# Are all atoms spherically symmetric? If so, why are atoms with half-filled/filled sub-shells often quoted as 'especially' spherically symmetric?

In my atomic physics notes they say

In general, filled sub-shells are spherically symmetric and set up, to a good approximation, a central field.

However sources such as here say that, even for the case of a single electron in Hydrogen excited to the 2p sub-shell, the electron is really in a spherically symmetric superposition $$\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$$ (which I thought made sense since there should be no preferred direction in space).

My question there now is, why is the central field approximation only an approximation if all atoms are really perfectly spherically symmetric, and why are filled/half-filled sub-shells 'especially spherically symmetric?

• It's worth noting that the reference here is a letter to the editor of a chemistry education journal. I think these typically are not peer-reviewed in the same way a journal article is, and so it is not particularly surprising IMO if the author is somewhat sloppy with some quantum mechanics notation, especially when this seems beside the point of the letter. – aquirdturtle Jan 26 at 19:37

In general, atoms need not be spherically symmetric.

• The source you've given is flat-out wrong. The wavefunction it mentions, $$\varphi=\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$$, is in no way spherically symmetric. This is easy to check: the wavefunction for the $$2p_z$$ orbital is $$\psi_{2p_z}(\mathbf r)=\frac {1}{\sqrt {32\pi a_0^5}}\:z \:e^{-r/2a_{0}}$$ (and similarly for $$2p_x$$ and $$2p_y$$), so the wavefunction of the combination is $$\varphi(\mathbf r)=\frac {1}{\sqrt {32\pi a_0^5}}\:\frac{x+y+z}{\sqrt 3} \:e^{-r/2a_{0}},$$ i.e., a $$2p$$ orbital oriented along the $$(\hat{x}+\hat y+\hat z)/\sqrt3$$ axis.

This is an elementary fact and it can be verified at the level of an undergraduate text in quantum mechanics (and it was also obviously wrong in the 1960s). It is extremely alarming to see it published in an otherwise-reputable journal.

• On the other hand, there are some states of the hydrogen atom in the $$2p$$ shell which are spherically symmetric, if you allow for mixed states, i.e., a classical probabilistic mixture $$\rho$$ of hydrogen atoms prepared in the $$2p_x$$, $$2p_y$$ and $$2p_z$$ states with equal probabilities. It is important to emphasize that it is essential that the mixture be incoherent (i.e. classical and probabilistic, as opposed to a quantum superposition) for the state to be spherically symmetric.

• As a general rule, if all you know is that you have "hydrogen in the $$2p$$ shell", then you do not have sufficient information to know whether it is in a spherically-symmetric or an anisotropic state. If that's all the information available, the initial presumption is to take a mixed state, but the next step is to look at how the state was prepared:

• The $$2p$$ shell can be prepared through isotropic processes, such as by excitation through collisions with a non-directional beam of electrons of the correct kinetic energy. In this case, the atom will be in a spherically-symmetric mixed state.
• On the other hand, it can also be prepared via anisotropic processes, such as photo-excitation with polarized light. In that case, the atom will be in an anisotropic state, and the direction of this anisotropy will be dictated by the process that produced it.

It is extremely tempting to think (as discussed previously e.g. here, here and here, and links therein) that the spherical symmetry of the dynamics (of the nucleus-electron interactions) must imply spherical symmetry of the solutions, but this is obviously wrong $$-$$ to start with, it would apply equally well to the classical problem! The spherical symmetry implies that, for any anisotropic solution, there exist other, equivalent solutions with complementary anisotropies, but that's it.

• The hydrogen case is a bit special because the $$2p$$ shell is an excited state, and the ground state is symmetric. So, in that regard, it is valid to ask: what about the ground states of, say, atomic boron? If all you know is that you have atomic boron in gas phase in its ground state, then indeed you expect a spherically-symmetric mixed state, but this can still be polarized to align all the atoms into the same orientation.

As a short quip: atoms can have nontrivial shapes, but the fact that we don't know which way those shapes are oriented does not make them spherically symmetric.

• So, given an atom (perhaps in a fixed excited state), what determines its shape? In short: its term symbol, which tells us its angular momentum characteristics, or, in other words, how it interacts with rotations.

• The only states with spherical symmetry are those with vanishing total angular momentum, $$J=0$$. If this is not the case, then there will be two or more states that are physically distinct and which can be related to each other by a rotation.
• It's important to note that this anisotropy could be in the spin state, such as with the $$1s$$ ground state of hydrogen. If you want to distinguish the states with isotropic vs anisotropic charge distributions, then you need to look at the total orbital angular momentum, $$L$$. The charge distribution will be spherically symmetric if and only if $$L=0$$.

A good comprehensive source for term symbols of excited states is the Levels section of the NIST ASD.

• @AlexGower It is definitely possible, and an everyday reality in a host of different experimental contexts. – Emilio Pisanty Jan 25 at 12:44
• (though note that to make a $2p_z$ state from $1s$ hydrogen, you need to use linearly polarized light along the $z$ axis.) – Emilio Pisanty Jan 25 at 12:44
• It depends on exactly what you mean by CFA. It looks like a rather technical term, and I haven't encountered it professionally. As defined by Wikipedia, the CFA describes the potential felt by each electron as produced by all the other electrons. Absent electron correlation (i.e. at Hartree-Fock level), this is accurate for the outermost $s$ electron, but not for the inner shells: the combined six electrons in the $2p$ shell produce a symmetric charge distribution, but each electron sees this symmetric distribution minus a shaped hole (corresponding to its own state). – Emilio Pisanty Jan 25 at 13:00
• I agree with the body of your text, but dislike the bold text at the top "atoms need not be spherically symmetric". Of course an atom can be prepared into a state where the electron orbitals are breaking the spherical symmetry. But that is like saying that this circular drum is not circularly symmetric because of its vibration mode. en.wikipedia.org/wiki/Vibration#/media/… Yes the vibration is not symmetric, but somehow the drum itself is, it could as easily vibrate any other way. I would prefer "atoms need not be in spherically symmetric states". – Dast Jan 25 at 21:13
• @Ilmari The corollary is mostly correct - those states do exist, including carbon. But the charge distribution is not anisotropic (the wording of that bullet point needs to be fixed). That state of carbon consists of P states in three different shapes, anticorrelated with three different spin states, such that the global charge distribution is isotropic. I hope this makes sense. Good catch! – Emilio Pisanty Jan 26 at 9:56

