# Proof that $L=S=0$ for filled electron subshells?

It is commonly stated that if you have a filled subshell, such as $$p^6$$ or $$d^{10}$$ that one must have $$L=S=0$$ implying $$J=0$$ and $$M_J=0$$ so that the atom is spherically symmetric.

Why is it clear that $$L=S=0$$?

It is clear to me that $$M_L=M_S=0$$. This is because for every electron with $$m_l$$ or $$m_s$$ there is another electron with $$-m_l$$ and $$-m_s$$ So clearly $$M_L=\sum_i m_{l_i} =0$$ and likewise for $$S$$.

But generally speaking, if you add up multiple spins and find $$M_L=M_S=0$$ this is NOT sufficient condition to have $$L=S=0$$.

This is clear with the two spin-1/2 triplet state which is well known to have total spin $$S=1$$:

$$|\uparrow \downarrow \rangle + |\downarrow \uparrow\rangle$$

I desire a proof that for filled shells we get the above property. I know that the answer is related to the requirement of antisymmetrization for the electron wavefunctions. For example, the triplet wavefunction above is not anti-symmetric. If I took the anti-symmetric combination I'd have the singlet which DOES satisfy $$S=0$$ as needed. However, I'd like a proof of how the requirement for anti-symmetrization leads to $$L=S=0$$ for any value of $$L$$ and $$2(2L+1)$$ electrons.

There are a number of similar questions on this site which I can link if desired. None of those provide satisfying proofs of the kind I'm desiring. Sometimes they point out that $$M_L=M_S=0$$ and leave it at that. Sometimes they point this out and wave their hands at anti-symmetrization and call it done. I'd like something more convincing.

The reason this whole situations concerns me is it seems like certain angular momentum states are inaccessible for fermions and bosons in a way that seems stronger to me than what is implied by Pauli-exclusion. Though perhaps I underestimate Pauli-exclusion. I guess this says that if I have, for example, 6 $$p$$ electrons then there is only ONE state they can occupy. But the general theory of angular momentum addition says that these 6 spin-$$1/2$$ particles with spin-1 orbital angular momentum should have like 36 angular momentum states they should occupy from $$J=0$$ to $$J=9$$.

Clearly I'm not thinking about this correct. I'd appreciate seeing the proof I'm looking for and having my intuition on Pauli anti-symmetrization clarified.

edit: the answer to this question Dimension of Hilbert space of spin $1/2$ identical particles? addresses my intuition about Pauli-exclusion. The short answer is that yes, a lot of angular momentum states that would be allowed for distinguishable particles are simply deleted when considering Fermions. I still seek a convincing proof for arbitrary $$L$$, that, after applying anti-symmetrization, the only state which remains in the $$S=L=J=0$$ state.

• When $m_l=m_s=0$, how does the atom pick the $z$-axis?
– JEB
Commented Apr 23, 2021 at 13:52
• @JEB for general angular momentum states m=0 does not imply the state doesn’t have an orientation. Look at spherical harmonics $Y_2^0(\theta, \phi)$. Somehow the additional constraint of anti-symmetrization makes it so. This is the fact for which I would like a proof. Commented Apr 23, 2021 at 15:49
• $m=0$ doesn't have an orientation (a vector direction), but it does have a tensor alignment with respect to $\pm \hat z$, so if $m=0$...how would the atom know which axis to pick?
– JEB
Commented Apr 23, 2021 at 17:21
• @JEB Perhaps it got to the state it’s in due to an interaction with a polarized electric field. Commented Apr 23, 2021 at 17:27
• An background ${\vec E}$ breaks the symmetry so that $Y_l^m$ aren't eigenstates anymore, so don't go there. The point I was making is that if $M=0$ for a filled shell, then that is independent of how you choose the axis, and that is only true if $L=0$.
– JEB
Commented Apr 23, 2021 at 21:55

I would like to share an answer, based on the proof (sketch) I found in L. Marchildon: Quantum Mechanics. From Basic Principles to Numerical Methods and Applications, chapter 18 (p. 403f.).

Since we are concerned with filled subshells only, let's first define the total spin and angular orbital operator for our subshell of interest, $$\mathbf L := \sum_{i\, \in \,\text{subshell}} \mathbf L_i , \qquad \mathbf S := \sum_{i\, \in \,\text{subshell}} \mathbf S_i .$$ The index $$i$$ runs over all electrons in that subshell.

