# Why do we assume that all electrons in an atom are in $m_s$ eigenstates, and that each one is in a simultaneous $m_l$ and $m_s$ eigenstate?

I have already read the following discussions and would like to point out that my question goes in a different direction:

If superposition is possible in QM, why do we often assume systems are already in their eigenstates?

Why is it often assumed that particles are found in energy eigenstates?

According to my current understanding, the labels that we assign to electrons in a multi-electron atom follow a hierarchy. The hierarchy is based on how precise and informative we want to make the description of the electrons. The simplest one is the electronic configuration, wherein we 'fill' the known eigenstates of a Hydrogen-like system with electrons, using a given set of rules (Pauli's exclusion principle, Hund's rule etc.) without considering any interactions among them. The next label in the hierarchy is a 'term', which is defined by considering spin-spin as well as orbit-orbit coupling resulting in the total spin and angular momentum quantum numbers $$\textit{S}$$ and $$\textit{L}$$ the system. This is followed by a 'level', which additionally specifies the quantum number $$\textit{J}$$, and finally the 'state', which specifies an $$\textit{m}_J$$ along with the first three.

My question is about the step in which we write electronic configurations. We start by making an energy level diagram of known eigenstates of hydrogen-like atoms and start to fill them up. We denote electrons as half-arrows whose direction represents their $$m_s$$ state. My first question lies here. The definition of $$m_s$$ relies on a well defined $$z$$-axis and hence a coordinate system. I understand that we can always choose the coordinate frame so as to have the first electron in an $$m_s$$ eigenstate. However, why do we assume that all the subsequent electrons would also be in $$m_s$$ eigenstates in the same coordinate frame?

My second question is about the act of 'placing' a half-arrow into one of the sketched out $$m_l$$ states, implying that they always have simultaneous eigenstates. Is there something fundamental that's preventing the electron from being in the superposition of $$m_l$$ states while being in an $$m_s$$ state and vice versa?

My first guess was that the spin of the first electron influences that of the others' to align with it. Similarly, that the electron 'prefers' to be in simultaneous $$m_l$$ and $$m_s$$ states because of energy considerations like the one discussed in L-S coupling. However, according to my current understanding, we only start incorporating those forms of interactions $$\textit{after}$$ having written the electronic configuration first, which is more of a symbolic (not representative of a physical reality) first step to get to the actual physical state (one specified by $$\textit{L}$$, $$\textit{S}$$, $$\textit{J}$$, $$\textit{m}_J$$). Is it that the electronic configuration is also a physical description and that coupling interactions are guiding it from the very beginning? Or is it also one of those cases where superposition is a genuine possibility that we neglect to consider the simplest case?

• There is simply no good reason. It is merely convenient within the Hartree-Fock scheme to do this. In reality, as long as you have more than one single electron, the actually occupied states are highly complicated molecular orbitals, not at all what Hartree-Fock simplifies them to be. It just so happens that Hartree-Fock gets close enough, and is predictive enough, that we can make do with arguing about its fictitious results as if they are gospel. Commented Apr 23 at 16:49
• Yeah, following on from the previous comment: This process of filling shells, making terms, and then making multiplets has two main purposes: determining degeneracy (i.e., developing a basis in terms of which we might calculate the real ground state) and determining total quantum numbers like $L$, $S$, and $J$ (in the case of $L$-$S$ coupling) for the ground state (and possibly, for excited states). The "putting arrows in states" is primarily book-keeping device, really, although it has vague physical interpretations if you're careful enough (see previous comment). Commented Apr 23 at 16:54
• Thank you for your inputs. @naturallyInconsistent, you mentioned that these results can be considered fictitious, which I interpret as 'not physical'. So whenever we talk about getting spectroscopic data out of atoms and molecules, does that mean that there is no rigorous theory / simulation as of now that predicts the experimental results precisely? As in, scientifically, are we in a place where we only make qualitative predictions and not quantitative ones about these spectra? Commented Apr 25 at 9:16
• That is the complete opposite. My last statement already said that Hartree-Fock is so good, that there are many of fields of science whereby we can pretend that the HF fictitious solutions are the actual real-world results. We also have rigorous post-Hartree-Fock simulations if you really need to access them. HF is just good enough that we can reliably use arguments based upon it to predict actual results, even if, ultimately, they are fictions. Commented Apr 25 at 9:28
• Okay, I understand. Thank you for the clarification. Commented Apr 25 at 10:01

OK, this question is too long to answer completely. Nevertheless, there are some factoids that may help:

1. The Slater determinant (https://en.wikipedia.org/wiki/Slater_determinant). No electron is in an eigenstate of $$|n, l, m\rangle$$ in a multi-electron atom. We generally ignore this in talking about atoms because it would otherwise be too cumbersome.

Consider a ground state lithium atom (ignoring nuclear-spin). The 3 lowest states are $$a = |1,0,+\rangle, b=|1,0,-\rangle, c=|2,0,m\rangle$$, where the $$m$$ is degenerates, so $$\pm \frac 1 2$$.

The 3 particle state is then:

$$\psi_{Li} = \frac 1 {\sqrt 6}\Big( |abc\rangle + |cab\rangle + |bca\rangle - |cba\rangle - |bac\rangle - |acb\rangle \Big)$$

So we use the language that in He, the two electrons are opposite spin in the $$1s$$-shell because of the Pauli Exclusion Principle, and then say a 3rd electron needs to go into the next state: $$2s$$. But what we really mean is, because of the Spin-Statistics Theorem, $$N$$-electrons need to be in at least $$N$$ states in order construct a wave function that is antisymmetric under the interchange of any pair of particle labels.

