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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?

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  • $\begingroup$ 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? $\endgroup$ – Brian Moths Feb 14 '14 at 20:41
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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 \begin{equation} \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 } \end{equation} So the momenta squared in invariant. Furthermore, if \begin{equation} V ( - x ) = V ( x ) \end{equation} then the potential is also invariant. Thus we have, \begin{equation} \hat{H} ( - x ) = \hat{H} ( x ) \end{equation} and the Hamiltonian must commute with the parity operator.

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  • $\begingroup$ Saying "it's trivially invariant" when it is exactly the problem of the OP is not answering the question... $\endgroup$ – Adam Feb 14 '14 at 18:47
  • $\begingroup$ I'm not exactly sure what the complaint was. But I updated the answer to make it more clear. $\endgroup$ – JeffDror Feb 14 '14 at 19:39

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