With respect to these 'total' operators we define raising and lowering operators as well, $$\mathbf L^- := \mathbf L_x -i \mathbf L_y \, \qquad \mathbf S^- := \mathbf S_x -i \mathbf S_y .$$ The raising operators are defined as the hermitian conjugate of the lowering operators, i.e. $$(\mathbf L^-)^\dagger = \mathbf L^+ = \mathbf L_x +i \mathbf L_y$$ and similarly, $$(\mathbf S^-)^\dagger = \mathbf S^+ = \mathbf S_x + i \mathbf S_y$$.

The effect of the lowering operator $$\mathbf L^-$$ on a quantum state $$\Psi$$ with quantum numbers $$M_L$$ and $$L$$ is to lower $$M_L$$ by 1, but keep $$L$$ fixed as long as $$M_L > -L$$. However, for minimal $$M_L$$ ($$M_L = -L$$) the lowering operator destroys the state $$\Psi$$. Analogously, we can describe the effect of the raising operator $$\mathbf L^+$$: If $$M_L < L$$, it acts on $$\Psi$$ by raising $$M_L$$ by 1 and keeping $$L$$ fixed. If $$M_L$$ is maximal, i.e. $$M_L = L$$, it destroys the state $$\Psi$$. (The same holds for the spin lowering and raising operators regarding the quantum numbers $$M_S$$ and $$S$$.)
So far, this is nothing new.

Remark

Let me just remark here that, in general, single electron raising and lowering operators $$\mathbf L^\pm_i$$ do not commute with particle permutations, but total raising and lowering operators (e.g. for a subshell) do. This means, after acting with $$\mathbf L^\pm = \sum_i \mathbf L^\pm_i$$ on a fully (anti-)symmetrized state it will remain (anti-)symmetric. However, the action of a single operator $$\mathbf L^\pm_i$$ will (in general) destroy any (anti-)symmetry.

For example, consider two electrons (with coordinates $$r_1$$ and $$r_2$$) in the states $$Y_{l=1}^{m=1}\otimes |\uparrow \rangle$$ and $$Y_{l=1}^{m=0}\otimes |\downarrow\rangle$$ (Of course, this is physically very unlikely; it just serves a demonstration purpose here.) A fully anti-symmetrized two-electron state is the Slater-determinant $$\Psi(r_1,r_2) = Y_{l=1}^{m=1}(r_1) Y_{l=1}^{m=0}(r_2) \otimes | \uparrow_1 \downarrow_2 \rangle - Y_{l=1}^{m=0}(r_1) Y_{l=1}^{m=1}(r_2) \otimes |\downarrow_1 \uparrow_2 \rangle .$$ Now, acting with $$\mathbf L^-$$ on $$\Psi(r_1, r_2)$$ we find, $$\mathbf L^- \Psi(r_1, r_2) = (\mathbf L^-_1 + \mathbf L^-_2 ) \, \Psi(r_1, r_2) = \sqrt{2} Y_{l=1}^{m=0}(r_1) Y_{l=1}^{m=0}(r_2) \otimes | \uparrow_1 \downarrow_2 \rangle - \sqrt{2} Y_{l=1}^{m=-1}(r_1) Y_{l=1}^{m=1}(r_2) \otimes |\downarrow_1 \uparrow_2 \rangle + \sqrt{2} Y_{l=1}^{m=1}(r_1) Y_{l=1}^{m=-1}(r_2) \otimes | \uparrow_1 \downarrow_2 \rangle - \sqrt{2} Y_{l=1}^{m=0}(r_1) Y_{l=1}^{m=0}(r_2) \otimes |\downarrow_1 \uparrow_2 \rangle .$$ (The appearance of the factor $$\sqrt{2}$$ can be seen from the identity $$\mathbf L^- \mathbf L^+ = \mathbf L^2 - \mathbf L_z^2 - \mathbf L_z$$.) Clearly, this state is again antisymmetric, although not a Slater determinant.

One can verify that a general anti-symmetric state-- which can be written as a sum of Slater determinants-- remains anti-symmetric after the action of $$\mathbf L^\pm$$. (If you believe this statement for any Slater determinant, then just use that particle permutation is a linear operation.)

Now, consider a state $$\Psi$$ with eigenvalues $$L,M_L, S, M_S$$ with respect to the total subshell operators we defined above. Furthermore, $$\Psi$$ should describe a filled subshell. In the following we will prove that $$M_L = L = 0$$ and $$M_S = S = 0$$ alike.

