1. Introduction
Recent progress in quantum gravity and black hole physics impresses on the fact that wormholes play important roles1(1 For an up-to-date review, see [1].). Many evidences suggest an appealing conjectural duality between a bulk gravitation theory and an ensemble theory on the boundary [2–54]. For example, the seminal work [2] shows that Jackiw-Teitelboim (JT) gravity is equivalent to a random matrix theory. On the other hand, this new conjectural duality is not compatible with our general belief about the AdS/CFT correspondence. A sharp tension is the puzzle of factorization [55, 56]. In [57], this puzzle is studied within a toy model introduced in [37], where they find that (approximate) factorization can be restored if other saddles which are called half-wormholes are included. Motived by this idea, in [58] a half-wormhole saddle is proposed in a 0-dimensional (0d) Sachdev–Ye–Kitaev (SYK) model, followed by further analyses in different models [59–66].
Another scenario where half-wormholes are crucial is in the example of the spectral form factor. Chaotic systems exhibit a linear ramp behavior in their spectral form factor at late times [67]. In gravity theory, the smooth portion of the ramp can be explained by the emergence of a ‘wormhole’ saddle [68]. The oscillation portion should also be captured by the gravity theory, and the half-wormhole saddle is a potential candidate. This phenomenon has been observed in other SYK-like models [66], even though their half-wormhole concept differs from ours [69].
In our previous works [44, 69], we pointed out the connection between the gravity computation in [57] and the field theory computation in [58] and tested the half-wormhole proposal in various models. The main difficulty of this proposal is the construction of the half-wormhole saddles. Furthermore, the ansatz proposed in [58, 62] seems to rely on the fact the ensemble is Gaussian with zero mean value. As a result, the 0d SYK model only has non-trivial cylinder wormhole amplitude. However for a generic gravity theory for example the JT gravity, disk and all kinds of wormhole amplitudes should exist. In our previous work [69], we find even turning on disk amplitude in 0d SYK model will change the half-wormhole ansatz dramatically.
In this work, we generalize the idea of [57] and propose a method of searching for half-wormhole saddles. In our proposal, the connection between [57] and [58] will manifest. One notable benefit of our approach is that it does not depend on the trick of introducing a resolution identity used in [58], the collective variables emerge automatically. More importantly, our proposal can be straightforwardly generalized to non-Gaussian ensemble theories.
2. Gaussian distribution or the CGS model
In [57], the main model is the Coleman and Giddings-Strominger (CGS) model. The CGS model is a toy model of describing spacetime wormholes and it is more suggestive to obtain it from the Marolf–Maxfield (MM) model [37] by restricting the sum over topologies to only include the disk and the cylinder [57].
Let the amplitudes of the disk and cylinder be μ and t2, i.e.
$\begin{eqnarray}\left\langle \hat{Z}\right\rangle =\mu ,\,\left\langle {\hat{Z}}^{2}\right\rangle -\left\langle {\hat{Z}}^{2}\right\rangle ={t}^{2},\end{eqnarray}$
where ∣⟩ = ∣HH⟩ denotes the no-boundary (Hartle-Hawking) state and $\hat{Z}$ denotes the boundary creation operator thus $\left\langle {\hat{Z}}^{n}\right\rangle $ computes the Euclidean path integral over all manifolds with n boundaries. For the CGS model the gravity amplitude or the ‘correlation function of the partition function’ $\left\langle {\hat{Z}}^{n}\right\rangle $ is a polynomial of μ and t2 and in particular its generating function is simply $\begin{eqnarray}\left\langle {{\rm{e}}}^{u\hat{Z}}\right\rangle =\exp \left(u\mu +\frac{{u}^{2}{t}^{2}}{2}\right).\end{eqnarray}$
Thus we can identify $\hat{Z}$ as a Gaussian random variable Z such that the gravity amplitude $\left\langle f\left(\hat{Z}\right)\right\rangle $ can be computed as the ensemble average ${\mathbb{E}}\left(f\left(Z\right)\right)\equiv \left\langle f\left(Z\right)\right\rangle .$ This equivalence is a baby version of gravity/ensemble duality.The crucial idea of [57] is that the correlation functions of partition function do not factorize in general but they factorize between α-states which are the eigenstates of $\hat{Z}$ 8 ) coincides with the trick used in [69] and [62] of rewriting ${Z}_{\alpha }^{n}$ as a formal average
$\begin{eqnarray}\left\langle \alpha \left|{\hat{Z}}^{2}\right|\alpha \right\rangle ={\left\langle \alpha \left|\hat{Z}\right|\alpha \right\rangle }^{2}={Z}_{\alpha }^{2}.\end{eqnarray}$
The α-state is also created by a generation operator acting on $\left|{\rm{HH}}\right\rangle $ $\begin{eqnarray}\left|\alpha \right\rangle ={\psi }_{\alpha }\left|{\rm{HH}}\right\rangle .\end{eqnarray}$
Note that ψ can be expressed in terms of $\hat{Z}$ in a very complicated way so ψ commutes with $\hat{Z}.$ Then (3) can be rewritten in a very suggestive way $\begin{eqnarray}{Z}_{\alpha }^{2}=\left\langle {\psi }_{\alpha }^{2}{\hat{Z}}^{2}\right\rangle =\left\langle {\hat{Z}}^{2}\right\rangle +{\left\langle {\psi }_{\alpha }^{2}{\hat{Z}}^{2}\right\rangle }_{c},\end{eqnarray}$
where we have assumed that α-state is normalized $\left\langle {\psi }_{\alpha }^{2}\right\rangle \mathrm{=1}.$ This rewriting is interesting because it separates out the self-averaged part $\left\langle {\hat{Z}}^{2}\right\rangle $ and non-self-averaged part ${\left\langle {\psi }_{\alpha }^{2}{\hat{Z}}^{2}\right\rangle }_{c}$. In the CGS model, since the eigenvalue of $\hat{Z}$ is continuous and supported on ${\mathbb{R}}$ so that we can express ψα in terms of $\hat{Z}$ schematically as $\begin{eqnarray}{\psi }_{\alpha }=\delta \left(\hat{Z}-{Z}_{\alpha }\right)=\int \frac{{\rm{d}}k}{2\pi }{{\rm{e}}}^{{\rm{i}}k\left(\hat{Z}-{Z}_{\alpha }\right)},\end{eqnarray}$
thus $\begin{eqnarray}\left\langle {Z}^{2}{\psi }_{\alpha }\right\rangle =\displaystyle \int \displaystyle \frac{{\rm{d}}k\,{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{2\pi }\left\langle {Z}^{2}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \end{eqnarray}$
$\begin{eqnarray}\to {Z}_{\alpha }^{2}=\displaystyle \int \displaystyle \frac{{\rm{d}}k\,{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{2\pi }\displaystyle \frac{\left\langle {Z}^{2}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }{\left\langle {\psi }_{\alpha }\right\rangle }\mathrm{}.\end{eqnarray}$
Noting that ⟨ψα⟩ = P(Zα), where P(Z) is the PDF of Z, we find that ( $\begin{eqnarray}{Z}_{\alpha }^{n}=\displaystyle \int {\rm{d}}Z\delta \left(Z-{Z}_{\alpha }\right)\displaystyle \frac{{Z}^{n}P\left(Z\right)}{P\left({Z}_{\alpha }\right)}=\displaystyle \int \displaystyle \frac{{\rm{d}}k}{2\pi }\displaystyle \frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P\left({Z}_{\alpha }\right)}\left\langle {Z}^{n}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \mathrm{}.\end{eqnarray}$
From which we can derive some useful approximation formula ${Z}_{\alpha }^{n}\approx \left\langle {Z}_{\alpha }^{n}\right\rangle +{\rm{\Phi }},$ where $\left\langle {Z}_{\alpha }^{n}\right\rangle $ and Φ are respectively recognized as the wormhole and half-wormhole contributions as shown in [62, 69]. So we can think of this trick as a refinement of the factorization proposal of [57]. We will elaborate on this below.2.1. Half-Wormhole in CGS-like model
In the CGS model, because Z satisfies the Gaussian distribution there is a more concrete expression for the half wormhole saddle as shown in [57]. The key point is the fact that when Z is Gaussian, it can be thought of as the position operator of a simple harmonic oscillator so there exists a natural orthogonal basis, the number basis {n} which is called the n-baby Universe basis in the context of the gravity model. If we insert the complete basis $\displaystyle {\sum }_{i}\left|i\right\rangle \left\langle i\right|$ into (9) we can get2(2 Note that our convention is Z = μ + t(a + a†).) 14 ) coincides with results in [69]. So we confirm the result that within the Gaussian approximation (only keep the first two cumulants), ${Z}_{\alpha }^{n}$ can be decomposed as (14 ) and it suggests that θi’s are the convenient building blocks of possible half-wormhole saddles. Some examples of the decomposition (14 ) are3(3 θ(i) is simply the (unnormalized) Hermite polynomial.)
$\begin{eqnarray}{Z}_{\alpha }^{n}=\int {\rm{d}}Z\delta \left(Z-{Z}_{\alpha }\right)\frac{{Z}^{n}P\left(Z\right)}{P\left({Z}_{\alpha }\right)}=\int \frac{{\rm{d}}k}{2\pi }\frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P\left({Z}_{\alpha }\right)}\left\langle {Z}^{n}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \end{eqnarray}$
$\begin{eqnarray}=\int \frac{{\rm{d}}k}{2\pi }\frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P({Z}_{\alpha })}\displaystyle \sum _{i=0}^{n}\left\langle {Z}^{n}| i\right\rangle \left\langle i| {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \end{eqnarray}$
$\begin{eqnarray}=\int \frac{{\rm{d}}k}{2\pi }\frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P({Z}_{\alpha })}\displaystyle \sum _{i=0}^{n}\left(\begin{array}{l}n\\ i\end{array}\right)\left\langle {Z}^{n-i}\right\rangle \left\langle {Z}^{i}| i\right\rangle \left\langle i| {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \end{eqnarray}$
$\begin{eqnarray}=\int \frac{{\rm{d}}k}{2\pi }\frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P({Z}_{\alpha })}\displaystyle \sum _{i=0}^{n}\left(\begin{array}{l}n\\ i\end{array}\right)\left\langle {Z}^{n-i}\right\rangle \sqrt{i!}{t}^{i}\left\langle i| {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \end{eqnarray}$
$\begin{eqnarray}=\int \frac{{\rm{d}}k}{2\pi }\frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P\left({Z}_{\alpha }\right)}\displaystyle \sum _{i=0}^{n}\left(\begin{array}{l}n\\ i\end{array}\right)\left\langle {Z}^{i}\right\rangle \left\langle {\left(at\right)}^{n-i}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \equiv \displaystyle \sum _{i=0}^{n}\left(\begin{array}{l}n\\ i\end{array}\right){\mu }_{i}{\theta }^{(n-i)},\end{eqnarray}$
where $\begin{eqnarray}{\theta }^{\left(n-i\right)}=\displaystyle \int \displaystyle \frac{{\rm{d}}k}{2\pi }\displaystyle \frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P\left({Z}_{\alpha }\right)}\left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle \displaystyle \frac{\left\langle {\left(at\right)}^{n-i}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }{\left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle },\,\displaystyle \frac{\left\langle {\left(at\right)}^{n-i}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }{\left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }\equiv {\phi }_{n-i}^{c}\mathrm{}.\end{eqnarray}$
Note that $\begin{eqnarray}\displaystyle \frac{\left\langle {\left(at\right)}^{n-i}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }{\left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }\equiv {\phi }_{n-i}^{c}={\left({\rm{i}}k{t}^{2}\right)}^{n-i}={\left({\phi }_{1}^{c}\right)}^{n-i},\end{eqnarray}$
$\begin{eqnarray}{\theta }^{\left(i\right)}=\displaystyle \int \left(\displaystyle \frac{{\rm{d}}k}{2\pi }\displaystyle \frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P\left({Z}_{\alpha }\right)}\right)\left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle {\left({\rm{i}}k{t}^{2}\right)}^{i}=\displaystyle \frac{{\left(-{t}^{2}{\partial }_{{Z}_{\alpha }}\right)}^{i}P\left({Z}_{\alpha }\right)}{P\left({Z}_{\alpha }\right)},\end{eqnarray}$
then ( $\begin{eqnarray}{Z}_{\alpha }^{1}={\theta }^{(1)}+\left\langle Z\right\rangle ,\end{eqnarray}$
$\begin{eqnarray}{Z}_{\alpha }^{2}={\theta }^{\mathrm{(2)}}+2\left\langle Z\right\rangle {\theta }^{\mathrm{(1)}}+\left\langle {Z}^{2}\right\rangle ,\end{eqnarray}$
$\begin{eqnarray}{Z}_{\alpha }^{3}={\theta }^{\mathrm{(3)}}+3\langle Z\rangle {\theta }^{\mathrm{(2)}}+3\langle {Z}^{2}\rangle {\theta }^{\mathrm{(1)}}+\langle {Z}^{3}\rangle ,\end{eqnarray}$
$\begin{eqnarray}{Z}_{\alpha }^{4}={\theta }^{\mathrm{(4)}}+4\left\langle Z\right\rangle {\theta }^{\mathrm{(3)}}+6\left\langle {Z}^{2}\right\rangle {\theta }^{\mathrm{(2)}}+4\left\langle {Z}^{3}\right\rangle {\theta }^{\mathrm{(1)}}+\left\langle {Z}^{4}\right\rangle ,\end{eqnarray}$
with $\begin{eqnarray}{\theta }^{\left(1\right)}=-\mu +{Z}_{\alpha },\,{\theta }^{\mathrm{(2)}}={\left(\mu -{Z}_{\alpha }\right)}^{2}-{t}^{2}={\theta }^{{\left(1\right)}^{2}}-{t}^{2},\end{eqnarray}$
