# 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)?

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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 $\mathrm{d}^3V = r^2\sin\theta\;\mathrm{d}r\;\mathrm{d}\theta\;\mathrm{d}\phi$..

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Thanks for the detailed answer. I thought I was going wrong because the plots of the PD would have been the same. My goal in calculating those PDs is plotting them in 3D, so two of the same plots wouldn't have been meaningful. –  nick_name Nov 7 '11 at 22:02
Indeed, it's redundant to draw both $nlm$ and $nl,-m$ because the probability densities are the same. That doesn't mean that the states are the same. The phase does contain physical information. Its change in space tells you about the momentum. So by looking at the phase, one may say that the electron is orbitting in one way (counter-clockwise) for $m=1$ and in the other way (clockwise) for $m=-1$. The two wave functions are orthogonal to each other - completely different, mutually exclusive states. –  Luboš Motl Nov 7 '11 at 22:20
@LubošMotl That was a solid explanation. Wavefunction orthogonality is a great way of tying it all together. –  nick_name Nov 7 '11 at 23:24
Thanks for your interest, @nick_name. I should have mentioned the simplest example: $C\exp(ipx)$ is a plane wave and the probability density is constant all over the space, independently of $p$. However, that doesn't mean that the functions are the same for all $p$: they're moving with different velocities. Their Fourier transforms, delta-function in momentum space, have different probability densities (for various momenta), located at different points. The angular-momentum case is just a wrapped, curved version of the same principle. –  Luboš Motl Nov 8 '11 at 7:32

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

$$dw=|\Psi(\vec{r})|^2 dV=|\Psi(\vec{r})|^2\cdot r^2dr \cdot sin\theta d\theta \cdot d\phi$$

You can enjoy the 3D visualizations and even rotate them with your mouse with this applet: http://www.falstad.com/qmatom/

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I love those visualisations from Falstad. In my on-going training as a physicist, they've all been useful at some time. –  Kasper Meerts Nov 7 '11 at 23:13