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Does there exist a proof/conjecture/counterexample, to the statement that the fermionic ground state, of the many-electron Schrödinger operator (including spin-orbit interaction) with spherically symmetric external potential, is unique modulo phase (think isolated atom in Born-Oppenheimer approximation).

Edit: Modified question to include spin-orbit based on dennismoore94’s counterexample.

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  • $\begingroup$ What about the free case? Or employ periodic boundary conditions and then consider the free case, i.e. constant external potential. $\endgroup$ Commented Mar 23 at 14:38
  • $\begingroup$ I’d be curious about that example too, but my intuition is that the gs of an isolated atom should be unique. $\endgroup$
    – phonon
    Commented Mar 23 at 14:44
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    $\begingroup$ What exactly do you mean by unique? If you mean non-degenerate, then the statement is not true (think of the $^3P_g$ ground state of carbon or oxygen). $\endgroup$ Commented Mar 23 at 16:31
  • $\begingroup$ Yes, non-degenerate = unique (modulo phase). In those examples, does spin-orbit interaction break the degeneracy leading to unique gs? $\endgroup$
    – phonon
    Commented Mar 23 at 17:00

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The non-relativistic Hamiltonian of the atom has rotational invariance, and it does not act on spins, which means that energy eigenstates $|\nu,L,M_L,\pi_P,S,M_S\rangle$ are $D=(2L+1)(2S+1)$-fold degenerate ($\pi_P$ standing for parity, and $\nu$ referring to other quantum numbers). This means that the non-relativistic ground state must be of ${^1}\hspace{-1mm}S$ symmetry (that is $L=S=0$) to be non-degenerate. In general, this does not hold: $$ \begin{aligned} \text{H:} & \ \ \ {^2}\hspace{-1mm}S && D=2 \ \ \ \ \rightarrow \ \ \ \ J=1/2 \\ \text{He:} & \ \ \ {^1}\hspace{-1mm}S && D=1 \ \ \ \ \rightarrow \ \ \ \ J=0\\ \text{Li:} & \ \ \ {^2}\hspace{-1mm}S && D=2 \ \ \ \ \rightarrow \ \ \ \ J=1/2 \\ \text{Be:} & \ \ \ {^1}\hspace{-1mm}S && D=1 \ \ \ \ \rightarrow \ \ \ \ J=0 \\ \text{B:} & \ \ \ {^2}\hspace{-1mm}P && D=6 \ \ \ \ \rightarrow \ \ \ \ J=1/2 \, , \, 3/2 \\ \text{C:} & \ \ \ {^3}\hspace{-1mm}P_g && D=9 \ \ \ \ \rightarrow \ \ \ \ J=0 \, , 1 \, , \, 2 \\ \text{N:} & \ \ \ {^4}\hspace{-1mm}S && D=4 \ \ \ \ \rightarrow \ \ \ \ J=3/2 \\ \text{O:} & \ \ \ {^3}\hspace{-1mm}P_g && D=9 \ \ \ \ \rightarrow \ \ \ \ J=0 \, , 1 \, , \, 2 \\ \text{F:} & \ \ \ {^2}\hspace{-1mm}P && D=6 \ \ \ \ \rightarrow \ \ \ \ J=1/2 \, , \, 3/2 \\ \text{Ne:} & \ \ \ {^1}\hspace{-1mm}S && D=1 \ \ \ \ \rightarrow \ \ \ \ J=0 \end{aligned} $$ The symmetry of the ground state can be guessed with the aufbau principle and the first two Hund rules.

Relativistic effects do break this degeneracy to some extent. The good quantum number is now $J$, and as long as relativistic effects are not too strong, we can use the $LS$ coupling to write $$ J=|L-S|,|L-S|+1,...,L+S \ . $$ The original $D$-fold degenerate manifold splits into smaller sets with different $J$ values, each of them being $(2J+1)$-fold degenerate. This means that if $L$ and $S$ are such that they can combine into $J=0$, and this state has the lowest energy in all sets, then the relativistic ground state is non-degenerate. Whether this is the case, you can guess with the third Hund rule:

among states arising from the same configuration with a half-filled or less than half-filled subshell, the lowest energy state is the one with the lowest $J$;

among states arising from the same configuration with a more than half-filled subshell, the lowest energy state is the one with the highest $J$.

This means that the ground state of the carbon atom (with $[(1s)^2(2s)^2(2p)^2]$) has $J=0$ and is non-degenerate, but the ground state of oxygen (with $[(1s)^2(2s)^2(2p)^4]$) has $J=2$ and is five-fold degenerate. Both results can be easily checked on NIST (C and O).

Of course, inclusion of QED and hyperfine effects would lead to further splittings in the spectrum.

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