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I am reading a paper arXiv:1606.01857. This paper expand the Schwarzian action (3.18) with a perturbation $$ \tau=u+\epsilon(u).\tag{4.26} $$ Then the linearized action becomes $$ I=\frac{C}{2}\int du[\epsilon^{\prime\prime2}-\epsilon^{\prime2}].\tag{4.27} $$

And my question is how to get the propagator $\langle\epsilon(u)\epsilon(0)\rangle$? In my mind, we already know the partition function of this system. In order to calculate the propagator, we need to identify the source for $\epsilon$. For this system, which quantity is the source for $\epsilon$? Another method is also welcomed.

Qmechanic
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1 Answers1

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  1. The linearized Schwarzian action is a quadratic Euclidean action$^1$ $$\begin{align} I[\epsilon]~=~&\frac{1}{2}\int_{-\pi}^{\pi}\!du~\left( \epsilon^{\prime\prime 2}-\epsilon^{\prime2}\right)\cr ~\stackrel{\text{IBP}}{=}\,&\frac{1}{2}\int_{-\pi}^{\pi}\!du~\epsilon~{\cal D}\epsilon\cr ~=~& \pi\sum_{n\in\mathbb{Z}}n^2(n^2-1)\epsilon_n\epsilon_{-n},\end{align}\tag{4.27} $$ with periodic boundary conditions $$ \epsilon(u)~=~\sum_{n\in\mathbb{Z}}\epsilon_ne^{inu}~=~\epsilon(u+2\pi),\qquad\epsilon_n^{\ast}~=~\epsilon_{-n},$$ has zero-modes $n\in\{-1,0,1\}$ and a differential operator $${\cal D}~:=~\partial_u^4 + \partial_u^2.$$

  2. The Green function/2-pt function/propagator $$ G(u)~=~\frac{1}{\hbar}\langle\epsilon(u)\epsilon(0)\rangle $$ satisfies $$\begin{align} {\cal D}G(u)~=~&\delta(u+2\pi\mathbb{Z})-\frac{1+2\cos u}{2\pi}\cr ~=~&\frac{1}{2\pi}\sum_{|n|\geq 2}e^{inu}~=~\frac{1}{\pi}\sum_{n\geq 2}\cos nu,\end{align}$$ where the relevant Dirac distribution is the Dirac comb (because the system is periodic) without the 3 zero-modes (because else the Green function $G$ does not exist, i.e. we exclude the 3 zero-modes from the dynamical variables of the system).

  3. The Green function is$^2$ [1,2] $$\begin{align} 2\pi G(u) ~=~&\sum_{|n|\geq 2}\frac{e^{inu}}{n^2(n^2-1)}~=~2\sum_{n\geq 2}\frac{\cos nu}{n^2(n^2-1)}\cr ~=~&2\sum_{n\geq 2}\left(\frac{1}{n^2-1}-\frac{1}{n^2}\right)\cos nu\cr ~=~&\sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n+1}-\frac{2}{n^2}\right)\cos nu\cr ~=~&1+\frac{5}{2}\cos u+\sum_{n\in\mathbb{N}}\left(\frac{\cos(n+1)u-\cos(n-1)u}{n}-\frac{2\cos nu}{n^2}\right)\cr ~=~&1+\frac{5}{2}\cos u-2\sum_{n\in\mathbb{N}}\left(\frac{\sin u\sin nu}{n}+\frac{\cos nu}{n^2}\right)\cr ~\stackrel{|u|<2\pi}{=}&1+\frac{5}{2}\cos u-(\pi{\rm sgn}(u)-u)\sin u-\frac{(\pi-|u|)^2}{2}+\frac{\pi^2}{6} .\end{align}\tag{3.128/4.28}$$ In the last equality we used the Fourier series $$\begin{align}\sum_{n\in\mathbb{N}}\frac{\cos nu}{n^2}~\stackrel{|u|<2\pi}{=}&\frac{(\pi-|u|)^2}{4}-\frac{\pi^2}{12} ~=~\frac{u^2}{4}-\frac{\pi|u|}{2}+\frac{\pi^2}{6} \cr\quad\Downarrow~&\quad\cr\sum_{n\in\mathbb{N}}\frac{\sin nu}{n}~\stackrel{|u|<2\pi}{=}&\frac{\pi{\rm sgn}(u)-u}{2}\quad\text{saw tooth profile}.\end{align}$$

References:

  1. J. Maldacena, D. Stanford & Z. Yang, Conformal symmetry and its breaking in 2D Nearly AdS space, arXiv:1606.01857; eq. (4.28).

  2. J. Maldacena & D. Stanford, Comments on the SYK model, arXiv:1604.07818; eq. (3.128).

  3. G. Sarosi, $AdS_2$ holography and the SYK model, arXiv:1711.08482; eq. (61).

$^1$ In the action (4.27) let us for simplicity set the normalization factor $C=1$. The propagator (3.128/4.28) should contain an overall normalization factor $C^{-1}$, cf. e.g. my Phys.SE answer here.

$^2$ Alternatively one may use the residue theorem [3] $$\begin{align} 2\pi G(u) ~=~&\sum_{|n|\geq 2}\frac{e^{inu}}{n^2(n^2-1)}\cr ~=~&\sum_{|n|\geq 2}\oint_n\frac{ds}{2\pi i(s-n)}\frac{e^{isu}}{s^2(s^2-1)}\cr ~=~&\sum_{|n|\geq 2}\oint_n\frac{ds}{e^{2\pi is}-1}\frac{e^{isu}}{s^2(s^2-1)}\cr \stackrel{0<u<2\pi}{=}&-\sum_{|n|\leq 1}\oint_n\frac{ds}{e^{2\pi is}-1}\frac{e^{isu}}{s^2(s^2-1)}\cr ~=~&\oint_0\frac{ds}{1-s^2}\frac{1+isu+\frac{(isu)^2}{2}+{\cal O}(s^3)}{2\pi is^3\left(1+\frac{2\pi i s}{2}+\frac{(2\pi i s)^2}{6}+{\cal O}(s^3)\right)} \cr ~-~&\sum_{\pm}\oint_{\pm 1}\underbrace{\frac{ds}{s^2}}_{=1+(s\pm 1)(s\mp 1)}\underbrace{\frac{1}{s\pm 1}}_{=\pm 2\left(1\pm\frac{s\mp 1}{2}\right)}\frac{e^{\pm i u}\left(1+iu(s\mp 1)+{\cal O}((s\mp 1)^2)\right)}{2\pi i(s\mp 1)^2 \left(1+\frac{2\pi i (s\mp 1)}{2}+{\cal O}((s\mp 1)^2)\right)} \cr ~=~&\oint_0\!ds\left(1+s^2+{\cal O}(s^4)\right)\frac{1+isu-\frac{(su)^2}{2}+{\cal O}(s^3)}{2\pi is^3}\left(1-\pi i s-\frac{(\pi s)^2}{3}+{\cal O}(s^3)\right) \cr ~+~&\sum_{\pm} \left(1+\frac{1}{4}\right)e^{\pm i u} -\sum_{\pm} \frac{1}{\pm 2}i(u-\pi)e^{\pm i u}\cr ~=~&\ldots. \end{align}\tag{61}$$

Qmechanic
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