Schwarzian quantum mechanics describes the collective IR mode of the SYK model
and captures key features of 2D black hole dynamics. Exact results for its correlation
functions were obtained in

Article funded by SCOAP3

0$ large with this combination $z$ held fixed. In the previous section we got this result from our exact formulas derived from 2d Liouville CFT between ZZ-branes. The purpose of this section is to rederive the semiclassical limit directly from the 2d picture without having to go through the details of the exact expressions. Instead of taking large $C$ with fixed $\beta$ we will consider units in which $C=1/2$. Since the dimensionless coupling is $2\pi C/\beta$ the semiclassical limit is equivalent to taking $\beta \to 0$ in these units. In the 2d picture, the inverse temperature $\beta$ of the Schwarzian gives the distance between the ZZ branes. Taking $\beta \to 0$ in the Schwarzian means sending the distance between the ZZ-branes to zero faster than the size of the circle in the extra dimension. Namely, $\beta$ goes to $0$ faster than $q \to 1$, where $q=e^{2 \pi i \tau}$ denoted the $q$-modulus of the 2d annulus. In this limit the Schwarzian becomes equivalent to Liouville between two infinite ZZ-branes, namely on a strip of width $\beta$ instead of an annulus. The upshot of the previous argument is that we can reproduce the semiclassical Schwarzian correlators from local operators between two infinite ZZ-branes. The Liouville one-point function, which corresponds to the Schwarzian two-point function, is easy to compute exactly from the 2d CFT perspective, since the system can be mapped to the upper-half-plane by a conformal transformation. The answer immediately has the required~form \beq \lb V \rb_{\rm strip} = \left(\frac{\pi }{\beta \sin \frac{\pi \tau}{\beta}} \right)^{2\ell}, \eeq where $\ell$ corresponds to the conformal dimension of the Liouville operator. This can be related to the real time answer~\eqref{gtwoclass} by analytic continuation. Now we compute the Liouville 2pt function/Schwarzian 4pt function using this approach. Again, we can map the infinite strip to the upper-half plane, and we take the positions of the two local vertex operator insertions to be $z_1$ and $z_3$, while the images of these operators will be denoted by $z_2$ and $z_4$ (even though they should strictly be given by $z_1^*$ and $z_3^*$ we will allow them to be generic). The two-point function can be written in two equivalent ways. First, we can take the OPE between the two insertions and between the two images, obtaining \bea \frac{\la V_1 V_2 W_3W_4 \ra}{\la V_1V_2\ra\la W_3 W_4\ra} \! \is\!\! \int \! dP\, \Psi_{\rm ZZ}(P)\, C_{VWP} \, \mathcal{F} \Big( {}^{V}_{W} \; {}^{V}_{W} , P, \eta \Big), \ \ \ \ \eea where $\eta=\frac{z_{13}z_{24}}{z_{14}z_{23}}$ is the cross-ratio, $\Psi_{\rm ZZ}$ is the ZZ-brane wavefunction, $C_{VWP}$ represents the Liouville OPE coefficient between the operators $V$, $W$ and an intermediate channel operator with Liouville momentum $P$. $ \mathcal{F} (P, \eta )$ denotes the conformal block in this channel. Another representation of this correlation function can be obtained by