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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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2 Answers 2

up vote 3 down vote accepted

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 '14 at 18:10

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