Let's simplify the situation a bit and start by looking at the angular momentum $$L$$ and $$M_L$$ and consider what happens when we act with $$\mathbf L^-$$ on $$\Psi$$. By def. of $$\mathbf L^-$$, the action of the subshell lowering operator is just the sum of the lowering operators $$\mathbf L^-_i$$ for each electron $$i$$, $$\mathbf L^- = \sum_{i\, \in \,\text{subshell}} \mathbf L^-_i \,.$$ (As we have seen, taking the sum is important as to keep the result anti-symmetrized).

Since $$\Psi$$ is a fully anti-symmetrized state, it may contain several terms/summands. We use an eigenbasis of the $$\mathbf L_i$$ (or rather $$\mathbf L_i^2$$ and $${\mathbf L_i}_z$$) as one-particle basis with respect to which we anti-symmetrize-- see example in the remark above. This means, on each summand the operator $$\mathbf L^-_i$$ acts on the state of electron $$i$$ with some well-defined quantum number $${m_l}_i$$.

• If $${m_l}_i > -l_i$$, then $$\mathbf L^-_i$$ simply lowers $${m_l}_i \rightarrow {m_l}_i -1$$. But because the subshell is fully occupied, another electron already occupies this state. So we have two electrons in the same state. Acting with the total subshell lowering operator $$\mathbf L^-$$ on $$\Psi$$ ensures anti-symmetry of the total final state $$\mathbf L^- \,\Psi$$, as we have seen in the remark above. Antisymmetrizing a state with two electrons occupying the same state yields zero (Pauli exclusion principle).
• If $${m_l}_i = -l_i$$, i.e. $${m_l}_i$$ is minimal, we cannot lower the quantum number $${m_l}_i$$ any further and this term/summand of $$\Psi$$ is destroyed.

In total, we find $$\mathbf L^- \,\Psi = 0$$.

By an analogous reasoning-- using the properties of the full subshell state $$\Psi$$ when acting with $$\mathbf L^+$$-- we can deduce $$\mathbf L^+ \,\Psi = 0$$.

Now comes the trick (you may already guess what comes next): Since the raising and lowering operators both destroy $$\Psi$$, its quantum number $$M_L$$ must be both minimal and maximal simultaneously. This only works if $$L=0$$ which restricts $$M_L$$ to be zero, too.

In the beginning I have also defined a raising and lowering operator $$\mathbf S^{\pm}$$ for the spin. Although I do not give the details here, the proof for $$S=M_S=0$$ works just the same as for the angular momentum that we have done here.

## Another approach

If you believe that empty shells have $$\mathbf L = \mathbf S = \mathbf 0$$, then you can convince yourself that filled shells have the exact same property by a particle-hole transformation. Basically, this amounts to mapping electron creation to hole annihilation operators (and electron annihilation to hole creation operators), while also assigning the corresponding hole the inverse magnetic angular momentum and spin quantum number than the electron, i.e. $$l \rightarrow l, \, m_l \rightarrow -m_l, s\rightarrow s, \, m_s \rightarrow -m_s$$. This assignment ensures that the (matrix) representation of the operators $$\mathbf L$$ and $$\mathbf S$$ are not changed by the transform, such that all properties regarding $$\mathbf L$$ and $$\mathbf S$$ applying to electronic empty shells applies to filled hole shells as well. (The details of this transformation can be found in atomic physics textbooks and online lecture notes as well.)

Of course, this is related to the fact that was elaborated on by JEB: Any object, that is invariant under rotations, i.e. actions under group generated by $$\mathbf L$$, transforms as rank-zero tensor, that is, as a scalar quantity in our example of angular momentum and space rotations. This is indeed independent of the Pauli exclusion principle and the space the group representation acts upon.

The general state with $$m=0$$ is:

$$|L=l,M=0\rangle = Y_l^0(\theta, \phi)= \sqrt{\frac{2l+1}{4\pi}}P_l(\cos\theta)$$

which describes a tensor polarization with a preferred alignment along the axis defining $$\theta$$. If you rotate that to different axes (primed), then it is a linear combination of spherical harmonics with the same degree $$l$$ but different magnetic quantum numbers:

$$Y_l^0(\theta, \phi)= \sum_{m'=-l}^lc_{lm'}Y_l^{m'}(\theta', \phi')$$

which is not a state with $$m'=0$$, unless $$l=0$$.

In the question you state that you know an atom with a full shell has $$M=0$$, but are uncertain about $$L$$. The point is, there is no preferred axis, so if any axis you pick must have $$M=0$$, then $$L$$ must be zero also.

