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  Is "vacuum entanglement" properly so called?

+ 4 like - 0 dislike

When I first read about vacuum entanglement, I understood it in exactly the same way as Ron wrote in this post. As can be very clearly seen, in a free scalar theory and in Schroedinger wavefunctional picture, vacuum state is a not a product state in terms of position-space wavefunction, but is a product state in momentum-space, and even more trivial(in terms of product structure) a state in the Fock-space representation. They should be all mathematically equivalent, I suppose. 

However, the question of whether "entanglement" is properly so called depends on how we identify subsystems that are in a sense "physical". Changing from position-space wavefunctional representation to the momentum-space one, or to the Fock-space one, is clearly not a change of basis of the Hilbert space, but a change of identification of subsystems. 

To elaborate my point,  consider a scenario in ordinary quantum mechanics, two spin-$\frac{1}{2}$ particles positioned at two points, are in a entangled state $|+-\rangle-|-+\rangle$. However, we can relabel the basis of the 4-dimensional Hilbert space as

\[|A\rangle:= |+-\rangle-|-+\rangle\\|B\rangle:=|+-\rangle+|-+\rangle\\|C\rangle:=|++\rangle\\|D\rangle:=|--\rangle .\] 

In this case, it is surely absurd to say our state is not entangled because it can be written as a single ket $|A\rangle$, because it is clear what must be identified as the physical subsystems.

The scalar field theory I discussed seems to be an infinite-dimensional analog mathematically, but in the field theory case I'm no longer sure what should be identified as the physical subsystem.

In fact, in the post I linked, twistor59 raised the same question under Xiao-Gang Wen's answer, I quote,

as you say, the vacuum isn't a product state, but I was curious about what the subsystems were in order even to discuss whether it is a product or not. 

but I see no satisfactory discussion in that post.

asked Jan 24, 2015 in Theoretical Physics by Jia Yiyang (2,640 points) [ revision history ]
edited Jan 24, 2015 by Jia Yiyang
To be a product, the states (subsystems) must be written in terms of separated variables. Then the subsystems look as "non-interacting" with each other although they may belong to one compound system. In the latter case they are still entangled due to conservation laws for the whole system, in my humble opinion.

3 Answers

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You're perhaps hitting problems partly because the operators in QFT are not $\hat\phi(x)$ and $\tilde\phi(k)$, which are operator-valued distributions — which we could say use the improper position and momentum basis sets for whatever test function space is used. For a scalar free field operator $\hat\phi_f$ constructed as a sum of creation and annihilation operators, without introducing distributions, $\hat\phi_f=a_{f^*}+a_f^\dagger$, we can define the vacuum as the zero eigenstate of the annihilation operators $a_f$, $a_f|0\rangle=0$, for all test functions $f\in\mathcal{S}$, a Schwartz space of test functions that decrease faster than exponentially in both position and momentum space, not including Dirac delta-functions. The Hilbert space structure is fixed by a choice of positive semi-definite commutator/inner product $[a_f,a_g^\dagger]=(f,g)$. There are many choices of countable basis for Schwartz space (including, say, sets of Hermite polynomials times a Gaussian centered on a chosen point), for each of which the vacuum state is a product state, but none is local in either position or momentum coordinates. Indeed, there is no such thing as an eigenstate of the momentum operator in the GNS-constructed Fock-Hilbert space that could be called a (Wigner) particle, there are only very close approximations (momentum eigenstates can be introduced as operator-valued distributions but they are not in the Hilbert space).

Insofar as entanglement refers to a state not being a product state in momentum coordinates, we might say that the concept of entanglement is not defined for the vacuum state in QFT, but insofar as we work carefully with operator-valued distributions in the momentum basis the vacuum is not entangled in the Wigner particle sense. If we take entanglement to refer to a state that cannot be expressed as a product state constructed using only localized test functions (in other words if we adopt a non-Wigner, non-momentum basis definition of particle in terms of localization, which often creeps silently into discussions), then the vacuum is entangled.

If we look at Stephen Summers "Yet More Ado About Nothing: The Remarkable Relativistic Vacuum State", http://arxiv.org/abs/0802.1854, which is linked to in the PDF you link to in your other question, we find a definition of a product state: "Given a pair (M,N) of algebras representing the observable algebras of two subsystems of a given quantum system, a state φ is said to be a product state across (M,N) if φ(MN) =φ(M)φ(N) for all M ∈ M, N ∈ N." An entangled state relative to (M,N) is any that cannot be written as sum of normal product states across (M,N). If M and N are local algebras (local in position space, an essential starting point for AQFT), then the free field QFT vacuum is entangled relative to (M,N). If P and Q are algebras that are local in momentum space, which are not an essential part of AQFT, then the free field QFT vacuum is not entangled relative to (P,Q).

