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In the usual SYK model described via

$$ H = -\frac{1}{N^{3/2}}\sum_{ijkl}^{N}J_{ij,kl}\psi_{i}\psi_{j}\psi_{k}\psi_{l}. $$

The normalization factor out front ($N^{-3/2}$) is chosen such that the energy is extensive, i.e. $$E \sim O(N).$$

However, I am a little confused as to how the above quantity scales like $N$. Since the sum is over four seperate variables that each range through $N$, shouldn't the sum scale like $N^{4}$? And, if so, then how is the $N^{-3/2}$ prefactor sufficient to make $H$ scale like $N$? I don't see any reason why the couplings themselves would depend on or scale with the system size, so I would think that the prefactor would have to be $N^{-3}$? Or is it just implicitly assumed that $J \sim N^{-3/2}$?

Qmechanic
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meer23
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1 Answers1

3

That the energy is extensive $E\sim N$ can e.g. be seen using [1]

  1. The free Lagrangian of $N$ Majorana fermions is $$ \underbrace{L_0^M(t)}_{\text{real}} ~=~\frac{\color{red}{i}}{2}\sum_{i=1}^N\underbrace{\psi^i(t)}_{\begin{matrix}\text{real}\cr\text{ferm}\end{matrix}}\partial_t\underbrace{\psi^i(t)}_{\begin{matrix}\text{real}\cr\text{ferm}\end{matrix}}\tag{108}$$ in Minkowskian 0+1D spacetime, and $$ \underbrace{L_0^E(\tau)}_{\text{imaginary}} ~=~\frac{1}{2}\sum_{i=1}^N\underbrace{\psi^i(\tau)}_{\begin{matrix}\text{real}\cr\text{ferm}\end{matrix}}\partial_{\tau}\underbrace{\psi^i(\tau)}_{\begin{matrix}\text{real}\cr\text{ferm}\end{matrix}} ~=~-L_0^M(t)\tag{108}$$ in Euclidean 1+0D spacetime with $\tau=\color{red}{i}t$. The corresponding free Greens functions are $$\begin{align} \underbrace{G^E_0(\tau,\tau^{\prime})}_{\begin{matrix}\text{real}\cr\text{antisym}\end{matrix}} ~=~(S^E_0)^{-1}(\tau,\tau^{\prime}) ~=~&\frac{1}{2}{\rm sgn}(\tau-\tau^{\prime}), \cr \underbrace{G^M_0(t,t^{\prime})}_{\begin{matrix}\text{imaginary}\cr\text{antisym}\end{matrix}} ~=~(S^M_0)^{-1}(t,t^{\prime}) ~=~&\frac{1}{2\color{red}{i}}{\rm sgn}(t-t^{\prime}). \end{align}\tag{108}$$

  2. The Hamiltonian is $$-H^E(\tau)~=~H^M(t)~=~ \frac{\color{red}{i}^{q/2}}{q!}\sum_{i_1,\ldots, i_q=1}^N\underbrace{\frac{J_{i_1\ldots i_q}}{N^{\frac{q-1}{2}}}}_{=:\widetilde{J}_{i_1\ldots i_q}}\psi^{i_1}(\tau)\ldots\psi^{i_q}(\tau),\tag{93+94}$$ where $q\in 2\mathbb{N}$ is an even natural number. The imaginary factors ensures that the Hamiltonian $H^M$ is Hermitian. Concerning the normalization, see also this related Phys.SE post.

  3. Average the partition function wrt. a Gaussian ensemble $$ \langle J_{i_1\ldots i_q}J_{j_1\ldots j_q}\rangle_{\psi=0}~=~ (q\!-\!1)!J^2 \sum_{\pi\in S_q}(-1)^{\sigma(\pi)}\delta_{i_1,j_{\pi(1)}}\ldots\delta_{i_q,j_{\pi(q)}}.\tag{101}$$

  4. Equivalently, introduce a Gaussian term in the action $$\begin{align} S^E[\psi,J]~=~&\int\! d\tau\left\{L_0^E(\tau) - H^E(\tau)\right\} \cr ~+~&\frac{1}{2J^2(q\!-\!1)!}\sum_{i_1,\ldots, i_q=1}^N \sum_{\pi\in S_q}(-1)^{\sigma(\pi)}J_{i_1\ldots i_q}J_{i_{\pi(1)}\ldots i_{\pi(q)}},\cr S^M[\psi,J]~=~&\int\! dt\left\{L_0^M(t) - H^M(t)\right\} \cr ~+~&\frac{\color{red}{i}}{2J^2(q\!-\!1)!}\sum_{i_1,\ldots, i_q=1}^N \sum_{\pi\in S_q}(-1)^{\sigma(\pi)}J_{i_1\ldots i_q}J_{i_{\pi(1)}\ldots i_{\pi(q)}}\cr ~=~&\color{red}{i}S^E[\psi,J] .\end{align} \tag{109}$$

