De Sitter space is sometimes not empty

Abstract

Abstract Multiple lines of evidence suggest that the Hilbert space of an isolated de Sitter universe is one dimensional but can appear larger when probed by a gravitating observer. To test this idea, we compute the von Neumann entropy of a field theory in a two-dimensional de Sitter universe which is entangled in a thermal-like state with the same field theory on a disjoint, asymptotically anti-de Sitter (AdS) black hole. Previously, it was shown that the replica trick for computing the entropy of such entangled gravitating systems requires the inclusion of a non-perturbative effect in quantum gravity — novel wormholes connecting the two spaces. Here we show that: (a) the expected wormholes connecting de Sitter and AdS universes exist, avoiding a no-go theorem via the presence of sources on the AdS boundary; (b) the entanglement entropy vanishes if the nominal entropy of the de Sitter cosmological horizon $$ \left({S}_{\textrm{dS}}={A}_{\textrm{horizon}}^{\textrm{dS}}/4{G}_{\textrm{N}}\right) $$ S dS = A horizon dS / 4 G N is less than the entropy of the AdS black hole horizon $$ \left({S}_{\textrm{BH}}={A}_{\textrm{horizon}}^{\textrm{AdS}}/4{G}_{\textrm{N}}\right) $$ S BH = A horizon AdS / 4 G N , i.e., SdS< SBH; (c) the entanglement entropy is finite when SdS > SBH. Thus, the de Sitter Hilbert space is effectively nontrivial only when SdS > SBH. The AdS black hole we introduce can be regarded as an “observer” for de Sitter space. In this sense, our result is a non-perturbative generalization of the recent perturbative argument that the algebra of observables on the de Sitter static patch becomes nontrivial in the presence of an observer.