# Mirror/Parity symmetry

I am trying to solve a problem of Griffiths' book.

$$\hat{\Pi} \psi(\vec{r}) = \psi(-\vec{r})$$ where $$\vec{r}$$= (x,y,z), eq. (1)

$$\hat{\Pi}$$ is the parity operator.

The problem says to show that this is equivalent to a mirror reflection followed by a rotation.

Rotation operator about z is given by $$\hat{R_z}(\varphi)\psi(r,\theta,\phi)=\psi(r,\theta, \phi-\varphi)$$

I don't understand this. The only way that eq(1) is equivalent to a mirror reflection followed by a rotation is only if I apply a mirror symmetry to z (z->-z) and then apply a $$\pi$$ (180º) rotation about z (x,y -> -x,-y).

So I would have to do something like:

$$\hat{R_z}(\varphi)\hat{\Pi}_z\psi(x,y,z) =\hat{R_z}(\varphi)\psi(x,y,-z) = \hat{R_z}(\varphi)\psi(r,-\theta,\phi) = \psi(r,-\theta,\phi-\pi) =\psi(-x,-y,-z)$$

This is ugly and I don't think is the answer. Could you help me?

• Looks all correct to me. Why is it ugly? Oct 15, 2020 at 3:50
• Well, then there is the second line of the exercise. And it says to use polar coordinates to show that $\hat{\Pi}\psi(r,\theta,\phi) = \psi(r, \pi-\theta, \phi+\varphi)$. I guess that in the first line I didn't need polar coordinates.
– AA10
Oct 15, 2020 at 8:58