# Action of Parity operator on Impulse representation

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

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

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

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

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

• Why wouldn't it be correct? Jan 26, 2014 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. Jan 26, 2014 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 $\tilde\psi(p)$. Jan 26, 2014 at 19:40
• @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. Jan 26, 2014 at 19:55
• @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 $\tilde\psi(-p)$. Jan 26, 2014 at 20:10