$\begin{eqnarray}{\theta }^{\left(3\right)}=-{\left(\mu -{Z}_{\alpha }\right)}^{3}+3\left(\mu -{Z}_{\alpha }\right){t}^{2}={\theta }^{{\left(1\right)}^{3}}-3{t}^{2}{\theta }^{\left(1\right)},\end{eqnarray}$
$\begin{eqnarray}{\theta }^{\mathrm{(4)}}\mathrm{=3}{t}^{4}-6{t}^{2}{\left(\mu -{Z}_{\alpha }\right)}^{2}+{\left(\mu -{Z}_{\alpha }\right)}^{4}={\theta }^{{\left(1\right)}^{4}}-6{t}^{2}{\theta }^{{\left(1\right)}^{2}}+3{t}^{4}\mathrm{}.\end{eqnarray}$
In general we have $\begin{eqnarray}{\theta }^{\left(i\right)}=\displaystyle \int \displaystyle \frac{{\rm{d}}k}{2\pi }\displaystyle \frac{{{\rm{e}}}^{-{\rm{i}}k{Z}_{\alpha }}}{P\left({Z}_{\alpha }\right)}\left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle {\left({\rm{i}}k{t}^{2}\right)}^{i}=\displaystyle \int \displaystyle \frac{{\rm{d}}k}{\sqrt{2\pi /{t}^{2}}}{{\rm{e}}}^{-\displaystyle \frac{{\left(k-\left({\rm{i}}\left(\mu -{Z}_{\alpha }\right)/{t}^{2}\right)\right)}^{2}}{\mathrm{2/}{t}^{2}}}{\left({\rm{i}}k{t}^{2}\right)}^{i},\end{eqnarray}$
so θ(i)/(it2)i is the ith moment of ‘Gaussian distribution’ ${\mathscr{N}}\left({\rm{i}}\left(\mu -{Z}_{\alpha }\right)/{t}^{2}\mathrm{,1/}{t}^{2}\right)$ and the generating function is $\begin{eqnarray}{\left\langle {{\rm{e}}}^{uk}\right\rangle }_{k}={{\rm{e}}}^{\displaystyle \frac{{\rm{i}}u\left(\mu -{Z}_{\alpha }\right)}{{t}^{2}}+\displaystyle \frac{{u}^{2}}{2{t}^{2}}}\mathrm{}.\end{eqnarray}$
Considering the following ensemble average $\begin{eqnarray}{\left\langle {\left\langle {{\rm{e}}}^{{u}_{1}{k}_{1}}\right\rangle }_{{k}_{1}}{\left\langle {{\rm{e}}}^{{u}_{2}{k}_{2}}\right\rangle }_{{k}_{2}}\right\rangle }_{{Z}_{\alpha }}={{\rm{e}}}^{-\displaystyle \frac{{u}_{1}{u}_{2}}{{t}^{2}}},\end{eqnarray}$
and expanding both sides into Taylor series of u1 and u2 one can find $\begin{eqnarray}{\left\langle {\theta }^{\left(i\right)}{\theta }^{\left(j\right)}\right\rangle }_{{Z}_{\alpha }}=i!{t}^{2i}{\delta }_{ij}\mathrm{}.\end{eqnarray}$
Due to this orthogonal condition we can directly tell which sector in the decomposition of ${Z}_{\alpha }^{n}$ is dominant by computing $\left\langle {Z}_{\alpha }^{n}{Z}_{\alpha }^{n}\right\rangle $ $\begin{eqnarray}{Z}_{\alpha }^{n}=\displaystyle \displaystyle \sum _{i}{c}_{i}{\theta }^{\left(i\right)},\,\left\langle {Z}_{\alpha }^{n}{Z}_{\alpha }^{n}\right\rangle =\displaystyle \displaystyle \sum _{i}{c}_{i}^{2}i!{t}^{2i}\mathrm{}.\end{eqnarray}$
In CGS model, since there is only a single random variable Z so it does not admit any approximation related to large N or small GN. Therefore the wormhole or half-wormhole are not true saddles in the usual sense. To breathe life into them we should consider a model with a large number N of random variables such as random matrix theory or SYK model which can be described by certain semi-classical collective variables like the G,∑ in SYK, which potentially have a dual gravity description. However, we find that it is illustrative to first apply the factorization proposal to some simple statistical models as we did in [69].2.2. Statistical model
Let us consider a function Y(Xi) of a large number N independent random variables Xi. Assuming that Xi’s are drawn from the Gaussian distribution then we have the decomposition
$\begin{eqnarray}\begin{array}{l}{Y}^{n}=\displaystyle \frac{1}{{\left(2\pi \right)}^{N}}\displaystyle \int \displaystyle \prod _{i}\left({\rm{d}}{k}_{i}\displaystyle \frac{{{\rm{e}}}^{-{\rm{i}}{k}_{i}{X}_{i}}}{P\left({X}_{i}\right)}\right)\left\langle {{\rm{e}}}^{{\rm{i}}\displaystyle {\sum }_{i}{k}_{i}{x}_{i}}\right\rangle \displaystyle \displaystyle \sum _{{n}_{1},\ldots ,{n}_{N}}\left\langle {Y}^{n}\left|{n}_{1},\ldots ,{n}_{N}\right.\right\rangle \displaystyle \frac{\left\langle {n}_{1},\ldots ,{n}_{N}\left|{{\rm{e}}}^{{\rm{i}}\displaystyle {\sum }_{i}{k}_{i}{x}_{i}}\right.\right\rangle }{\left\langle {{\rm{e}}}^{{\rm{i}}\displaystyle {\sum }_{i}{k}_{i}{x}_{i}}\right\rangle }\\ =\,\displaystyle \displaystyle \sum _{k=\displaystyle {\sum }_{i}{n}_{i}}{{\rm{\Gamma }}}_{k},\end{array}\end{eqnarray}$
where Γk denotes different sectors, in particular Γ0 = ⟨Yn⟩. This kind of model can be also thought of as the CGS model with species [57].2.2.1. Simple observables
The simplest operator is
$\begin{eqnarray}Y=\displaystyle \sum _{i\mathrm{=1}}^{N}{X}_{i}\mathrm{}.\end{eqnarray}$
Apparently for n = 1 there are only two sectors $\begin{eqnarray}Y=\displaystyle \sum _{i\mathrm{=1}}^{N}\mu +\displaystyle \sum _{i\mathrm{=1}}^{N}{\theta }_{i}^{\left(1\right)}={{\rm{\Theta }}}_{0}+{{\rm{\Theta }}}_{1},\left\langle {{\rm{\Theta }}}_{0}^{2}\right\rangle ={N}^{2}{\mu }^{2},\,\left\langle {{\rm{\Theta }}}_{1}^{2}\right\rangle =N{t}^{2},\end{eqnarray}$
and for n = 2 there are three sectors $\begin{eqnarray}{Y}^{2}={{\rm{\Phi }}}_{0}+{{\rm{\Phi }}}_{1}+{{\rm{\Phi }}}_{2},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{0}=\left\langle {Y}^{2}\right\rangle ,\,{{\rm{\Phi }}}_{1}\mathrm{=2}N\mu \displaystyle \displaystyle \sum _{i}{\theta }_{i}^{\mathrm{(1)}},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{2}=\displaystyle \displaystyle \sum _{ij}\left({\theta }_{i}^{\left(1\right)}{\theta }_{j}^{\left(1\right)}+{\delta }_{ij}\left({\theta }_{i}^{\left(2\right)}-{\theta }_{i}^{{\left(1\right)}^{2}}\right)\right)=\displaystyle \displaystyle \sum _{ij}\left({\theta }_{i}^{\left(1\right)}{\theta }_{j}^{\left(1\right)}-{\delta }_{ij}{t}^{2}\right).\end{eqnarray}$
In general the parameters μ and t2 are N independent therefore Yn is self-averaged Yn ≈ ⟨Yn⟩ in the large N limit. This is also true even Xi are not Gaussian because of the central limit theorem. But we also know in the literature that in order to have well-defined semi-classical approximation, the parameters μ and t2 should depend on N in a certain way like in SYK model. Interestingly in this case if t2 ∼ μ2N, the self-averaged part and non-self-averaged part are comparable and we should keep them both. This is exactly what we have encountered in the 0-SYK model. But a crucial difference is that for this simple choice of observables, all the non-self-averaged sectors are also comparable so it is not fair to call any of them the half-wormhole saddle and to restore factorization we have to include all the non-self-averaged sectors. The extremal case is t2 ≫ μ2N. In this limit we find that the sector with highest level dominates. For example, $\begin{eqnarray}Y\approx {{\rm{\Theta }}}_{1},\,{Y}^{2}\approx \left\langle {Y}^{2}\right\rangle +{{\rm{\Phi }}}_{2},\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Theta }}}_{1}^{2}\right\rangle \approx \left\langle {Y}^{2}\right\rangle ,\,\left\langle {{\rm{\Theta }}}_{2}^{2}\right\rangle \approx 2{\left\langle {Y}^{2}\right\rangle }^{2},\end{eqnarray}$
then it is reasonable to identify Θ1 with half-wormhole and identify Φ2 with the 2-linked half-wormhole. Similarly we can introduce n-linked half-wormholes. For example, in this extremal case, we can approximate Y3 with $\begin{eqnarray}{Y}^{3}\approx 3\left\langle {Y}^{2}\right\rangle {{\rm{\Theta }}}_{1}+{{\rm{\Lambda }}}_{3},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Lambda }}}_{3}=\displaystyle \displaystyle \sum _{i\ne j\ne k}\left({\theta }_{i}^{\left(1\right)}{\theta }_{j}^{\left(1\right)}{\theta }_{k}^{\left(1\right)}\right)+3\displaystyle \displaystyle \sum _{i\ne j}\left({\theta }_{i}^{\left(2\right)}{\theta }_{j}^{\left(1\right)}\right)+\displaystyle \displaystyle \sum _{i}\left({\theta }_{i}^{\left(3\right)}\right),\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \displaystyle \sum _{i,j,k}\left({\theta }_{i}^{\left(1\right)}{\theta }_{j}^{\left(1\right)}{\theta }_{k}^{\left(1\right)}\right)-3{t}^{2}N\displaystyle \displaystyle \sum _{i}\left({\theta }_{i}^{\left(1\right)}\right),\end{eqnarray}$
where the sector Λ3 should describe the 3-linked half-wormhole. We will consider a similar construction in the 0-SYK model.2.2.2. Exponential observables
In the Random Matrix Theory or quantum mechanics, the most relevant observable is the exponential operator Tr(eβH) since it relates to the partition function. So it may be interesting to consider a similar exponential operator
$\begin{eqnarray}Y=\displaystyle \displaystyle \sum _{i}{{\rm{e}}}^{\beta {X}_{i}},\end{eqnarray}$
in the toy statistical model. By a Taylor expansion of the exponential operator we find the following decomposition $\begin{eqnarray}{{\rm{e}}}^{\beta X}=\left\langle {{\rm{e}}}^{\beta X}\right\rangle \displaystyle \displaystyle \sum _{k}\displaystyle \frac{{\beta }^{k}{\theta }^{\left(k\right)}}{k!},\,{\theta }^{\left(0\right)}\equiv \mathrm{1,}\end{eqnarray}$
thus $\begin{eqnarray}Y=\displaystyle \displaystyle \sum _{k}{{\rm{\Theta }}}_{k},\,{{\rm{\Theta }}}_{k}={{\rm{e}}}^{\mu \beta +\displaystyle \frac{{\beta }^{2}{t}^{2}}{2}}\displaystyle \displaystyle \sum _{i}\displaystyle \frac{{\beta }^{k}{\theta }_{i}^{\left(k\right)}}{k!},\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle =N{{\rm{e}}}^{2\mu \beta +{t}^{2}{\beta }^{2}}\displaystyle \frac{{\left(\beta t\right)}^{2k}}{k!},\,\displaystyle \frac{\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle }{\left\langle {Y}^{2}\right\rangle }={{\rm{e}}}^{-{t}^{2}{\beta }^{2}}\displaystyle \frac{{\left(\beta t\right)}^{2k}}{k!}\equiv {r}_{k}\mathrm{}.\end{eqnarray}$
Interestingly the ratio rk follows the Poisson distribution Pois (βt2), some examples are in figure 1. When βt ≪ 1 the dominant sector is Θ0 while for βt ≫ 1 the Poisson distribution approaches Gaussian distribution N(β2t2, β2t2) so we have to include all the sectors in the peak k ∈ (β2t2 − βt, β2t2 + βt) to have a good approximation. We can decompose Y2 in a similar way $\begin{eqnarray}{Y}^{2}=\displaystyle \displaystyle \sum _{k}{{\rm{\Phi }}}_{k},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{k}={{\rm{e}}}^{2\mu \beta +{\beta }^{2}{t}^{2}}{\beta }^{k}\displaystyle \displaystyle \sum _{i\ne j}\displaystyle \displaystyle \sum _{n}\displaystyle \frac{{\theta }_{i}^{\left(n\right)}}{n!}\displaystyle \frac{{\theta }_{j}^{\left(k-n\right)}}{\left(k-n\right)!}+{{\rm{e}}}^{2\mu \beta +2{\beta }^{2}{t}^{2}}{\left(2\beta \right)}^{k}\displaystyle \displaystyle \sum _{i}\displaystyle \frac{{\theta }_{i}^{\left(k\right)}}{k!},\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle }{\left\langle {Y}^{4}\right\rangle }={{\rm{e}}}^{-2{\beta }^{2}{t}^{2}}\displaystyle \frac{{\left(2{\beta }^{2}{t}^{2}\right)}^{k}}{k!}\displaystyle \frac{2\left(N-1\right)+{2}^{k}{{\rm{e}}}^{2{\beta }^{2}{t}^{2}}+4\left(N-1\right){{\rm{e}}}^{{\beta }^{2}{t}^{2}}}{2\left(N-1\right)+{{\rm{e}}}^{4{\beta }^{2}{t}^{2}}+4\left(N-1\right){{\rm{e}}}^{{\beta }^{2}{t}^{2}}}\mathrm{}.\end{eqnarray}$
The behavior is similar. When βt ≪ 1, the dominant sector is the self-averaged sector Φ0. When 2β2t2 > log N (47 ) approaches the Gaussian N(4β2t2, 4β2t2). On the other hand, when 1 ≪ 2β2t2 ≪ log N (47 ) approaches the Gaussian N(2β2t2, 2β2t2). In the end when 2β2t2 ∼ log N, (46 ) will have two comparable peaks, see figure 2 as an example. However the half-wormhole ansatz proposed in [62, 69] which can be written as
$\begin{eqnarray}{\rm{\Phi }}=\displaystyle \sum _{k\mathrm{=0}}^{\infty }{\phi }_{k},\end{eqnarray}$
$\begin{eqnarray}{\phi }_{k}={{\rm{\Phi }}}_{k}+\left({{\rm{e}}}^{-{\beta }^{2}{t}^{2}}-1\right){{\rm{e}}}^{2\mu \beta +2{\beta }^{2}{t}^{2}}{\left(2\beta \right)}^{k}\displaystyle \displaystyle \sum _{i}\displaystyle \frac{{\theta }_{i}^{\left(k\right)}}{k!},\end{eqnarray}$
only works for small value of βt.To summarize our proposal, by introducing the basis {θi} which is the generalization of n-baby Universe basis [57] we can decompose the observables or partition functions into a single self-averaged sector and many non-self-averaged sectors. These sectors are independent in the sense of (28 ). The contributions from each sector have interesting statistics: in the large N limit leading contributing sectors may condense to peaks. This condensation is a signal that the observable potentially has a bulk description (or semi-classical description) in the large N limit. If the self-averaged sector survives then it means the observable is approximately self-averaging. The surviving non-self-averaged sectors in the large N limit are naturally interpreted as the (n-linked) half-wormholes which are the results of sector condensation. In the extremal case, only one non-self-averaging survives reminiscing the famous Bose–Einstein condensation.
