0
$\begingroup$

I need to show the following:

$$P x P^{-1} = -x, \ P p P^{-1} = -p, \ P L P^{-1} = L$$

where $P$ is the parity operator and $x$, $p$ and $L$ are the position, momentum and angular momentum operators. How can I prove that? Maybe someone could show me an example on one of these and I would try to make the others by myself.

Improve this question
$\endgroup$
6
  • $\begingroup$ These are operators - maps - something that maps an element of the Hilbert space on another one. So you should verify that the identities hold whenever multiplied by any "psi" on the right. The product of operators acting on a "psi" is obtained by acting by them one by one, starting from the right ones, closest operators to the "psi". $\endgroup$
    – Luboš Motl
    Commented Nov 11, 2015 at 9:15
  • $\begingroup$ Thank you. But what does the $P^{-1}$ do? Is it changing the sign to the opposite again? $\endgroup$
    – Alex
    Commented Nov 11, 2015 at 9:54
  • $\begingroup$ Sorry for the extra comment, but what exactly does the position operator do? Is it just returning the argument of the wave function? I don't really understand that. $\endgroup$
    – Alex
    Commented Nov 11, 2015 at 10:01
  • $\begingroup$ Dear Alex, I think you should know that before your homework but because there's an answer here, anyway, let me say that the position operator acts on $\psi(x)$ as $[\hat x \psi](x) = x\cdot \psi(x)$. It maps a function to another function that is the original one multiplied by the function $g(x)=x$. Also, the operator $P^{-1}$ may be seen to be basically the same as $P$ in this non-relativistic context because $P^2=1$ and therefore $P=P^{-1}$. Multiply the second equation by $P$ to get the first. $\endgroup$
    – Luboš Motl
    Commented Nov 11, 2015 at 11:02
  • $\begingroup$ But you don't need to have any explicit form for $P^{-1}$ because $PxP^{-1}=-x$ is equivalent to $Px=-xP$ and then you only need to know how $P$, and not $P^{-1}$, acts on $\psi$. $\endgroup$
    – Luboš Motl
    Commented Nov 11, 2015 at 11:03

1 Answer 1

Reset to default
2
$\begingroup$

I think the first two things that is transformation of position and momentum operators are defined from definition...because by parity transformation sign of the position coordinates changes and time coordinate remain unchanged...so accordingly we get change in sign for position and momentum...once you do that then angular momentum should remain unchanged since it's given by cross product of position and momentum..these relations are valid in classical and quantum mechanics both..but here you can,as @Lubos Motl already pointed out the above identities are operator identities..so it should be true for any state of the hilbert space...e.g. $$p^{-1}\hat{x}p\psi=\sum_{x'}p^{-1}\hat{x}p|x'\rangle\psi(x')\\=\sum_{x'}p^{-1}\hat{x}|-x'\rangle\psi(x')=\sum_{x'}p^{-1}-x'|-x'\rangle\psi(x')=\sum_{x'}(-\hat{x})|x'\rangle\psi(x')=(-\hat{x})\psi$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Sorry, I don't really understand: why does the position operator change the sign? $\endgroup$
    – Alex
    Commented Nov 11, 2015 at 10:46
  • $\begingroup$ where exactly you had the problem. $\newcommand\ket[1]{\left|{#1}\right>} \hat{x}\ket{-x}=-x\ket{-x}$ This is just how position operator works on it...property of eigenstate..In the end I had something like $\sum_{x'}-x'\ket{x'}\psi(x')$Now I can write it as$\sum_{x'}-\hat{x'}\ket{x'}\psi(x')$ since $\hat{x}$ acting on \ket{x} will give me x\ket{x}..and then the completeness relation has been used..By $\ket{-x}$ I mean it's an eigenstate of $\hat{x}$ with eigenvalue $-x$. $\endgroup$
    – kau
    Commented Nov 11, 2015 at 10:55

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.