Is there an intuitive description of vacuum entanglement? People often refer to the fact that the vacuum is an entangled state (It's even  described as a maximally entangled state).  
I was trying to get a feeling for what that really means.  The problem is that most descriptions of this are done in the formalism of AQFT, which I'm not very familiar with.  The entanglement definitions which I have some feeling for are those of the form 

System S Hilbert space $\mathcal{H}$ factorizes as
  $\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$ where A and B are
  two subsystems of S.  An entangled state can't be written in the form
  $\phi_A \otimes \phi_B$

There are then various measures of this, such as entanglement entropy.
So my question is - is it possible to describe the entanglement of the QFT vacuum in these more familiar terms?
Can such a description be given for a simple QFT example, say a Klein Gordon field on Minkowski space?
 A: If you have a harmonic oscillator in x, the ground state wave function is a Gaussian;
$$ H = {p^2\over 2} + {\omega^2 x^2\over 2} $$
$$ \psi_0(x) = e^{ - {\omega x^2\over 2}} $$
If you have two independent oscillators x,y;
$$ H = {p_x^2\over 2} + {p_y^2\over 2} + {\omega_1^2 x^2\over 2} + {\omega_2^2 y^2\over 2} $$
the ground state is a product:
$$ \psi_0(x,y) = e^{-{\omega_1 x^2\over 2}} e^{-{\omega_2 y^2\over 2}} $$
So there is no entanglement in the ground state between x and y. But if you look at it in a rotated basis (and $\omega_1 \ne \omega_2$), there is entanglement.
For a scalar quantum field in a spatial lattice in finite volume (time is still continuous), you have (if you Fourier transform in space) a bunch of decoupled harmonic oscillators (the sum on k is over non-redundant k's for a real scalar field, this is half the full space $k_x>0$):
$$ H = \sum_k {1\over 2} \dot{\phi_k}^2 + {k^2+m^2\over 2} \phi^2 $$
Which is a bunch of decoupled oscillators, so the ground state is;
$$ \psi_0(\phi_k) = \prod_k e^{-{\sqrt{k^2+m^2} |\phi_k|^2\over 2}} $$
That's not entangled in terms of $\phi_k$, but in terms of the $\phi_x$ (on the lattice), it is entangled. The vacuum wave-function Gaussian can be expressed here as:
$$ \psi_0(\phi) = e^{-\int_{x,y} \phi(x) J(x-y) \phi(y)} $$
Where $J(x-y) = {1\over 2} \sqrt{\nabla^2 + m^2} $ is not the propagator, it is this weird nonlocal square-root operator.
The vacuum for bosonic field theories is a statistical distribution, it is a probability distribution, which is the probability of finding a field configuration $\phi$ in a Monte Carlo simulation at any one imaginary time slice in a simulation (when you make the t-coordinate long). This is one interpretation of the fact that it is real and positive. The correlations in this probability distribution are the vacuum correlations, and for free fields they are simple to compute.
The axiomatic field theory material is not worth reading in my opinion. It is obfuscatory and betrays ignorance of the foundational ideas of the field, including Monte Carlo and path-integral.
General vacuum wave function for bosonic fields
In any path integral for bosonic fields with a real action (PT invariant theory), and this includes pure Yang-Mills theory and theories with fermions integrated out, the vacuum wave-function is the exact same thing as the probability distribution of the field values in the Euclidean time formulation of the theory. This is true outside of perturbation theory, and it makes it completely ridiculous that the rigorous mathematical theory doesn't exist. The reason is that the limits of probability distributions on fields as the lattice becomes fine are annoying to define in measure theory, since they become measures on distributions.
To see this, note that at t=0, neither the imaginary time nor the real time theory has any time evolution factors, so they are equivalent. So in an unbounded imaginary box in time, the expected values in the Euclidean theory at one time slice are equal to the equal time vacuum expectation values in the Lorentzian theories.
This gives you a Monte Carlo definition of the vacuum wave function of any PT invariant bosonic field theory, free or not. This is the major insight on ground states due to Feynman, described explicitly in the path integral and in the work on the ground state of liquid He4 in the 1950s (this is also a bosonic system, so the ground state is a probability distribution). It is used to describe the 2+1 Yang-Mills vacuum in the 1981 by Feynman (his last published paper), and this work is extended to compute the string tension by Karbali and Nair about a decade ago.
A: To answer "Is there an intuitive description of vacuum entanglement?",
we would like to point out to define  entanglement in a quantum theory
(defined by a Hilbert space and a Hamiltonian), we need to assume that the total
Hilbert space is a direct-product of local Hilbert spaces: $\cal{H}_{tot}=\otimes_i \cal{H}_i$.
