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Physically, can quantum-states which are a superposition of states of different numbers of fermions exist? i.e. states of the form $\vert \psi \rangle = a \vert N\rangle + b \vert N' \rangle$ where $N \neq N'$.

An example state might be the superposition $\vert \psi \rangle = (1/\sqrt{2})(\vert \rm vac \rangle + \vert \uparrow \downarrow \rangle)$ where $\rm \vert vac \rangle$ is the vacuum state and $\vert \uparrow \downarrow \rangle$ is the state with two fermions of different spin. Another example might be the superposition of the ground states of a fermionic many-body Hamiltonian at two different fillings (say the fermionic Hubbard model at half and quarter filling).

I understand that $N$ and $N'$ must have the same parity due to the parity superselection rule for fermions but are there any restrictions beyond that? I also know there is a charge superselection rule but considering not all fermions have charge this would seemingly not apply here.

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  • $\begingroup$ One obvious example is the BCS ground state. $\endgroup$ Jun 14, 2021 at 13:54
  • $\begingroup$ The BCS ground state is a superposition of different numbers of Cooper-paired electrons, not a superposition of different numbers of total electrons. The unpaired electrons are still there in the material, just not paired. If they weren't there at all, then we'd have a superposition of different values of the total electric charge, violating the charge superselection rule that was noted in the question. $\endgroup$ Jun 14, 2021 at 13:58
  • $\begingroup$ Some people call this "fluffy bunny entanglement" when considering such things with a particle-number superselection rule arxiv.org/abs/quant-ph/0309046 - for bosonic systems these superpositions only make sense with a phase reference $\endgroup$ Jun 14, 2021 at 14:09

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Yes, such superpositions are allowed. Relativistic quantum field theory wouldn't work without them.

Consider an initial state $|n\rangle$ with just a single isolated neutron. That's a single-fermion state. Over the next several minutes, in the Schrödinger picture, the single-neutron component of the state smoothly declines from $1$ toward $0$, and the proton-plus-electron-plus-neutrino $|p,e,\nu\rangle$ component of the state smoothly grows from $0$ toward $1$: $$ \alpha(t)|n\rangle+\beta(t)|p,e,\nu\rangle $$ with $|\alpha(t)|=1$ initially, $|\beta(t)|\to 1$ as $t\to\infty$, and $|\alpha(t)|^2+|\beta(t)|^2\approx 1$ always. (I wrote this an approximation because there are actually other terms in the sum, too.) After a few hours, the single-neutron component is negligible: if we tried to measure the number of neutrons remaining in the state, we would get zero with high probability, because the state is $\approx |p,e,\nu\rangle$.

For several minutes during the transition period, we had a quantum superposition of a one-fermion state and a three-fermion state in which both $\alpha(t)$ and $\beta(t)$ have non-negligible magnitudes.

For some cartoons, see https://en.wikipedia.org/wiki/Free_neutron_decay, but beware that those cartoons don't convey the all-important intermediate quantum superposition. If that weren't allowed, then neutrons wouldn't be able to decay.

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