Figure 1. Poisson distribution ( |
2.3. 0-SYK model
In this section we apply our proposal to the 0-SYK model which has the ‘action’54 ) is nothing but the hyperpfaffian Pf(J). According to (30 ), we can similarly decompose it as56 ) can be derived by a combinatorial method used in [69] or by using the G,∑ trick as follows. First we expand z into series of θ(1)
$\begin{eqnarray}z=\displaystyle \int {{\rm{d}}}^{N}\psi \exp \left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle \sum {J}_{{i}_{1}\ldots {i}_{q}}{\psi }_{{i}_{1}\ldots {i}_{q}}\right),\end{eqnarray}$
where ${\psi }_{{i}_{1}\ldots {i}_{q}}={\psi }_{{a}_{1}}{\psi }_{{a}_{2}}\ldots {\psi }_{{a}_{q}}$ and ψi are Grassmann numbers. The random couplings ${J}_{{i}_{1}\ldots {i}_{q}}$ is drawn from a Gaussian distribution $\begin{eqnarray}\left\langle {J}_{{i}_{1}\ldots {i}_{q}}\right\rangle =u,\,\left\langle {J}_{{i}_{1}\ldots {i}_{q}}{J}_{{j}_{1}\ldots {j}_{q}}\right\rangle ={t}^{2}{\delta }_{{i}_{1}{j}_{1}}\ldots {\delta }_{{i}_{q}{j}_{q}},\,{t}^{2}={\tau }^{2}\displaystyle \frac{\left(q-1\right)!}{{N}^{q-1}},\end{eqnarray}$
where we found in [69] in order to have a semi-classical description u should also have a proper dependence $\begin{eqnarray}u={\left(-{\rm{i}}\right)}^{q\mathrm{/2}}\mu \displaystyle \frac{\left(q\mathrm{/2}-1\right)!}{2{N}^{q\mathrm{/2}-1}}\mathrm{}.\end{eqnarray}$
We sometimes use the collective indies A, B to simplify the notation $\begin{eqnarray}A=\left\{{a}_{1}\lt \ldots \lt {a}_{q}\right\},\,{J}_{A}{\psi }_{A}\equiv {J}_{{a}_{1}\ldots {a}_{q}}{\psi }_{{a}_{1}\ldots {a}_{q}}\mathrm{}.\end{eqnarray}$
Integrating out the Grassmann numbers directly gives4(4 Here we choose the measure of Grassmann integral to be $\int {{\rm{d}}}^{N}\psi {\psi }_{1\ldots N}={{\rm{i}}}^{-N\mathrm{/2}}$.): $\begin{eqnarray}z=\displaystyle \int {{\rm{d}}}^{N}\psi \exp \left({{\rm{i}}}^{q\mathrm{/2}}{J}_{A}{\psi }_{A}\right)=\displaystyle \sum _{{A}_{1}\lt \ldots \lt {A}_{p}}^{^{\prime} }{\rm{sgn}}\left(A\right){J}_{{A}_{1}}\ldots {J}_{{A}_{p}},\,p=N/q,\end{eqnarray}$
where the expression ( $\begin{eqnarray}z=\displaystyle \displaystyle \sum _{i}{{\rm{\Theta }}}_{i},\,{{\rm{\Theta }}}_{0}=\left\langle z\right\rangle ,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Theta }}}_{k}={u}^{p-k}\displaystyle \sum _{{I}_{1}\lt \ldots \lt {I}_{p-k}}^{^{\prime} }{\rm{Pf}}\left({\theta }_{A}^{\left(1\right)\left({I}_{1},\ldots ,{I}_{p-k} \ \right)}\right),\end{eqnarray}$
where the tensor ${\theta }_{A}^{\left(1\right)\left({I}_{1},\ldots ,{I}_{p-k} \ \right)}$ means that the index A is not in the set (I1, …, Ip-k). The expression ( $\begin{eqnarray}z=\displaystyle \int {{\rm{d}}}^{N}\psi {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle \displaystyle \sum _{A}{J}_{A}{\psi }_{A}}=\displaystyle \int {{\rm{d}}}^{N}\psi {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}u{\psi }_{A}}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{\psi }_{A}}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {{\rm{d}}}^{N}\psi \displaystyle \displaystyle \sum _{k\mathrm{=0}}\displaystyle \frac{{\left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\mathrm{(1)}}{\psi }_{A}\right)}^{k}}{k!}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}u\displaystyle {\sum }_{A}{\psi }_{A}},\end{eqnarray}$
thus by matching the power of θ(1) we get a integral expression of Θk $\begin{eqnarray}{{\rm{\Theta }}}_{k}=\displaystyle \int {{\rm{d}}}^{N}\psi \displaystyle \frac{{\left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\mathrm{(1)}}{\psi }_{A}\right)}^{k}}{k!}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}u\displaystyle {\sum }_{A}{\psi }_{A}}\mathrm{}.\end{eqnarray}$
Next following [69] we can introduce G,∑ variables directly as $\begin{eqnarray}G=\displaystyle \frac{1}{N}\displaystyle \displaystyle \sum _{i\lt j}{\psi }_{i}{\psi }_{j},\end{eqnarray}$
$\begin{eqnarray}z=\displaystyle \int {{\rm{d}}}^{N}\psi \displaystyle {\int }_{{\mathbb{R}}}{\rm{d}}G\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}\displaystyle \frac{{\rm{d}}{\rm{\Sigma }}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{u{{\rm{i}}}^{q\mathrm{/2}}\displaystyle \frac{{N}^{q\mathrm{/2}}}{\left(q\mathrm{/2}\right)!}{G}^{q\mathrm{/2}}}{{\rm{e}}}^{-N{\rm{\Sigma }}G}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{\psi }_{A}}{{\rm{e}}}^{{\rm{\Sigma }}\displaystyle {\sum }_{i\lt j}{\psi }_{i}{\psi }_{j}}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {{\rm{d}}}^{N}\psi \displaystyle {\int }_{{\mathbb{R}}}{\rm{d}}G\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}\displaystyle \frac{{\rm{d}}{\rm{\Sigma }}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{u{{\rm{i}}}^{q\mathrm{/2}}\displaystyle \frac{{N}^{q\mathrm{/2}}}{\left(q\mathrm{/2}\right)!}{G}^{q\mathrm{/2}}}{{\rm{e}}}^{-N{\rm{\Sigma }}G}\displaystyle \displaystyle \sum _{k}\displaystyle \frac{{\left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\mathrm{(1)}}{\psi }_{A}\right)}^{k}}{k!}{{\rm{e}}}^{{\rm{\Sigma }}\displaystyle {\sum }_{i\lt j}{\psi }_{i}{\psi }_{j}},\end{eqnarray}$
and $\begin{eqnarray}{{\rm{\Theta }}}_{k}=\displaystyle \int {{\rm{d}}}^{N}\psi \displaystyle {\int }_{{\mathbb{R}}}{\rm{d}}G\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}\displaystyle \frac{{\rm{d}}{\rm{\Sigma }}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{u{{\rm{i}}}^{q\mathrm{/2}}\displaystyle \frac{{N}^{q\mathrm{/2}}}{\left(q\mathrm{/2}\right)!}{G}^{q\mathrm{/2}}}{{\rm{e}}}^{-N{\rm{\Sigma }}G}\displaystyle \frac{{\left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\mathrm{(1)}}{\psi }_{A}\right)}^{k}}{k!}{{\rm{e}}}^{{\rm{\Sigma }}\displaystyle {\sum }_{i\lt j}{\psi }_{i}{\psi }_{j}},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{c}=\int {{\rm{d}}}^{N}\psi {\int }_{{\mathbb{R}}}{\rm{d}}G{\int }_{{\rm{i}}{\mathbb{R}}}\frac{{\rm{d}}{\rm{\Sigma }}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{u{{\rm{i}}}^{q/2}\frac{{N}^{q/2}}{(q/2)!}{G}^{q/2}}{{\rm{e}}}^{-N{\rm{\Sigma }}G}{\frac{({{\rm{i}}}^{q/2}\displaystyle \sum _{A}{\theta }_{A}^{(1)}{\psi }_{A})}{k!}}^{k}\frac{{(q/2!)}^{p-k}{{\rm{\Sigma }}}^{\displaystyle \frac{N-{qk}}{2}}{(\displaystyle {\sum }_{A}{\psi }_{A})}^{p-k}}{(N/2-{qk}/2)!}\\ ={\int }_{{\mathbb{R}}}{\rm{d}}G\displaystyle \sum _{{\rm{i}}{\mathbb{R}}}\frac{{\rm{d}}{\rm{\Sigma }}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{u{{\rm{i}}}^{q/2}\frac{{N}^{q/2}}{(q/2)!}{G}^{q/2}}{{\rm{e}}}^{-N{\rm{\Sigma }}G}{({\rm{i}}{\rm{\Sigma }})}^{\frac{N-{qk}}{2}}\frac{{(q/2!)}^{p-k}(p-k)!}{(N/2-{qk}/2)}\\ \times \int {{\rm{d}}}^{N}\psi \displaystyle \sum _{{A}_{1}\lt \ldots \lt {A}_{k}}{\theta }_{{A}_{1}}^{(1)}\ldots {\theta }_{{A}_{k}}^{(1)}{\psi }_{{A}_{1}}\ldots {\psi }_{{A}_{k}}\times \displaystyle \sum _{{I}_{1}\lt \ldots \lt {I}_{p-k}}{\psi }_{{I}_{1}}\ldots {\psi }_{{I}_{p-k}}\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{c}={\int }_{{\mathbb{R}}}{\rm{d}}G{\int }_{{\rm{i}}{\mathbb{R}}}\frac{{\rm{d}}{\rm{\Sigma }}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{u{{\rm{i}}}^{q/2}\frac{{N}^{q/2}}{\left(q/2\right)!}{G}^{q/2}}{{\rm{e}}}^{-N{\rm{\Sigma }}G}{\left({\rm{i}}{\rm{\Sigma }}\right)}^{\frac{N-{qk}}{2}}\frac{\left(p-k\right)!{\left(q/2!\right)}^{p-k}}{\left(N/2-{qk}/2\right)!}\\ \times \displaystyle \sum _{{I}_{1}\lt \ldots \lt {I}_{p-k}}^{\mbox{'}}{\rm{PF}}\left({\theta }_{A}^{\left(1\right)\left({I}_{1},\ldots ,{I}_{p-k} \ \right)}\right),\\ ={u}^{p-k}\displaystyle \sum _{{I}_{1}\lt \ldots \lt {I}_{p-k}}^{\mbox{'}}{\rm{PF}}\left({\theta }_{A}^{\left(1\right)\left({I}_{1},\ldots ,{I}_{p-k} \ \right)}\right),\end{array}\end{eqnarray}$
where the tensor ${\theta }_{A}^{\left(1\right)\left({I}_{1},\ldots ,{I}_{p-k} \ \right)}$ means that the index A is not in the set (I1, …, Ip-k). To figure out which one is dominant let us compute $\begin{eqnarray}\left\langle {z}^{2}\right\rangle =\displaystyle \displaystyle \sum _{i}\left\langle {{\rm{\Theta }}}_{i}{{\rm{\Theta }}}_{i}\right\rangle \mathrm{}.\end{eqnarray}$
The expression of ⟨z2⟩ is derived in [69] $\begin{eqnarray}\left\langle {z}^{2}\right\rangle =\displaystyle \sum _{k\mathrm{=0}}^{p}{c}_{k}{m}_{p-k}^{2}{t}^{2k}{u}^{2p-2k}\equiv \displaystyle \displaystyle \sum _{k}{z}_{2}^{\left(k\right)},\end{eqnarray}$
where $\begin{eqnarray}{c}_{k}=\displaystyle \frac{1}{k!}\left(\begin{array}{c}N\\ q\end{array}\right)\left(\begin{array}{c}N-q\\ q\end{array}\right)\ldots \left(\begin{array}{c}N-\left(k-1\right)q\\ q\end{array}\right)=\displaystyle \frac{N!}{k!{\left(q!\right)}^{k}\left(N-kq\right)!},\end{eqnarray}$
$\begin{eqnarray}{m}_{p}=\displaystyle \frac{\left(pq\mathrm{/2}\right)!}{p!{\left(\left(q\mathrm{/2}\right)!\right)}^{p}}\mathrm{}.\end{eqnarray}$
By matching the power of t2 we can identify $\begin{eqnarray}{z}_{2}^{\left(k\right)}=\left\langle {{\rm{\Theta }}}_{k}{{\rm{\Theta }}}_{k}\right\rangle ={c}_{k}{m}_{p-k}^{2}{t}^{2k}{u}^{2p-2k}\mathrm{}.\end{eqnarray}$
The coefficient is very involved so let us first consider some simple cases. If p = 2, then there are only three sectors $\begin{eqnarray}z=\left\langle z\right\rangle +{{\rm{\Theta }}}_{1}+{{\rm{\Theta }}}_{2},\end{eqnarray}$
$\begin{eqnarray}{z}_{2}^{\mathrm{(0)}}=\displaystyle \frac{{\left(q!\right)}^{2}}{4{\left(\displaystyle \frac{q}{2}!\right)}^{4}}{u}^{4},\,{z}_{2}^{\left(1\right)}=\displaystyle \frac{\left(2q\right)!}{{\left(q!\right)}^{2}}{u}^{2}{t}^{2},\,{z}_{2}^{\left(2\right)}=\displaystyle \frac{\left(2q\right)!}{2{\left(q!\right)}^{2}}{t}^{4}\mathrm{}.\end{eqnarray}$
Taking the large N limit, we find $\begin{eqnarray}{z}_{2}^{\left(1\right)}\sim \sqrt{N}\displaystyle \frac{{t}^{2}}{{u}^{2}}{z}_{2}^{\left(0\right)},\,{z}_{2}^{\left(2\right)}\sim \sqrt{N}\displaystyle \frac{{t}^{4}}{{u}^{4}}{z}_{2}^{\left(0\right)},\end{eqnarray}$
and $\begin{eqnarray}\displaystyle \frac{{t}^{2}}{{u}^{2}}\approx \displaystyle \frac{1}{N}\displaystyle \frac{{\tau }^{2}}{4{\mu }^{2}}\displaystyle \frac{\displaystyle \frac{N}{2}!}{{\left(\displaystyle \frac{N}{4}!\right)}^{2}}\sim \displaystyle \frac{{\tau }^{2}}{{\mu }^{2}}\displaystyle \frac{{2}^{N\mathrm{/2}}}{N\sqrt{N}},\end{eqnarray}$
which implies that $\begin{eqnarray}{z}_{2}^{\left(1\right)}\sim \displaystyle \frac{{2}^{N\mathrm{/2}}}{N}{z}_{2}^{\left(0\right)},\,{z}_{2}^{\left(2\right)}\sim \displaystyle \frac{{2}^{N}}{{N}^{2}\sqrt{N}}{z}_{2}^{\left(0\right)}\sim \displaystyle \frac{{2}^{N\mathrm{/2}}}{N\sqrt{N}}{z}_{2}^{\left(1\right)},\end{eqnarray}$
so that we have the approximation $\begin{eqnarray}z\sim {{\rm{\Theta }}}_{2}\mathrm{}.\end{eqnarray}$
Similarly when p = 3, we can find $\begin{eqnarray}z=\left\langle z\right\rangle +{{\rm{\Theta }}}_{1}+{{\rm{\Theta }}}_{2}+{{\rm{\Theta }}}_{3},\end{eqnarray}$
and $\begin{eqnarray}{z}_{2}^{\left(3\right)}\sim \displaystyle \frac{{t}^{2}}{{u}^{2}}\displaystyle \frac{1}{3}{z}_{2}^{\left(2\right)},\,{z}_{2}^{\left(2\right)}\sim \displaystyle \frac{{t}^{2}}{{u}^{2}}\sqrt{N}{z}_{2}^{\left(1\right)},\,{z}_{2}^{\left(1\right)}\sim \displaystyle \frac{{t}^{2}}{{u}^{2}}\sqrt{N}{z}_{2}^{\left(0\right)},\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{{t}^{2}}{{u}^{2}}\sim \displaystyle \frac{{2}^{N\mathrm{/3}}}{N\sqrt{N}},\end{eqnarray}$
thus $\begin{eqnarray}z\approx {{\rm{\Theta }}}_{3}\mathrm{}.\end{eqnarray}$
This turns out be general: when p ≪ N the dominant term is Θp. Therefore, the self-averaged ⟨z⟩ will not survive. This behavior is the same as we found in the simple statistical model in the regime when the cylinder amplitude is much larger than the disk amplitude.
