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.

  • $\begingroup$ One obvious example is the BCS ground state. $\endgroup$
    – Roger V.
    Commented 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. $\endgroup$ Commented 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 $\endgroup$ Commented Jun 14, 2021 at 14:09

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.