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In field theory, by quantizing a dirac field, we can obtain a creation operator for a single electron of definite momentum, of definite spin up or down, these respectively are: $$a^\dagger_{+}(p)|0\rangle, {a^\dagger}_{-}(p)|0\rangle$$ Where we've defined the former to create a spin +1/2 electron, the latter to create a spin -1/2 electron. By addition and repeated-application of these creation operators we can write down a state of any number of particles, each having any superposition of spins. : $$ \int dp f(p)\prod_{i=0}^n(\alpha_ia^\dagger_{i+}(p)+\beta_i{a^\dagger}_{i-}(p))|0 \rangle $$ Where of course $i$ labels the particle, and $a,b,f$ are some distributions.

Question: For a given field theory, how does one write down a creation operator for a pair of entangled particles? (say electrons in a Dirac theory of spinors)

In quantum mechanics, an entangled state is one which lives in a tensor product Hilbert space, but does not have a tensor product decomposition. Since Fock space is essentially built up with a bunch of tensor products of Hilbert spaces, it doesn't seem unreasonable to demand that it contains such entangled states. But how does one explicitly write down such a state?

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  • $\begingroup$ I would like to note that this is not a duplicate of any previous questions on measurement in QFT or entanglement in QFT, for I am asking for something specific and explicit that has not been answered by any previous questions. $\endgroup$
    – zzz
    Commented Jul 25, 2014 at 17:36
  • $\begingroup$ 1007.1569 appears to be a relevant reference, maybe I will read through it and write an answer. $\endgroup$
    – zzz
    Commented Jul 25, 2014 at 17:51
  • $\begingroup$ this previous question gives an example of what's not an answer. $\endgroup$
    – zzz
    Commented Jul 25, 2014 at 17:55

2 Answers 2

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Let $f(x,y)\in L^2(\mathbb{R}^{2d})$ and $\Omega$ the vacuum of the symmetric Fock space $\Gamma_s(L^2(\mathbb{R}^d))$. Suppose there is no $f_1,f_2\in L^2(\mathbb{R}^d)$ such that $f(x,y)=f_1(x)f_2(y)$: then $f_s$ (the symmetrized of $f$) is an "entangled" two particle state of $\Gamma_s(L^2(\mathbb{R}^d))$. This is created by $$\frac{1}{\sqrt{2}}\int f(x,y) a^*(x)a^*(y)dxdy\Omega\; .$$ For antisymmetric particles and/or more degrees of freedom the reasoning is the same.

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  • $\begingroup$ Thank you very much. This looks right to me, but can you give a link of where this form is used explicitly in literature? $\endgroup$
    – zzz
    Commented Jul 26, 2014 at 2:59
  • $\begingroup$ @bechira Unluckily I don't know of papers utilizing this type of operators in the context of entanglement (mainly because I don't know much about entanglement itself), however as you can see it is a fairly easy construction. Trying to remember the literature I know better, I found a state similar to what you look for in this paper (page 10); but I don't know if this may be useful to you...there are surely other better examples that I am not aware of ;-) $\endgroup$
    – yuggib
    Commented Jul 26, 2014 at 9:37
  • $\begingroup$ Okay thanks, I will dig through literature as well, and I'll accept this answer as soon as I find some confirmation in literature. $\endgroup$
    – zzz
    Commented Jul 26, 2014 at 16:24
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A state like $ \frac{1}{\sqrt{2}}(a^\dagger_+(\vec p)a^\dagger_+(-\vec p) + a^\dagger_-(\vec p)a^\dagger_-(-\vec p))|0\rangle$ would be an example. It is both entangled in spin and entangled in momenta.

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    $\begingroup$ It is an entangled state. If you measure the spin of one of the $2$ particles relatively to an axis $Oz$ and find $+1$, then a measure of the spin of the other particle relatively to this axis $Oz$ will give you $+$ too. The measurement will project the initial state, into the state $a^\dagger_+(\vec p)a^\dagger_+(-\vec p)$ $\endgroup$
    – Trimok
    Commented Jul 27, 2014 at 11:41
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    $\begingroup$ this is actually right now that I think about it, it has this weird form because Fock space is already equipped with a tensor product structure, but the state you give here would indeed give an entangled system. $\endgroup$
    – zzz
    Commented Aug 11, 2014 at 3:52

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