(For example, in a lattice model, $\cal{H}_i$ can be the Hilbert space on site-$i$.)
Such a direct-product structure can be viewed as an UV completion of a quantum field theory.
Therefore, in order to discuss vacuum entanglement, we need to assume that the total Hilbert space of our universe to have the structure $\cal{H}_{tot}=\otimes_i \cal{H}_i$.
The following discussion is based on such an assumption where the "vacuum" is simply
the ground-state vector in the total Hilbert space $\cal{H}_{tot}$.
The ground states of almost all Hamiltonians are entangled (since those ground states
are in general not product states). So, the vacuum, like a generic ground state, is also an entangled state.
However, the vacuum of our universe is very spectial: our vacuum is actually a 
long-range entangled state, or in other words,
a topologically ordered state.
This is because only long-range entangled states
are known to produce electromagnatic wave that satisfy Maxwell equation and
fermions that satisfy Dirac equations (as collective excitations above the ground state). I wrote an article to discribe this in detail. See also the PE 
question.
So the fact that our vacuum supports photons and fermions (as quasiparticles)
implies that our vacuum is a long-range entangled state.
A: The slides by Summers misuse the conventional terminology (though for a formally justified reason explained below), thereby introducing confusion. 
Entangled states are, by the conventional definition (as given,e.g., by Wikipedia), defined in a tensor product with more than one factor of dimension $>1$.
On the other hand, the vacuum state of a free theory and of any asymptotic representation of an interacting theory is a state defined in a Fock space, which is a direct sum of all tensor product spaces $H_N$ representing the $N$-particle sector ($N=0,1,2,\dots$). By definition, the vacuum state spans the $0$-particle sector, which is a 1-dimensional space and not part of any of the tensor product spaces inside the Fock space. 
Thus it is meaningless (i.e., not backed up by consistent formal definitions) to call the vacuum state entangled in the conventional sense.
To disentangle things further, it may be a good exercise to consider nonrelativistic QM in the second quantization formalism used in statistical mechanics. There the above is seen neatly displayed and interpretable in terms of ordinary multiparticle wave functions, and it becomes clear that Summer's application of the conventional entanglement concept to the vacuum state is spurious.
However, Summers introduces in slide 12 a different entanglement concept adapted to states in a quantum field theory, which applies to the vacuum state. It is loosely related to ordinary entanglement in that the $N=1$ sector of a QFT is represented by 2-point vacuum correlation functions, though none of the states with $N=1$ is a vacuum state. Therefore one can imitate the usual Bell inequality stuff in this framework.
According to this definition, the statements of Summers about the vacuum state make sense. But they should not be confused with ordinary entanglement, as they represent, translated to ordinary QM, statements about pairs of 1-particle states rather than statements about the vacuum. 
Edit: The analogy in which things should be regarded is that in the QFT case, the tensor product is not on the space of states but on a suitably chosen space of operators. This is why the formal Bell-type machinery can be adapted to this situation.
A: For a non-interacting quantum field, the whole mathematical structure of purely Gaussian VEVs that is the vacuum state is contained in the 2-point VEV, which for the KG field is the distribution
$$\left<0\right|\hat\phi(x+y)\hat\phi(y)\left|0\right>=\frac{m\theta(x^2)}{8\pi\sqrt{x^2}}\left[Y_1(m\sqrt{x^2})+\epsilon(x_0)iJ_1(m\sqrt{x^2})\right]-\frac{\epsilon(x_0)i}{4\pi}\delta(x^2)$$
  $$\hspace{7em}+\frac{m\theta(-x^2)}{4\pi^2\sqrt{-x^2}}K_1(m\sqrt{-x^2}).$$
The second line gives the correlation function at space-like separation, where joint measurements are always possible, whereas at time-like or light-like separation the imaginary component of the first line causes measurements to be incompatible. Measurement incompatibility introduces issues that are not easily given an intuitive gloss, of course, but the above shows the nature of the correlations for the free field case.
The Bessel function term at space-like separation is $\frac{1}{4\pi^2(-x^2)}$ at small $x$, whereas it is asymptotically becomes $\sqrt{\frac{2m}{\pi^3\sqrt{-x^2}^3}}\,\frac{\exp{\left(-m\sqrt{-x^2}\right)}}{8}$ for large $x$.
For interacting fields, the 2-point function is always of a comparable form, smeared by a mass density, the Källén–Lehmann representation, but higher order VEVs are relatively nontrivial.
