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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?

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  • $\begingroup$ Looks all correct to me. Why is it ugly? $\endgroup$ Oct 15, 2020 at 3:50
  • $\begingroup$ 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. $\endgroup$
    – AA10
    Oct 15, 2020 at 8:58

1 Answer 1

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Here's a easy way to do that,you can rotate the axis(passive rotation,it's physically equivalent with rotating the wavefunction).Staring with keeping the z axis standing still and rotating the x-y plane 180 degrees,then applying the mirror reflection to z axis,you'll get the new axis with x',y',z' = -x,-y,-z

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