Extension from Excitons to Trions

I am curious about the fundamentals of how a trion state is defined. An exciton state is considered a bound electron-hole pair. This can be formed by is we have a ground state which consists of a full valence band and a photon comes along with momentum $$\vec{Q}$$ and excites an electron to the conduction band, from which we leave a hole (absence of an electron) in itself place which is bound together through an interaction term (i.e Coulomb). Therefore such a state could be written as: $$c^\dagger(\vec{k}+\vec{Q})b(\vec{k})|GS\rangle$$

If we then look at a trion which can be considered as a charged exciton for this case. We have a three-body particle that either contains 2 holes and 1 electron (positive trion) or 2 electrons and 1 hole (negative trion).

Therefore we can consider a trion to be formed when a photo-excited exciton (as mentioned above binds with either an extra electron or a hole. My question here is how does one write this state? Does it have an impact on the extra momentum $$\vec{Q}$$ mentioned for the exciton? Because in theory how in this set-up would one have an extra electron in the conduction band to bind with the exciton system to form a trion? Hence, for a negative trion would this be:

$$c^\dagger(\vec{k}_1)c^\dagger(\vec{k}_2+\vec{Q})b(\vec{k}_1+\vec{k}_2)|GS\rangle$$ or would both of the electrons now in the system have equal extra momentum i.e $$c^\dagger(\vec{k}_1+\vec{Q}/2)c^\dagger(\vec{k}_2+\vec{Q}/2)b(\vec{k}_1+\vec{k}_2)|GS\rangle$$ In turn, both of these states can give the same results when going to calculation trion energies, but I am wondering what the correct formalism is