That is unrelated to the Pauli Exclusion Principle (PEP). The PEP results from the spin-statistics theorem, which states that the wave function of identical fermions must be antisymmetric under the interchange of any two particles:

If:

$$\psi_{nm}(x_1, x_2) = \frac 1 {\sqrt 2}[\psi_n(x_1)\psi_m(x_2) - \psi_m(x_1)\psi_n(x_2)]$$

and

$$\psi_{nm}(x_2, x_1) = \frac 1 {\sqrt 2}[\psi_n(x_2)\psi_m(x_1) - \psi_m(x_2)\psi_n(x_1)] = -\psi_{nm}(x_1, x_2)$$

where $$(n,m)$$ label all the quantum numbers defining a state, then if $$n=m$$:

$$\psi_{nn}(x_1, x_2) = 0$$

which is the PEP. Note that it holds for all quantum numbers, not just $$L=S=0$$.

If you're considering six electrons in a single $$P$$-shell, then the wave function is described by a Slater determinant:

$$\psi(1,2,3,4,5,6) = \frac 1 {\sqrt{6!}} \times$$ $$\begin{vmatrix}Y_1^{1}(\theta_1, \phi_1)|\uparrow_1\rangle&Y_1^{1}(\theta_2, \phi_2)|\uparrow_2\rangle&Y_1^{1}(\theta_3, \phi_3)|\uparrow_3\rangle&Y_1^{1}(\theta_4, \phi_4)|\uparrow_4\rangle&Y_1^{1}(\theta_5, \phi_5)|\uparrow_5\rangle&Y_1^{1}(\theta_6, \phi_6)|\uparrow_6\rangle\\Y_1^{0}(\theta_1, \phi_1)|\uparrow_1\rangle&Y_1^{0}(\theta_2, \phi_2)|\uparrow_2\rangle&Y_1^{0}(\theta_3, \phi_3)|\uparrow_3\rangle&Y_1^{0}(\theta_4, \phi_4)|\uparrow_4\rangle&Y_1^{0}(\theta_5, \phi_5)|\uparrow_5\rangle&Y_1^{0}(\theta_6, \phi_6)|\uparrow_6\rangle\\Y_1^{-1}(\theta_1, \phi_1)|\uparrow_1\rangle&Y_1^{-1}(\theta_2, \phi_2)|\uparrow_2\rangle&Y_1^{-1}(\theta_3, \phi_3)|\uparrow_3\rangle&Y_1^{-1}(\theta_4, \phi_4)|\uparrow_4\rangle&Y_1^{-1}(\theta_5, \phi_5)|\uparrow_5\rangle&Y_1^{-1}(\theta_6, \phi_6)|\uparrow_6\rangle\\Y_1^{1}(\theta_1, \phi_1)|\downarrow_1\rangle&Y_1^{1}(\theta_2, \phi_2)|\downarrow_2\rangle&Y_1^{1}(\theta_3, \phi_3)|\downarrow_3\rangle&Y_1^{1}(\theta_4, \phi_4)|\downarrow_4\rangle&Y_1^{1}(\theta_5, \phi_5)|\downarrow_5\rangle&Y_1^{1}(\theta_6, \phi_6)|\downarrow_6\rangle\\Y_1^{0}(\theta_1, \phi_1)|\downarrow_1\rangle&Y_1^{0}(\theta_2, \phi_2)|\downarrow_2\rangle&Y_1^{0}(\theta_3, \phi_3)|\downarrow_3\rangle&Y_1^{0}(\theta_4, \phi_4)|\downarrow_4\rangle&Y_1^{0}(\theta_5, \phi_5)|\downarrow_5\rangle&Y_1^{0}(\theta_6, \phi_6)|\downarrow_6\rangle\\Y_1^{-1}(\theta_1, \phi_1)|\downarrow_1\rangle&Y_1^{-1}(\theta_2, \phi_2)|\downarrow_2\rangle&Y_1^{-1}(\theta_3, \phi_3)|\downarrow_3\rangle&Y_1^{-1}(\theta_4, \phi_4)|\downarrow_4\rangle&Y_1^{-1}(\theta_5, \phi_5)|\downarrow_5\rangle&Y_1^{-1}(\theta_6, \phi_6)|\downarrow_6\rangle\end{vmatrix}$$

From that you can calculate the probability density versus angular coordinate, and it will look like:

$$P(\theta, \phi) \propto |Y_1^1{\theta, \phi}|^2+|Y_1^0{\theta, \phi}|^2+|Y_1^{-1}{\theta, \phi}|^2$$ $$\propto |\sin\theta e^{+i\phi}|^2+|\sqrt 2\cos\theta|^2+|\sin\theta e^{-\phi}|^2 = 2(\sin^2{\theta}+\cos^2{\theta}) =2$$

that is, it is spherically symmetric. Spherical symmetry means $$L=0$$.

