# 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

• if you are referring to the global phase factor $e^{i\phi}$ multiplying the whole wavefunction, that has probably simply been taken to be 1, as it does not affect any of the physics (remember that both the "blue" and the "red" areas are equally multiplied by it, so no problems on that regard) – glS Jan 24 '15 at 21:12
• So are you saying that for the sake of these diagrams, you can basically take the exp(+nφ) term (that multiplies the whole equation) to be 1, then compute the wavefunction and if it is negative color it red, and if positive, color it blue? – Lars Jan 24 '15 at 21:19
• No, this is incorrect. See my answer for details. – Emilio Pisanty Jan 24 '15 at 21:26

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.

• Thanks for your answer. Could you explain what $\psi_e$ and $\psi_o$ are? Specifically, what do the subscripts refer to? – Lars Jan 24 '15 at 21:43
• They refer to 'even' and 'odd'. – Emilio Pisanty Jan 24 '15 at 21:44
• So does this basically boil down to re-writing the eigenfunction in the "cartesian" form, because then it will have no imaginary parts to it? – Lars Jan 24 '15 at 23:06
• Yes, that is about it. – Emilio Pisanty Jan 24 '15 at 23:15

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".

• This is mostly misleading, I'm afraid. The picture shown has amplitude changes as a function of $\phi$, which means that the $\phi$ dependence goes beyond a pure phase factor $e^{i\phi}$. Instead, it depicts an (equivalent) real-valued wavefunction with $\phi$ dependence proportional to $\sin(\phi)$ (or $\cos(\phi)$). – Emilio Pisanty Jan 24 '15 at 22:39
• @EmilioPisanty I guess it all depends on what we mean by $\phi$. I meant a global constant phase factor, which can always be added to a wave function. I actually realized after having answered that the OP may be referring to the $\phi$ usually appearing in the $e^{im\phi}$ factor, with $m$ labeling eigenstates of $L_z$. This is of course another matter, as the $\phi$ in this factor is the azimuthal angle upon which the wave-function depends, and which thus makes a lot of difference – glS Jan 24 '15 at 22:48
• The OP makes a clear reference to this being the $|3,1,1\rangle$ hydrogenic state, so the $\phi$ is quite clearly the azimuthal angle. – Emilio Pisanty Jan 24 '15 at 22:53
• @EmilioPisanty yes I see that now. I edited the question to make the difference clear. I think there may be some (maybe small) value in pointing out that the wave function is defined up a global phase factor anyway. – glS Jan 24 '15 at 23:00