Figure 2. Plot of ( |
On the other hand, if q ≪ N then
$\begin{eqnarray}\displaystyle \frac{{t}^{2}}{{u}^{2}}\sim \displaystyle \frac{{\tau }^{2}}{{\mu }^{2}}\sim \displaystyle \frac{1}{N},\end{eqnarray}$
the situation is very different. As a simple demonstration let us consider the case of q = 2 $\begin{eqnarray}{z}_{2}^{\left(p\right)}\sim {N}^{N\mathrm{/2}}\displaystyle \frac{{t}^{N}}{{u}^{N}}{z}_{2}^{\left(0\right)}=\displaystyle \frac{1}{{N}^{N\mathrm{/2}}}{z}_{2}^{\left(0\right)},\end{eqnarray}$
$\begin{eqnarray}{z}_{2}^{\left(1\right)}\sim {N}^{2}\displaystyle \frac{{t}^{2}}{{u}^{2}}{z}_{2}^{\left(0\right)}=N{z}_{2}^{\left(0\right)},\,{z}_{2}^{\left(2\right)}\sim {N}^{4}\displaystyle \frac{{t}^{4}}{{u}^{4}}{z}_{2}^{\left(0\right)}={N}^{2}{z}_{2}^{\left(0\right)},\end{eqnarray}$
$\begin{eqnarray}{z}_{2}^{\left(3\right)}\sim {N}^{6}\displaystyle \frac{{t}^{6}}{{u}^{6}}{z}_{2}^{\left(0\right)}={N}^{3}{z}_{2}^{\left(0\right)},\ldots ,\end{eqnarray}$
$\begin{eqnarray}{z}_{2}^{\left(k\right)}\sim {s}_{k}{z}_{2}^{\left(0\right)},\,{s}_{k}=\displaystyle \frac{1}{{2}^{k}{N}^{k}}\displaystyle \frac{N!}{k!\left(N-2k\right)!}\mathrm{}.\end{eqnarray}$
The dominant term is neither ⟨z⟩ nor Θp but some intermediate term Θk as argued in [69]. With this detailed analysis we find that we should also include some ‘sub-leading’ sectors. The distribution of the surviving sectors in the large N limit has a peak centered at the ‘dominant’ sector with a width roughly √N. One possible interpretation of this result is the surviving sectors are only approximate saddles or constrained saddles with some free parameters. Even though each approximate saddle contribution is as tiny as 1/√N but after integrating over the free parameters the total contribution is significant. Note that similar approximate saddles are also found for the spectral form factor in the SYK model [68]. We plot the ratio ${z}_{2}^{\left(k\right)}/{z}_{2}^{\left(0\right)}$ as function of k in figure 3. With increasing q or equivalently decreasing p, the peak moves to the left (small k) and becomes sharper and sharper. This is consistent with our analysis of limit of small p where there is only one dominant saddle, Θp. So our result shows that the wormhole (actually disk in this case) does not persist but the half-wormhole appears. As we found in [69] ⟨z2⟩ can be computed by a trick of introducing the collective variables
$\begin{eqnarray}{G}_{LR}=\displaystyle \frac{1}{N}\displaystyle \displaystyle \sum _{i}{\psi }_{i}^{L}{\psi }_{i}^{R},\,{G}_{L}=\displaystyle \frac{1}{N}\displaystyle \displaystyle \sum _{i\lt j}{\psi }_{i}^{L}{\psi }_{j}^{L},\,{G}_{R}=\displaystyle \frac{1}{N}\displaystyle \displaystyle \sum _{i\lt j}{\psi }_{i}^{R}{\psi }_{j}^{R},\end{eqnarray}$
and doing the path integral. The final expression is $\begin{eqnarray*}\begin{array}{c}\left\langle {z}^{2}\right\rangle =\displaystyle {\int }_{R}{{\rm{d}}}^{3}{G}_{i}\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}{{\rm{d}}}^{3}{\sum }_{i}\,{{\rm{e}}}^{\displaystyle \frac{N}{q}\left({\tau }^{2}{G}_{LR}^{q}+\mu {G}_{L}^{q\mathrm{/2}}+\mu {G}_{R}^{q\mathrm{/2}}\right)-N\left({{\rm{\Sigma }}}_{i}{G}_{i}\right)}\displaystyle \frac{1}{2}\left({\left({{\rm{\Sigma }}}_{LR}+{\rm{i}}\sqrt{{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}}\right)}^{N}+{\left({{\rm{\Sigma }}}_{LR}-{\rm{i}}\sqrt{{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}}\right)}^{N}\right)\\ =\displaystyle {\int }_{R}{{\rm{d}}}^{3}{G}_{i}\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}{{\rm{d}}}^{3}\displaystyle \displaystyle \sum _{i}\displaystyle \sum _{m=0}^{N/2}\left(\begin{array}{c}N\\ 2m\end{array}\right){\left({\sum }_{LR}\right)}^{2m}{\left({\rm{i}}{\sum }_{L}{\sum }_{R}\right)}^{\displaystyle \frac{N}{2}-m}{{\rm{e}}}^{\displaystyle \frac{N}{q}\left({\tau }^{2}{G}_{LR}^{q}+\mu {G}_{L}^{q\mathrm{/2}}+\mu {G}_{R}^{q\mathrm{/2}}\right)}{{\rm{e}}}^{-N\left({{\rm{\Sigma }}}_{i}{G}_{i}\right)},\end{array}\end{eqnarray*}$
Figure 3. The ratio ${z}_{2}^{\left(k\right)}/{z}_{2}^{\left(0\right)}$ in ( |
In [69] we indeed find a new non-trivial saddle point whose saddle contribution is larger than the saddle contribution of the trivial disk saddle and wormhole saddle. The new non-trivial saddle should correspond to $\displaystyle {\sum }_{k}\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle $ with k in the peak. The expression (87 ) of ⟨z2⟩ leads to a G, ∑ expression of each ${z}_{2}^{\left(k\right)}$ 57 ) factorizes88 ) can be decomposed into
$\begin{eqnarray}\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle ={z}_{2}^{\left(k\right)}=\left(\begin{array}{c}N\\ kq\end{array}\right)\displaystyle {\int }_{R}{{\rm{d}}}^{3}{G}_{i}\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}{{\rm{d}}}^{3}{{\rm{\Sigma }}}_{i}\,{\left({{\rm{\Sigma }}}_{LR}\right)}^{kq}{\left({{\rm{i}}}^{2}{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}\right)}^{\displaystyle \frac{N-kq}{2}}{{\rm{e}}}^{\displaystyle \frac{N}{q}\left({\tau }^{2}{G}_{LR}^{q}+\mu {G}_{L}^{q\mathrm{/2}}+\mu {G}_{R}^{q\mathrm{/2}}\right)}{{\rm{e}}}^{-N\left({{\rm{\Sigma }}}_{i}{G}_{i}\right)}\mathrm{}.\end{eqnarray}$
Actually we can derive a different G, ∑ expression from Θi directly in a more enlightening way. Because ψi are Grassmann numbers and q is even then the exponential in ( $\begin{eqnarray}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}}=\displaystyle \prod _{A}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}{J}_{A}{\psi }_{A}}\mathrm{}.\end{eqnarray}$
Using Tyler expansion the definition of θ(i) one can derive a useful identity $\begin{eqnarray}{{\rm{e}}}^{\alpha X}=\left\langle {{\rm{e}}}^{\alpha X}\right\rangle \displaystyle \sum _{i\mathrm{=0}}^{\infty }\displaystyle \frac{{\alpha }^{n}}{n!}{\theta }^{\left(n\right)},\end{eqnarray}$
where X is the random variable. With the help of this identity and ${\psi }_{A}^{2}=0,$ ( $\begin{eqnarray}\begin{array}{c}\displaystyle \prod _{A}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}{J}_{A}{\psi }_{A}}=\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}}\right\rangle \left(1+\displaystyle \displaystyle \sum _{A}{\theta }_{A}^{\left(1\right)}\left({{\rm{i}}}^{q\mathrm{/2}}{\psi }_{A}\right)+\displaystyle \frac{1}{\mathrm{2!}}\displaystyle \displaystyle \sum _{A,B}{\theta }_{A}^{\left(1\right)}\left({{\rm{i}}}^{q\mathrm{/2}}{\psi }_{A}\right){\theta }_{B}^{\left(1\right)}\left({{\rm{i}}}^{q\mathrm{/2}}{\psi }_{B}\right)+\ldots \right)\\ =\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}}\right\rangle {{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{\psi }_{A}}.\end{array}\end{eqnarray}$
Thus the we can express $\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle $ as
$\begin{eqnarray}\begin{array}{c}\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle =\displaystyle \int {{\rm{d}}}^{2N}{\psi }^{L\left(R\right)}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}^{L}}\right\rangle \left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}^{R}}\right\rangle \displaystyle \frac{1}{k{!}^{2}}{\left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle \displaystyle \sum _{A}{\psi }_{A}^{L}{\theta }_{A}^{\left(1\right)}\right)}^{k}{\left({{\rm{i}}}^{q\mathrm{/2}}\displaystyle \displaystyle \sum _{A}{\psi }_{A}^{R}{\theta }_{A}^{\left(1\right)}\right)}^{k}\\ =\displaystyle \int {{\rm{d}}}^{2N}{\psi }^{L\left(R\right)}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}^{L}}\right\rangle \left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}{\psi }_{A}^{R}}\right\rangle \displaystyle \frac{{t}^{2k}}{k!}{\left(\displaystyle \displaystyle \sum _{A}{\psi }_{A}^{L}{\psi }_{A}^{R}\right)}^{k}\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{c}=\displaystyle \int {{\rm{d}}}^{2N}{\psi }^{L\left(R\right)}{\rm{d}}{G}_{{LR}}{\rm{d}}{{\rm{\Sigma }}}_{{LR}}\,{{\rm{e}}}^{{{\rm{i}}}^{q/2}u\displaystyle {\sum }_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}{{\rm{e}}}^{-N{{\rm{\Sigma }}}_{{LR}}\left({G}_{{LN}}-\displaystyle {\sum }_{i}{\psi }_{i}^{L}{\psi }_{i}^{R}\right)}\displaystyle \frac{1}{k!}{\left(\displaystyle \frac{N{\tau }^{2}}{q}{G}_{{LR}}^{q}\right)}^{k}\\ =\displaystyle {\int }_{R}{{\rm{d}}}^{3}{G}_{i}\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}{{\rm{d}}}^{3}{\sum }_{i}\displaystyle \frac{1}{2}\left({\left({{\rm{\Sigma }}}_{{LR}}+{\rm{i}}\sqrt{{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}}\right)}^{N}+{\left({{\rm{\Sigma }}}_{{LR}}-{\rm{i}}\sqrt{{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}}\right)}^{N}\right)\\ {{\rm{e}}}^{\displaystyle \frac{N}{q}\left(\mu {G}_{L}^{q/2}+\mu {G}_{R}^{q/2}\right)}{{\rm{e}}}^{-N\left({\sum }_{i}{G}_{i}\right)}\displaystyle \frac{1}{k!}{\left(\displaystyle \frac{N{\tau }^{2}}{q}{G}_{{LR}}^{q}\right)}^{k}.\end{array}\end{eqnarray}$
The integral (92 ) is not convergent but we can introduce the generating function
$\begin{eqnarray}F\left(v\right)=\displaystyle {\int }_{R}{{\rm{d}}}^{3}{G}_{i}\displaystyle {\int }_{{\rm{i}}{\mathbb{R}}}{{\rm{d}}}^{3}{{\rm{\Sigma }}}_{i}\,\displaystyle \frac{1}{2}\left({\left({{\rm{\Sigma }}}_{{LR}}+{\rm{i}}\sqrt{{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}}\right)}^{N}+{\left({{\rm{\Sigma }}}_{{LR}}-{\rm{i}}\sqrt{{{\rm{\Sigma }}}_{L}{{\rm{\Sigma }}}_{R}}\right)}^{N}\right){{\rm{e}}}^{\displaystyle \frac{N}{q}\left(v{\tau }^{2}{G}_{{LR}}^{q}+\mu {G}_{L}^{q/2}+\mu {G}_{R}^{q/2}\right)}{{\rm{e}}}^{-N\left({\sum }_{i}{G}_{i}\right)},\end{eqnarray}$
which can be computed with a saddle point approximation and the $\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle $ is given by $\begin{eqnarray}\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle =\displaystyle \frac{1}{k!}\displaystyle \frac{{{\rm{d}}}^{k}F{\left(v\right)}_{{\rm{saddle}}}}{{\rm{d}}{v}^{k}}{| }_{v\mathrm{=0}}\mathrm{}.\end{eqnarray}$
As a simple test, we know that the exact result of F(v) is just $\begin{eqnarray}F\left(v\right)={\left\langle {z}^{2}\right\rangle }_{{t}^{2}\to {t}^{2}v}=\displaystyle \displaystyle \sum _{k}{c}_{k}{m}_{p-k}^{2}{t}^{2k}{u}^{2p-2k}{v}^{k},\end{eqnarray}$
which indeed leads to $\begin{eqnarray}\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle =\displaystyle \frac{1}{k!}\displaystyle \frac{{{\rm{d}}}^{k}F{\left(v\right)}_{{\rm{saddle}}}}{{\rm{d}}{v}^{k}}{| }_{v\mathrm{=0}}={c}_{k}{m}_{p-k}^{2}{t}^{2k}{u}^{2p-2k}\mathrm{}.\end{eqnarray}$
2.3.1. Half-wormhole in z2
To make the half-wormhole saddle manifest below we will set u = 0. In this case ‘Bose–Einstein’ condensation happens. As found in [58] for the square of partition function z2 the wormhole persists and there is only one dominant non-self-averaged sector. Applying (30) directly leads to the decomposition89 ) first
$\begin{eqnarray}{z}^{2}=\displaystyle \displaystyle \sum _{i}{{\rm{\Phi }}}_{2i},\end{eqnarray}$
with $\begin{eqnarray}{{\rm{\Phi }}}_{0}=\left\langle {z}^{2}\right\rangle =\displaystyle \sum _{{A}_{1}\left({B}_{1}\right)\lt \ldots \lt {A}_{p}\left({B}_{p}\right)}^{^{\prime} }{\rm{sgn}}\left(A\right){\rm{sgn}}\left(B\right){t}^{2}{\delta }_{{A}_{1}{B}_{1}}\ldots \ldots {t}^{2}{\delta }_{{A}_{p}{B}_{p}},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{2}=\displaystyle \displaystyle \sum _{k}\displaystyle \sum _{{A}_{1}\left({B}_{1}\right)\lt \ldots \lt {A}_{p}\left({B}_{p}\right)}^{^{\prime} }{\rm{sgn}}\left(A\right){\rm{sgn}}\left(B\right){t}^{2}{\delta }_{{A}_{1}{B}_{1}}\ldots \left({\theta }_{{A}_{k}}^{\left(1\right)}{\theta }_{{B}_{k}}^{\left(1\right)}+{\delta }_{{A}_{k}{B}_{k}}\left({\theta }_{{A}_{k}}^{\left(2\right)}-{\theta }_{{A}_{k}}^{\left(1\right)}2\right)\right)\ldots {t}^{2}{\delta }_{{A}_{p}{B}_{p}},\ldots \end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{2p}=\displaystyle \sum _{{A}_{1}\left({B}_{1}\right)\lt \ldots \lt {A}_{p}\left({B}_{p}\right)}^{{\rm{^{\prime} }}}{\rm{sgn}}\left(A\right){\rm{sgn}}\left(B\right)\left({\theta }_{{A}_{1}}^{\left(1\right)}{\theta }_{{B}_{1}}^{\left(1\right)}+{\delta }_{{A}_{1}{B}_{1}}\left({\theta }_{{A}_{1}}^{\left(2\right)}-{\theta }_{{A}_{1}}^{\left(1\right)}{2}^{}\right)\right)\ldots \left({\theta }_{{A}_{p}}^{\left(1\right)}{\theta }_{{B}_{p}}^{\left(1\right)}+{\delta }_{{A}_{p}{B}_{p}}\left({\theta }_{{A}_{p}}^{\left(2\right)}-{\theta }_{{A}_{p}}^{\left(1\right)}{2}^{}\right)\right)\end{eqnarray}$
where Φ2p is the half-wormhole saddle which is found in [58, 62] by noticing ${\theta }_{A}^{\left(1\right)}={J}_{A}$ and ${\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{\left(1\right)2}=-{t}^{2}.$ Actually the connection between the half-wormhole proposed in [58] and factorization proposal introduced in [57] has been pointed out in [44]. A useful way to derive the expression of Φi is to use ( $\begin{eqnarray}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}=\displaystyle \prod _{A}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}{J}_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}\end{eqnarray}$
$\begin{eqnarray}=\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}\right\rangle \displaystyle \prod _{A}\left(1+{{\rm{i}}}^{q\mathrm{/2}}{\theta }_{A}^{\left(1\right)}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)+{{\rm{i}}}^{q}{\theta }_{A}^{\left(2\right)}{\psi }_{A}^{L}{\psi }_{A}^{R}\right)\end{eqnarray}$
$\begin{eqnarray}=\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}\right\rangle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)+{{\rm{i}}}^{q}\displaystyle {\sum }_{A}\left({\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{\left(1\right)2}\right){\psi }_{A}^{L}{\psi }_{A}^{R}}\end{eqnarray}$
and then to substitute it into the integral form of z2 $\begin{eqnarray}{z}^{2}=\int {{\rm{d}}}^{2N}{\psi }^{L\left(R\right)}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}{\sum }_{A}{J}_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}\right\rangle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}{\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)+{{\rm{i}}}^{q}{\sum }_{A}\left({\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{\left(1\right)2}\right){\psi }_{A}^{L}{\psi }_{A}^{R}}\end{eqnarray}$
$\begin{eqnarray}=\int {{\rm{d}}}^{2N}{\psi }^{L\left(R\right)}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}{\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)+{{\rm{i}}}^{q}{\sum }_{A}\left[\left({\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{\left(1\right)2}\right)+{t}^{2}\right]{\psi }_{A}^{L}{\psi }_{A}^{R}}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{c}={{\rm{i}}}^{N}\displaystyle \sum _{k\mathrm{=0}}^{p}\displaystyle \int {{\rm{d}}}^{2N}{\psi }^{L\left(R\right)}\displaystyle \frac{{\left(\displaystyle {\sum }_{A}\left[\left({\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{{\left(1\right)}^{2}}\right)+{t}^{2}\right]{\psi }_{A}^{L}{\psi }_{A}^{R}\right)}^{k}}{k!}\displaystyle \frac{{\left(\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{\psi }_{A}^{L}\right)}^{p-k}}{\left(p-k\right)!}\displaystyle \frac{{\left(\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{\psi }_{A}^{R}\right)}^{p-k}}{\left(p-k\right)!}\\ =\displaystyle \sum _{A\left(B\right)}^{^{\prime} }{\rm{sgn}}\left(A\right){\rm{sgn}}\left(B^{\prime} \right)\displaystyle \prod _{i}\left({\theta }_{{A}_{i}}^{\left(1\right)}{\theta }_{{B}_{i}}^{\left(1\right)}+{\delta }_{{A}_{i}{B}_{i}}\left({\theta }_{{A}_{i}}^{\left(2\right)}-{\theta }_{{A}_{i}}^{{\left(1\right)}^{2}}+{t}^{2}\right)\right).\end{array}\end{eqnarray}$
By matching the power of t2 we can extract the expression of Φi. Note that the expressions of Φi have been derived in [62] based on the proposal of [58]. In [62] the non-dominant sectors are derived as fluctuations of the dominant saddle Φ2p with the help of introducing G,∑ variables. Because our derivation here does not rely on G,∑ trick so it can be used to derive possible n-linked half-wormholes in zn. First we notice that ${\left\langle {z}^{2}\right\rangle }^{2}=\left\langle {{\rm{\Phi }}}_{0}^{2}\right\rangle $ is in the same order of ⟨z4⟩ ≈ ⟨z2⟩2 as proved in [58] so the wormhole saddle persists. To confirm that Φ2p is the only dominant non-self-averaged saddle we only need to show104 ) is that we can introduce G,∑ variable directly if needed because the appearance of $\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{L}+{\psi }_{A}^{R}\right)}\right\rangle $ instead of introducing them ‘by hand’ by inserting an identity as proposed in [58]. As we argued in [69] when u ≠ 0, Φ2p will not be the dominant sector anymore. Instead, there will be a package of surviving non-self-averaged sectors.
