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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.

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  • $\begingroup$ When $m_l=m_s=0$, how does the atom pick the $z$-axis? $\endgroup$ – JEB Apr 23 at 13:52
  • $\begingroup$ @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. $\endgroup$ – Jagerber48 Apr 23 at 15:49
  • $\begingroup$ $m=0$ 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? $\endgroup$ – JEB Apr 23 at 17:21
  • $\begingroup$ @JEB Perhaps it got to the state it’s in due to an interaction with a polarized electric field. $\endgroup$ – Jagerber48 Apr 23 at 17:27
  • $\begingroup$ 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 $L=0$. $\endgroup$ – JEB Apr 23 at 21:55
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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.

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  • $\begingroup$ 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... $\endgroup$ – Jagerber48 Apr 24 at 17:42
  • $\begingroup$ 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? $\endgroup$ – Jagerber48 Apr 24 at 17:44
  • $\begingroup$ 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. $\endgroup$ – Jagerber48 Apr 24 at 17:47
  • $\begingroup$ 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. $\endgroup$ – Jagerber48 Apr 25 at 20:04
  • $\begingroup$ 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. $\endgroup$ – JEB Apr 27 at 4:34

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