# [1911.12333] [Almheiri, Hartman, Maldacena, Shaghoulian, Tajdini] Replica Wormholes & Information Paradox

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## Refs

- JT
  - Harlow & Jafferis (H&J), 1804.01081
  - Gábor Sárosi, 1711.08482
- Island & Wormholes: 2006.06872
- Calculations:
  - 1911.12333

## $\mathrm{AdS}_2$ Basics

- Euclidean Poincaré:

  $$
    \dd{s}^2
    = \frac{\dd{x} \dd{\bar{x}}}{z^2},
  \quad
    z = \frac{x + \bar{x}}{2},
  \quad
    z < 0
  $$

  Note that we choose to look at the $z < 0$ region for future convenience. Poincaré horizon:

  $$
    0> z \to -\infty,
  \quad
    \mcal{I}^\pm
    \colon\ x^\mp \to -\infty
  $$

- Rindler:

  $$
    x = \tanh ky,
  \quad
    x \to -1,
  \quad
    y = \sigma + i\tau \to -\infty
  $$

  $y = \sigma + i\tau$ is similar to some cylindrical coordinate.
  $\tau\colon$ Euclidean time, $\tau = it$, Lorentzian $y^\pm = \sigma \pm t$. Note that $y^+ = \bar{y},\ y^- = y$.

  $k$ is related to the temperature. To see this explicitly, compute the induced metric in Euclidean signature, and we get:

  $$
    \dd{s}^2
    = \frac{
        k^2 \dd{y} \dd{\bar{y}}
      }{
        \pqty{
          \tfrac{1}{2}\,
          \sinh 2k\sigma
        }^2
      },
  \quad
    \sigma = \frac{y + \bar{y}}{2}
  $$

  - $\sigma \to -\epsilon$, $\dd{s}^2 = \frac{1}{\epsilon} \dd{y} \dd{\bar{y}}$, we see a cutoff boundary with flat metric.

  - $\sigma \to -\infty$, i.e. near the Euclidean "horizon", after some trial and error, we see that if we set $\rho = e^{2k\sigma} \to 0$, then $\dd{\rho} = 2k \rho \dd{\sigma}$,

    $$
      \dd{s}^2
      \sim \frac{
          (2k)^2
        }{
          \pqty{
            \tfrac{1}{2}\,
            e^{-2k\sigma}
          }^2
        }
        \pqty{
          \frac{\dd{\rho}^2}{(2k)^2\rho^2}
          + \dd{\tau}^2
        }
      = 4\,(
        \dd{\rho}^2 + \rho^2 (2k)^2 \dd{\tau}^2
      )
    $$

    i.e. the Euclidean geometry is smooth around $\rho \to 0$ if $2k\beta = 2\pi$, or $k = \frac{\pi}{\beta}$. In fact, one can get the standard Poincaré disk with:

    $$
      w = e^{\frac{2\pi}{\beta} y},
    \quad
      \dd{s}^2
      = \dd{w}\dd{\bar{w}}
        \bigg/
        \pqty{
          \frac{1 - \abs{w}^2}{2}
        }^2
    $$

- The Rindler patch is in fact exactly the "right half" of a two sided "black hole". To see this, consider:

  $$
    x = \tan u,
  \quad
    \begin{aligned}
      x &\to -1,
    \quad u \to -\tfrac{\pi}{4},\\
      x &\to -\infty,
    \quad u \to -\tfrac{\pi}{2},
    \end{aligned}
  $$

  $$
    \dd{s}^2
    = \frac{
        \dd{u} \dd{\bar{u}}
      }{
        \pqty{
          \tfrac{1}{2}\,
          \sin \,(u + \bar{u})
        }^2
      },
  \quad
    \sigma = \frac{y + \bar{y}}{2}
  $$

  This is, in fact, global $\mathrm{AdS_2}$, with two asymptotic boundaries at $\frac{u + \bar{u}}{2} = 0, -\frac{\pi}{2}$.









## JT Gravity

AdS/CFT: too strong, not realistic! $\leadsto$ holographic systems. How do we see that some theory of quantum gravity is holographic or not? Case study: JT gravity.

Different UV completions of JT gravity (H&J):

### Canonical quantization

JT gravity is renormalizable!

$$
  S \sim \frac{1}{16\pi G_N}
    \int \dd[2]{x}
      \Phi\,(R + 2) + \cdots
$$

Classical power counting: mass dim $
  [g] = 0, [x] = -1
$, $
  [\Phi] = 0,
  \ [R] = +2,
  \ [G_N] = 0
$.

<img
  src="img/JTblackhole.png"
  class="center"
  style="width: 180px; max-width: 80%;"
/>

$$
  0 = \var{S}
  = \int_M \dd[2]{x} \sqrt{-g}
      \ (\mathrm{EOM})
    + \int_{\pd M} \dd{x} \sqrt{\abs{\gamma}}
      \ (\mathrm{BC})
$$

Boundary conditions (BC) at the cutoff $r_c$ --- asymptotic $\mathrm{AdS}_ 2$. Classical solution: two-sided black hole. Singularities emerge at the $\infty$-ly blue-shifted surface. Given such BC, the solution to the EOM is _unique!_ With only one free parameter.

Details: H&J, Section 2. Conformal compactification: $x = \tan z$,

$$
  \dd{s}^2
  = (\sec{z})^2 \pqty{
      -\dd{\tau}^2 + \dd{z}^2
    },
  \quad z\in (-\tfrac{\pi}{2}, +\tfrac{\pi}{2}),
\\
  \Phi = \Phi_h \sec z \cos \tau
$$

Alternatively, use coordinates in the right patch: horizon $r_s = \Phi_h / \phi_b$,

$$
  T^{tt}_\mathrm{CFT}
  = H = \frac{2\Phi_h^2}{\phi_b}
$$

**Lorentzian path integral: no sum of topologies!** Classical solution is completely fixed by $\Phi_h$, or $r_s$, or equivalently, $H$. One can thus canonically quantized such system like any QM system: find dynamical d.o.f. and promote the Poisson brackets to commutators. Here the natural conjugate variables are $(t,H)$, which can be transformed to $(L,P)$ where $L$ is the cutoff-independent "renormalized bulk size"; see H&J (2.29). We have:

$$
  H = \frac{P^2}{2\phi_b}
      + \frac{2}{\phi_b}\,e^{-L}
$$

**Note:** it is kind of weird to include $t$ in the phase space. However, note that what we are actually doing is trying to assign a natural weight to each configuration (labeled by $\Phi_h$ or $H = E$). One finds the result is equivalent to **Hartle--Hawking construction / Euclidean path integral, if we include only the trivial (disk) topology!** [H&J]

**Information paradox:**

Couple the right patch to a reservoir to let it evaporates! $S_{\text{res}}$ can be computed by a RT surface

$$
  S_{\text{res}}
  \gg S_{BH}
  \sim 8\pi\Phi_0 + 8\pi^2\phi_b T_0
$$

Interpretation: there are remnants not counted by $S_{BH}$! H&S: _this renormalizable bulk theory can have an arbitrarily large number of low-energy excitations near black hole horizon._ [Details? NOT shown / cited by H&S.]

### Recommendations

- 1804.01081: to better understand JT
- 1905.08762: contains most of the calculations that H&S relies on
- 1211.3494: Aron Wall on maximin & HRT.
