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


3 Answers 3


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.


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

The parity operator therefore commutes with Hamiltonian.

  • 2
    $\begingroup$ 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. $\endgroup$
    – user121330
    Oct 28, 2014 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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