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Cogswell
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Homework Statement
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) ##.
Homework Equations
Equation [1.0]:
## E \psi = -\dfrac{\hbar ^2}{2m} \dfrac{d^2 \psi}{dx^2} + V(x) ##
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?
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