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Stationary states, if V(x) is an even function proof

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove the following theorum:

    If V(x) is an even function (that is, ##V(-x) = V(x)##) then ## \psi (x) ## can always be taken to be either even or odd.
    Hint: If ## \psi ## satisfies equation [1.0] for a given E, so too does ## \psi (-x) ## and hence also the odd and even linear combinations ## \psi (x) \pm \psi (-x) ##.

    2. Relevant equations
    Equation [1.0]:

    ## E \psi = -\dfrac{\hbar ^2}{2m} \dfrac{d^2 \psi}{dx^2} + V(x) ##

    3. The attempt at a solution
    I'm guessing if V(x) is an even function then it means an even function plus an odd function ##E \psi = \underbrace{ -\dfrac{\hbar ^2}{2m} \dfrac{d^2 \psi}{dx^2}}_{\text{even or odd function}} + \underbrace{[V(x)]}_{\text{even function}}## will just be a similar even or odd function?

    But if V(x) is a function that isn't zero, then wouldn't solving the time independent schrodinger equation be much more complicated?
    Last edited: May 6, 2013
  2. jcsd
  3. May 6, 2013 #2


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    It is important to note that this theorem applies to eigenfunctions of the energy only. The wavefunction of a particle does not have to be either even or odd.

    With the given hint, you can convert every set of eigenfunctions in a set where all functions are either even or odd. You do not have to consider the Schrödinger equation.
  4. May 7, 2013 #3


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    V(x) should be multiplied by ψ.

    I'm not sure what you're trying to get at here.

    Consider the change of variables u=-x.
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