$\begin{eqnarray}\left\langle {z}^{4}\right\rangle \approx \left\langle {{\rm{\Phi }}}_{0}^{2}\right\rangle +\left\langle {{\rm{\Phi }}}_{2p}^{2}\right\rangle ,\end{eqnarray}$
which also has been proved in [58, 62]. Another benefit of the rewriting (2.3.2. Half-wormhole in z3
As we argued in the statistical toy model, there should exist n-linked half-wormholes. For simplicity let us focus on 3-linked half-wormholes and z3. Similar to (104 ), z3 can be rewritten as112 ) and (113 ) it is obvious to show ⟨Λ3p⟩ = ⟨Λp⟩ = 0 as they should be. To confirm that they are dominant let us compute $\langle {{\rm{\Lambda }}}_{3p}^{2}\rangle $ and $\langle {{\rm{\Lambda }}}_{p}^{2}\rangle $ 38 ). We believe that this analogy persists for all other higher moments zn. Recall that θ(i) can be thought of as moments thus it is reasonable to introduce the connected moments or the cumulants ${\tilde{\theta }}^{\left(i\right)}$ with those z3 can be cast into122 ) leads to zn directly as it should be since (122 ) is nothing but a rewriting of zn in a convenient way of extracting contributions from different sectors and it is a direct generalization of the trick introduced in [58]. In particular the highest level sector of zn can be expressed as
$\begin{eqnarray}{z}^{3}=\displaystyle \int {{\rm{d}}}^{3N}{\psi }^{i}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{1}+{\psi }_{A}^{2}+{\psi }_{A}^{3}\right)}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {{\rm{d}}}^{3N}{\psi }^{i}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q/2}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{1}+{\psi }_{A}^{2}+{\psi }_{A}^{3}\right)}\right\rangle {{\rm{e}}}^{{{\rm{i}}}^{q/2}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({\psi }_{A}^{1}+{\psi }_{A}^{2}+{\psi }_{A}^{3}\right)}\times {{\rm{e}}}^{{{\rm{i}}}^{q}\displaystyle {\sum }_{A}\left({\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{\left(1\right)}{2}^{}\right)\left({\psi }_{A}^{1}{\psi }_{A}^{2}+{\psi }_{A}^{1}{\psi }_{A}^{3}+{\psi }_{A}^{2}{\psi }_{A}^{3}\right)}{{\rm{e}}}^{{{\rm{i}}}^{3q/2}\displaystyle {\sum }_{A}\left({\theta }_{A}^{\left(3\right)}-3{\theta }_{A}^{\left(2\right)}{\theta }_{A}^{\left(1\right)}+2{\theta }_{A}^{\left(1\right)}{3}^{}\right){\psi }_{A}^{1}{\psi }_{A}^{2}{\psi }_{A}^{3}},\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \sum _{i\mathrm{=0}}^{3p}{{\rm{\Lambda }}}_{i}\mathrm{}.\end{eqnarray}$
Again the expression of Λi can be extracted by matching the power of t2. Since ⟨z3⟩ = 0, so the self-averaged sector does not exist and z3 is only dominated by non-self-averaged sectors which we expect are Λ3p: $\begin{eqnarray}{{\rm{\Lambda }}}_{3p}=\displaystyle \int {{\rm{d}}}^{3N}{\psi }^{i}{{\rm{e}}}^{{{\rm{i}}}^{q/2}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({\psi }_{A}^{1}+{\psi }_{A}^{2}+{\psi }_{A}^{3}\right)}\times {{\rm{e}}}^{{{\rm{i}}}^{q}\displaystyle {\sum }_{A}\left({\theta }_{A}^{\left(2\right)}-{\theta }_{A}^{\left(1\right)2}\right)\left({\psi }_{A}^{1}{\psi }_{A}^{2}+{\psi }_{A}^{1}{\psi }_{A}^{3}+{\psi }_{A}^{2}{\psi }_{A}^{3}\right)}{{\rm{e}}}^{{{\rm{i}}}^{3q/2}\displaystyle {\sum }_{A}\left({\theta }_{A}^{\left(3\right)}-3{\theta }_{A}^{\left(2\right)}{\theta }_{A}^{\left(1\right)}+2{\theta }_{A}^{\left(1\right)3}\right){\psi }_{A}^{1}{\psi }_{A}^{2}{\psi }_{A}^{3}},\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {{\rm{d}}}^{3N}{\psi }^{i}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{1}+{\psi }_{A}^{2}+{\psi }_{A}^{3}\right)}{{\rm{e}}}^{-{{\rm{i}}}^{q}{t}^{2}\displaystyle {\sum }_{A}\left({\psi }_{A}^{1}{\psi }_{A}^{2}+{\psi }_{A}^{1}{\psi }_{A}^{3}+{\psi }_{A}^{2}{\psi }_{A}^{3}\right)},\end{eqnarray}$
and Λp: $\begin{eqnarray}{{\rm{\Lambda }}}_{p}=\displaystyle \int {{\rm{d}}}^{3N}{\psi }^{\mathrm{1,2,3}}\displaystyle \displaystyle \sum _{\left(i,j,k\right)=\left(\mathrm{1,2,3}\right),\left(\mathrm{1,3,2}\right),\left(\mathrm{2,3,1}\right)}\left({{\rm{e}}}^{{{\rm{i}}}^{q}{t}^{2}\displaystyle {\sum }_{A}{\psi }_{A}^{i}{\psi }_{A}^{j}}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(k\right)}{\psi }_{A}^{k}}\right)\end{eqnarray}$
$\begin{eqnarray}=3\left\langle {z}^{2}\right\rangle z,\end{eqnarray}$
where we have substituted the explicit expressions of ${\theta }_{A}^{\left(i\right)}.$ The term of triple product ${\psi }_{A}^{1}{\psi }_{A}^{2}{\psi }_{A}^{3}$ drops out in Λ3p because of JA is Gaussian so that there is no tri-linear interactions. From ( $\begin{eqnarray}\langle {{\rm{\Lambda }}}_{3p}^{2}\rangle =\langle {{\rm{\Lambda }}}_{3p}^{L}{{\rm{\Lambda }}}_{3p}^{R}\rangle \end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {{\rm{d}}}^{6N}{\psi }^{{L}_{i}\left({R}_{i}\right)}{{\rm{e}}}^{{{\rm{i}}}^{q}{t}^{2}\displaystyle {\sum }_{A}\left(\displaystyle {\sum }_{i,j\mathrm{=1}}^{3}{\psi }_{A}^{{L}_{i}}{\psi }_{A}^{{R}_{j}}\right)}\end{eqnarray}$
$\begin{eqnarray}\langle {{\rm{\Lambda }}}_{p}^{2}\rangle =\langle {{\rm{\Lambda }}}_{p}^{L}{{\rm{\Lambda }}}_{p}^{R}\rangle =9{\left\langle {z}^{2}\right\rangle }^{3},\end{eqnarray}$
which give $\begin{eqnarray}\langle {{\rm{\Lambda }}}_{3p}^{2}\rangle +\langle {{\rm{\Lambda }}}_{p}^{2}\rangle \approx 15{\langle {z}^{2}\rangle }^{3}\approx \langle {z}^{6}\rangle \mathrm{}.\end{eqnarray}$
Therefore the approximation $\begin{eqnarray}{z}^{3}\approx {{\rm{\Lambda }}}_{p}+{{\rm{\Lambda }}}_{3p},\end{eqnarray}$
is the analogue of ( $\begin{eqnarray}{z}^{3}=\displaystyle \int {{\rm{d}}}^{3N}{\psi }^{i}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left({\psi }_{A}^{1}+{\psi }_{A}^{2}+{\psi }_{A}^{3}\right)}\right\rangle \displaystyle \prod _{k}{{\rm{e}}}^{{{\rm{i}}}^{kq\mathrm{/2}}\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(k\right)}\displaystyle \frac{{\left(\displaystyle {\sum }_{i}{\psi }_{A}^{i}\right)}^{k}}{k!}}\mathrm{}.\end{eqnarray}$
In general, we expect $\begin{eqnarray}{z}^{n}=\displaystyle \int {{\rm{d}}}^{nN}{\psi }^{i}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left(\displaystyle {\sum }_{i\mathrm{=1}}^{n}{\psi }_{A}^{i}\right)}\right\rangle \displaystyle \prod _{k\mathrm{=1}}^{n}{{\rm{e}}}^{{{\rm{i}}}^{kq\mathrm{/2}}\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(k\right)}\displaystyle \frac{{\left(\displaystyle {\sum }_{i}{\psi }_{A}^{i}\right)}^{k}}{k!}},\end{eqnarray}$
which is simple to check for small n by a direct calculation. Since θ(i) is Gaussian so the only non-vanishing cumulants are ${\tilde{\theta }}^{\left(1\right)}$ and ${\tilde{\theta }}^{\left(2\right)}$ thus $\begin{eqnarray}{z}^{n}=\displaystyle \int {{\rm{d}}}^{nN}{\psi }^{i}\left\langle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{J}_{A}\left(\displaystyle {\sum }_{i\mathrm{=1}}^{n}{\psi }_{A}^{i}\right)}\right\rangle {{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left(\displaystyle {\sum }_{i}{\psi }_{A}^{i}\right)}{{\rm{e}}}^{{{\rm{i}}}^{q}\displaystyle \frac{1}{2}\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{\left(\displaystyle {\sum }_{i}{\psi }_{A}^{i}\right)}^{2}}\mathrm{}.\end{eqnarray}$
As a consistency check, substituting the explicit expressions ${\theta }_{A}^{\left(1\right)}={J}_{A}$ and ${\tilde{\theta }}_{A}^{\left(2\right)}=-{t}^{2}$ into ( $\begin{eqnarray}{\rm{\Theta }}=\displaystyle \int {{\rm{d}}}^{nN}{\psi }^{i}{{\rm{e}}}^{{{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left(\displaystyle {\sum }_{i}{\psi }_{A}^{i}\right)}{{\rm{e}}}^{{{\rm{i}}}^{q}\displaystyle \frac{1}{2}\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{\left(\displaystyle {\sum }_{i}{\psi }_{A}^{i}\right)}^{2}},\end{eqnarray}$
which is expected to be one of the dominant non-self-averaged sector in the large N limit.2.4. 0+1 SYK model
Now let us apply our proposal to the 1-SYK model. The partition function is defined as51 ). We will assume that (122 ) is approximately valid at least semi-classically. In other words, the saddle point can be derived from (122 ). The possible problem of (122 ) in the one-dimensional SYK model is that the fermions are not Grassmann numbers but Majorana fermions. As a result, ψA does not commute with ψB if there are odd number of common indexes in the collective indexes A and B. Therefore (89 ) is not exact anymore. The reason why we expect such subtlety is negligible in the large N limit is that when we introduce standard G,∑ variables in the SYK model we already ignore this fact and it is shown in [68] this approximation is correct in the large N limit.
$\begin{eqnarray}z\left(\beta \right)=\displaystyle \int {\rm{D}}\psi \,\exp \left\{-\displaystyle {\int }_{0}^{\beta }{\rm{d}}\tau \left({\psi }_{i}{\partial }_{\tau }{\psi }_{i}+{{\rm{i}}}^{q\mathrm{/2}}{J}_{A}{\psi }_{A}\right)\right\},\end{eqnarray}$
with JA’s satisfy (2.4.1. Half-wormhole in z and complex coupling
First let us consider z(β + iT)123 ) in the 0d SYK model. It can also be written as ${\left\langle {{\rm{e}}}^{{\sum }_{A}{\theta }_{A}^{\left(1\right)}{{\mathscr{O}}}_{A}}\right\rangle }_{\bar{\mathrm{SYK}}}$, where $\bar{\mathrm{SYK}}$ can be thought of as the anti-SYK model which is an SYK model but with an opposite bi-linear coupling or it can be thought of as an SYK model with purely imaginary random coupling ${\rm{i}}{\tilde{J}}_{A}.$ The relation between factorization and complex couplings in the SYK model was also proposed in [62]. To confirm this approximation, let us compute130 ) is the same saddle point solution of ⟨z2⟩ with GLL = GRR = 0. Such solutions are found in [68]. To be more precise, these solutions found in [68] are time-dependent and only in the ramp region we have GLL, GRR → 0. This is why we stress that only in the ramp region our approximation is good. Away from this region, we have to include other sectors which can be obtained by the expansion (125 ) as
$\begin{eqnarray}z\left(\beta +{\rm{i}}T\right)=\displaystyle \int {\rm{D}}\psi {{\rm{e}}}^{-\displaystyle {\int }_{0}^{\beta }{\rm{d}}\tau \left({\psi }_{i}{\partial }_{\tau }{\psi }_{i}\right)+{t}^{2}\displaystyle {\sum }_{A}{{O}}_{A}{{O}}_{A}}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{{O}}_{A}}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{{O}}_{A}{{O}}_{A}},\end{eqnarray}$
where we have defined the operator $\begin{eqnarray}{{\mathscr{O}}}_{A}\left(\beta +{\rm{i}}T\right)\equiv {{\rm{i}}}^{q\mathrm{/2}}\displaystyle {\int }_{0}^{\beta +{\rm{i}}T}{\rm{d}}{\tau }_{A}{\psi }_{A}\mathrm{}.\end{eqnarray}$
The reason we consider z(β + iT) is that its square ⟨z(β + iT)z(β − iT)⟩ is the spectral form factor (SFF) which has universal behaviors for chaotic systems like SYK model and random matrix theories. When T is small, SFF is self-averaged so it is dominated by disconnected piece ⟨z(β + iT)z(β − iT)⟩ ≈ ⟨z(β + iT)z(β − iT)⟩. Because the one point function decays with respect to time and so is SFF. This decay region of SFF is called the slope. Because of the chaotic behavior SSF should not vanish in the late time. It will be the non-self-averaged sector dominates which are responsible for the ramp of the SFF. Therefore, in the ramp region we expect the approximation $\begin{eqnarray}z\approx {\rm{\Theta }}\left(\beta +{\rm{i}}T\right)\equiv \displaystyle \int {\rm{D}}\psi {{\rm{e}}}^{-\displaystyle {\int }_{0}^{\beta +{\rm{i}}T}\left[{\rm{d}}\tau {\psi }_{i}{\partial }_{\tau }{\psi }_{i}\right]}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{{\mathscr{O}}}_{A}}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{{\mathscr{O}}}_{A}{{\mathscr{O}}}_{A}}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {\rm{D}}\psi {{\rm{e}}}^{-\displaystyle {\int }_{0}^{\beta +{\rm{i}}T}{\rm{d}}\tau {\psi }_{i}{\partial }_{\tau }{\psi }_{i}}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{{\mathscr{O}}}_{A}}{{\rm{e}}}^{-\displaystyle \frac{{t}^{2}}{2}\displaystyle {\sum }_{A}{{\mathscr{O}}}_{A}{{\mathscr{O}}}_{A}},\end{eqnarray}$
which is the analog of the highest level sector ( $\begin{eqnarray}\langle {\rm{\Theta }}\left(\beta +{\rm{i}}T\right){\rm{\Theta }}\left(\beta -{\rm{i}}T\right)\rangle =\displaystyle \int {\rm{D}}{\psi }^{L}{\rm{D}}{\psi }^{R}{{\rm{e}}}^{-\displaystyle {\int }_{0}^{\beta \pm {\rm{i}}T}{\rm{d}}\tau {\psi }_{i}^{L\left(R\right)}{{\rm{\partial }}}_{\tau }{\psi }_{i}^{L\left(R\right)}}{{\rm{e}}}^{-\displaystyle \frac{{t}^{2}}{2}\left({{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}{{\mathscr{O}}}_{A}^{R}\right)}\times \left\langle {{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{{\mathscr{O}}}_{A}^{L}}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{{\mathscr{O}}}_{A}^{R}}\right\rangle \end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {\rm{D}}{\psi }^{L}{\rm{D}}{\psi }^{R}{{\rm{e}}}^{-\displaystyle {\int }_{0}^{\beta \pm {\rm{i}}T}{\rm{d}}\tau {\psi }_{i}^{L\left(R\right)}{\partial }_{\tau }{\psi }_{i}^{L\left(R\right)}}{{\rm{e}}}^{{t}^{2}\displaystyle {\sum }_{A}{{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{R}},\end{eqnarray}$
which describes the wormhole saddle considering that we can introduce the GLR as $\begin{eqnarray}{t}^{2}\displaystyle \displaystyle \sum _{A}{{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{R}\approx \displaystyle \frac{{t}^{2}}{q!}\displaystyle \int {\rm{d}}{\tau }_{L}\displaystyle \int {\rm{d}}{\tau }_{R}{\left(\displaystyle \displaystyle \sum _{i}{\psi }_{i}^{L}{\psi }_{i}^{R}\right)}^{q}\equiv \displaystyle \frac{N{\tau }^{2}}{q}\displaystyle \int {\rm{d}}{\tau }_{L}\displaystyle \int {\rm{d}}{\tau }_{R}{G}_{LR}^{q}\mathrm{}.\end{eqnarray}$
so the saddle point solution of ( $\begin{eqnarray}{{\rm{\Theta }}}_{k}=\displaystyle \int {\rm{D}}\psi \,{{\rm{e}}}^{-\displaystyle {\int }_{0}^{\beta }{\rm{d}}\tau \left({\psi }_{i}{\partial }_{\tau }{\psi }_{i}\right)}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}{{O}}_{A}}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{{O}}_{A}{{O}}_{A}}\left(\displaystyle \frac{{\left({t}^{2}\displaystyle {\sum }_{A}{{O}}_{A}{{O}}_{A}\right)}^{k}}{k!}\right),\end{eqnarray}$
$\begin{eqnarray}\approx \frac{1}{k!}{\left\langle {{\rm{e}}}^{{\sum }_{A}{\theta }_{A}^{\left(1\right)}{O}_{A}}{\left(\frac{N{\tau }^{2}}{q}{G}_{LL}^{q}\right)}^{k}\right\rangle }_{\bar{\mathrm{SYK}}}.\end{eqnarray}$
