Symmetric potential and the commutator of parity and hamiltonian

Question

In one dimension -

How can one prove that the Hammiltonian and the parity operator commute in the case where the potential is symmetric (an even function)?

i.e. that [H, P] = 0 for V(x)=V(-x)

Answer 1

You prove the equality of operators by applying them to a function, we have

$$ H = - \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x) $$ Ergo: $$ HP f(x) = H f(-x) = (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(-x) = - \frac{\hbar^2}{2 m} f''(-x) + V(x) f(-x) $$ and $$ PH f(x) = P (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(x) = P (- \frac{\hbar^2}{2 m} f''(x)) + P (V(x) f(x)) ... $$

$$ ... = - \frac{\hbar^2}{2 m} f''(-x) + V(-x) f(-x) $$ When you use $$ V(-x) = V(x) $$ you see that both expressions are equal.