# Why do the X, Y, Z operators switch parity?

1. Aug 7, 2012

### VortexLattice

I'm reading about selection rules, and the book is talking about how if you have a parity switching operator in between two wave vectors of opposite (definite) parity, the result is 0. For example, we have

$\left\langle2,0,0 \right|\hat{X}\left|2,0,0\right\rangle = 0$ because $\left|2,0,0\right\rangle$ is of even parity, and X switches its parity (where these kets are the hydrogen wave functions). Then, we have an even parity bra with an odd parity ket, and the result is 0.

My question is, why do these operators switch parity? I'd love to have both a physical and mathematical reason.

Thanks!

2. Aug 7, 2012

### Ben Niehoff

Think about the representation of these operators in the position basis.

3. Aug 7, 2012

### VortexLattice

Hmm...I mean, we have the radial function, and the spherical harmonics...sorry, I'm not seeing it :(

4. Aug 7, 2012

### Ben Niehoff

Use Cartesian coordinates instead of spherical. What does an even-parity wavefunction look like? What is the X operator in this basis?

5. Aug 10, 2012

### VortexLattice

I guess an even-parity wavefunction is symmetric over the x axis? And, as far as I know, the X operator in Cartesian coordinates is just x.

I mean, I think I have the basic idea. I know that that on some level it's essentially integrating an odd function from -a to a, which will always be 0.

Actually, I think I have it. These wave functions are always even or odd. So x turns an odd function into an even one, or an even one into an odd one.