# What is the QFT state with two distinguishable fermions present?

I want to describe a system with two non-interacting and definitely different fermions, say an electron neutrino, $$\nu_e$$, and an electron, $$e^-$$.

The state describing a single electron is given using the Dirac creation operator $$c^\dagger_e|0\rangle$$ where $$|0\rangle$$ is the vacuum state.

Similarly the state describing a single neutrino is $$c_\nu^\dagger|0\rangle$$

My question is what is the state with both fermions present? Is it simply $$c^\dagger_ec^\dagger_\nu|0\rangle$$ If so, does it matter which way around the two creation operators are? If not, surely that means different fermion operators commute, since I could have just as well have chosen to write $$c^\dagger_\nu c^\dagger_e|0\rangle$$ Other questions on this topic (e.g. this question) suggest to me that they should anticommute, and in general I've come to believe that the whole point of these operators and what makes them so special is they anticommute with just about anything.

• Yes, they anticommute. – knzhou Feb 20 at 16:49
• In response to the edited question: yes, they anticommute. This is not only not weird, but also completely necessary to get correct results, e.g. see my answer here. – knzhou Feb 20 at 18:06
• You can define the state with two fermions present as either $c_e^\dagger c_\nu^\dagger |0 \rangle$ or $c_\nu^\dagger c_e^\dagger |0 \rangle$, but the two definitions will differ by a minus sign. To avoid sign errors, you need to check what convention your source uses. – knzhou Feb 20 at 18:11
• "...they anticommute with just about anything." Not quite. All fermion creation operators anticommute with each other, but they commute with boson creation operators. – Chiral Anomaly Feb 21 at 3:12