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Wave Functions With Same Energies Are the Same (only differ by a complex phase)

  1. May 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Assume a particle with a wave function ##\psi(x)## such that ##-\infty < x < \infty##, that move under some potential ##V(x)##.

    Show that:
    a) two wave functions with same energies can only differ by a complex phase;
    b) if the potential is real, then you can choose the wave function to be real as well;
    c) the wave function of the ground state (with real potential) doesn't change sign.

    2. Relevant equations
    a) Schrodinger's time independent equation.

    3. The attempt at a solution
    I'm stuck at (a). Need a push in the right direction for the very start.
    I want to show that if two wave functions ## \psi_1, \psi_2## satisfy
    $$ \psi_{1/2}''(x) + \frac{2m}{\hbar^2}\left(E-V(x)\right)\psi_{1/2}(x)=0$$
    then I can find an equation that ties them in a phase relation.
    But aside from writing this statement down, I don't know how to proceed. Thanks.
     
  2. jcsd
  3. May 12, 2016 #2

    blue_leaf77

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  4. May 12, 2016 #3
    If the Hamiltonian be ##\hat H## and the two normalised energy eigen vectors be ##|\psi_1> ## and ##|\psi_2>##, then from Schroedinger's time independendent equation ##\hat H |\psi_1> = E|\psi_1>## and ##\hat H |\psi_2> = E|\psi_2>##. From these two equation you can conclude that ##|\psi_2> = c|\psi_1>##. Where ##c## is a complex number. Now as the wave vectors are normalised, you can write ##<\psi_2|\psi_2> = <\psi_1|c^* c|\psi_1> \Rightarrow <\psi_2|\psi_2> = c^* c <\psi_1|\psi_1> \Rightarrow c^*c =1##. Naturally ##c = exp (i \theta)##.
     
  5. May 12, 2016 #4

    blue_leaf77

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    It does not explain why in 1D system like the one considered in the problem, there can be no degeneracy. Your method can be applied, for example, to a hydrogen atom. We know that for quantum numbers ##n>1##, the eigenstates are degenerate and those eigenstates are not merely related by a complex constant.
     
  6. May 12, 2016 #5
    Thanks for your replies!
    I managed both (a) and (b) and understood its underlying principles. I'm now left with (c).
    We were suggested to define ##\phi = |\psi_0|## where ##\psi_0## is the wave function of the ground state, and then express ##\phi## in terms of the Hamiltonian eigenvectors and find its energy. Then, use (b) to finish. But how can I express ##\phi## in terms of the eigenvectors of the Hamiltonian when I don't know what it is?

    Ok got it... sorry, I should have given it more thought before throwing the question in the air.
     
  7. May 12, 2016 #6

    blue_leaf77

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    (c) asks you to prove that the ground state never crosses the x axis.
     
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