Why are general QFT states locally similar to the vacuum?

Question

Recently, when reading about the Black Hole Information Paradox, many authors have said that any state in a quantum field theory "looks" like the vacuum at sufficiently short distances. One example from a recent review by Marolf is

First recall that at sufficiently short distances any state of our quantum field will be well approximated by the vacuum. We must then also recall that the vacuum of quantum field theory contains divergent ultraviolet correlations between spacelike separated points.

I have not managed to find a detailed explanation why this should be so, either by doing a quick search on google or looking at my QFT textbooks, so I would appreciate if either a detailed explanation (with all the maths) or at least a heuristic (with maybe a link to some book/article) could be given.

Answer 1

First recall that at sufficiently short distances any state of our quantum field will be well approximated by the vacuum. We must then also recall that the vacuum of quantum field theory contains divergent ultraviolet correlations between spacelike separated points.

Given that Marolf's review concerns quantum fields in the strongly curved space-times of black holes, I think this statement may concern the fact that physically reasonable field states $\omega$ are Hadamard states satisfying the Hadamard condition. That is, the singularity structure of their 2-point correlation function $\omega(\phi(x) \phi(y))$ has the general form $$ \omega(\phi(x) \phi(y)) = \frac{u}{\sigma} + v\log\sigma + w \tag{*} $$ where $\sigma(x, y)$ is the square geodesic distance between points $x$ and $y$, and $u$, $v$, $w$ are smooth functions.

The reason is that if the Fock space is constructed starting from a Hadamard vacuum state $\omega_0$, with a 2-point function $\omega_0(\phi(x) \phi(y))$ satisfying (*), then any allowed physical state $\omega$ will inherit the same singularity structure for its 2-point correlations and the same type of short range (ultraviolet) behavior as the vacuum state. In other words, at short distances physically allowed states "look like the vacuum" and share its divergent ultraviolet correlations.

In fact, the Hadamard condition is imposed precisely to ensure this behavior, because this is what is required to construct a meaningful (finite) stress-energy tensor. Specifically, finite stress-energy requires a finite $\omega(\phi^2(x))$, which is renormalized by "subtracting the vacuum" $$ \omega(\phi^2(x)) \sim \lim_{x\rightarrow y} \Big(\omega(\phi(x) \phi(y)) - \omega_0(\phi(x) \phi(y)) \Big) < \infty $$ If the Fock states would have a different behavior at small scales compared to the vacuum, they would compromise the energy spectrum.

Source: A. Wipf, Quantum fields near black holes

Note: According to an argument by Srednicki, for a field in a vacuum state the entropy of entanglement of the degrees of freedom in a bounded region of space with the degrees of freedom outside that region is proportional both to the ultraviolet cutoff of the theory and to the area of the boundary of said region (no, no black hole involved). A similar argument may be construed for two spacelike separated bounded regions and their mutual entanglement entropy. But Marolf's reference is to spacelike separated points, which in this context would mean vanishing boundary areas, etc. So I think it's much more probable that his reference to divergent ultraviolet correlations concerns instead the obviously divergent 2-point function.

Answer 2

I'm not sure I agree with the claim. But let's pretend that the claim is correct, and let's try to understand what it means, at least from a heuristic point of view.

The first step is to properly characterise the vacuum. What do we mean by $|0\rangle$, the vacuum state of the theory? In flat space-time, a state is called the vacuum if and only if it satisfies $$ P^\mu|0\rangle=M^{\mu\nu}|0\rangle=0 $$ where $P^\mu$ are the generators of translations, and $M^{\mu\nu}$ are the generators of Lorentz transformations. The states $|0\rangle$ that satisfy the conditions above are typically highly degenerate, and that is not really a problem (contrary to what one may think). But this is irrelevant here, so let's move on.

Given an arbitrary (normalised) state $|\psi\rangle$, we can construct two tensors that give us the characteristic energy and angular momentum of the state, $$ p^\mu=\langle \psi|P^\mu|\psi\rangle\qquad\text{and}\qquad m^{\mu\nu}=\langle \psi|M^{\mu\nu}|\psi\rangle $$ the first having units of energy, and the second being dimensionless.

If we probe the state $|\psi\rangle$ with a very energetic particle, say, with energy $E\gg p^\mu$, then we can certainly neglect the energy of $|\psi\rangle$, to within an accuracy $\mathcal O(p/E)$. In this sense, when we look at $|\psi\rangle$ from a very short distance, it looks like it has $p^\mu\sim 0$. In fact, any state of finite-energy state may me taken to have zero energy if we are probing it with a high enough energy.

This means that the condition $$ P^\mu|\psi\rangle\approx0 $$ is approximately satisfied for any state of finite energy, as long as we look at it from a very short distance.

On the other hand, no matter how large the energy of our probe is, the tensor $m^{\mu\nu}$ will never look like $m^{\mu\nu}\sim 0$. For one thing, the eigenvalues of $M^{ij}$ are half-integers, which are not continuously connected to zero. Therefore, $$ M^{\mu\nu}|\psi\rangle\not\approx0 $$ in general.

But, and there is a big but, the condition $M^{\mu\nu}|0\rangle=0$ need not hold in general relativity: in curved space-time, the vacuum may have non-zero angular momentum! This fact is discussed in Strominger's notes here. I'll try to find a more specific reference, but for now this is the best one I can find.

The punchline is that any state $|\psi\rangle$ approximately satisfies the conditions for it to be a vacuum state, at least if we look at it from a very short distance. This is more or less what we wanted to understand. I'd love to see a more technical discussion of the claim in then OP, but I hope that this heuristic argument is more or less convincing.