The reference paper is actually not really wrong, just poorly communicated for this physics audience. If you read the paper this letter is responding to, it becomes clear that the authors are referring to the probability distribution of an incoherent sum of the referenced orbitals. The letter to the editor sloppily writes (emphasis mine)

We therefore must describe the electron as (to use chemically familiar language) a “resonance hybrid” of $$2p_x, 2p_y,$$ and $$2p_z$$. In more detail, we write if $$\psi = \frac{1}{\sqrt{3}}(2p_x + 2p_y + 2p_z)$$ and this is exactly a spherical distribution, as Johnson and Rettew have shown.

Where, out of context, every physicist assumes that the author specifically means a coherent sum of wavefunctions, and where both the wavefunction and the probability distribution resulting from it are not spherically symmetric. However, I believe (I'm no chemist) to a chemistry audience the common phrase "resonance hybrid" would immediately imply an incoherent superposition of the given states, as there's nothing particularly coherent about normal chemistry. The word "distribution" also hints that something is funny, as it's not typical to call the wavefunction itself a "distribution". Specifically, Johnson and Rettew showed that $$\psi_{2p_x}^2 + \psi_{2p_y}^2 + \psi_{2p_z}^2$$ is spherically symmetrical, which it is. Since there is basically only one equation in the referenced article, this is clearly what Cohen was referring to. The phrasing of the letter to the editor could clearly should have been clearer here, but good communication does take effort from both sides, especially when the two sides come from different fields where notation is not so well standardized or understood.

For completeness, if a shell is partially filled then there's the possibility of there being a specific angle between the orbits of electrons in the inner shell and the electrons in the outermost shell (think, e.g. of two concentric donuts rotating independently). Even in a mixed state, the outermost electron would be in a mixed state of interacting with various orientations of inner electrons, none of which are independently centric, which suggests that assuming a central field approximation will miss some important physics.

• Thank you for putting some thought into correctly interpreting the paper and digging into the Johnson and Rettew reference. – d_b Jan 26 at 5:48
• Thanks for this. Alkali's are often used as the prime example of where to use the central field approximation, and my lecture notes say the central field approximation is 'excellent' in this case. What factors make it not perfect for alkalis? Is it the fact that eventhough the valence electron sees a perfectly spherically symmetric set of inner electrons, the inner electrons themselves do not as they do not interact with themselves and thus see a 'hole' in the distribution? – Alex Gower Jan 26 at 11:33
• @AlexGower Hmm, I'm not honestly super confident about the details at this level myself tbh, but I can think of a couple things. Even for a single pair of electrons, one of which is in a perfectly spherical distribution, the interaction in the hamiltonian is still $~e^2/|(\vec{r}_1-\vec{r}_2)|$ which is clearly not central. So the question boils down to when is $e^2/|(\vec{r}_1-\vec{r}_2)|\sim a/r_1$ for some effective interaction strength $a$. So you see, just interacting with a spherically symmetric distribution is not enough. – aquirdturtle Jan 26 at 19:26
• Perhaps part of the confusion is in specifying the difference between it being a central interaction vs. being a spherically symmetric interaction. I think interacting with a filled shell should be to a very good approximation very spherically symmetric (although as pointed out there might be some subtleties with the phases of the inner electrons, not sure). – aquirdturtle Jan 26 at 19:31
• Determining when it is effectively central should boil down to when $r_2<<r_1$ or if r_2 fluctuates around zero fast enough that the dynamics are well-approximated by the time-average of $r_2$, or if averaging multiple atoms in different orbits makes it approximately central. I think it's not obvious when these various approximations should work. Perhaps there's a good intuitive picture for them, but I wouldn't be surprised if the most honest answer is just "it turns out given the numbers that they work pretty well". But all this is a bit speculative on my part, so take it with some salt. – aquirdturtle Jan 26 at 19:32

No coherent superposition of 2p orbitals is spherically symmetric. Your example $$\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$$ is a 2p orbital pointing in the 111 direction and is not spherical. The proper description is by a diagonal density matrix, which states that the atom is in an incoherent superposition of the three states.

• Ah okay, my link was a letter from the 1960s so was probably wrong about that but yes that does make sense from usual density matrix logic. Do you have any thoughts about what makes the central field approximation only an approximation? – Alex Gower Jan 25 at 12:09
• After this I would take Chem Ed letters with a grain of salt :-) – my2cts Jan 25 at 12:15

Your quote references "filled sub-shells". You then write in your question "all atoms are really perfectly spherically symmetric".

Not all atoms have only filled sub-shells. Noble gasses are strong examples of atoms with only filled sub-shells and the small ones behave approximately spherically symmetrically.

Most atoms do not have all of their sub-shells filled. It is rather common for the "last" one or two sub-shells to be partially filled. Look at the electron configuration of the transition metals for many, many examples. As one such example, Chromium's configuration ends with $$\mathrm{[Ar] 3d^5 4s^1}$$, so neither of the last two sub-shells is filled and we should not expect (and do not find) that Chromium is spherically symmetric.