performing the OPE between an operator and its image. In this case it was shown that only the vacuum block appears (see section 6 of~\cite{Zamolodchikov-ml-2001ah} and also~\cite{LeFloch-ml-2017lbt} for a different perspective on this result). Defining the new cross ratio via $x=1-\eta$ and using the exponential map $z_i =e^{\frac{2\pi}{\beta} \tau_i}$, the ZZ identity~gives \beq \frac{\lb V_1 V_2 W_3 W_4\rb }{\lb V_1 V_2\rb \lb W_3 W_4\rb} = x^{2\Delta_V} \mathcal{F} \Big( {}^{V}_{V} \; {}^{W}_{W} , {\rm vac},x\Big),~~~~x= -\frac{\sinh \frac{\pi t_{12}}{\beta} ~\sinh \frac{\pi t_{34}}{\beta} }{\sinh \frac{\pi t_{32}}{\beta} ~\sinh \frac{\pi t_{41}}{\beta} }. \eeq For fixed $t_1,\ldots, t_4 \sim \mathcal{O}(1)$ and $c\to \infty$, the cross-ratio is finite and the vacuum block becomes trivial, implying that $\lb V_1V_2W_3W_4\rb \sim \lb V_1V_2\rb \lb W_3W_4 \rb$. For the time-ordered four-point function this is the final answer. The out-of-time ordered four-point function is equal to the vacuum block evaluated on the second sheet. It turns out this indeed exactly reproduces the shockwave calculation. The vacuum block on the second sheet is found by performing a monodromy operation on the block. As observed in~\cite{Kaplan}, this monodromy remains non-trivial in the combined $x\to 0$ and $c\to \infty$ limit, with the product $c \cdot x$ is held fixed and finite. The exact formula for the identity block in this limit was found to be~\cite{Kaplan} \bea \frac{\lb V_1 W_3 V_2 W_4\rb }{\lb V_1 V_2\rb \lb W_3 W_4\rb}&=&\lim_{c\to \infty, cx~{\rm fixed}} x^{2\Delta_V} \mathcal{F}_{2^{nd}{\rm sheet}} \Big( {}^{V}_{V} \; {}^{W}_{W} , {\rm vac},x\Big) \nonumber \\[3mm] &=& z^{-2\ell_1} U(2\ell_1,1+2\ell_1-2\ell_2,1/z), \ea where the right hand side involves the cross ratio $z$ defined in equation~\eqref{xratio}. Here we used the precise relation between the Virasoro central charge $c$ and the Schwarzian coupling $2\pi C/\beta$. This matches exactly with the shockwave calculation in equation~\eqref{eq:MSYsw}. ]]>

0$.} From this, we indeed identify the Kruskal content of the Unruh modes as $p^{-i\nu-1/2}\theta(p)$ and $\left|p\right|^{-i\nu-1/2}\theta(-p)$ as in~\eqref{pmmodes}: \begin{equation} h^1_\nu(U) = \frac{1}{\sqrt{2\pi}} \int_{0}^{+\infty} dp\, p^{-i\nu-1/2} \psi_p(U), \qquad h^2_\nu(U) = \frac{1}{\sqrt{2\pi}} \int_{0}^{+\infty} dp\, p^{i\nu-1/2} \psi_p(U). \end{equation} Inspecting the modes~\eqref{unruhmodes1},~\eqref{unruhmodes2}, one can write the Unruh modes more economically by defining: \begin{equation} \mathbf{h}_\nu (U) \equiv \begin{cases} h^1_{\nu}(U), & \nu > 0 \\[2mm] h^2_\nu(U) = h^1_{-\nu}(U),& \nu < 0 \end{cases}, \qquad \mathbf{c}(\nu) \equiv \begin{cases} c^1(\nu), & \nu > 0 \\[2mm] c^2(\nu) = c^1(-\nu),& \nu < 0 \end{cases}. \end{equation} Then the Unruh creation operator can be expanded into the Kruskal creation operators as Mellin transforms: \begin{equation} \label{Unruexpa} \mathbf{c}^{\dagger}(\nu) = \frac{1}{\sqrt{2\pi}} \int_{0}^{+\infty} dp\, p^{-i\nu-1/2} a^\dagger_p, \qquad a^{\dagger}_p = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} d\nu\, p^{i\nu-1/2} \mathbf{c}^\dagger(\nu), \end{equation} consistent