This holds for any order:

$$\sum_{m=-l}^l |Y_l^m(\theta, \phi)|^2$$

does not depend on $$\theta$$ nor $$\phi$$. Hence, filled shells are always spherically symmetric with total angular momentum $$L=0$$.

Nevertheless, there is a deep connection between antisymmetry and rotational invariance. For example: the antisymmetric Levi Civitta symbol, $$\epsilon_{ijk}$$ , is an isotropic tensor. This arises through Schur-Weyl duality, which describes the rotationally closed subspaces of tensors via the representations of the permutation group and Young diagrams. The dimensions of the subspaces can be calculated with "the remarkable" Hook Length Formula.

The antisymmetric permutation corresponds to a subspace of dimension 1, which is a scalar (thus, spherically symmetric). The simplest example is included in your question, where you combined two 2D representations (spinors) and get a symmetric triplet and an antisymmetric singlet:

$${\bf 2}\otimes{\bf 2} ={\bf 3}_S \oplus {\bf 1}_A$$

The triplet transforms like a vector, and the singlet as a scalar.

Likewise if you combine 3 vectors (say, 3 P-orbitals), you get: $${\bf 3}\otimes{\bf 3} \otimes{\bf 3}={\bf 10}_S \oplus {\bf 8}_M \oplus {\bf 8}_M\oplus{\bf 1}_A$$

where the symmetric $${\bf 10}$$ is $$L=3$$ and $$L=1$$, the octets are $$L=2$$ and $$L=1$$, and the fully antisymmetric singlet is $$L=0$$. You can verify this by laboriously working through the Clebsch Gordon coefficients by hand.

• I have some takeaways/questions. First, you show explicitly that the 6-electron state filling the $p$-shell is isotropic. You then claim that this holds for any value of $L$. I'm curious to hear a fleshing out of the proof of this latter statement, that is essentially my whole question. You next sections about the relationship between antisymmetry and rotational invariance seems like the key. Espcially that the antisymmetric permutation is dimension one. But clearly not all anti-symmetric states are spherically symmetric. If I just have two electrons in $p$-orbitals... Commented Apr 24, 2021 at 17:42
• They still form a antisymmetric slater determinant state but it is not rotationally symmetric. So it is something like, there are 3 orbital states available and 2 spin states available. So the single particle Hilbert space is 6 dimensional. If you now make 6 copies of that space (6 electrons) but restrict to antisymmetric permutations you're saying it has to be 1 dimensional. Can that be easily proved? Commented Apr 24, 2021 at 17:44
• I guess the proof is something like, since it's antisymmetric you know you have to have one copy of each state and there's only one way to choose 6 unique objects from a group of 6 objects. Commented Apr 24, 2021 at 17:47
• How were you able to assign the labels S, M, and A to the various subspaces at the end of the answer? I also aksed physics.stackexchange.com/questions/631984/… which may more directly get to what it is I'm trying to figure out. If you're able, perhaps you could answer there to clarify if it makes more sense. Commented Apr 25, 2021 at 20:04
• You get the symmetries from something called "The Young Symmetrizer", so for each standard tableaux you have a prescription for getting the rotationally closed permutations.
– JEB
Commented Apr 27, 2021 at 4:34

CONTEXT

The question here asks for a proof that for a filled subshell the total angular momentum, $$\mathbf{L}$$, and total spin angular momentum, $$\mathbf{S}$$, are such that $$\mathbf{L}=\mathbf{S}=\mathbf{0}.$$

Above, @JEB provides an answer to the question. I have no issues with that answer, except that there is no mention (as far as I can tell) of $$\mathbf{S}$$. Therefore, as far as I can tell, the original question remains unanswered. in my answer, I do not provide a proof, which is what is requested; rather, I provide a demonstration indicating my rational. I imagine that such rational could be extended to a proof.

DEMONSTRATION

Magnesium ($$Z=12$$) is a divalent metal with ground state electronic configuration [Ne]3s$$^2$$. What is the angular momentum of the ground state?

