Suppose that [tex]\psi (x)[/tex] is some solution to the time-independent Schrödinger equation;(adsbygoogle = window.adsbygoogle || []).push({});

[tex]-\frac{h^2}{2m}\frac{\partial^2\psi(x)}{\partial x} + V(x)\psi(x) = E\psi(x)[/tex].

I want to show that if the potential V(x) is anevenfunction, then [tex]\psi(-x)[/tex] is also a solution to same equation (same E and V).

I know I'm supposed to combine the facts that [tex]\psi(x)[/tex] is a solution and that V(x) = V(-x), but I can't see how. I've noted that

[tex]\frac{\partial^2\psi(-x)}{\partial x} = \frac{\partial^2\psi(x)}{\partial x}[/tex],

but that's pretty much it

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# Basic qm derivation

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