with the Minkowski commutation relations: \begin{equation} \left[a_p,a^{\dagger}_{p'}\right] = \delta(p-p'), \quad \text{all others vanish}. \end{equation} The TFD state is, by definition, annihilated by all positive Kruskal frequency modes: \begin{equation} c^1(\nu)\left|\text{TFD}\right\rangle = c^2(\nu)\left|\text{TFD}\right\rangle = 0. \end{equation} To link the first and second quantized formalism, we should identify states through \begin{equation} a^{\dagger}_{{\!}_{\rm R}}(\nu) \left|0 \right\rangle = \frac{1}{\sqrt{2\pi\nu}} \left|\nu\right\rangle, \end{equation} leading to \begin{equation} \label{tbog} \sqrt{2\pi\nu} \, a^{\dagger}_{{\!}_{\rm R}}(\nu) \left|\text{TFD}\right\rangle \,\, \sim \,\, e^{\pi\nu/2}\Gamma(1+i\nu) \, c^{1\dagger}(\nu) \left|\text{TFD}\right\rangle, \end{equation} where the resulting states can finally be expanded into Kruskal eigenstates using~\eqref{Unruexpa}~as: \begin{equation} c^{1\dagger}(\nu) \left|\text{TFD}\right\rangle = \frac{1}{\sqrt{2\pi}} \int_{0}^{+\infty} dp\, p^{-i\nu-1/2} \left|p\right\rangle. \end{equation} At this point, one can make contact with the 't Hooft-Dray shockwave $\mathcal{S}$-matrix computation done in~\eqref{smatf}. The result~\eqref{tbog} means there is an extra factor of $e^{\pi\nu/2}\Gamma(1+i\nu)$ when going from modes that are localized within either the $L$- or $R$-wedge, to a mode with positive Kruskal momentum. An interesting example correlator to compute using the above formulas, is: \begin{equation} \left\langle \text{TFD}\right| a_{{\!}_{\rm L}}(-\nu_1)a^{\dagger}_{{\!}_{\rm R}}(\nu_2)\left|\text{TFD} \right\rangle = \frac{1}{2\sinh(\pi \nu_1)} \delta(\nu_1-\nu_2), \end{equation} which is non-zero. ]]>

] (0,2.7) arc (120:150:2); \draw[thick,->] (-2,0.7) arc (120:150:2); \draw[thick,<-] (2.3,3) arc (60:30:2); \draw[thick,->] (-2.3,-3) arc (240:210:2); \draw (1,2.5) node {\small \color{red}$k_2$}; \draw (-1,-2.5) node {\small \color{red}$k_5$}; \draw (1.5,-1.5) node {\small \color{red}$k_6$}; \draw (2.5,1) node {\small \color{red}$k_1$}; \draw (-1.5,1.5) node {\small \color{red}$k_{3}$}; \draw (-2.5,-1) node {\small \color{red}$k_{4}$}; \end{tikzpicture} \end{align} The arrows depict the choice of sign of the $\omega$'s, so we define new redundant variables: \begin{align} k_2^2 - k_1^2 &= \omega_1, \quad k_6^2-k_1^2 = \omega_4, \quad k_5^2- k_6^2 = \omega_6, \nonumber \\ k_{4}^2-k_5^2 &= \omega_2, \quad k_{4}^2-k_{3}^2 = \omega_5, \quad k_3^2 - k_2^2 = \omega_3, \quad k_1^2 = M, \end{align} satisfying energy conservation $\omega_1+\omega_3+\omega_5 = \omega_2+\omega_4+\omega_6$, with as usual $M \gg \omega_i$ as $C \gg t_{ij}$. Defining again $\nu_i = \frac{\omega_i}{2\sqrt{M}}$, we find two final Gamma's associated to the two shockwave processes, written as \begin{equation} \Gamma\left(-i\nu_3+i\nu_4\right)\Gamma\left(-i\nu_5+i\nu_6\right), \end{equation} whereas the remainder gives precisely the required Schwarzschild wavefunctions~\eqref{wvsch}: \begin{align} \sim \,\, &e^{i\nu_1 t_1}\frac{\Gamma(\ell_1 - i\nu_1)}{\sqrt{\Gamma(2\ell_1)}} \times e^{- i\nu_2 t_2}\frac{\Gamma(\ell_1 + i\nu_2)}{\sqrt{\Gamma(2\ell_1)}} \times