# Boundary conditions on azimuthal component of Hydrogen atom eigenstates (Schrödinger)

I have a question relating to quantum mechanics that keeps coming back in one form or another, but it can be summed up most concisely in the context of the Hydrogen eigenstates.

When solving S.E. for hydrogen and separating variables, one arrives at a 2nd-order diff. eq. in the azimuthal component, where $$f'' = -m^2f$$. This has the general solution $$Ae^{im\phi} + Be^{-im\phi}$$ . $$m$$ has to be real and integer in order for the function to be continuous and smooth at the meridian, otherwise the exponentials would spiral and/or have a phase discontinuity at the meridian. $$A$$ and $$B$$ are complex in general.

Every textbook and online derivation I have seen so far, simply says the solution is $$Ae^{im\phi}$$, and then sets $$A$$ to real (argument from global $$U(1)$$ symmetry and/or time translational symmetry in light of the time-dependent oscillating term). And then the magnitude of $$A$$ is found to be $$\frac{1}{\sqrt{2\pi}}$$ in order to normalize the function's contribution to the $$|\psi|^2$$ integral. So I get all that.

What I don't get is how they got rid of the $$B$$ term, i.e. the term which depends on $$e^{-im\phi}$$. I would love if someone can tell me why that term must be zero.

Some thoughts:

• $$A$$ and $$B$$ are in general complex, so together they have four degrees of freedom ($$a+bi$$, $$c+di$$)
• Normalization provides a constraint which reduces this to three degrees of freedom
• Global $$U(1)$$ symmetry lets us fix the phase and (I think) reduces this to two degrees of freedom
• I don't know how to remove those other two degrees of freedom. Boundary conditions, symmetry arguments?
• $$m$$ is in $$\{...,-3,-2,-1,0,1,2,3,...\}$$ so that probably has some implications for the $$e^{im\phi}$$ terms vs. $$e^{-im\phi}$$, right?

Anyway, I hope this question makes sense, and I would be very grateful for your help.

Your last bullet point captures the spirit of it. Before imposing B.C.'s, the solutions for a Helmholtz differential equation are:

$$$$f(\varphi) = A_{\nu} \, e^{i\nu \varphi} + B_{-\nu} \, e^{-i\nu \varphi}$$$$

where $$\nu \in \mathbb{R}^{+}_{0}$$ as to not double-count certain contributions. You can expand the field of definition of $$\nu$$ to the entirety of the real number line $$\mathbb{R}$$ and define new constants $$\tilde{A}_{\nu}$$ such that there is agreement with the first equation. As a result you have:

$$$$f(\varphi) = \tilde{A} _{\nu} e^{i\nu \varphi}$$$$

Upon imposing periodicity for revolutions in the azimuthal direction, $$\nu$$ is quantized to $$m \in \mathbb{Z}$$, like you state in your question.

Now it might seem you have lost some degrees of freedom due to having just 1 complex constant, but this is simply illusory. For the complete time-dependent wave function solution, you'll have an infinite sum with respect to $$m$$ based on your field of definition for it. Since the new field of definition of $$m$$ is "twice as big" (not quite in a strict mathematical sense, but intuitive in a more restricted numerical sense), the sum has twice as many terms. By corresponding each term in the two sums one-to-one, there will be a set of $$\tilde{A}_{\pm m}$$ constants for each $$A_{m}$$ and $$B_{-m}$$.

From a physical standpoint, this choice makes sense even for a single time-independent state with a specific set of quantum numbers. That is because you expect such wave functions to also be eigenstates of the angular momentum operator $$\hat{L}_{\hat{n}} \rightarrow -i\hbar \frac{\partial}{\partial \varphi}$$, where $$\hat{n}$$ is the unit vector in the direction around which the rotations happens (usually just the $$z$$ axis). Had you kept the original solution, a single state would not be an eigenstate of this operator.