2.4.2. Half-wormhole in z2 and factorization
Let us consider z(iT)z(-iT) and apply our decomposition proposal (122 )
$\begin{eqnarray}z\left({\rm{i}}T\right)z\left(-{\rm{i}}T\right)=\int {\rm{D}}{\psi }^{L\left(R\right)}\,{{\rm{e}}}^{{S}_{\mathrm{SYK}}^{L}+{S}_{\mathrm{SYK}}^{R}}{{\rm{e}}}^{{\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}{{\rm{e}}}^{\frac{1}{2}{\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}^{2}}.\end{eqnarray}$
Motivated by the result of 0-SYK model, we expect that there is also a ramp region where the dominant non-self-averaged sector is given by the 2-linked half-wormhole5(5 Note that we have normalized the fermionic integral such that $\int {\rm{d}}\psi \mathrm{=0}$ thus ⟨Φ⟩ = 0.) 137 ) of the 2-linked half-wormhole is very close to the one proposed in [62] which has two more bi-linear terms ${{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}{{\mathscr{O}}}_{A}^{R}$ in the second exponent. It seems that our proposal is more proper considering that in ⟨Φ2⟩ there are only bi-linear correlations between L(R) and L′(R′)
$\begin{eqnarray}{\rm{\Phi }}=\displaystyle \int {\rm{D}}{\psi }^{L\left(R\right)}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}{{\rm{e}}}^{\displaystyle \frac{1}{2}\displaystyle {\sum }_{A}{\tilde{\theta }}_{A}^{\left(2\right)}{\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}^{2}}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {\rm{D}}{\psi }^{L\left(R\right)}\,{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}{{\rm{e}}}^{-\displaystyle \frac{{t}^{2}}{2}\displaystyle {\sum }_{A}{\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}^{2}}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \int {\rm{D}}{\psi }^{L\left(R\right)}{{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left({{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}\right)}{{\rm{e}}}^{-\displaystyle \frac{{t}^{2}}{2}\displaystyle {\sum }_{A}\left({{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{L}+{{\mathscr{O}}}_{A}^{R}{{\mathscr{O}}}_{A}^{R}+{{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{R}+{{\mathscr{O}}}_{A}^{R}{{\mathscr{O}}}_{A}^{L}\right)}\end{eqnarray}$
Our proposal ( $\begin{eqnarray}\langle {{\rm{\Phi }}}^{2}\rangle =\displaystyle \int {\rm{D}}{\psi }^{L}{\rm{D}}{\psi }^{R}{{\rm{e}}}^{-\displaystyle \int {\rm{d}}\tau {\psi }_{i}^{L\left(R\right)}{{\rm{\partial }}}_{\tau }{\psi }_{i}^{L\left(R\right)}}\displaystyle \int {\rm{D}}{\psi }^{{L}^{{\rm{{\prime} }}}}{\rm{D}}{\psi }^{{R}^{{\rm{{\prime} }}}}{{\rm{e}}}^{-\displaystyle \int {\rm{d}}\tau {\psi }_{i}^{{L}^{{\rm{{\prime} }}}\left({R}^{{\rm{{\prime} }}}\right)}{{\rm{\partial }}}_{\tau }{\psi }_{i}^{{L}^{{\rm{{\prime} }}}\left({R}^{{\rm{{\prime} }}}\right)}}{{\rm{e}}}^{{t}^{2}\displaystyle {\sum }_{A}\left({{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{{L}^{{\rm{{\prime} }}}}+{{\mathscr{O}}}_{A}^{L}{{\mathscr{O}}}_{A}^{{R}^{{\rm{{\prime} }}}}+{{\mathscr{O}}}_{A}^{R}{{\mathscr{O}}}_{A}^{{L}^{{\rm{{\prime} }}}}+{{\mathscr{O}}}_{A}^{R}{{\mathscr{O}}}_{A}^{{R}^{{\rm{{\prime} }}}}\right)},\end{eqnarray}$
as shown in figure 4. Thus it implies the approximate factorization $\begin{eqnarray}{\rm{Error}}={z}^{2}-\left\langle {z}^{2}\right\rangle -{\rm{\Phi }},\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{Error}}}^{2}\right\rangle \approx \left\langle {z}^{4}\right\rangle -{\left\langle {z}^{2}\right\rangle }^{2}-2\left\langle {\rm{\Phi }}{z}^{2}\right\rangle +\left\langle {{\rm{\Phi }}}^{2}\right\rangle \approx \left(3-1-4+2\right){\left\langle {z}^{2}\right\rangle }^{2}\approx \mathrm{0,}\end{eqnarray}$
where we have assumed in the regime where the wormhole dominates the partition function z approximates a Gaussian random variable. The bulk point of view of the factorization is also interesting. The insertion of ${{\rm{e}}}^{\displaystyle {\sum }_{A}{\theta }_{A}^{\left(1\right)}\left(\displaystyle {\sum }_{i\mathrm{=1}}^{n}{{\mathscr{O}}}_{A}^{i}\right)}$ can be thought of as inserting spacetime branes in the gravity path integral and the opposite bi-linear coupling means the wormhole amplitudes connecting the branes are opposite to the usual spacetime wormhole amplitudes such that including all the effects of wormholes and branes factorization is achieved. In [65], it is proposed that JT gravity can be factorized by inserting such spacetime branes.
Figure 4. The illustration of ( |
2.5. Random matrix theory
In this section, let us apply our proposal to the Random Matrix Theory: the GUE ensemble which can also be thought of the CGS model with end-of-world (EOW) branes. The random matrix element Hij is identified with an EOW brane $\left({\hat{\psi }}_{i},{\hat{\psi }}_{j}\right)$ in the notation of [37] or the topological complex matter field ${Z}_{{\psi }^{\dagger }\psi }$ in the notation of [44] with the restriction that the disk amplitude of $\left({\hat{\psi }}_{i},{\hat{\psi }}_{i}\right)$ vanishes i.e. $\left\langle \left({\hat{\psi }}_{i},{\hat{\psi }}_{i}\right)\right\rangle \mathrm{=0}.$ The equivalence between these two models can be understood as the following. The correlation functions of Hij are computed by the Wick contractions which exactly describe how to connect different EOW branes $\left({\hat{\psi }}_{i},{\hat{\psi }}_{j}\right)$ with spacetime wormholes in the Disk-Cylinder approximation. Therefore the correlation functions of the random matrix theory are equal to the gravity path integral as we have seen in the CGS model. In this theory, we are interested in the observable
$\begin{eqnarray}z\left(\beta \right)={\rm{T}}r\left({{\rm{e}}}^{-\beta H}\right),\end{eqnarray}$
whose ensemble average is given by $\begin{eqnarray}\left\langle z\right\rangle =\displaystyle \int {\rm{d}}H\,{{\rm{e}}}^{-\displaystyle \frac{1}{2{t}^{2}}{\rm{T}}r{H}^{2}}z,\end{eqnarray}$
where t2 is usually taken to be 1/N.2.5.1. Half-wormhole
First let us consider the non-self-averaged sector in z. It is useful to study a simpler observable TrHn to get some intuitions about the non-self-averaged sector of matrix functions. For the random variable Hij we can not use the decomposition (30 ) directly. One possible way of adapting to (30 ) is to rewrite Hij as a linear combination of the Gaussian random variables. However this rewriting is not very convenient. Alternatively, we can transfer the matrix integral into the integral over eigenvalues146 )
$\begin{eqnarray}{\mathscr{Z}}\left(H\right)=\displaystyle \int {\rm{d}}H{{\rm{e}}}^{-\displaystyle \frac{1}{2{t}^{2}}{\rm{tr}}{H}^{2}}=\displaystyle \int \displaystyle \prod _{i}{\rm{d}}{\lambda }_{i}{{\rm{e}}}^{-\displaystyle \frac{1}{2{t}^{2}}\displaystyle {\sum }_{i}{\lambda }_{i}^{2}}{\rm{\Delta }}{\left(\lambda \right)}^{2},\end{eqnarray}$
where Δ(λ) is the Vandermonde determinant $\begin{eqnarray}{\rm{\Delta }}\left(\lambda \right)=\displaystyle \prod _{i\lt j}^{L}\left({\lambda }_{i}-{\lambda }_{j}\right)\mathrm{}.\end{eqnarray}$
Then the simple single-trace observable translates to $\begin{eqnarray}{\rm{Tr}}{H}^{n}=\displaystyle \displaystyle \sum _{i}{\lambda }_{i}^{n}\,\mathrm{}.\end{eqnarray}$
However those eigenvalues are not Gaussian random variables. As a result, even though we can still do the sector decomposition but the resulting different sectors are not orthogonal anymore. Although when the level is finite, we can obtain a new orthogonal basis by a direct diagonalization but it is still very cumbersome. We will make some preliminary analysis beyond Gaussian distribution in next section. Here we will take a similar approach as before. Considering the non-vanishing correlator ⟨HijHji⟩ = t2 we should define $\begin{eqnarray}{\theta }_{ij}^{\left(1\right)}={H}_{ij},\,{\theta }_{ij,ji}^{\left(2\right)}={\theta }_{ij}^{\left(1\right)}{\theta }_{ji}^{\left(1\right)}-{t}^{2},\end{eqnarray}$
thus we have $\begin{eqnarray}{H}_{ij}{H}_{km}=\left\langle {H}_{ij}{H}_{km}\right\rangle +{\theta }_{ij}^{\left(1\right)}{\theta }_{km}^{\left(1\right)}+{\delta }_{jk}{\delta }_{im}\left({\theta }_{ij,ji}^{\left(2\right)}-{\theta }_{ij}^{\left(1\right)}{\theta }_{ji}^{\left(1\right)}\right)\end{eqnarray}$
$\begin{eqnarray}\equiv \left\langle {H}_{ij}{H}_{km}\right\rangle +\left[{H}_{ij}{H}_{km}\right]\end{eqnarray}$
and $\begin{eqnarray}{H}_{ij}{H}_{kl}{H}_{mn}=\displaystyle \sum _{i\mathrm{=0}}^{3}{{\rm{\Theta }}}_{i},\,{{\rm{\Theta }}}_{0}=\left\langle {H}_{ij}{H}_{kl}{H}_{mn}\right\rangle ,\,{{\rm{\Theta }}}_{2}=0,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Theta }}}_{1}={\theta }_{ij}^{\left(1\right)}\left\langle {H}_{kl}{H}_{mn}\right\rangle +{\theta }_{kl}^{\left(1\right)}\left\langle {H}_{ij}{H}_{mn}\right\rangle +{\theta }_{mn}^{\left(1\right)}\left\langle {H}_{kl}{H}_{ij}\right\rangle ,\end{eqnarray}$
( $\begin{eqnarray}{{\rm{\Theta }}}_{3}=\left(\begin{array}{l}{\theta }_{ij}^{\left(1\right)}{\theta }_{kl}^{\left(1\right)}{\theta }_{mn}^{\left(1\right)},\,{\rm{no}}\,{\rm{pairs}}\,{\rm{like}}\\ {\theta }_{ab,ba}^{\left(2\right)}{\theta }_{ji}^{\left(1\right)}={\theta }_{ab}^{\left(1\right)}{\theta }_{ba}^{\left(1\right)}{\theta }_{ji}^{\left(1\right)}-{t}^{2}{\theta }_{ji}^{\left(1\right)},\,{\rm{there}}\,{\rm{is}}\,{\rm{only}}\,{\rm{one}}\,{\rm{pair}}\,{\rm{like}}\\ {\theta }_{ab,ba,ab}^{\left(3\right)}={\theta }_{ab}^{\left(1\right)}{\theta }_{ba}^{\left(1\right)}{\theta }_{ab}^{\left(1\right)}-2{t}^{2}{\theta }_{ab}^{\left(1\right)},\,a\ne b\\ {\theta }_{aa,aa,aa}^{\left(3\right)}={\theta }_{aa}^{\left(1\right)3}-3{t}^{2}{\theta }_{aa}^{\left(1\right)}\mathrm{}.\end{array}\right.,\end{eqnarray}$
$\begin{eqnarray}=\,{\theta }_{ij}^{\left(1\right)}{\theta }_{kl}^{\left(1\right)}{\theta }_{mn}^{\left(1\right)}-\left({\rm{all}}\,{\rm{the}}\,{\rm{possible}}\,{\rm{contractions}}\right)\end{eqnarray}$
So the highest level sector can also be understood as the observable in the ‘normal order’. Applying this rule of decomposition to the single-trace observables we get $\begin{eqnarray}{\rm{Tr}}H={\rm{Tr}}{\theta }^{\left(1\right)},\end{eqnarray}$
$\begin{eqnarray}{\rm{Tr}}{H}^{2}=\left\langle {\rm{Tr}}{H}^{2}\right\rangle +\displaystyle \displaystyle \sum _{ij}{\theta }_{ij}^{\left(2\right)}=\left\langle {\rm{Tr}}{H}^{2}\right\rangle +\left({\rm{Tr}}{\theta }^{{\left(1\right)}^{2}}-{N}^{2}{t}^{2}\right),\end{eqnarray}$
$\begin{eqnarray}{\rm{Tr}}{H}^{3}=3\left\langle {\rm{Tr}}{H}^{2}\right\rangle {\rm{Tr}}{\theta }^{\left(1\right)}+\left({\rm{Tr}}\left({\theta }^{{\left(1\right)}^{3}}\right)-3N{t}^{2}{\rm{Tr}}{\theta }^{\left(1\right)}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\rm{Tr}}{H}^{4}=\left\langle {\rm{Tr}}{H}^{4}\right\rangle +\left(4{N}^{2}{t}^{2}{\rm{Tr}}\left({\theta }^{\left(2\right)}\right)+2{t}^{2}\left[{\rm{Tr}}{\theta }^{\left(1\right)}{\rm{Tr}}{\theta }^{\left(1\right)}\right]\right)\\ \,+\,\left[{\rm{Tr}}{\theta }^{{(1)}^{4}}\right],\ldots \end{array}\end{eqnarray}$
where the normal ordered terms are explicitly given by $\begin{eqnarray}\left[{\rm{Tr}}{\theta }^{\left(1\right)}{\rm{Tr}}{\theta }^{\left(1\right)}\right]={\rm{Tr}}{\theta }^{\left(1\right)}{\rm{Tr}}{\theta }^{\left(1\right)}-N{t}^{2},\end{eqnarray}$
$\begin{eqnarray}\left[{\rm{Tr}}{\theta }^{\left(1\right)4}\right]={\rm{Tr}}{\theta }^{{\left(1\right)}^{4}}-\left(\left\langle {\rm{Tr}}{H}^{4}\right\rangle +\left(4{N}^{2}{t}^{2}{\rm{Tr}}\left({\theta }^{\mathrm{(2)}}\right)+2{t}^{2}\left[{\rm{Tr}}{\theta }^{\mathrm{(1)}}{\rm{Tr}}{\theta }^{\mathrm{(1)}}\right]\right)\right).\end{eqnarray}$
Like the Wick transformation in quantum field theory, the normal order or the highest level sector can be defined as $\begin{eqnarray}\left[f\left(H\right)\right]={{\rm{e}}}^{-\displaystyle \frac{{t}^{2}}{2}{\rm{Tr}}\left(\displaystyle \frac{\delta }{\delta {H}_{ij}}\displaystyle \frac{\delta }{\delta {H}_{ji}}\right)}f\left(H\right),\end{eqnarray}$
or we can introduce the formal integral $\begin{eqnarray}\left[f\left(H\right)\right]=\displaystyle \int {\rm{d}}\tilde{H}\,{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}{\rm{Tr}}{\tilde{H}}^{2}}{{\rm{e}}}^{-\displaystyle \frac{1}{{t}^{2}}{\rm{Tr}}\left({\theta }^{\left(1\right)}\tilde{H}\right)}{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}{\rm{Tr}}{\theta }^{{\left(1\right)}^{2}}}f\left(\tilde{H}\right)-f\left(0\right),\end{eqnarray}$
which is more convenient sometimes. Therefore we can rewrite the decomposition as $\begin{eqnarray}f\left(H\right)=\left[f\left(H\right)\right]+\displaystyle \sum _{k}{\mathrm{Con}}_{k},\end{eqnarray}$
where the Conk means choosing all possible k pairs of matrix elements from [f(H)] and replacing each pair HabHcd with its expectation value ⟨HabHcd⟩. It implies the identification $\begin{eqnarray}\left[f\left(H\right)\right]={\Theta }_{p},\,{\mathrm{Con}}_{k}={\Theta }_{p-2k}.\end{eqnarray}$
For these single-trace observables, in the large N limit their correlation functions factorize so the dominant sector is always the self-averaged sector. The more interesting observable is z(iT) whose expectation value is $\begin{eqnarray}\mathop{\mathrm{lim}}\limits_{N\to \infty }\left\langle z\left({\rm{i}}T\right)\right\rangle =\displaystyle \displaystyle \sum _{k\mathrm{=0}}\displaystyle \frac{{\left({\rm{i}}T\right)}^{2k}}{\left(2k\right)!}\left\langle {\rm{T}}r{H}^{2k}\right\rangle =N\displaystyle \displaystyle \sum _{k\mathrm{=0}}\displaystyle \frac{{\left({\rm{i}}T\sqrt{N{t}^{2}}\right)}^{2k}}{\left(2k\right)!}{C}_{k}\end{eqnarray}$
$\begin{eqnarray}=N\displaystyle \frac{{J}_{1}\left(2\alpha T\right)}{\alpha T}\sim \mathrm{0,}\,{\rm{when}}\,T\gg \mathrm{1,}\end{eqnarray}$
where Ck is a famous Catalan number and $\alpha =\sqrt{N{t}^{2}}.$ So in the late time, the non-self-averaged sector becomes important. The lowest sector can be simply obtained by expanding z and picking the term with θ(1)6(6 There is a 1/N in front because one of the summation of indexes gives the trace of θ(1) instead of a factor of N. For example $\displaystyle {\sum }_{a,i,j,k,m}{\theta }_{ai}^{\left(1\right)}\left\langle {H}_{ij}{H}_{jk}{H}_{km}{H}_{ma}\right\rangle ={\rm{Tr}}{\theta }^{\left(1\right)}.$): $\begin{eqnarray}{{\rm{\Theta }}}_{1}={\rm{Tr}}{\theta }^{\left(1\right)}\displaystyle \frac{1}{N}\left(N{\rm{i}}T+\displaystyle \frac{{\left({\rm{i}}T\right)}^{3}}{\mathrm{3!}}3\times \left\langle {\rm{Tr}}{H}^{2}\right\rangle +\displaystyle \frac{{\left({\rm{i}}T\right)}^{5}}{\mathrm{5!}}5\times \left\langle {\rm{Tr}}{H}^{4}\right\rangle +\ldots \right)\end{eqnarray}$
