# Parity of wavefunction.

1. Feb 6, 2012

### humanist rho

1. The problem statement, all variables and given/known data
The wavefunction describing state of a system is,
$$\psi (r,\theta ,\phi )=\frac{1}{8\sqrt{\pi }}\left( \frac{1}{a_{0}}% \right) ^{3/2}\frac{r}{a_{0}}e^{-4/2a_{0}}\sin \theta e^{-i\phi }$$
Find the parity of system in this state.

3. The attempt at a solution

$$\psi (-r,\theta ,\phi )=\frac{1}{8\sqrt{\pi }}\left( \frac{1}{a_{0}}% \right) ^{3/2}\frac{-r}{a_{0}}e^{-4/2a_{0}}\sin \theta e^{-i\phi } =-\psi (r,\theta ,\phi )$$
odd parity.

I'm wrong.But donno where. Pls help.

2. Feb 6, 2012

### vela

Staff Emeritus
In spherical coordinates, the parity operator takes
\begin{align*}
r &\to r \\
\theta &\to \pi-\theta \\
\phi &\to \phi+\pi
\end{align*}

3. Feb 6, 2012

### humanist rho

you mean r →-r ?

Then that gives me the perfect answer.
Thank you very much dear friend.

4. Feb 6, 2012

### vela

Staff Emeritus
No, r goes to +r.

5. Feb 6, 2012

### humanist rho

How is that possible?
is it bcz r^2 = x^2+y^2+z^2?
But the reflection changes direction of vector also na?