# Is there an intuitive description of vacuum entanglement?

+ 8 like - 0 dislike
4515 views

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?

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59

recategorized Mar 24, 2014

+ 5 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Xiao-Gang Wen
answered Jul 19, 2012 by (3,485 points)
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?

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59
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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Xiao-Gang Wen
Is "vacuum entanglement" properly so called?Test
+ 4 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
answered Jul 15, 2012 by (7,720 points)
@ArnoldNeumaier: I also take objection to calling the Feynman integral ill defined, or saying that continuation to imaginary time "turns" it into anything. Feynman was doing a slight continuation (an i\epsilon rotation) in the path integral right from the beginning, and he and Kac understood the connection to Weiner stochastic path integration from the beginning. The path integral was always defined using this rotation (at least infinitesimally) and so was always well defined.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
@RonMaimon: You probably refer to the free functional integral, whereas I refer to the one with a nonquadratic action in the exponent. If this functional integral were well-defined in 4D, QED, QCD, and the standard model would exist with mathematical rigor. But many physicists think that QED does not exist rigorously, and there is a one million dollar price for proving existence just for Yang-Mills.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Arnold Neumaier
@RonMaimon: Note that Summers makes in his slides no assumption that the QFT is free. Your explanation only addresses the free case, though. This is inadequate as the interacting correlation functions are determined by the noninteracting ones only through a power series whose construction involves renormalization with all its difficulties, and whose convergence is almost certainly not the case. Summers works nonperturbatively, hence isn't troubled by that.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Arnold Neumaier
@ArnoldNeumaier: The vacuum probability distribution exists in pure gauge theory, and in any real bosonic theory, free or not. It exists in QED after you integrate out the Fermions, but then you have nonlocal action. The identity of the ground state wavefunction with the probability distribution in Euclidean space is general, it does not require perturbation theory, it is exploited by Feynman and later Karabali and Nair to describe the 2+1 pure Yang-Mills gauge vacuum. You are right that given this identity, it is completely ridiculous that there is no mathematical theory of QCD.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
let us continue this discussion in chat

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Arnold Neumaier
@twistor59: This is one of the fundamental tools of path-integrals. I said it slightly wrong in the comments (the thing I said is true when you add a gradient term to make the Euclidean theory time-derivatives pure forward derivatives, it isn't true when the time-derivatives are centered derivatives, then the ground state wavefunction is just the same as the probability distribution for a field in Euclidean space, not the square root of this, but these details are silly to go into at this level).

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
The formula doesn't work for interacting field theories (obviuosly), but you don't need a formula to calculate the correlations in the interacting vacuum. The interacting vacuum entanglement in a bosonic theory is still determined by the probability distribution on a slice in Euclidean space. The fact that AQFT people pretend to talk about this quantity in high-falutin' terms full of big words and little brains without ever mentioning the one and only exact solution they are mining to get all their results shows me that they are obscurantists frauds, not worth paying attention to.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Arnold Neumaier
answered Jul 15, 2012 by (15,787 points)
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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59
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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
@RonMaimon: But it is what Summers says vacuum entanglement is. As I said, there are multiple notions.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Arnold Neumaier
@twistor59: I added a remark at the end to showin which sense one can reconcile the two notions.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Arnold Neumaier
@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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon
+ 2 like - 0 dislike

Regarding your question of an intuitive description of vacuum entanglement, I would like to mention Maldacena's ads-cft-correspondence.

The vacuum entanglement yields spacetime and gravity (intuitively spoken), see e.g. Mark van Raamsdonk, Gravity and Entanglement, (especially around minute 19).

answered Dec 5, 2016 by (90 points)
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Peter Morgan
answered Jul 15, 2012 by (1,230 points)
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?

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59
@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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Peter Morgan
@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).

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Peter Morgan
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?

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59
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").

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Peter Morgan
@twistor59: The rigorous theorem is not so surprising, and this answer is too formal.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user Ron Maimon

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysi$\varnothing$sOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification