# 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?

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I accidentally realized this site. Maybe, for the answer one shall take a glance on the following manuscript. Mutual emergence of noncausal optical response and nonclassicality in an optomechanical system arxiv.org/abs/1502.01294 Best regards. –  Mehmet Emre Tasgin Feb 16 at 18:05

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.

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Thanks for your answer. In the OP, I was really trying to understand the sense in which the term "entanglement" was being applied to the vacuum of conventional relativistic QFT - 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. After the answers above I'm OK with this now. Am I right in saying that your description of the vacuum here, in terms of topological ordering etc. is a very special one, specific to a particular model - string-net theory? –  twistor59 Jul 19 '12 at 7:11
You made a very good point in your comment about the subsystem. I update my answer to make it more precise. I did not answer your original question directly since I feel that the issue about the subsystem (or the direct-product structure of the total Hilbert space) is more important. My description of the vacuum is not a special one. I simply assume the total Hilbert space to have a direct-product structure and the Hamiltonian to be local respect to the direct-product structure. Under such two general assumptions, the vacuum must be long-range entangled to have emergent photons and fermions. –  Xiao-Gang Wen Jul 19 '12 at 12:27

If you have a harmonic oscillator in x, the ground state wavefunction 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 nonredundant 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 wavefunction 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 wavefunction 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 probaility 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.

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I have never seen anyone measure probabilities for finding something in imaginary time. It is purely formal analogy. - Your low esteem for axiomatic field theory is not well-grounded, but just an indication that you don't know its language well enough to translate its insights into ordinary QFT terms. –  Arnold Neumaier Jul 16 '12 at 7:38
@ArnoldNeumaier: I know the language of axiomatic field theory, I can translate every statement. It is stupid, it is formal, and it is trivial. You should know that the ground state wavefunction of a bosonic field (or any bosonic variable) is equal to the square-root of the probability density for finding this field configuration in imaginary time along a time slice. This is an exact identity. This is what allows Schwinger to write the vacuum wave-functional of QED in some physical gauge. –  Ron Maimon Jul 16 '12 at 7:57
How then do you explain that I learnt a lot from AQFT that I couldn't discern from the traditional QFT treatment a la Peskin/Schroeder or Weinberg? Lots of interesting stuff is going on in AQFT, it just doesn't match your interests. –  Arnold Neumaier Jul 16 '12 at 8:16
@ArnoldNeumaier: I think this is incorrect. When I look at AQFT, I see results which are obfuscated to be maximally unclear, because of a phony attention to rigor which is unwarranted at this stage, because the construction isn't there. The relation to statistical theory is most important, because it is a practical tool for rigorously defining the theory and also because it makes relations which are obscure in the AQFT language obvious. Can you tell me what's contained in AQFT now? I haven't been following recent developments. –  Ron Maimon Jul 16 '12 at 9:21
Interesting - I hadn't heard of this technique. I guess it's the imaginary time propagation method described (for the Schroedinger eqn) here? –  twistor59 Jul 16 '12 at 10:19

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.

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Thanks, that's very illuminating. So the answer to my original question is "no" it's not possible to phrase the vacuum entanglement referred to by Werner and Summers in conventional ($\mathcal{H}_A \otimes \mathcal{H}_B$) entanglement terms. –  twistor59 Jul 15 '12 at 18:26
This answer is not describing what people mean by vacuum entanglement--- they mean the entaglement between different field states in the Schrodinger wave-functional. This is something which doesn't refer to asymptotic particle states, and you can work it out in free field theory easily. –  Ron Maimon Jul 15 '12 at 21:34
@RonMaimon: But it is what Summers says vacuum entanglement is. As I said, there are multiple notions. –  Arnold Neumaier Jul 16 '12 at 7:28
@twistor59: I added a remark at the end to showin which sense one can reconcile the two notions. –  Arnold Neumaier Jul 16 '12 at 7:34
@ArnoldNeumaier: No it isn't. I looked at what Summers says, he just says that the vacuum is entangled with respect to the local observables at two separated regions. For a free field, this is the same as the entanglement when you rotate the unentangled k-state into the x-state field basis. It's a not so enlightening statement. –  Ron Maimon Jul 16 '12 at 8:08

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.

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Thanks for the response! So the states $\phi(x)|0\rangle$ (localized at x) and $\phi(x+y)|0\rangle$ (localized at x+y) are correlated to the extent given by the formula. My understanding of the maximal entanglement of the vacuum was that you could take any state localized at x+y and construct it just by applying operators local to x on the vacuum (OK we should really be talking regions and smearing, but take that as read). Is there any way to demonstrate that explicitly in the KG case? –  twistor59 Jul 15 '12 at 16:47
@twistor59 Apologies, but I'm going to split some hairs. $\hat\phi(x)|0\rangle$ is a vector-valued distribution, not a state. A state is a positive map such as $\hat A\mapsto\langle 0|\hat\phi_f^\dagger\hat A\hat\phi_f|0\rangle$. A state is not a local object in the sense that it tells you what results you would expect if you make a local measurement, such as $\hat A$, wherever that measurement is made. –  Peter Morgan Jul 15 '12 at 17:11
@twistor59 The Reeh-Schlieder theorem, of which you speak, is subtle. It says that if we construct vectors using only local operator-valued distributions $\hat\phi(x)$ acting on the vacuum vector, with $x$ in some region $\mathcal{O}$, that vector space is a dense subspace of the whole Hilbert space. So for any given vector in the Hilbert space, we can approximate it as well as we like using only resources we can construct in $\mathcal{O}$. To do this, one "must judiciously exploit the small but nonvanishing long distance correlations" in the vacuum state (Haag, Local Q Physics, II.5.3). –  Peter Morgan Jul 15 '12 at 17:23
Thanks Peter. I think I need to read up a bit on Wightman-style rigorous QFT to get more used to the terminology! Yes, it was the Reeh-Schlieder theorem I was talking about. I'm getting the expression, though, that its statements remain in the realm of existence proofs - i.e. they can't be made explicit in terms of the "chug and plug" calculations of elementary QFT? –  twistor59 Jul 15 '12 at 17:31
The only fields in 3+1 that are known to be Wightman fields are free quantum fields, so to that extent yes. It would be nice to be able to construct regularization and renormalization with enough precision to decide whether a Reeh-Schlieder-like theorem is satisfied for "chug and plug" interacting QFTs (also known as Lagrangian QFT, I hear, but I would never call them "elementary"). –  Peter Morgan Jul 15 '12 at 17:51