4
$\begingroup$

In large N strong coupling limit of AdS/CFT, We know that not every CFT state has a bulk dual described by a classical metric. What about the converse? Does any asymptotically AdS classical bulk metric have a dual CFT state?

$\endgroup$
2
  • $\begingroup$ Generic bulk metrics do not have the same assymptotic symmetries as AdS space and even if they were dual to some QFT, it would most likely not be a CFT. $\endgroup$
    – Heidar
    Jan 19, 2020 at 16:56
  • $\begingroup$ @Heidar Sorry. I mean AlAdS metric. $\endgroup$
    – Shadumu
    Jan 19, 2020 at 17:15

1 Answer 1

2
$\begingroup$
  1. Given a state in the CFT, one can either causally reconstruct a portion of the bulk using HKLL reconstruction, or obtain the Einstein equations from the CFT using RT formula. In both cases we have a bulk description. So the first statement in your question is not correct.

  2. The basic premise of AdS-CFT is that upon correct matching of symmetries, spectrum and operators in the theory according to the holographic dictionary, there is an equivalence between observables computed in both theories, say correlation functions. This follows from the fact that the partition function on both sides upon using the dictionary match, and by its very nature this duality is at a quantum level. You can see that the theory on the AdS side cannot be just classical gravity, as is raised in your question.

  3. There are deeper questions however on whether the CFT potentially contains all possible bulk states in the full quantum gravity description, i.e. whether all states in the AdS side map to those on the CFT side (I think this was probably what you wanted to ask). A claimed aspect of this question in the language of AMPSS (note that this is the second paper by the firewall people, the first one is just plain wrong) is that the states generated by the action of operators behind the horizon of a black hole are not present in the CFT. However it was shown by PR (also see preceding papers) that there are state-dependent maps from the bulk to the CFT which can describe the behind horizon operators (which I consider is most likely correct).

  4. Again for black holes, there can be bags-of-gold paradox that there can be too many excitations in the AdS bulk. So again there is a question on whether all these excitations are actually present in the CFT description or not (I personally think they are actually present in the CFT description, one does need to cleverly describe them though).

$\endgroup$
5
  • $\begingroup$ Thanks. I am not asking about full QG in the bulk, but rather if all classical geometry has a corresponding state on the boundary. What is your answer to that? Regarding your point 1, it is true that you can do reconstruction, but it is certainly not true that every state has a CLASSICAL bulk dual. A superposition of two BHs with different masses is a counterexample. Do you agree? $\endgroup$
    – Shadumu
    Jan 26, 2020 at 17:41
  • $\begingroup$ @Shadumu 1. I stated in point 1 two different ways how corresponding to a state in the boundary you get the classical geometry in the bulk. The map from bulk states description to the boundary is not a settled business yet for arbitrary geometries (See points 3 and 4). To get the classical limit in the bulk, you need to to perform a semi-classical quantization of metric perturbations and construct coherent states in the Hilbert space, and the expectation value of the metric operator on this coherent state will give you a classical value $\langle \hat{g}_{\mu \nu}\rangle = g^{cl}_{\mu \nu}$ . $\endgroup$
    – Bruce Lee
    Jan 27, 2020 at 0:58
  • $\begingroup$ But as I said the complete map is not a settled business yet, so there is no general notion. 2. The dual description of a Euclidean black hole is that of a CFT living on a manifold, whose temperature determines the manifold in question. The basic problem with superposing 2 BHs with different masses here is that they would belong to two different CFTs, and you cannot add vectors from two different Hilbert spaces like that. $\endgroup$
    – Bruce Lee
    Jan 27, 2020 at 1:10
  • $\begingroup$ The two ways you mentioned in point 1 worked under the assumption of existence of a geometric bulk dual. Surely not every CFT state has a classical bulk dual, right? Why do you say the first statement in my question is not correct? $\endgroup$
    – Shadumu
    Jan 27, 2020 at 12:29
  • $\begingroup$ The dual of BH is a thermal state in the CFT. BHs with different masses are thermal states with different temperatures, why would they belong to two different CFTs? Surely one can have ensembles at different temperatures in one CFT. $\endgroup$
    – Shadumu
    Jan 27, 2020 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.