$\begin{eqnarray}={\rm{iTr}}{\theta }^{\left(1\right)}\displaystyle \frac{{J}_{1}\left(2\alpha T\right)}{\alpha }.\end{eqnarray}$
Similarly we find that the next sector is7(7 The factor 6 × 2 comes from the adjacent terms like ${\theta }_{ab,ij}^{\left(2\right)}\left\langle {H}_{ij}{H}_{ik}{H}_{km}{H}_{mn}\right\rangle $ and factor 3 comes from the pairs like ${\theta }_{ab,cd}^{\left(2\right)}\left\langle {H}_{bi}{H}_{ic}{H}_{dk}{H}_{ka}\right\rangle .$) Appendix that
$\begin{eqnarray}{{\rm{\Theta }}}_{2}={\rm{Tr}}{\theta }^{\left(2\right)}\displaystyle \frac{1}{{\alpha }^{2}}\left(\displaystyle \frac{{\left({\rm{i}}T\alpha \right)}^{2}}{2}+\displaystyle \frac{{\left({\rm{i}}T\alpha \right)}^{4}}{\mathrm{4!}}\times 4+\displaystyle \frac{{\left({\rm{i}}T\alpha \right)}^{6}}{\mathrm{6!}}\times \left(6\,\ast \,2+3\right)+\ldots \right),\end{eqnarray}$
$\begin{eqnarray}=-{\rm{Tr}}{\theta }^{\left(2\right)}\displaystyle \frac{{J}_{2}\left(\left(2T\alpha \right)\right)}{{\alpha }^{2}},\end{eqnarray}$
where we have dropped the terms Trθ(1)Trθ(1) because they are suppressed by 1/N. Comparing with the known results8(8 For example see [65].) of the wormhole contribution to ⟨z(iT)z(iT2)⟩c $\begin{eqnarray}{\left\langle z\left({\rm{i}}{T}_{1}\right)z\left({\rm{i}}{T}_{2}\right)\right\rangle }_{c}=\displaystyle \sum _{l\mathrm{=0}}^{\infty }\left(l+1\right){\left(-1\right)}^{l+1}{J}_{l+1}\left(2\alpha {T}_{1}\right){J}_{l+1}\left(2\alpha {T}_{2}\right),\end{eqnarray}$
we will show in the $\begin{eqnarray}{{\rm{\Theta }}}_{k}={\left({\rm{i}}\right)}^{k}\displaystyle \frac{{J}_{k}\left(2\alpha T\right)}{{\alpha }^{k}}{\rm{Tr}}{\theta }^{\left(k\right)},\,k\gt 0.\end{eqnarray}$
We plot $\left\langle {{\rm{\Theta }}}_{0}^{2}\right\rangle +\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle $ in figure 6(a) and $\left\langle {{\rm{\Theta }}}_{k}^{2}\right\rangle $ in figure 6(b). The result is very interesting. We see that every curve has the typical slop, ramp and plateau regimes. Another interesting fact is that only the first few sectors contribute to the slop and ramp regions. For example, adding the first 20 sectors we find that the ramp region is roughly located at [2.5/α, 4/α] and we plot the contribution of each sector in figure 5. Actually including the first 10 sectors is a very good approximation
$\begin{eqnarray}\displaystyle \frac{\displaystyle {\sum }_{i\mathrm{=1}}^{10}\left\langle {{\rm{\Theta }}}_{i}^{2}\right\rangle }{\displaystyle {\sum }_{i\mathrm{=1}}^{20}\left\langle {{\rm{\Theta }}}_{i}^{2}\right\rangle }\mathrm{=0.999974}.\end{eqnarray}$
Figure 5. Contributions of different sectors (170), α= 1, T = 3. The horizontal axis represents different sectors, and the vertical axis is the proportion of each sector. |
Therefore if we only focus on the ramp which is supposed to relate to wormholes we only need to include the first 10 non-self-averaged sectors. In this sense, we may call ${\rm{\Theta }}=\displaystyle {\sum }_{i\mathrm{=1}}^{10}{{\rm{\Theta }}}_{i}$ the half-wormhole of z. This is similar to the half-wormhole of the simple exponential observable (41 ) in the regime βt2 < 1. We can follow the same procedure to study the decomposition and the half-wormhole of z2. But it is very cumbersome and we expect the its behavior is similar to the exponential observable.
3. Beyond Gaussian distribution or the generalized CGS model
One of simplest way to go beyond CGS model is again starting from the MM model but including connected spacetimes with other topologies in the Euclidean path integral. So the next simplest case beyond CGS model is the Disk-Cylinder-Pants model. Let the amplitudes of the disk, cylinder and pants to be9 ) we can decompose ${Z}_{\alpha }^{n}$ into different sectors which are exactly like (18 )–(21 ). In other words, the number basis or {θ(i)} is still the basis for decomposition. But {θ(i)} should be determined from the recursion relations (18 )–(21 ). For example, in the Disk-Cylinder-Pants model the first few θ(i) are
$\begin{eqnarray}\left\langle \hat{Z}\right\rangle ={\kappa }_{1},\,\left\langle {\hat{Z}}^{2}\right\rangle -{\left\langle \hat{Z}\right\rangle }^{2}={\kappa }_{2},\,\left\langle {\hat{Z}}^{3}\right\rangle -3\left\langle \hat{Z}\right\rangle \left\langle {\hat{Z}}^{2}\right\rangle +2{\left\langle \hat{Z}\right\rangle }^{2}={\kappa }_{3}\mathrm{}.\end{eqnarray}$
The generating function is $\begin{eqnarray}\left\langle {{\rm{e}}}^{u\hat{Z}}\right\rangle =\exp \left(u{\kappa }_{1}+\displaystyle \frac{{u}^{2}{\kappa }_{2}}{\mathrm{2!}}+\displaystyle \frac{{u}^{3}{\kappa }_{3}}{\mathrm{3!}}\right),\end{eqnarray}$
so we can also identify $\hat{Z}$ as a random variable albeit with a very complicated PDF. We can simply think of the distribution is defined by the same generating function. In [69] we introduce the connected correlators to decompose ⟨ZneikZ⟩ for example $\begin{eqnarray}\left\langle Z{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle =\left\langle Z\right\rangle \left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle +{\left\langle Z{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }_{{\rm{c}}},\end{eqnarray}$
$\begin{eqnarray}\left\langle {Z}^{2}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle =\left\langle {Z}^{2}\right\rangle \left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle +2\left\langle Z\right\rangle {\left\langle Z{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }_{c}+{\left\langle {Z}^{2}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }_{{\rm{c}}},\end{eqnarray}$
$\begin{eqnarray}\left\langle {Z}^{3}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle =\left\langle {Z}^{3}\right\rangle \left\langle {{\rm{e}}}^{{\rm{i}}kZ}\right\rangle +3\left\langle {Z}^{2}\right\rangle {\left\langle Z{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }_{c}+3\left\langle Z\right\rangle {\left\langle {Z}^{2}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }_{c}+{\left\langle {Z}^{3}{{\rm{e}}}^{{\rm{i}}kZ}\right\rangle }_{{\rm{c}}},\ldots \end{eqnarray}$
such that using the trick ( $\begin{eqnarray}{\theta }^{\left(1\right)}=Z-{\kappa }_{1},\,{\theta }^{\left(2\right)}={\theta }^{{\left(1\right)}^{2}}-{\kappa }_{2},\,{\theta }^{\left(3\right)}={\theta }^{{\left(1\right)}^{3}}-3{\kappa }_{2}{\theta }^{\left(1\right)}-{\kappa }_{3}\mathrm{}.\end{eqnarray}$
Because of the inclusion of new wormholes, the pants, the basis is not orthogonal anymore in the sense $\begin{eqnarray}\left\langle {\theta }^{\left(i\right)}{\theta }^{\left(j\right)}\right\rangle \ne {\delta }_{ij}\mathrm{}.\end{eqnarray}$
It is easy to find that $\begin{eqnarray}\left\langle {\theta }^{\left(1\right)}{\theta }^{\left(2\right)}\right\rangle ={\kappa }_{3},\,\left\langle {\theta }^{\left(2\right)}{\theta }^{\left(3\right)}\right\rangle \mathrm{=6}{\kappa }_{2}{\kappa }_{3}\mathrm{}.\end{eqnarray}$
Moreover the matrix ${M}_{ij}^{\left(3\right)}=\left\langle {\theta }^{\left(i\right)}{\theta }^{\left(j\right)}\right\rangle ,i,j=1,2,3$ is $\begin{eqnarray}{M}^{\left(3\right)}=\left(\begin{array}{ccc}{\kappa }_{2} & {\kappa }_{3} & 0\\ {\kappa }_{3} & 2{\kappa }_{2}^{2} & 6{\kappa }_{2}{\kappa }_{3}\\ 0 & 6{\kappa }_{2}{\kappa }_{3} & 6{\kappa }_{2}^{3}+9{\kappa }_{3}^{2}\end{array}\right)\mathrm{}.\end{eqnarray}$
Figure 6. Behaviors of the sectors (170) in z(iT) (141), the horizontal axis is the time T and the vertical axis is the value of sectors. (a) The sum of two sectors. (b) An individual sector. |
Figure 7. Illustrations of the metric ( |
Figure 8. Illustrations of the metric ( |
3.1. Toy statistical model
We start from the simplest operator36 ).
$\begin{eqnarray}Y=\displaystyle \displaystyle \sum _{i}\left({X}_{i}-\langle {X}_{i}\rangle \right)\mathrm{}.\end{eqnarray}$
The modification starts to show up in $\begin{eqnarray}{Y}^{3}={{\rm{\Delta }}}_{0}+{{\rm{\Delta }}}_{1}+{{\rm{\Delta }}}_{3},\end{eqnarray}$
where $\begin{eqnarray}{{\rm{\Delta }}}_{0}=\left\langle {Y}^{3}\right\rangle ,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{1}=3N{\kappa }_{2}\displaystyle \sum {\theta }_{i}^{\left(1\right)},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{3}=\displaystyle \displaystyle \sum _{i}{\theta }_{i}^{\left(3\right)}+3\displaystyle \displaystyle \sum _{i\ne j}{\theta }_{i}^{\left(2\right)}{\theta }_{j}^{\left(1\right)}+\displaystyle \displaystyle \sum _{i\ne j\ne k}{\theta }_{i}^{\left(1\right)}{\theta }_{j}^{\left(1\right)}{\theta }_{k}^{\left(1\right)}\mathrm{}.\end{eqnarray}$
In this special case since ⟨θ(1)θ(3)⟩ = 0, there is no cross terms in ⟨Y6⟩ $\begin{eqnarray}\left\langle {Y}^{6}\right\rangle =\displaystyle \displaystyle \sum _{i}\left\langle {{\rm{\Delta }}}_{i}^{2}\right\rangle ,\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{0}^{2}\right\rangle \sim {N}^{2}{\kappa }_{3}^{2},\,\left\langle {{\rm{\Delta }}}_{1}^{2}\right\rangle \sim {N}^{3}{\kappa }_{2}^{3},\,\left\langle {{\rm{\Delta }}}_{3}^{2}\right\rangle \sim {N}^{3}{\kappa }_{2}^{3}+{N}^{2}{\kappa }_{3}^{2},\end{eqnarray}$
where we only keep the possible leading terms. When ${\kappa }_{2},{\kappa }_{3}\sim {\mathscr{O}}\left(1\right),$ the operator Y3 is not self-averaged and the effect of κ3 is negligible. The interesting case is when ${N}^{2}{\kappa }_{3}^{2}\gt \gt {N}^{3}{\kappa }_{2}^{3}$ so that we have the approximation $\begin{eqnarray}{Y}^{3}\approx \left\langle {Y}^{3}\right\rangle +{{\rm{\Delta }}}_{3},\end{eqnarray}$
which is the analog of (3.2. 0-SYK model
Let us reconsider the 0-SYK model but assume the random couplings satisfying69 ) and
$\begin{eqnarray}\left\langle {J}_{{i}_{1}\ldots {i}_{q}}\right\rangle \mathrm{=0,}\,\left\langle {J}_{{i}_{1}\ldots {i}_{q}}{J}_{{j}_{1}\ldots {j}_{q}}\right\rangle ={\kappa }_{2}{\delta }_{{i}_{1}{j}_{1}}\ldots {\delta }_{{i}_{q}{j}_{q}},\,{\kappa }_{2}={\tau }^{2}\displaystyle \frac{\left(q-1\right)!}{{N}^{q-1}},\end{eqnarray}$
$\begin{eqnarray}\left\langle {J}_{A}{J}_{B}{J}_{C}\right\rangle ={\kappa }_{3}{\delta }_{ABC},\end{eqnarray}$
we will determine the scaling of κ3 in a moment. Then the averaged quantity is $\begin{eqnarray}\left\langle {z}^{3}\right\rangle =\displaystyle \int {{\rm{d}}}^{3N}\psi {{\rm{e}}}^{{\kappa }_{2}\displaystyle {\sum }_{A}\left({\psi }_{A}^{1}{\psi }_{A}^{2}+{\psi }_{A}^{1}{\psi }_{A}^{3}+{\psi }_{A}^{2}{\psi }_{A}^{3}\right)}{{\rm{e}}}^{{{\rm{i}}}^{3q}{\kappa }_{3}\displaystyle {\sum }_{A}{\psi }_{A}^{1}{\psi }_{A}^{2}{\psi }_{A}^{3}},\end{eqnarray}$
which can be computed by introducing the collective variables $\begin{eqnarray}{G}_{ab}=\displaystyle \frac{1}{N}\displaystyle \displaystyle \sum _{i}{\psi }_{i}^{a}{\psi }_{i}^{b},\,\left(a,b\right)=\left(\mathrm{1,2}\right),\left(\mathrm{1,3}\right),\left(\mathrm{2,3}\right),\end{eqnarray}$
$\begin{eqnarray}{G}_{3}=\displaystyle \frac{1}{N}\displaystyle \displaystyle \sum _{i\lt j}{\psi }_{i}^{1}{\psi }_{j}^{1}{\psi }_{i}^{2}{\psi }_{j}^{2}{\psi }_{i}^{3}{\psi }_{j}^{3},\end{eqnarray}$
$\begin{eqnarray}\displaystyle \displaystyle \sum _{A}{\psi }_{A}^{a}{\psi }_{A}^{b}=\displaystyle \frac{{N}^{q}}{q!}{G}_{ab}^{q},\,\displaystyle \displaystyle \sum _{A}{\psi }_{A}^{1}{\psi }_{A}^{2}{\psi }_{A}^{3}=\displaystyle \frac{{N}^{q\mathrm{/2}}}{\left(q\mathrm{/2}\right)!}{G}_{3}^{q\mathrm{/2}},\end{eqnarray}$
to rewrite ⟨z3⟩ as $\begin{eqnarray}\left({z}^{3}\right)=\displaystyle \int \displaystyle \frac{{\rm{d}}{G}_{{ab}}{\rm{d}}{{\rm{\Sigma }}}_{{ab}}}{2\pi {\rm{i}}/N}\displaystyle \int \displaystyle \frac{{\rm{d}}{G}_{3}{\rm{d}}{{\rm{\Sigma }}}_{3}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{\displaystyle \frac{N}{q}\left({\tau }^{2}\sum {G}_{{ab}}^{q}+2{\gamma }^{3}{G}_{3}^{q/2}\right)}{{\rm{e}}}^{-N\left(\sum {{\rm{\Sigma }}}_{{ab}}{G}_{{ab}}+{{\rm{\Sigma }}}_{3}{G}_{3}\right)}\displaystyle \int {{\rm{d}}}^{3N}\psi {{\rm{e}}}^{{{\rm{\Sigma }}}_{{ab}}\displaystyle {\sum }_{i}{\psi }_{i}^{a}{\psi }_{i}^{b}+{{\rm{\Sigma }}}_{3}\displaystyle {\sum }_{i\lt j}{\psi }_{i}^{1}{\psi }_{j}^{1}{\psi }_{i}^{2}{\psi }_{j}^{2}{\psi }_{i}^{3}{\psi }_{j}^{3}}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{c}=\displaystyle \int \displaystyle \frac{{\rm{d}}{G}_{ab}{\rm{d}}{{\rm{\Sigma }}}_{ab}}{2\pi {\rm{i}}/N}\displaystyle \int \displaystyle \frac{{\rm{d}}{G}_{3}{\rm{d}}{{\rm{\Sigma }}}_{3}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{\displaystyle \frac{N}{q}\left({\tau }^{2}\displaystyle \sum {G}_{ab}^{q}+2{\gamma }^{3}{G}_{3}^{q\mathrm{/2}}\right)}{{\rm{e}}}^{-N\left(\displaystyle \sum {{\rm{\Sigma }}}_{ab}{G}_{ab}+{{\rm{\Sigma }}}_{3}{G}_{3}\right)}{\sum }_{3}^{N/2}\\ =\displaystyle \int \displaystyle \frac{{\rm{d}}{G}_{3}{\rm{d}}{{\rm{\Sigma }}}_{3}}{2\pi {\rm{i}}/N}{{\rm{e}}}^{\displaystyle \frac{N}{q/2}{\gamma }^{3}{G}_{3}^{q/2}}{{\rm{e}}}^{-N{{\rm{\Sigma }}}_{3}{G}_{3}}{\sum }_{3}^{N/2}={\gamma }^{3p}{m}_{p},\end{array}\end{eqnarray}$
where mp is defined in ( $\begin{eqnarray}{\gamma }^{3}\equiv {{\rm{i}}}^{3q}{\kappa }_{3}\displaystyle \frac{{N}^{q\mathrm{/2}-1}}{\left(q\mathrm{/2}-1\right)!},\,\gamma \sim {\mathscr{O}}\left(1\right),\end{eqnarray}$
thus $\begin{eqnarray}{\kappa }_{3}\sim \displaystyle \frac{\left(q\mathrm{/2}-1\right)!}{{N}^{q\mathrm{/2}}-1}\mathrm{}.\end{eqnarray}$
Recall that $\begin{eqnarray}{z}^{3}=\displaystyle \sum _{A,B,C}^{^{\prime} }{\rm{sgn}}\left(A\right){\rm{sgn}}\left(B\right){\rm{sgn}}\left(C\right){J}_{{A}_{1}}{J}_{{B}_{1}}{J}_{{C}_{1}}\ldots {J}_{{A}_{p}}{J}_{{B}_{p}}{J}_{{C}_{p}}.\end{eqnarray}$
In general decomposing z3 is still very complicated. Let us consider some simple examples. If p = 2, then we have18 )–(21 ):213 ), the three-mouth-wormhole amplitude is favored thus the possible dominate sectors are Δ0, Δ3p-3 and Δ3p:2.3 . In the same limit, z is not self-averaged neither while the half-wormhole emerges.