1. The $$z$$-axis for spin up/down. This is a good question and a common point of confusion, since: atoms don't care about your choice of coordinates. For this, just forget the atom and look at a two particle spin $$\frac 1 2$$ state

$$|s=0,s_z=0\rangle = \frac 1 {\sqrt 2} \Big( |\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle \Big)$$ with some random $$z$$-axis. There are several ways to approach it, but if you rotate that state, say to the $$x$$-axis, you get: $$|s=0,s_x=0\rangle = \frac 1 {\sqrt 2} \Big( |\rightarrow\leftarrow\rangle - |\leftarrow\rightarrow\rangle \Big)$$

so nothing has changed. This works for any orientation because for one electron:

$$|s=\frac 1 2, m_{\theta, \phi}=+\frac 1 2\rangle = \cos(\theta/2)e^{+i\sin(\phi/2)}|\uparrow\rangle + \sin(\theta/2)e^{-i\sin(\phi/2)}|\downarrow\rangle$$

is a pure spin-up state along the axis defined by the angles.

And: the $$|s=0, s_3=0\rangle$$ two particle state is spherically symmetric, it does not care about our choice of $$z$$-axis.

1. On the various $$m$$'s at fixed $$l$$: This is just a property of spherical harmonics, which are the $$(2l+1)$$ irreducible representations of $$SO(3)$$ in cartesian space. What that means is that if you rotate a state $$|l, m\rangle$$ onto new axes, the result is a sum:

$$Y_{l,m'}(\theta', \phi') = \sum_{m=-l}^{+l} c_{m'm}Y_{l, m}(\theta, \phi)$$ And that means: your choice of a quantization axis doesn't matter, because all the states at fixed $$l$$ are degenerate. (I just want add something that made it click for me: degenerate means the time evolution is just an increasing phase:

$$\phi(t) = \frac{E_{nlm}}{\hbar}t$$

and since $$E_{nlm} = E_{nlm'}$$, the axis on which you choose to define $$m$$ or $$m'$$ or $$m''$$ truly does not matter because it does not change phase differences, just an unmeasurable global phase).

Of course, an external field, or one from nuclear spin, lifts that degeneracy.

Wigner D-matrices treat this in matrix form, where $$Y_{l, m}$$ are rotation eigenstates about the axis defining $$\phi$$:

$$R_z(\phi)Y_{l,m}(\theta, \phi) = e^{im\phi}Y_{l,m}(\theta, \phi)$$

and the $$c_{m'm}$$ are the matrix elements for theta rotations, and the general rotations are $$(2l+1) \times (2l+1)$$ block-diagonal for polar angle ($$\theta$$) rotations.

Also note that:

$$\sum_{m=-l}^{+l} |Y_{l, m}(\theta, \phi)|^2 = \frac 1 {4\pi}$$

so a filled shell is spherically symmetric, and changing the quantization axis is the same as multiplying "$$\times 1$$", which is the same a doing nothing: coordinates don't matter.

The above was all for the Coulomb hamiltonian, which provides approximate states to work with. Once you include fine-structure and beyond, it gets more complicated. The approximation works well because the "fine-structure" constant is small.

Note that in the nuclear shell model, this not true, spin-orbit is large, inter particle interactions are large, and it's a whole mess.

• It is necessarily worse than Slater determinant because we can prove that Slater determinant has no electron correlation, whereas the true wavefunction must have it. We are extremely lucky that electron correlation effects are tiny. Commented Apr 24 at 0:45
• Thank you for the detailed answer. I have a couple of follow up questions. In 2. you talk about the invariance of a singlet state under rotations, and this would explain my doubt if we assumed that all paired electrons always exist as singlets. Is this an assumption that we actively make when we write the electronic configuration? It comes with this connotation that spin-spin couplings are taking place only within one orbital. When we write the total spin S, we consider spin couplings not just within an orbital but also among different orbitals in a subshell (the outermost one). Commented Apr 25 at 9:31
• I can intuitively understand why we would do that, but is there a more rigorous explanation for when to consider couplings and when not to? Also, it seems that the reasons for spherical symmetry of orbitals and spins are fundamentally different. For spins, we arrive at symmetry after considering coupling for a two spin system, and for orbitals, it comes directly from the solution to the SE. It does make sense, of course, but I felt it would be interesting to make a note of that. Commented Apr 25 at 9:51
• @UVcatastrophe to get at the spherical symmetry, you need to get into the representation of Lie groups: SO(3) for orbitals, and SU(2) for spin. For spin, then tensor product of two 2D irreps says: 2 x 2 = 3 + 1 (triplet and singlet), both are closed under rotations, and from a singlet that means: invariant. For SO(3), you have 2l+1 dimensional irreps (scalar, vector, tensor, etc...) that are all closed under rotations. So, e.g, if you have 5 L=2 states with equal weight and different "m" and do a rotation, you still have 5 states with m's on a new axis. All that changes is their phases.
– JEB
Commented Apr 25 at 19:09
• @JEB Thank you for the explanation. Commented Apr 28 at 19:36