  5. Completing the square produces a bi-local Hamiltonian density $$\begin{align} \underbrace{{\cal H}^E(\tau,\tau^{\prime})}_{\begin{matrix}\text{real}\cr\text{sym}\end{matrix}}~=~& \frac{(-1)^{q/2}}{2q}\underbrace{\frac{J^2}{N^{q-1}}}_{=:\widetilde{J}^2}\sum_{i_1,\ldots, i_q=1}^N\psi^{i_1}(\tau)\ldots\psi^{i_q}(\tau)\psi^{i_1}(\tau^{\prime})\ldots\psi^{i_q}(\tau^{\prime})\cr ~=~&\color{red}{i}\underbrace{{\cal H}^M(t,t^{\prime})}_{\begin{matrix}\text{imaginary}\cr\text{sym}\end{matrix}} \end{align}\tag{110}$$ inside a bi-local Lagrangian density.

  6. Introduce a bi-local master field $$\color{red}{i}\underbrace{G^M(t,t^{\prime})}_{\begin{matrix}\text{real}\cr\text{antisym}\end{matrix}}~=~\underbrace{G^E(\tau,\tau^{\prime})}_{\begin{matrix}\text{imaginary}\cr\text{antisym}\end{matrix}} ~\approx~\frac{1}{N}\sum_{i=1}^N\psi^i(\tau)\psi^i(\tau^{\prime}) ,\tag{112}$$ and a bi-local Lagrange multiplier $$\color{red}{i}\underbrace{\Sigma^M(t,t^{\prime})}_{\begin{matrix}\text{imaginary}\cr\text{antisym}\end{matrix}}~=~\underbrace{\Sigma^E(\tau,\tau^{\prime})}_{\begin{matrix}\text{real}\cr\text{antisym}\end{matrix}},$$ which are both antisymmetric in $\tau\leftrightarrow\tau^{\prime}$. The extended bi-local Lagrangian density becomes $$\begin{align} \widetilde{\cal L}^M(t,t^{\prime}) ~=~&\delta(t\!-\!t^{\prime})L_0^M(t)-{\cal H}^M(t,t^{\prime})\cr ~+~&\frac{\color{red}{i}N}{2}\Sigma^M(t,t^{\prime})\left\{G^M(t,t^{\prime}) -\frac{1}{N\color{red}{i}}\sum_{i=1}^N\psi^i(t)\psi^i(t^{\prime})\right\},\cr \widetilde{\cal L}^E(\tau,\tau^{\prime}) ~=~&\delta(\tau\!-\!\tau^{\prime})L_0^E(\tau)-{\cal H}^E(\tau,\tau^{\prime})\cr ~+~&\frac{N}{2}\Sigma^E(\tau,\tau^{\prime})\left\{G^E(\tau,\tau^{\prime}) -\frac{1}{N}\sum_{i=1}^N\psi^i(\tau)\psi^i(\tau^{\prime})\right\}\cr ~=~&\color{red}{i}\widetilde{\cal L}^M(t,t^{\prime}) .\end{align}\tag{113}$$ See also e.g. this related Math.SE post.

  7. On-shell the bi-local Hamiltonian density becomes $${\cal H}^E(\tau,\tau^{\prime}) ~\approx~ \frac{NJ^2}{2q} G^E(\tau,\tau^{\prime})^q.\tag{113}$$ The bi-local field $G^E$ is intensive wrt. $N$, so that ${\cal H}^E \sim N$ is extensive, cf. OP's title question. $\Box$

  8. Performing the Gaussian path integral over the Majorana fermions produces a bi-local effective action $$\begin{align}-\frac{S^E_{\rm eff}[G^E,\Sigma^E]}{N}~=~&\ln{\rm Pf}\frac{(G^E_0)^{-1}-\Sigma^E}{\text{ref}}\cr ~-~&\frac{1}{2}\iint\!d\tau~d\tau^{\prime}\left\{\Sigma^E(\tau,\tau^{\prime})G^E(\tau,\tau^{\prime})-\frac{J^2}{q}G^E(\tau,\tau^{\prime})^q\right\}.\end{align}\tag{116}$$ Using the formula $$\delta\ln{\rm Pf}(A) ~=~\frac{1}{2}\iint\!d\tau~d\tau^{\prime}~{\rm Tr}\left\{A^{-1}(\tau^{\prime},\tau)\delta A(\tau,\tau^{\prime})\right\}, $$ we can derive the corresponding Euler-Lagrange (EL) equations $$\begin{align}\delta\Sigma^E:\quad (G^E)^{-1}~=~&(G^E_0)^{-1}-\Sigma^E\quad\text{self-energy},\cr \delta G^E:\quad\quad \Sigma^E~=~&J^2(G^E)^{q-1}\quad\text{Schwinger-Dyson (SD) eq}. \end{align}\tag{107}$$ This eq. implies the dominance of melon diagrams in the large $N$ limit.

References:

  1. G. Sarosi, $AdS_2$ holography and the SYK model, arXiv:1711.08482.
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