How much freedom is there in a quantum field? Let's imagine we have a free scalar quantum field, and that it has 2 particles in a specific momentum eigenstate only. Does this information completely fix the quantum field, or is there additional information needed, like correlations / entanglements between the particles or something?
There could be some additional subtlety to this question and I can imagine more than 2 possible answers, for example:


*

*There is just one mathematical state that corresponds to a field with 2 particles in a specific momentum eigenstate only.

*There are multiple formal states that correspond to this but they have identical phenomenology / the freedom is in the model only.

*There are multiple states that have this interpretation and they exhibit different phenomenology.


Edit:
To be a bit clearer about the motivation for my question: if I were to talk about a Fock space in a state containing 2 particles at 2 positions, this is not enough to uniquely specify the state. It could mean equal chance of both particles at each point, or certainty of a single particle at each point (ie. 
$\frac{1}{\sqrt{2}} a_{x1}^\dagger a_{x1}^\dagger |0\rangle + \frac{1}{\sqrt{2}} a_{x2}^\dagger a_{x2}^\dagger |0\rangle$ or $a_{x1}^\dagger a_{x2}^\dagger |0\rangle$ I believe) This sort of idea is what I was thinking of when I wrote "correlations / entanglements". I'm not necessarily referring to any specific meaning of "correlations" or "entanglements".
 A: The state of a quantum field is fully described by a state in Fock-Space.
Thus in general the number of particles in a given set of modes is not enough information as you gave in you example of:
$
\frac{1}{\sqrt{2}} a_{x1}^\dagger a_{x1}^\dagger |0\rangle + \frac{1}{\sqrt{2}} a_{x2}^\dagger a_{x2}^\dagger |0\rangle
\neq
a_{x1}^\dagger a_{x2}^\dagger |0\rangle
$
In otherwords, the state of your field is not described by just the single particle correlations $\left<a_{x2}^\dagger a_{x1}\right>$.  You need to specify all n particle correlations to give a full description of the state.  If your state has $N$ particles in it then all correlations up to the $N$ particle correlation function will be enough.
If your state only has exactly n particles in one mode, then you  have specified all the information.  There is only one state of this situation:
$
(a^{\dagger})^n|0\rangle
$
but you could have a coherent state where on average you have $n$ particles in one mode, but the number particles is not fixed:
$e^{\alpha a^\dagger-\alpha^* a}\left|0\right> $ with $\left<a^\dagger a\right>=\alpha^2$
