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Background:

The following question is from page 131 of Tom Hartman's notes on Quantum Gravity and Black Holes.

The GKPW dictionary states that

$$Z_{\text{grav}}[\phi_{0}^{i};\partial M] = \langle \exp \left( - \frac{1}{\hbar} \sum_{i} \int d^{d}x\ \phi_{0}^{i}(x)O^{i}(x) \right) \rangle_{\text{CFT on } \partial M},$$

where

  1. the LHS is the gravitational path integral in the bulk $M$ for fields $\phi^{i}(x)$ with boundary conditions $\phi^{i}(\partial M) = \phi_{0}^{i}(x)$, and

  2. the RHS is the CFT partition function on the boundary for operators $O^{i}$ sourced by fields $\phi^{i}_{0}(x).$


Question:

The usual form for the partition function $Z[J]$ of a quantum field theory with a source $J$ for a scalar field $\phi$ is given by

$$Z[J] = \int \mathcal{D}\phi \exp\left( -\frac{1}{\hbar}\left[ S[\phi] + \int d^{d}x\ J(x)\phi(x) \right] \right).$$

How would you then prove that, for fields $O^{i}(x)$ with sources $\phi^{i}_{0}(x)$, we have that

$$\langle \exp \left( - \frac{1}{\hbar} \sum_{i} \int d^{d}x\ \phi_{0}^{i}(x)O^{i}(x) \right) \rangle = \int \mathcal{D}O \exp\left( - \frac{1}{\hbar} \left[ S_{\text{CFT}}[O^{i}(x)] + \int d^{d}x\ \phi^{i}_{0}(x)O^{i}(x) \right] \right)?$$

Furthermore, is the quantum state over which you are taking the expectation value in the LHS given by the boundary conditions on the Euclidean path integral on the RHS?


Attempt:

My understanding is that the LHS is just the usual way of writing the partition function (in statistical mechanics) as the trace over all the quantum states of the exponential of source times operator, while the RHS is just the usual way of writing the partition function in quantum field theory.

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  • $\begingroup$ $e^{A+B} = e^Ae^B$ $\endgroup$ – user1504 Jun 27 '17 at 0:42
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    $\begingroup$ It is a definition of the expectation brackets $<\dots>$. Also, what user1504 said. $\endgroup$ – Prof. Legolasov Jun 27 '17 at 3:51
  • $\begingroup$ 1. You'll have to give your definition of $\langle\dots\rangle$ for this question to make sense. 2. Even if you do, it's still just asking about a computation - please a) give references for why you think the claims you want to prove are true, b) show some effort to show the statement yourself and c) try to ask a more specific question about a particular step or problem that arises during your attempt. $\endgroup$ – ACuriousMind Jun 27 '17 at 8:45
  • $\begingroup$ I've edited my question. $\endgroup$ – nightmarish Jun 27 '17 at 11:47
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    $\begingroup$ You have still not defined $\langle \dots \rangle$. The point is that, in the path integral formalism, it is usually defined to be $\langle O \rangle = \int O \exp(-\mathrm{i}S[\phi])\mathcal{D}\phi$ for any operator $O$, which trivially yields the equation you want to prove as per user1504's comment. It's not clear why you think the actual relation should be what you write; it appears your problem is with how one takes expectation values of operators in the path integral formalism in general, and has nothing to do with the specific AdS/CFT theories you are ostensibly asking about. $\endgroup$ – ACuriousMind Jun 27 '17 at 13:25

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