foundations - 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".

Answer

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 $\left$. 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 $\left=\alpha^2$