How do you determine the "phase" of a hydrogen eigenfunction? I've been reading the wikipedia article on the atomic orbitals of hydrogen. They have a nice collection of diagrams, such as this one for n,l,m = 3,1,1

This is apparently showing the wavefunction, not the probability density, and the blue area represents positive phase, red represents negative.
My problem is this: this particular wavefunction contains a term $\exp(+i\phi)$, so how has this graph been drawn taking into account the complex part? And what are they referring to when they talk about phase?

Edit, this is the 311 wavefunction containing the $\exp(+i\phi)$ term

 A: Hydrogenic wavefunctions (as well as anything with well-defined angular momentum about a given axis) come in two flavours. 


*

*The first set is 'cylindrical', and has wavefunctions $\psi\sim e^{\pm i|m|\phi}$.

*The second set is 'cartesian', and has wavefunctions $\psi_\text{even}\sim\cos(m\phi)$ and $\psi_\text{odd}\sim\sin(m\phi)$.


Both sets are perfectly valid bases for the subspace with a given $l$ and $|m|$ (or, alternatively, $m^2$, so all functions are eigenfunctions of $L_z^2$). It should be clear that either set can be obtained as linear combinations of the other set, so they are equivalent. However:


*

*The first set shares (more explicitly) the cylindrical symmetry of the problem, which is reflected by the fact that its wavefunctions are eigenfunctions of $L_z$.

*The second set has the advantage that its wavefunctions are real and that the $\phi$ variation is explicitly encoded in the amplitude, and not in a hard-to-recover phase factor, so they make it easier to display the structure in plots. Additionally, they are eigenfuctions of the parity operator, and they are simpler to deal with in code, which means that quantum chemistry software often works with them.
The second set has a flat phase except for $\pi$ jumps - changes of sign - at azimuthal nodes. These are the phase changes shown in your diagram.
A: If you are referring to a global phase factor $e^{i\phi_0}$ (with $\phi_0 \in \mathbb{R}$ a real number, NOT the azimuthal angle $\phi$ upon which the hydrogenic wavefunction depends) multiplying the whole wavefunction, then that has probably been simply taken to be 1, as this choice does not affect any of the physics. This is because the probability density of finding a particle in some location is given the square modulus of the wavefunction, and with such an operation the phase factor vanishes.
You could equally well take the phase factor to be $−1$, and then red and blue in the picture would be exchanged. Indeed, you should keep in mind that the only thing that matters is the difference in sign between the two regions, not that one is "positive" and the other "negative".
