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Halfwormholes in nearly AdS$_2$ holography
by Antonio M. GarcíaGarcía, Victor Godet
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Authors (as registered SciPost users):  Victor Godet 
Submission information  

Preprint Link:  scipost_202109_00031v2 (pdf) 
Date submitted:  20220115 07:51 
Submitted by:  Godet, Victor 
Submitted to:  SciPost Physics 
Ontological classification  

Academic field:  Physics 
Specialties: 

Approaches:  Theoretical, Computational 
Abstract
We find halfwormhole solutions in JackiwTeitelboim gravity by allowing the geometry to end on a spacetime Dbrane with specific boundary conditions. This theory also contains a Euclidean wormhole which leads to a factorization problem. We propose that halfwormholes provide a gravitational picture for how factorization is restored and show that the Euclidean wormhole emerges from averaging over the boundary conditions. The wormhole is known to be dual to a SachdevYeKitaev (SYK) model with random complex couplings. We find that the free energy of the halfwormhole is strikingly similar to that of a single realization of this SYK model. These results suggest that the gravitational path integral computes an average over spacetime Dbrane boundary conditions.
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minor changes detailed in the replies to the referees
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Reports on this Submission
Report #1 by Anonymous (Referee 3) on 2022122 (Invited Report)
 Cite as: Anonymous, Report on arXiv:scipost_202109_00031v2, delivered 20220122, doi: 10.21468/SciPost.Report.4220
Report
What I wanted to say is that $z$ is an overall shift in the SYK definition of energy: for a generic tensor $J_{ijkl}$, the sum $\sum_{ijkl} J_{ijkl} \psi_i \psi_k \psi_k \psi_l$ can be separated into two parts. The antisymmetric part of $J_{ijkl}$ gives the standard SYK hamiltonian.
The symmetric part of $J_{ijkl}$ yields a constant, since fermions anticommute: $\{\psi_i, \psi_j\} = \delta_{ij}$. Since all fermion operators disappear, the residual sum over $ijkl$ yields $z$. This is a purely analytic argument. Interestingly, eq. (4.8) is indeed the shift in the ground state energy, after cancelling $\arctan$ and $\tan$ and factors of $T$. So one does not need to perform a numerical fit to obtain a precise relation between $j_0$ and $z$(modulo subtleties of taking a $\log $ of a complex $Z$, presumably this is why you have $\arctan(\
tan)$ in eq. (4.8) ). I strongly believe that Section 6 should emphasize this point.
Author: Victor Godet on 20220202 [id 2143]
(in reply to Report 1 on 20220122)Thanks for your comment. I agree that $z$ can be interpreted as a couplingdependent ground state energy. I have modified the end of paragraph below 6.2 to reflect this, reproduced here:
The pattern in the imaginary part of F is indeed a consequence of the branch cut in the argument. This effect results from the fact that the partition is a sum of two exponentials, so we have schematically
Anonymous on 20220206 [id 2161]
(in reply to Victor Godet on 20220202 [id 2143])I thank Authors for the quick response. I recommend the paper for publication.