M. wave function?

  1. Apr 22, 2016 #1
    In cylindrical coordinates could Ψ=f(r)exp[-i(Et-pz+Φ/2)] be a valid quantum mechanical wave-function for the right boundary conditions and with the right choice of f(r)?

    Thanks!
     
  2. jcsd
  3. Apr 22, 2016 #2
    Your suggestion doesn't appear to be normalizable if you allow ##z\in(-\infty,\infty)##. Consider ##\int |\Psi|^2 \mathrm{d}zr\mathrm{d}r\mathrm{d}\theta##. You are forced to do the integral ##\int\limits_{-\infty}^{\infty}1\mathrm{d}z##.

    EDIT: I suppose the answer to your exact question is that, yes, it could be a wavefunction with an appropriate restriction on the domain in the z-direction.
     
    Last edited: Apr 22, 2016
  4. Apr 22, 2016 #3

    blue_leaf77

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    Judging from the expression of the wavefunction, there can be two possibilities about the condition of the potential along the z-direction:1) The potential is constant along this direction, for which reason ##p## is continuous or 2) there is an infinite square well confining the space along z direction, in this case ##p## must be discrete.
    Either way, the function you have there is bounded at infinities and therefore it can act as a basis function, despite being not normalizable for the case of continuous ##p##, for a physically realizable state.

    EDIT: The only possible potential form in the z direction in this case is a constant potential. Therefore ##p## must be continuous.
     
    Last edited: Apr 24, 2016
  5. Apr 24, 2016 #4
    Thanks for pointing out things I should of worried about. My main concern was the angular coordinate Φ divided by two by which I was naively trying to get something like spin 1/2 angular momentum.

    Thanks!
     
  6. Apr 24, 2016 #5

    blue_leaf77

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    Spin? That was not mentioned in your original post.
     
  7. Apr 25, 2016 #6
    Yes, but but my function depends on angular coordinates (there is angular momentum?). My function only returns to itself after a 4π rotation and I'm not sure that is acceptable.

    Thanks!
     
  8. Apr 25, 2016 #7

    blue_leaf77

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    Rotation operator due to spin angular momentum can only rotate spin state, it does not act on spatial wavefunction.
     
  9. May 1, 2016 #8
    So two problems with my function above, Ψ=f(r)exp[-i(Et-pz+Φ/2)]. The first is the function is undefined along the cylindrical axis (though I guess I could set f(r) to zero there). I thought I could get away from that problem by using the function in some annular region where the azis was excluded.

    The second more worrisome problem is the fact that the function only returns to itself after two complete rotations,

    Ψ=f(r)exp[-i(Et-pz+Φ/2)] = Ψ=f(r)exp[-i(Et-pz+[Φ+4π]/2)]

    What trouble do we get in if Ψ(r,z,Φ) ≠ Ψ(r,z,Φ+2π) but Ψ(r,z,Φ) = Ψ(r,z,Φ+4π)?

    Thanks Blue_leaf for your help!
     
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