My question is closely related to this and this questions. However, instead of asking about the commutation relation between the operators, I would like to ask about their action of the Fock states.
For definiteness, let's assume that we have two relativistic fermionic fields of different species whose annihilation operators are denoted $b$, $d$, $\widetilde{b}$, $\widetilde{d}$. Let $\xi$ be the multi-index denoting a mode in a free theory, i.e. $\xi$ includes momentum, spin, and whatever else quantum numbers are present in the theory: $\xi=\{\vec{p}, s,\ldots\}$.
When writing the Fock state, I will only include modes with non-zero occupancies. I will first list fermions of type $b$, then anti-fermions of type $d$, then fermions of type $\widetilde{b}$, then fermions of type $\widetilde{d}$. Consider a the state of the form $$ |\Psi\rangle= |\cdots\xi_1\xi_2\cdots;\cdots\xi_3\cdots;\cdots\xi_4\cdots;\cdots\xi_5\cdots\rangle =b^\dagger_{\xi_1}b^\dagger_{\xi_2} d^\dagger_{\xi_3} \widetilde{b}^\dagger_{\xi_4} \widetilde{d}^\dagger_{\xi_5} |0\rangle \ . $$ In the equation above I assumed that if I'm filling a mode in a state "on the left" from all the existing modes, the sign remains unchanged. The sign $\cdots$ in this equation stands for the absence of any other excited modes (so, this is a 5-particle state).
What I would like to know is the results of the action of various operators on this state. For example, according to my definition above, $$ b_{\xi_1}|\Psi\rangle = b^\dagger_{\xi_2} d^\dagger_{\xi_3} \widetilde{b}^\dagger_{\xi_4} \widetilde{d}^\dagger_{\xi_5} |0\rangle \ . $$
I would appreciate if someone could explain how to determine the sign in the following expressions: \begin{alignat}{8} b_{\xi_2}|\Psi\rangle &= (?) b_{\xi_1}^\dagger d^\dagger_{\xi_3} \widetilde{b}^\dagger_{\xi_4} \widetilde{d}^\dagger_{\xi_5} |0\rangle \ ,\\ d_{\xi_3}|\Psi\rangle &= (?) b_{\xi_1}^\dagger b_{\xi_2}^\dagger \widetilde{b}^\dagger_{\xi_4} \widetilde{d}^\dagger_{\xi_5} |0\rangle \ ,\\ {\widetilde{b}}_{\xi_4}|\Psi\rangle &= (?) b_{\xi_1}^\dagger b_{\xi_2}^\dagger d_{\xi_3}^\dagger \widetilde{d}^\dagger_{\xi_5} |0\rangle \ ,\\ {\widetilde{d}}_{\xi_5}|\Psi\rangle &= (?) b_{\xi_1}^\dagger b_{\xi_2}^\dagger d_{\xi_3}^\dagger \widetilde{b}^\dagger_{\xi_4} |0\rangle \ . \end{alignat}
Here is my guess:
Since I have specified from the very beginning that the fields are relativistic, any two creation/annihilation operators whose type and mode index don't match have to anticommute. For example: $$ \{b^\dagger_{\xi_1}, \ \widetilde{d}_{\xi_5}\} = 0 \ . $$
Therefore (here the actual guess begins), I have to treat all the modes in an exactly same way as if all of them represented the same species. Thus, I expect the following rule to hold: $$ \widetilde{d}_{\xi_5} |\Psi\rangle = (-1)^{\sum_\alpha n_\alpha} b^\dagger_{\xi_1}b^\dagger_{\xi_2} d^\dagger_{\xi_3} \widetilde{b}^\dagger_{\xi_4} |0\rangle = (+1) b^\dagger_{\xi_1}b^\dagger_{\xi_2} d^\dagger_{\xi_3} \widetilde{b}^\dagger_{\xi_4} |0\rangle \ , $$ where $(-1)^{\sum_\alpha n_\alpha} $ is the sum of occupancies of modes of all types standing to left of $\xi_5$ within my ordering of modes. In the case above, ${\sum_\alpha n_\alpha} = 4$.
UPDATE:
(Assuming that my guess is correct.) Quite interestingly, it looks like in the practical applications, one does not have to keep track of the sign in front of the wave function. This is owing to the fact that all known interactions preserve "fermionic parity". By this I mean that all the interaction vertices have an even number of fermionic lines.
Consider the action of some reasonable second-quantized Hamiltonian onto a Fock state:
$$ \hat{H} |\Psi\rangle = \sum_i c_i|\Phi_i\rangle \ , $$
where $|\Psi\rangle$ and $|\Phi_i\rangle$ are the Fock states, containing $m$ and $m_i$ fermions, correspondingly. For any reasonable interaction: $$ (m - m_i) \mod 2 = 0 \ . $$ Therefore, one doesn't really have to keep track of the Fock state sign arising due to the fermionic commutation relations.
