Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 $V(x)=V(-x)$

share|cite|improve this question

3 Answers 3

up vote 4 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.

share|cite|improve this answer

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

The parity operator therefore commutes with Hamiltonian.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.