e^{i\nu_3 t_3}\frac{\Gamma(\ell_2 + i\nu_3)}{\sqrt{\Gamma(2\ell_2)}} \nonumber \\ &\times e^{-i\nu_4 t_6}\frac{\Gamma(\ell_2 - i\nu_4)}{\sqrt{\Gamma(2\ell_2)}} \times e^{i\nu_5 t_4}\frac{\Gamma(\ell_3 + i\nu_5)}{\sqrt{\Gamma(2\ell_3)}} \times e^{-i\nu_6 t_5}\frac{\Gamma(\ell_3 - i\nu_6)}{\sqrt{\Gamma(2\ell_3)}}. \end{align} Finally, the $M$-integral, readily generalized to such an $n$-point OTO function, can be done as usual by saddle point methods, giving for any correlator the same saddle $M_0 = 2\pi^2 C/\beta^2$ as found before. As alluded to already several times, the structure of this computation is immediately generalized to arbitrary $n$-point OTO crossed diagrams of this specific graph topology. ]]>

}_1}{2}$. The boundary conditions at $\tau_1$ and $\tau_2$ are read off from~\eqref{eqmot}. Using the notation $T(u) = -\frac{\ell}{F'_1}\delta'(u-\tau_1) - \frac{2}{F_1-F_2}\delta(u-\tau_1)$, we can integrate~\eqref{eqmot} into: \beq \frac{F''}{F'} = \int^t \!\! du \, (F(\tau)-F(u)) T(u), \qquad \qquad F' = \int^t\!\! du \, \frac{(F(\tau)-F(u))^2}{2} T(u). \eeq A $\delta$-function insertion in $T(u)$ requires a jump for $F'''$, whereas $\delta'$-insertions require a jump already for $F''$. In this case, the gluing conditions at $\tau=\tau_1, \tau_2$ are \beq F, F' \text{ continuous}, \qquad \quad \Delta F'' = -\frac{\ell}{C}F'. \eeq We set $\tau_1=0$ without loss of generality. For the combined solution, we make the following~Ansatz \begin{alignat}{2} F(\tau) & = \, \frac {2C} {k_2}\tan\Bigl(\frac{k_2\spc \tau} {2C}\Bigr),\qquad \quad &&\tau<0, \nonumber \\[-4mm] & & & \qquad \qquad \qquad \qquad k_2 \, = \, \sqrt{2E_2 C},\nonumber \\[-2mm] &= \, \frac{\tan\Bigl(\frac{k_1\spc \tau} {2C}\Bigr)}{\frac{\ell}{2C}\tan\Bigl(\frac{k_1\spc \tau} {2C}\Bigr)+\frac{k_1}{2C}},\qquad \quad &&0 <\tau<\tau_2, \label{zerotsolution}\\[-2mm] & & & \qquad \qquad \qquad \qquad k_1 \, = \, \sqrt{2E_1 C}, \nonumber \\[-3.3mm] &= \, \frac{a \tan\left(\frac{k_2} {2C}(\tau-\tau_2)\right) + b \, \frac{k_2}{2C}}{ c \tan\left(\frac{k_2} {2C}(\tau-\tau_2)\right) + d\, \frac{k_2} {2C}} ,\qquad \quad &&\tau_2 < \tau.\nonumber \end{alignat} Setting $d=1$ and imposing the continuity conditions gives that \begin{equation} b = F_2, \quad c = -\frac{1}{2}\frac{F''_2}{F_2} = -\frac{1}{2}\frac{F''^{<}_2}{F_2'} + \frac{\ell}{2C}, \quad a = F'_2 + F_2 c. \end{equation} From this point onwards, it is a straightforward calculation to show that the continuity conditions and the requirement that the distance between the asymptotes of $f(\tau)$ are fixed by the inverse temperature $\beta$ are precisely equivalent to the conditions~\eqref{finitetsaddle}. An example of a classical solution is depicted in figure~\ref{ClassicalHeavyLimitT}. The finite temperature on-shell action is \begin{align} S_{0} &= -\frac{\tau k_1^{2}}{2C} - \frac{(\beta-\tau) k_2^{2}}{2C} - \ell \ln \frac{F'_2}{F_2^2}, \end{align} where \begin{equation} \frac{F'_2}{F_2^2} = \frac{1}{(4C\ell)^2}((k_1+k_2)^2+\ell^2)((k_1-k_2)^2+\ell^2) \end{equation} agreeing with~\eqref{sadintT}. \begin{figure} \centerline{ \includegraphics[width=0.45\textwidth]{ClassicalHeavyLimitT.pdf} } ]]>