9.2.3 Error correction

^[1]
Take the same SYK problem from section SYK. \begin{align} &H =i^{p / 2} !!!!!!!!\sum_{1 \leq i_{1} <\cdots < i_{p} \leq N} !!!!!!J_{i_{1} \ldots i_{p}} \psi_{i_{1}} \cdots \psi_{i_{p}},
&\left{\psi_{i}, \psi_{j}\right}=2 \delta_{i j},
&\left\langle J_{i_{1} \ldots i_{p}}^{2}\right\rangle =\frac{\mathcal{J}^2}{ {N \choose p} } \end{align}

Now, instead of fixing $p$ if we fix $\lambda := 2p^2/N$ then we get DSSYK. We switched the discrete parameter $p$ that is always even (as $H$ is Hermitian) with the continuous $\lambda$. The parameter $\lambda$ is somewhat similar to $G_N$ or $l_P/l_{AdS}$. As $\lambda\to 0$, we get back the original SYK but with large $p$. This limit is a semi-classical limit. When $\lambda$ is not near $0$, quantum gravity effects become significant. So, compared to SYK, this model 1) Probes quantum gravity instead of just semi-classical gravity and 2) Is exactly solvable for large $N$ at all energy scales; SYK is only exactly solvable for large $N$ in the IR limit.

Normally SYK model is solved by introducing the master fields $G,\Sigma$ and then using a saddle point approximation. This method can also be used for DSSYK when $\lambda \to 0$ (large $p$ SYK). But to solve for arbitrary $\lambda$, we need to use “chord diagrams”.

Chord diagrams: Let $q := \exp(-\lambda)$. Holography is defined by $Z_{\mathrm{boundary}}=Z_{\mathrm{bulk}}$ and $\mathcal{H}{\mathrm{boundary}}=\mathcal{H}{\mathrm{bulk}}$. $Z_{\mathrm{boundary}}$ is the most important quantity to calculate.

By Taylor expansion we get, $\displaystyle Z =\langle\langle T r ~ e^{-\beta H} \rangle\rangle=\sum_{k}\frac{(-\beta)^k}{k!}m_k$ where $m_k=\langle\langle T r ~ H^k\rangle\rangle$. We normalize such that $Tr(\mathbb{I})=1$. Let $\mathcal{J}=1$ from now.

Let’s consider $m_4$. From now on, we follow $I \equiv{i_{1}, i_{2},… i_{p} }$ notation $\begin{aligned} m_4 &= \sum\limits_{I_1,I_2,I_3,I_4} \langle\langle J_{I_1}J_{I_2}J_{I_3}J_{I_4}\rangle\rangle Tr(\psi_{I_1}\psi_{I_2}\psi_{I_3}\psi_{I_4})\\ =& \frac{\mathcal{J}^2}{N\choose p}^2\sum\limits_{I_1 I_2} \left\lbrace Tr(\psi_{I_1}\psi_{I_1}\psi_{I_2}\psi_{I_2}) + Tr(\psi_{I_1}\psi_{I_2}\psi_{I_1}\psi_{I_2}) \right. \\ & \left. + Tr(\psi_{I_1}\psi_{I_2}\psi_{I_2}\psi_{I_1})\right\rbrace \end{aligned}$

Here, we are only considering pairwise matching and this can be diagrammatically represented using chord diagrams as shown below.

\resizebox{.9\hsize}{!}{ \begin{tikzpicture} \node[black] at (-2.75,0) {$m_4 = \dfrac{1}{ {N \choose p}^2}\sum\limits_{I_1,I_2}$}; \draw[thick] (0,0) circle(1 cm); \draw[thick,color=black] (-1,0) to[bend right] (0,1); \draw[thick,color=black] (0,-1) to[bend left] (1,0); \node[black] at (1.5,0) {$+$}; \draw[thick] (3,0) circle(1 cm); \draw[thick,color=black] (3,1) to[bend right] (4,0); \draw[thick,color=black] (3,-1) to[bend right] (2,0); \node[black] at (4.5,0) {$+$}; \draw[thick] (6,0) circle(1 cm); \draw[thick,color=black] (5,0) to[bend right] (7,0); \draw[thick,color=black] (6,-1) to[bend right] (6,1); \end{tikzpicture} }

There will be higher-order corrections, such as some pairs and some four $\psi_I$s matching, etc., but those are suppressed in the large $N$ limit. Notice that $\psi_{I}^2=(-1)^{\frac{p}{2}}\mathbb{I}$. So, the first two terms just give $1$ and $\sum\limits_{I_1,I_2}$ cancels ${N \choose p}^2$. $\psi_I\psi_J = (-1)^{|I\cap J|} \psi_J\psi_I$ In the above identity, $|I\cap J|$ is the number of sites in common to $I$ and $J$. In the formula for $m_4$ as we have $\sum\limits_{I_1,I_2}$ we only need to find the average value $(-1)^{|I\cap J|}$ for 3rd diagrams contribution.