answered Jan 24, 2015 by Peter Morgan (1,230 points) [ no revision ]
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@JiaTiyang: E and B are strongly interacting subsystems. Their equations are coupled. It is like two constituent particles connected with a spring. But you can introduce separated variables - the center of mass coordinate and the relative distance coordinate. Equations for the latter are not coupled, they describe collective normal modes of the total system whose QM states make a product.
@VK, Yes it is a system of infinite coupled harmonic oscillators, but we call it a free theory if there's no terms with order higher than 2. It's just semantics, and it doesn't change my question, the free EM ground state, as well as free scalar ground state, are Gaussian in wavefuncional picture. BTW the downvote is not from me.
@JiaYiyang Free field EM is still translation invariant, crucially, so the positive semi-definite inner product is still diagonal in momentum space. The presentation can be just the same as for the free scalar field, four equations, $$\hat F_f=a_{f^*}+a_f^\dagger, a_f|0\rangle=0,\langle 0|0\rangle=1, [a_f,a_g^\dagger]=(f,g),$$ as it can be for any free real boson field, but with the test function space being antisymmetric tensors and with the positive semi-definite inner product being $$(f,g)=-\hbar\int \tilde f^{\alpha\mu*}(k)k_\alpha k_\beta g_{\mu\nu}2\pi\delta(k^2)\tilde g^{\beta\nu}(k)\frac{\mathrm{d}^4k}{(2\pi)^4}$$(with the metric $g_{\mu\nu}$ and the test function $\tilde g^{\beta\nu}(k)$ hopefully being distinguishable). This is positive semi-definite because $k_\alpha\tilde f^{\alpha\mu*}(k)$ and $k_\beta\tilde g^{\beta\nu}(k)$ are both orthogonal to the null 4-vector $k$ and space-like. We can do something similar for the nonobservable EM potential, but there are some long-winded details.
@PeterMorgan, I understand that in momentum space the free field vacuum is a product state, both for EM and scalar. My mentioning of EM is to argue there seems to be at least a merit to consider position space wavefunctional to be more "natural" than momentum space functional or Fock space: when we measure the local E-field strength(you may smear it bit if you prefer rigor) out of vacuum, the functional $\Psi[E(x)]$ gives the probability amplitude to obtain a value $E(x)$ at the point $x$. This argument wouldn't work for scalar since it's hard to say if the (smeared) field operator is an observable or not.
@JiaYiyang I take position space to be natural insofar as the symmetry group of the equations is the Poincaré group, including translations in position space. Translations in momentum space are not a symmetry. I don't see the EM and scalar fields as so very much different, at least not in the way you suggest.
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Thanks, so your stand is basically that for QFT vacuum there's no preferred way of identifying subsystems?

@JiaYiyang There's no preferred countable basis for the test function space, but the (positive semi-definite) inner product on the test function space is diagonal in the momentum space coordinates (as it has to be to ensure translation invariance), $$(f,g)=\hbar\int \tilde f^*(k)2\pi\delta(k^2-m^2)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4},$$ so I think I'd say that the Wigner definition of a particle is natural, for the free field, in the sense implied by this equation, but limited by the difficulties that arise if we use the improper momentum basis carelessly.

+ 3 like - 0 dislike
One sense in which the vacuum is entangled is in position space. Take a scalar field in 1+1 dimensions for simplicity, so that the spatial dimension is just a line. We have operators \phi(x) on each spatial lattice site, that commute with the operators on all other sites. So we can divide the Hilbert space up into two Hilbert spaces, one of which consists of the operators \phi(x) for x>0 and the other one with x<0. The two subsystems are the positive and negative half-lines. Now one can trace out the left subsystem, to get a density matrix $\rho$ for the right subsystem. The statement that the vacuum is entangled is equivalent to saying that $\rho$ is not a pure state, and has some entanglement entropy. This entanglement entropy is in fact UV divergent due to contributions from modes very close to x=0 on the left and right. For example for a CFT it's log divergent (and determined in terms of the central charge).

All of this is equivalent to the statement that the Rindler vacuum is an excited state in terms of the Minkowski vacuum.
answered Jan 29, 2015 by Matthew [ no revision ]
Hi Matthew, thanks for the new perspective!
+ 1 like - 0 dislike
One sense in which the vacuum is entangled is in position space. Take a scalar field in 1+1 dimensions for simplicity, so that the spatial dimension is just a line. We have operators \phi(x) on each spatial lattice site, that commute with the operators on all other sites. So we can divide the Hilbert space up into two Hilbert spaces, one of which consists of the operators \phi(x) for x>0 and the other one with x
answered Jan 29, 2015 by Matthew [ no revision ]

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