Hydrogen 2P Probability Density Question I'd like to calculate the probability density for Hydrogen in the $|2,1,1\rangle$ and $|2,1,-1\rangle$ states. There is an $\exp(\pm i\phi)$ term attached to the wave function for these states. It seems to me that $|\Psi|^2$ in this case would be the same for $|2,1,1\rangle$ and $|2,1,-1\rangle$  since $|\exp(i\phi)| = 1$ with the same being true for $|\exp(-i\phi)|$. Where am I going wrong here?
Also, since I'm working in spherical coordinates the PD should be $r^2|\Psi|^2$, right (this will result in those classic energy orbital diagram functions)?
 A: You are right, the $\phi$-dependence disappears from the probability of this state. The probability is symmetric with respect to the reflections.
The differential probability in spherical coordinates is determined as
$$dw=|\Psi(\vec{r})|^2 dV=|\Psi(\vec{r})|^2\cdot r^2dr \cdot sin\theta d\theta \cdot d\phi$$
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A: Why did you think you were doing something wrong? The phase factor does indeed become irrelevant when you calculate the probability density.
As for the factor of $r^2$: the proper way to interpret $|\Psi|^2$ is that, when integrated over some region, it gives the probability of the electron being found in that region:
$$P(\text{e in }V) = \iiint_V|\Psi|^2\mathrm{d}^3V$$
This definition means that $|\Psi|^2$ matches up with the mathematical definition of a probability density function. So your probability density is just $|\Psi|^2$, pretty much by definition. However, if you do this in spherical coordinates, you will get a factor of $r^2$ (and $\sin\theta$) from the measure of integration, namely $\mathrm{d}^3V = r^2\sin\theta\;\mathrm{d}r\;\mathrm{d}\theta\;\mathrm{d}\phi$..
