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In one dimension -

How can one prove that the Hamiltonian 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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up vote 5 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

While the accepted answer is very clear, I'll write an operator proof. The $\hat{p^2}$ in $\hat{H}$ commutes with $\hat{\mathbb{P}}$ (the parity operator). So, to show that $\hat{H}$ and $\hat{\mathbb{P}}$ commute, we have to show this:


Note that since $V(x)$ is a symmetric function i.e. even function, it is an eigenfunction of $\hat{\mathbb{P}}$.


$\Rightarrow V(x)\hat{\mathbb{P}}-V(-x)\hat{\mathbb{P}}=0$ (QED)

I did the last step keeping in mind that when you have a product of functions on which the parity operator needs to be applied, you can apply at one (i.e. change the $+x$ to $-x$) and transfer the Parity to the right.

P.S. As a consequence of this commutation, in one dimension, whenever you have a symmetric potential, the eigenstates are either even or odd, since, only even and odd functions are the eiegenstates of the parity operator.

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