# Commutator of parity and Hamiltonian operators under even potential function

I need to show what is $[H,P]$ where $H$ is the Hamiltonian and $P$ the parity operator. $V(\underset{\sim}x) = V(-\underset{\sim}x)$ in this case.

I start off with

$$\langle \underset{\sim}x|HP|\psi\rangle = \langle \underset{\sim}x|(\frac{p^2}{2m}+V)P|\psi\rangle = \langle \underset{\sim}x|(\frac{p^2}{2m}+V)P|\psi\rangle = \langle \underset{\sim}x|\frac{p^2}{2m}P|\psi\rangle+\langle \underset{\sim}x|VP|\psi\rangle$$

and since $\langle \underset{\sim}x|V = V(\underset{\sim}x)\langle \underset{\sim}x|$ (is this step valid?) and $\langle \underset{\sim}x|P = \langle -\underset{\sim}x|$, the above equation becomes

$$\langle \underset{\sim}x|HP|\psi\rangle = \langle \underset{\sim}x|\frac{p^2}{2m}P|\psi\rangle + V(\underset{\sim}x)\psi(-\underset{\sim}x)$$

Similarly I have

$$\langle \underset{\sim}x|PH|\psi\rangle = \langle \underset{\sim}x|P\frac{p^2}{2m}|\psi\rangle + V(-\underset{\sim}x)\psi(-\underset{\sim}x) = \langle \underset{\sim}x|P\frac{p^2}{2m}|\psi\rangle + V(\underset{\sim}x)\psi(-\underset{\sim}x)$$

Taking the difference of the two, I find that

$$\langle \underset{\sim}x|HP-PH|\psi\rangle = \langle \underset{\sim}x|\frac{p^2}{2m}P-P\frac{p^2}{2m}|\psi\rangle = -\langle \underset{\sim} x|\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_i^2}P|\psi \rangle +\langle -\underset{\sim} x|\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_i^2}|\psi\rangle$$

which I had trouble evaluating. Any hints?

• I don't see how you can't just plug in the definitions. How are things defined? Also have you learned about the wavevector basis? – Brian Moths Feb 14 '14 at 20:41

I think there is an easier way to go about this. Acting with the Parity operator on the Hamiltonian we have: \begin{align} P \hat{H} P & = \hat{H} ( - x ) \\ \Rightarrow P \hat{H} & = \hat{H} ( - x ) P \end{align} So the Hamiltonian commutes with the Parity operator if $\hat{H} ( x ) = \hat{H} ( - x )$. Now $$\frac{ p ^2 }{ 2 m } = - \frac{1}{ 2m} \frac{ \partial ^2 }{ \partial x ^2 } \xrightarrow{P} - \frac{1}{ 2m} \frac{ \partial ^2 }{ \partial (-x) ^2 } =- \frac{1}{ 2m} \frac{ \partial ^2 }{ \partial x ^2 }$$ So the momenta squared in invariant. Furthermore, if $$V ( - x ) = V ( x )$$ then the potential is also invariant. Thus we have, $$\hat{H} ( - x ) = \hat{H} ( x )$$ and the Hamiltonian must commute with the parity operator.