From the Penrose diagram of de Sitter space, we see it has a future and past conformal boundary, and they are both spacelike. So, does de Sitter space admit an asymptotic S-matrix? Sure, in the usual coordinates which only cover half of the full space, we don't have an S-matrix because it's not causally complete, but in a coordinate system which covers the entire space, why not? Just because we don't have a globally timelike Killing vector field doesn't mean we can't have an S-matrix.

# [Physics] Does de Sitter space admit an asymptotic S-matrix

de-sitter-spacetimequantum-field-theorys-matrix-theory

#### Related Solutions

You're right that the definition of a black hole based on an event horizon is non-local - you need to examine the entire future evolution to check if a light ray makes it to $\mathcal{I}^+ $. However, more local alternative definitions have been discussed, such as those involving *apparent horizons*: You basically define a closed two-surface to be trapped if, when you look at the pair of future-oriented null directions which are orthogonal to it, you find that they're converging as the surface evolves in time. A point on a timeslice is trapped if it lies on a trapped surface in the timeslice. The apparent horizon is then the boundary of the union of the trapped points.

This just makes use of local quantities - the test for convergence of the null directions is computed using derivatives only. No need to wait for future null infinity! More discussion and refinements here.

A simple and direct answer to the question would be to say that the timelike boundary can't be the issue, because the same paradox can be constructed using a combination of past and future spacelike boundaries:

The full answer is that a timelike boundary is perfectly reasonable, and neither of these situations are paradoxes, as long as you're careful.

The path integral is a function of the boundary conditions. When you're sewing / composing the two regions, the boundary $\Sigma$ between those two regions stops being a boundary, and becomes just more points internal to the region. So to compose them you need to integrate over all the possible field states on $\Sigma$, with the integrand being the product of the amplitudes for each of the two regions:

$$ \langle\phi_2|U_{1\to2}|\phi_1\rangle = \int \mathcal D \phi(\Sigma)\langle\phi_2|U_{\Sigma\to2}|\phi_\Sigma\rangle\langle\phi_\Sigma|U_{1\to\Sigma}|\phi_1\rangle $$

That's how the influence of $J$ on $\Sigma$ interacts with the influence of $\Sigma$ on $\Sigma_2^L$. This does involve the influence going "backwards" from the second region to the first region, but that's fine. Something like that *has* to happen given the spacetime diagram, and as you said, it works fine formally if you just think of it in terms of "fields in, amplitudes out".

To take a concrete situation, say that $|\phi_1\rangle$ is a vacuum state, and $J$ produces some photons which will pass through $\Sigma_2^L$. So $U_{\Sigma\to2}$ tells us that there is a large amplitude for $\phi_\Sigma$ to have photons passing through it (informally speaking). If instead $J$ were 0, there would be a large amplitude for $\phi_\Sigma$ to be undisturbed. And if $U_{1\to\Sigma}$ has a boundary condition where photons are coming through $\Sigma$, it will give large amplitudes for photons passing through $\Sigma_2^L$. Otherwise, if $\Sigma$ is undisturbed, there will be large amplitudes for vacuum. So when you perform the integral over all the possible boundary conditions, whether there's a large value for vacuum or for photons at $\Sigma_2^L$ depends on whether there is a large amplitude vacuum or for photons at $\Sigma$. Which in turn depends on $J$.

The timelike part of the boundary is acting like a weird mixture of initial-value and final-value. But the integral of the product of amplitudes is symmetrical, so it doesn't really care which one comes before the other, and doesn't care whether they have a causally messy relationship, like in this case.

One thing to note is that $U_{1\to\Sigma}$ and $U_{\Sigma\to2}$ aren't necessarily unitary, despite the naming. One other thing to note is that you can specify fields on $\Sigma$ that are actually inconsistent with each other, which should cause the path integral to spit out infinitesimal amplitudes in those cases.

## Best Answer

Dear inflation, the answer is No, one can't define an asymptotic S-matrix in de Sitter space. See e.g. section V.A. in this paper by Bousso:

The problem is that the Penrose diagram of a de Sitter space starts and ends with a horizontal line. So at the beginning, the observer is forbidden - by causality - to set up the initial state on the initial horizontal line (because he can only affect its future light cone, a small triangle); and at the end, for a mirror reason, he can't measure what happened.

So at most, if such an S-matrix existed, it would dictate the evolution of states that no one can prepare to states that no one can observe, as Bousso says. This S-matrix would not be an observable, it would be a "meta-observable".

There are good reasons to think that this is not just an aesthetic conflict with the "testability" criteria in science. It's usually the case in physics that if one can prove that something can't be measured, not even in principle, it doesn't make sense. Attempts to measure/define the values of $x,p$ with a high accuracy that violates the uncertainty principle is a canonical example. We know that it's not just because of our technological limitations; accurate values of $x,p$ simply can't exist at the same moment. The operators don't commute with one another, after all.

In the case of the hypothetical de Sitter S-matrix "meta-observable", the experience with the uncertainty principle suggests that e.g. the "arbitrary" initial state indeed can't be prepared - surely not by an observer, as argued above, but not even by God or Nature - because the degrees of freedom that belong to vastly separated regions on the infinite initial line are not really independent of each other. This notion is known as the complementarity - and while it's usually talked about in the case of black holes, de Sitter space offers a straightforward generalization.

De Sitter space may be thought of as a black hole but the "inner visible part" of the Universe should be identified with the black hole

exteriorwhile the black holeinterioris the infinite region behind the cosmic horizon of the de Sitter space. Much like the microscopic information about the black hole interior is ultimately imprinted in the Hawking radiation, one should believe that all the information about physics beyond the cosmic horizon is imprinted to detailed properties of particles in the visible patch.For similar reasons, the fact that the de Sitter horizon has a finite entropy - because the cosmic horizon has a finite area - should be taken pretty seriously and it arguably applies to the whole space and not just the visible patch. Because of the finiteness of the entropy, it seems that one can't define

anyobservable that is arbitrarily accurate and that depends on continuous parameters such as momenta, not even in principle.