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I Confused about degeneracy

  1. Mar 12, 2017 #1

    dyn

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    Hi. I'm confused about degeneracy in I-D As far as I understand it ( and please tell me if i'm wrong ) ; a free particle is doubly degenerate with a continuous energy spectrum with eigenfunctions eikx and e-ikx. A particle on a ring in I-D is doubly degenerate but this time the energy is quantized.
    My main question concerns a particle in a 1-D infinite well with infinite walls at x=0 and x=a. The eigenfunction is given by ψ = sin (nπx/a) where n = 1,2,3,.... These eigenfunctions are non-degenerate as n only takes positive values but sin kx can be written as sin kx = ( eikx - e-ikx) / 2i which is a superposition of 2 waves travelling in opposite directions. Would this not make the particle in a box doubly degenerate ?
     
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  3. Mar 12, 2017 #2

    hilbert2

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    There's not much room for choice of ##A## and ##B## if you want the function ##\psi (x) = Ae^{ikx}+Be^{-ikx}## to be zero at x=0 and x=a. For most values of k it's not possible at all (except if A and B are both zero).
     
  4. Mar 12, 2017 #3

    Nugatory

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    Neither of those functions are solutions to the eigenvalue equation for the particle-in-a-box Hamiltonian.
     
  5. Mar 12, 2017 #4

    dyn

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    The solution to the particle in a box Hamiltonian is sin kx where k = nπ /a and n takes positive integer values. This shows that the energy eigenvalues are non-degenerate. But the part that confuses me is that sin kx can be written as a linear combination of eikx and e-ikx which implies doubly degenerate ?
     
  6. Mar 12, 2017 #5

    hilbert2

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    The function ##e^{ikx}## is not zero for any value of x. Its absolute value is 1 everywhere.
     
  7. Mar 12, 2017 #6

    PeterDonis

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    No, because, as Nugatory said, ##e^{ikx}## and ##e^{-ikx}## are not solutions of the eigenvalue equation, whereas ##\sin kx## is. Writing a function that is a solution as a linear combination of functions that are not tells you nothing at all about degeneracy; degeneracy implies that you are talking only about functions which are solutions. For each ##k## there is only one such function, hence no degeneracy.
     
  8. Mar 12, 2017 #7

    hilbert2

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    Yeah, even the harmonic oscillator ground state ##\psi (x) = Ae^{-kx^2}## can be written as ##\psi (x) = A\left(\frac{1}{2}e^{-kx^2}+\sin kx\right) + A\left(\frac{1}{2}e^{-kx^2}-\sin kx\right)## if you want to, but neither of the functions

    ##\psi (x) = \frac{1}{2}e^{-kx^2}+\sin kx##, or
    ##\psi (x) =\frac{1}{2}e^{-kx^2}-\sin kx##,

    is a solution to the SHO Schrödinger equation.
     
  9. Mar 13, 2017 #8

    dyn

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    Thanks everyone. Much appreciated !
     
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