Poisson distribution: Let $I$ be of size $p$ and $J$ be of size $p’$. For fixed $I$, the probability of $|I\cap J|=k$ is $\displaystyle\frac{\binom{p} {k} \binom{N-p} {p^{\prime}-k}} {\binom{N} {p^{\prime}}}\approx \frac{1} {k!} \left( \frac{p p^{\prime}} {N} \right)^{k} e^{-p p^{\prime} / N} $. That is a Poisson distribution with parameter $p p^{\prime} / N$.

For our case $p’=p$ $\frac{1}{ {N \choose p}^2}\sum\limits_{I_1,I_2}(-1)^{|I\cap J|}=\sum\limits_{k=0}^{p}(-1)^{k}\frac{(\lambda/2)^{k}}{k!}e^{-\lambda/2}=e^{-\lambda}=q$

So, $m_4=1+1+q$. This method can be generalized to $m_k\neq m_4$. We just need to 1) Find all chord diagrams corresponding to $m_k$ and 2) In each chord diagram, multiply by $q$ for every chord intersection. Now, using these 2 rules and the transfer-matrix method, we will get

\[m_k = \int\limits_0^\pi \frac{d\theta}{2\pi} (q,e^{\pm 2i\theta};q)_{\infty}\left(\frac{2\cos\theta}{\sqrt{1-q}}\right)^k\]

where the q-Pochhammer symbol is defined as $(q,e^{\pm 2i\theta};q)_{\infty}=(q;q)_\infty(e^{ 2i\theta};q)_\infty(e^{-2i\theta};q)_\infty$ $(a;q)_{\infty} = \prod\limits_{k=0}^\infty (1-aq^{k}).$

Transfer-matrix method: At a random point on the chord diagrams, we cut them open. We can think of the number of chords $l$ between 2 nodes as making a state $

l\rangle$ in some auxiliary Hilbert space. Once we know the state before a node, then the state after this node has 1 way of creating a chord and $l$ ways of annihilating a chord as shown in Fig fig:TransferMatrix. If you think of these nodes as some discrete time, then the corresponding Hamiltonian is defined as the transfer matrix $T

l \rangle=

l+1 \rangle+\left( 1+q+\cdots+q^{l-1} \right)

l-1 \rangle=

l+1 \rangle+\frac{1-q^{l}} {1-q}

l-1 \rangle\,$. Now we can solve $Z$ using $\displaystyle\langle\langle T r \, e^{-\beta H} \rangle\rangle=\langle0

e^{-\beta T}

0 \rangle$. Equivalently $T$ can be also be defined using the $q$ deformed harmonic oscillator

$T=A_{q}+A_{q}^{\dagger}, \ A_{q} A_{q}^{\dagger}-q A_{q}^{\dagger} A_{q}=1$ $[ N, A_{q} ]=-A_{q}, \ [ N, A_{q}^{\dag} ]=A_{q}^{\dag}.$

Now that we know how to solve $Z$, we can see how this can be generalized to correlation functions $\mathrm{t r} ( e^{-\tau_{1} H} M_{s} e^{-\tau_{2} H} M_{s} \cdots)$ of matter operators defined as

\[M_{s}=i^{s / 2} \sum_{I} K_{I} \Psi_{I}^{s}.\]

where $K_{I}$ are independent Gaussian random variables defined so that $TrM_{s}^2=1$. In the Double Scaled limit, we let $s\to \infty$ by fixing $\Delta = s/p$. So we should now call it $M_{\Delta}$. Now we can proceed similarly using chord diagrams. We will now have 2 different chords corresponding to $H$ and $M_{\Delta}$. When 2 $M_{\Delta}$ chords intersect we get a factor of $q_{M}=e^{-2 s^{2} / N}=e^{-\lambda\Delta^{2}}$ and when a $H$ chord and $M_{\Delta}$ chord intersect we get a factor of $q_{HM}=e^{-2 ps/ N}=e^{-\lambda\Delta}$. Similarly, we can introduce multiple matter operators $M_1, M_2\cdots$ with $\Delta_1,\Delta_2,\cdots$.

Quantum groups and noncommutative geometry:~^[2]

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