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.


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".

  • 1
    $\begingroup$ It is not clear what you're asking - a quantum field is an operator, not a state, but your entire question sounds as if you think the quantum field should somehow be part of the state information. $\endgroup$
    – ACuriousMind
    Jul 29, 2018 at 10:10
  • $\begingroup$ Thank you, could you please help me with the terminology? What should I call the domain/set of states upon which the operator acts, ie. the "stuff" that the universe (or in this case toy universe) is made of? $\endgroup$
    – user183966
    Jul 29, 2018 at 10:13
  • $\begingroup$ ACM is correct but being a bit strict. The object we call a quantum field is indeed an operator field, a mathematical object, but it is describing a real object i.e. the free scalar field. The eigenstates of this object are called Fock states. The Fock states cannot become entangled since in a free field there are no interactions to entangle them. The overall state of the field is a product of the Fock states. $\endgroup$ Jul 29, 2018 at 10:19
  • $\begingroup$ @ACuriousMind Regarding the terminology, I would love to know the precise answer so I can be clearer in my questions, but instead of clearing it up here I have moved to a whole question so any answer can have proper status: physics.stackexchange.com/questions/420015/… $\endgroup$
    – user183966
    Jul 29, 2018 at 10:47
  • 1
    $\begingroup$ I see, I was think two particles in two different modes. Two particles in one mode completely specifies the state $\endgroup$ Jul 29, 2018 at 13:20

1 Answer 1


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 $\left<a^\dagger a\right>=\alpha^2$

  • 1
    $\begingroup$ Thank you so much, very helpful answer. Good to know there is no extra 'magic', and the extra info was spot on. For anybody else reading "correlation function" is apparently a standard term and you can read more at en.wikipedia.org/wiki/… . $\endgroup$
    – user183966
    Jul 29, 2018 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.