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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".

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    $\begingroup$ 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 '18 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 '18 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. $\endgroup$ – John Rennie Jul 29 '18 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 '18 at 10:47
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    $\begingroup$ I see, I was think two particles in two different modes. Two particles in one mode completely specifies the state $\endgroup$ – Shane P Kelly Jul 29 '18 at 13:20
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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 $\left<a^\dagger a\right>=\alpha^2$

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    $\begingroup$ 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 '18 at 14:39

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