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Is my derivation of the action of the parity operator $\mathbb{P}$ on the $|p\rangle$ representation correct?

$$\left( \mathbb{P}\tilde\psi \right)(p)= - \tilde\psi (p).$$

Obtained from

$$\left( \mathbb{P}\tilde\psi \right)(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}px}(\mathbb{P}\psi)(x)= $$

$$ = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}px}\psi(-x)=$$

$$= -\frac{1}{\sqrt{2\pi\hbar}}\int_{\infty}^{-\infty} dx \, e^{-\frac{i}{\hbar}(-p)x}\psi(x)=-\tilde\psi(p).$$

$$= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}(-p)x}\psi(x)=\tilde\psi(-p).$$

PhilipV
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    Why wouldn't it be correct? – David Z Jan 26 '14 at 19:03
  • @DavidZ Isn't the definition of parity $(\mathbb P\psi)(x) = \psi(-x)$? In that case there is an error when the OP writes $(\mathbb P\psi)(x) = -\psi(x)$. Also, the dummy variable $r$ is being used as the argument of the momentum space representation of the wavefunction, but it does not appear in the integral expressions for the Fourier transform. – joshphysics Jan 26 '14 at 19:16
  • @joshphysics I used $\tilde\psi(r)$ when I meant $\tilde\psi(p)$, which is present on the exponential inside the integral.

    I know the definition of parity is $\left( \mathbb{P}\tilde\psi \right)(x)=\tilde\psi (-x)$, and that's why I'm not sure what it's action would be on $\tilde\psi(p)$.

     – PhilipV Jan 26 '14 at 19:40
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    @PhilipV Your approach is exactly right, you've simply mis-applied the action of parity in the second line as far as I can tell. – joshphysics Jan 26 '14 at 19:55
  • Note also that since $p=m,dx/dt$, then if $\mathbb{P}x=-x$ it ought to be the case that $\mathbb{P}p=-p$. – Kyle Kanos Jan 26 '14 at 19:56
  • @joshphysics That's (another) error in the formula I wrote. I made a mess at writing it! But the next line should be correct, no? I send $x\rightarrow -x$. – PhilipV Jan 26 '14 at 20:06
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    @PhilipV I think it's a good practice not to edit so much because then the comments above become difficult to follow unless we constantly refer to which version we were commenting on. You do need to send $x\to -x$, then we careful in the end because the sign of the exponential changes when you do this, and the result should be $\tilde\psi(-p)$. – joshphysics Jan 26 '14 at 20:10
  • @joshphysics I'd say it's good to edit even if it invalidates comments. In fact that's the whole point of comments, to be incorporated into the post. Also, just to clarify, my purpose in posting the first comment was not to suggest that it was correct (honestly, I didn't even read it), but to get PhilipV to identify a specific problem and ask about it. Questions where someone just asks for us to check their work generally aren't permitted under the homework policy. – David Z Jan 26 '14 at 23:54
  • @DavidZ I think you make a good point. I'll take it upon myself to refer to versions more religiously. – joshphysics Jan 27 '14 at 01:23
  • @DavidZ I'm sorry. I knew that I had made an error, but did not realized where. I solved it, now. – PhilipV Jan 27 '14 at 20:22
  • @joshphysics this is a very legitimate rather advanced technical question that should not be closed: voted to reopen. – Dilaton Jan 27 '14 at 21:59

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