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

[Cogswell]

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?

 

[mfb]

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.

 

[vela]

Homework Equations

Equation [1.0]:

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

V(x) should be multiplied by ψ.

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?

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

Consider the change of variables u=-x.