In the LS-scheme (https://en.wikipedia.org/wiki/Angular_momentum_coupling), the rules are that $$\mathbf{L}_{[Ne]3s ^2 }= \boldsymbol{\ell}_1 +\boldsymbol{\ell_2} \quad~\text{and}\quad\mathbf{S}_{[Ne]3s ^2 }= \boldsymbol{s}_1 +\boldsymbol{s_2} .$$

For the 1s$$^2$$,2 s$$^2$$, 2p$$^6$$,3s$$^2$$ ground state, we have two valence electrons in the $$s$$ shell. Each and every electron in the $$s$$ shell has the quantum number $$\ell=0$$. Therefore $$\ell_1=\ell_2=0$$. Thus, the possible vales of the quantum numbers for $$\mathbf{L}_{[Ne]3s ^2 }= \boldsymbol{\ell}_1 +\boldsymbol{\ell_2} \quad~\text{with}\quad ~\ell_1=0\quad\text{and}\quad \ell_2=0,$$ are $$L _{[Ne]3s ^2 } =\left\{ (0+0),\ldots,(|0-0|) \right\} =\left\{ 0 \right\}.$$ Hence, $$|\mathbf{L} _{[Ne]3s ^2 }|=\sqrt{0\,(1+0)}\,\hbar=0,$$ and $$$$\mathbf{L} _{[Ne]3s ^2 } = \mathbf{0}. \label{eq:mag-2}$$$$

For the 1s$$^2$$,2 s$$^2$$, 2p$$^6$$,3s$$^2$$ ground state, we have two valence electrons in the $$s$$ shell. Each and every electron has the quantum number $$s=1/2$$. Therefore $$s_1=s_2=1/2$$. Thus, the possible vales of the quantum numbers for $$\mathbf{S}_{[Ne]3s ^2 }= \boldsymbol{s}_1 +\boldsymbol{s_2} \quad~\text{with}\quad ~s_1=1/2\quad\text{and}\quad s_2=1/2,$$ appear to be $$S _{[Ne]3s ^2 } =\left\{ (1/2+1/2),\ldots,(|1/2-1/2|) \right\} =\left\{1, 0 \right\}.$$ It further appears that $$|\mathbf{S} _{[Ne]3s ^2 }|=\left\{\sqrt{0\,(1+0)}\,\hbar, \sqrt{1\,(1+1)}\,\hbar\right\} .$$ That said, we must take the following into account.

In the special case of ground configurations of equivalent electrons, the spin and orbital angular momentum of the lowest-energy term follow some emipircal rules called Hund's rules: the lowest-energy term has the largest value of $$S$$ consistent with the Pauli exclusion priniple. (Atomic Physics, Foot, 2005, p.81)''

In this example, we are clearly dealing the with ground configuration and the lowest-energy term. According to the Pauli exclusion principle we can not have two electrons with the quantum state $$|n,\ell,m_\ell,s,m_s> = |3,0,0,1/2,+1/2>;$$ similarly, we can not have two electrons with the quantum state $$|n,\ell,m_\ell,s,m_s> = |3,0,0,1/2,-1/2>.$$ Thus, for ground configuraion, one each (but not both) of the electrons must be quantised as follows: \begin{align*} |\psi_1 > =|n,\ell,m_\ell,s,m_s> = |3,0,0,1/2,+1/2> \\ |\psi_2 > = |n,\ell,m_\ell,s,m_s> = |3,0,0,1/2,-1/2> \end{align*} The value of $$S=1$$, which would lead to a non-zero value $$|\mathbf{S} _{[Ne]3s ^2 }|$$, is not physically realizable since the spin we would need two electrons whose individual spin angular momenta are not equal and opposite. This would be inconsistent with the Pauli exclusion principle (i.e. the two spin angular momenta must be equal and opposite and thus result in no net spin angular momentum). In accordance with Hund's rule, we find that $$S _{[Ne]3s ^2 } = \left\{ 0 \right\}.$$ Hence, $$|\mathbf{S} _{[Ne]3s ^2 }|=\sqrt{0\,(1+0)}\,\hbar=0,$$ and $$$$\mathbf{S} _{[Ne]3s ^2 } =\mathbf{0}. \label{eq:mag-1}$$$$

In summary, I have demonstrated a particular instance such that for a filled subshell the total angular momentum, $$\mathbf{L}$$, and total spin angular momentum, $$\mathbf{S}$$, are identically zero. Namely, I find $$\mathbf{L} _{[Ne]3s ^2 } =\mathbf{S} _{[Ne]3s ^2 } =\mathbf{0}.$$