# Symmetric potential and the commutator of parity and hamiltonian

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)

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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.

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$$[P,H]f(x)=(PH-Hp)f(x)$$ But $$H=P^2/2m+E(x)$$ $$=PE(x)-Hf(x)$$ $$=E(-x)-E(-x)$$ $$=0$$

The parity operator therefore commutes with Hamiltonian.

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Is there a conflation of parity operators and momentum operators? The germ of truth is very powerful with this proof, but the lack of clarity makes it un-decipherable. –  user121330 Oct 28 at 18:10