• Wonderful, thanks so much for taking the time to respond, and that makes a lot of sense. One thing I'm confused about though is why A_nu and B_-nu are real numbers. Can't they be complex, for the general solution, before B.C.s are imposed? Edit: sorry, I misread, nu is real and it has to be because otherwise the thing would spiral. Got it. But I'm confused as to how you get from the first equation to the second. Commented Apr 10, 2023 at 23:51
• The second equation is not equivalent to the first one. It is just a different way to write the solutions by making different assumptions about $\nu$. However, the constant coefficients have to be consistent, because the two expressions have to agree once you sum over all possible $m$ (after imposing B.C.'s). Commented Apr 11, 2023 at 6:52
• Hmm. I see what you're saying, but if we are left with only one complex constant, and just the exp(im phi) term, then the probability density would be invariant with respect to rotations in phi, wouldn't it? Because psi would be a real radial function times a real polar function times a complex number with magnitude 1. But the probability density should actually vary in phi, except for the m = 0 case. So shouldn't the function contain both an exp(im phi) term as well as an exp(-im phi) term? Commented Apr 11, 2023 at 16:37
• Why do you expect probability density to be $\varphi$-dependent? Even with the alternative choice, integrating the probability density over $\varphi$ will "kill" all cross terms that are $\varphi$-dependent, leaving only the independent ones. The actual states remain $\varphi$-dependent in either case, so azimuthal observables (e.g. angular momentum) are still consistent. Commented Apr 11, 2023 at 18:30
• There should be some preferred directions in phi where the probability density is higher, right? In the hydrogen orbital plots, all but the ones with m = 0 have some lobes which stick out in various directions as you go around phi. For that kind of structure to arise, there would have to be something more than just an exp(im phi) term, since that just swings around in the complex plane but does not change in length. But, something like a sum of real sines and cosines (or linear combination of +im phi and -im phi exponentials) could produce those lobes in the probability density. Commented Apr 11, 2023 at 20:05

The reason is that only the eigenstates of the time-independent Schrödinger equation of the hydrogen atom are sought. As part of the determination of these eigenstates by separation of variables, only a complete set of eigenfunctions of the Sturm-Liouville eigenvalue problem represented by the azimuthal homogeneous second order ordinary differential equation with periodic boundary condition are necessary, not the general solution. A simple complete set is given by $$e^{im\phi}$$ with $$m$$ is in $$\{...,-3,-2,-1,0,1,2,3,...\}$$ and the appropriate normalization constant.

• Thanks for your response. I am confused though because if the azimuthal function only contains an exp(im phi) term, then the probability density would be invariant under rotations in phi, assuming the radial and polar components are both real. In order to give the probability density its lobes around phi for the m != 0 cases, don't we need a linear combination of exp(im phi) and exp(-im phi) terms? I.e. the magnitude of the azimuthal component should vary around phi. Commented Apr 11, 2023 at 16:43
• @Richard B. The square of the magnitude of the azimuthal component and also of the total wave functions |ψ|2 have always azimuthal symmetry. There are no "lobes around phi". Commented Apr 11, 2023 at 20:38
• Interesting. I might be completely wrong then. But when you see pictures of the probability density for the hydrogen orbitals, there is some phi dependence which I thought came from the spherical harmonics. Example: google.com/… Commented Apr 11, 2023 at 21:10
• @Richard.B The "lobes" you see in the probability density orbitals are only lobes in theta, not in phi. Please look at the formula for the hydrogen wave function, where the angular dependence on theta and phi are in the spherical harmonics. The only phi dependence of the spherical harmonics is in the factor exp(i m phi) whose magnitude squared is always 1. Thus the probability density orbitals (in contrast to the real and imaginary parts of psi) are always independent of phi, i.e., invariant for rotations around the z-axis. Commented Apr 12, 2023 at 7:38
• @Richard.B You'll find a very instructive interactive 3d visualization of the probability density orbitals with the Wolfram Player here: demonstrations.wolfram.com/HydrogenOrbitals Commented Apr 12, 2023 at 7:38