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  CFT Entanglement Entropy - relation between translations and the stress-energy tensor

+ 1 like - 0 dislike

In a recent paper on CFT entanglement entropy, I want to understand the defintion of a certain partition function. They consider a metric space $S^1 \times \mathbb{H}^{d-1}_q$ with metric:

$$ ds^2_{H_q^{d-1}} = d\tau^2 + du^2 + \sinh^2 u \; d\Omega_{d-2}^2 $$

Here $d\tau$ is probably a Wick-rotated time, $u$ is a radial variable and $d\Omega$ is the spherical area measure.

Then they define a partition function $ Z_q = \mathrm{tr}(e^{-2\pi q H_\tau}) $ where $H_\tau$ "generates translations along the $S^1$". What does that mean? Could it mean this?

$$ H_\tau = \frac{d}{d\tau}$$

This is the generator for translations along the $\tau$-axis.

However, they also say this is related to the stress-energy tensor: $H_\tau = \int_{\mathbb{H}^{d-1}} dx^{d-1} \sqrt{g} T_{\tau\tau}$ This seems like a very complicated way of describing translations. Could there be another meaning for the phrase "generates translations along $S^1$?

  1. Jeongseog Lee, Aitor Lewkowycz, Eric Perlmutter, Benjamin R. Safdi
    Renyi entropy, stationarity, and entanglement of the conformal scalar
This post imported from StackExchange Physics at 2014-08-10 19:42 (UCT), posted by SE-user john mangual
asked Jul 31, 2014 in Theoretical Physics by john mangual (310 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

Everything is standard here. Think simply to a classical field on a Minkowski space $M = R_t \times R^3$.

The time translation generator, the hamiltonian, is defined by $H_t = P_0=\int T_{0i} d\sigma^i$, where $d\sigma^i = \epsilon^{ijkl} dx^j \wedge dx^k \wedge dx^l$.

Now, we may, in fact, consider only the $d\sigma^o= d\sigma^t$ component which is equals to $dx^1 \wedge dx^2 \wedge dx^3$, so finally :

$H_t = \int d^3xT_{00} = \int d^3xT_{tt}$

This post imported from StackExchange Physics at 2014-08-10 19:42 (UCT), posted by SE-user Trimok
answered Aug 1, 2014 by Trimok (955 points) [ no revision ]
I think what I am missing is that you still need an action to define the CFT, and your equation computes the time evolution Hamiltonian from this action.

This post imported from StackExchange Physics at 2014-08-10 19:42 (UCT), posted by SE-user john mangual

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