$\begin{eqnarray}{z}^{3}=\displaystyle \sum _{A}^{{\rm{^{\prime} }}}{\rm{s}}{\rm{g}}{\rm{n}}\left(A\right){J}_{{A}_{1}}^{3}{J}_{{A}_{2}}^{3}+3\displaystyle \sum _{A,B,{A}_{i}\ne {B}_{i}}^{{\rm{^{\prime} }}}{\rm{s}}{\rm{g}}{\rm{n}}\left(B\right){J}_{{A}_{1}}^{2}{J}_{{A}_{2}}^{2}{J}_{{B}_{1}}{J}_{{B}_{2}}+\displaystyle \sum _{A,B,C,{A}_{i}\ne {B}_{i}\ne {C}_{i}}^{{\rm{^{\prime} }}}{\rm{s}}{\rm{g}}{\rm{n}}\left(A\right){\rm{s}}{\rm{g}}{\rm{n}}\left(B\right){\rm{s}}{\rm{g}}{\rm{n}}\left(C\right){J}_{{A}_{1}}{J}_{{A}_{2}}{J}_{{B}_{1}}{J}_{{B}_{2}}{J}_{{C}_{1}}{J}_{{C}_{2}},\end{eqnarray}$
and there are seven different sectors. A simple way to derive the explicit expression of each sector is to first decompose each ${J}_{A}^{n}$ as ( $\begin{eqnarray}{J}_{A}={\theta }_{A}^{\left(1\right)},\,{J}_{A}^{2}={\theta }_{A}^{\left(2\right)}+{\left({\kappa }_{2}\right)}_{A},\,{J}_{A}^{3}={\theta }_{A}^{\left(3\right)}+{\left({\kappa }_{3}\right)}_{A}+3{\left({\kappa }_{2}\right)}_{A}{\theta }_{A}^{\left(1\right)},\end{eqnarray}$
then collect the terms in the same sector: $\begin{eqnarray}{{\rm{\Delta }}}_{0}={\kappa }_{3}^{2}\sum ^{^{\prime} }{\rm{sgn}}\left(A\right)={m}_{p}{\kappa }_{3}^{2}=\left\langle {z}^{3}\right\rangle ,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{1}\mathrm{=3}{\kappa }_{3}\displaystyle \sum _{A}^{^{\prime} }{\rm{sgn}}\left(A\right){\theta }_{{A}_{1}}^{\left(1\right)}{\left({\kappa }_{2}\right)}_{{A}_{2}}+{\left({\kappa }_{2}\right)}_{{A}_{1}}{\theta }_{{A}_{2}}^{\left(1\right)},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{2}=\left(6+3{M}_{p}\right){\kappa }_{2}^{2}\displaystyle \sum _{A}^{^{\prime} }{\rm{sgn}}\left(A\right){\theta }_{{A}_{1}}^{\left(1\right)}{\theta }_{{A}_{2}}^{\left(1\right)},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{3}=\displaystyle \sum _{A}^{^{\prime} }{\rm{sgn}}\left(A\right)\left({\theta }_{{A}_{1}}^{\left(3\right)}{\left({\kappa }_{3}\right)}_{{A}_{2}}+{\left({\kappa }_{3}\right)}_{{A}_{1}}{\theta }_{{A}_{2}}^{\left(3\right)}\right),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{4}\mathrm{=3}{\kappa }_{2}\displaystyle \sum _{A}^{^{\prime} }{\rm{sgn}}\left(A\right)\left({\theta }_{{A}_{1}}^{\left(3\right)}{\theta }_{{A}_{2}}^{\left(1\right)}+{\theta }_{{A}_{1}}^{\left(1\right)}{\theta }_{{A}_{2}}^{\left(3\right)}+\displaystyle \displaystyle \sum _{B,B\ne {A}_{1},{A}_{2}}{\theta }_{B}^{\left(2\right)}{\theta }_{{A}_{1}}^{\left(1\right)}{\theta }_{{A}_{2}}^{\left(1\right)}\right),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Delta }}}_{5}=0,\,{{\rm{\Delta }}}_{6}=\displaystyle \sum _{A}^{{\rm{^{\prime} }}}{\rm{s}}{\rm{g}}{\rm{n}}\left(A\right){\theta }_{{A}_{1}}^{\left(3\right)}{\theta }_{{A}_{2}}^{\left(3\right)}+3\displaystyle \sum _{A,B,{A}_{i}\ne {B}_{i}}^{{\rm{^{\prime} }}}{\rm{s}}{\rm{g}}{\rm{n}}\left(B\right){\theta }_{{A}_{1}}^{\left(2\right)}{\theta }_{{A}_{2}}^{\left(2\right)}{\theta }_{{B}_{1}}^{\left(1\right)}{\theta }_{{B}_{2}}^{\left(1\right)}+\displaystyle \sum _{A,B,C,{A}_{i}\ne {B}_{i}\ne {C}_{i}}^{{\rm{^{\prime} }}}{\rm{s}}{\rm{g}}{\rm{n}}\left(A\right){\rm{s}}{\rm{g}}{\rm{n}}\left(B\right){\rm{s}}{\rm{g}}{\rm{n}}\left(C\right){\theta }_{{A}_{1}}^{\left(1\right)}{\theta }_{{A}_{2}}^{\left(1\right)}{\theta }_{{B}_{1}}^{\left(1\right)}{\theta }_{{B}_{2}}^{\left(1\right)}{\theta }_{{C}_{1}}^{\left(1\right)}{\theta }_{{C}_{2}}^{\left(1\right)},\end{eqnarray}$
where ${M}_{p}=\displaystyle \frac{\left(pq\right)!}{p!{\left(q!\right)}^{p}}.$ Now we are ready to compute ⟨ΔiΔj⟩ using the relation (180). It turns out that different sectors are still orthogonal for this case: $\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{0}^{2}\right\rangle ={m}_{p}^{2}{\kappa }_{3}^{4},\,\left\langle {{\rm{\Delta }}}_{1}^{2}\right\rangle \mathrm{=18}{M}_{p}{\kappa }_{3}^{2}{\kappa }_{2}^{3},\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{2}^{2}\right\rangle ={\left(6+3{M}_{p}\right)}^{2}{M}_{p}{\kappa }_{2}^{6},\,\left\langle {{\rm{\Delta }}}_{3}^{2}\right\rangle \mathrm{=2}{M}_{p}{\kappa }_{3}^{2}\left(6{\kappa }_{2}^{3}+9{\kappa }_{3}^{2}\right),\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{4}^{2}\right\rangle \mathrm{=18}{M}_{p}{\kappa }_{2}^{3}\left(6{\kappa }_{2}^{3}+9{\kappa }_{3}^{2}\right)+18{M}_{p}\left(2{M}_{p}-2\right){\kappa }_{2}^{6},\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{6}^{2}\right\rangle ={M}_{p}{\left(6{\kappa }_{2}^{3}+9{\kappa }_{3}^{2}\right)}^{2}+9{M}_{p}\left({M}_{p}-1\right)4{\kappa }_{2}^{6}+9\left({m}_{p}^{2}-{M}_{p}\right){\kappa }_{3}^{4}\end{eqnarray}$
$\begin{eqnarray}+6{M}_{p}\left({M}_{p}-1\right)\left({M}_{p}-2\right){\kappa }_{2}^{6}\mathrm{}.\end{eqnarray}$
In large N limit the relevant parameters have the following asymptotic behaviors $\begin{eqnarray}{\kappa }_{2}\sim \displaystyle \frac{\left(N\mathrm{/2}-1\right)!}{{N}^{N\mathrm{/2}-1}},\,{\kappa }_{3}\sim \displaystyle \frac{\left(N\mathrm{/4}-1\right)!}{{N}^{N\mathrm{/4}-1}},\,{\kappa }_{3}^{2}\sim {\kappa }_{2}^{3}{m}_{p}{{\rm{e}}}^{N},\,{M}_{p}\sim {m}_{p}^{2}\sqrt{N},\end{eqnarray}$
then the approximation can be given as $\begin{eqnarray}{z}^{3}\sim {{\rm{\Delta }}}_{3}+{{\rm{\Delta }}}_{6}\mathrm{}.\end{eqnarray}$
In general we find that when p ≪ N(or q ≫ 1), z3 is not self-averaged, i.e. the wormhole does not persists, but the (three-linked) half-wormhole emerges. This fact can be intuitively understood as the following. In this limit because of the scaling ( $\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{0}^{2}\right\rangle ={m}_{p}^{2}{\kappa }_{3}^{2p},\,\left\langle {{\rm{\Delta }}}_{3p}^{2}\right\rangle \gt \left\langle {\left(\displaystyle \sum _{A}^{^{\prime} }{\rm{sgn}}\left(A\right){\theta }_{{A}_{1}}^{\left(3\right)}\ldots {\theta }_{{A}_{p}}^{\left(3\right)}\right)}^{2}\right\rangle \sim {M}_{p}{\kappa }_{3}^{2p},\end{eqnarray}$
$\begin{eqnarray}\left\langle {{\rm{\Delta }}}_{3p-3}^{2}\right\rangle \gt \left.{\left(\displaystyle \sum _{A}^{^{\prime} }{\rm{sgn}}\left(A\right)\displaystyle \displaystyle \sum _{i}{\theta }_{{A}_{1}}^{\left(3\right)}\ldots {\left({\kappa }_{3}\right)}_{{A}_{i}}\ldots {\theta }_{{A}_{p}}^{\left(3\right)}\right)}^{2}\right\rangle \sim {M}_{p}{\kappa }_{3}^{2p},\end{eqnarray}$
and since Mp ≫ m2p we conclude that z3 ≈ Δ3p−3 + Δ3p. This is similar to the result obtained in section 4. Discussion
In this paper, we have generalized the factorization proposal introduced in [57]. The main idea is to decompose the observables into the self-averaging sector and non-self-averaging sectors. We find that the contributions from different sectors have interesting statistics in the semi-classical limit. When the self-averaging sector survives in this limit, the observable is self-averaging. An interesting phenomenon is the sector condensation, meaning the surviving non-self-averaging trend to condense, and in the extreme case, only one non-self-averaging sector is left-over, resembling the Bose–Einstein condensation. Then the half-wormhole saddle is naturally understood as the condensed sectors. We apply this proposal to a simple statistical model, a 0-SYK model and a random matrix model. Half-wormhole saddles are identified and they are in agreement with the known results. With our proposal, we also show the equivalence between the results in [57] and [58]. We also studied multi-linked-half-wormholes and their relations. There are some future directions.
4.1. Sector condensation
It is interesting to understand the sector condensation better. We expect that it is some criterion for an ensemble theory or a statistical observable to potentially have a bulk description, and so it deserves to be studied in other gravity/ensemble theories. Definitely, the extreme case mimicking the Bose–Einstein condensation is the most interesting one. We have not understood when it will happen and could it be used as some order parameter. We expect by studying the ‘phase diagram’ in the sector space we can obtain more information about the observables and systems.
4.2. Complex coupling and half-wormholes
In [62], it shows that factorization is related to the complex couplings. In our approach, the complex coupling emerges as an auxiliary parameter to obtain the half-wormhole saddle. The trick here is similar to the one used by Coleman, Giddings and Strominger [70–72], where the non-local effect of spacetime wormhole is ‘localized’ with the price of introducing random couplings. But the current analysis shows that this is only possible when ‘Bose–Einstein’ happens such that the dominant sector can be obtained from this trick. So it would be interesting to explore the relation between complex coupling and half-wormhole further using our approach.
4.3. Relations to other factorization proposal
Besides the half-wormhole proposal, there exists other proposals of factorization. For example, in [65] it shows two-dimensional gravity can be factorized by including other branes in the gravitational path integral. These new branes correspond to specific operators in the dual matrix model. From the point of view of our approach, inserting operators may be related to adding back the contributions from non-self-averaging sectors. In [73], it is argued that factorization can be restored by adding other kinds of asymptotic boundaries corresponding to the degenerate vacua. It is clear that from our approach, this is equivalent to introducing new random variables. It would be interesting to see how this changes the statistic of contribution form different sectors.
Acknowledgments
We thank Cheng Peng for valuable discussions and comments on an early version of the draft. We thank many of the members of KITS for interesting related discussions. JT is supported by the National Youth Fund No.12105289 and funds from the UCAS program of special research associates. YY is supported by the Fundamental Research Funds for the Central Universities, by funds from the University of Chinese Academy of Science (UCAS), and NSFC NO. 12175237.
Appendix: Details of 2.5.1
First let us rederive the non-self-averaged sectors of z in a more systematic way. For simplicity let us set t2 = 1/N. Defining218 ) Hba has to contract with other Hab or by a argument of symmetry it is obvious160 ) the generating function of the normal-ordered operator is27 ) we have226 ) we get (224 ) as promised. So the task is to compute the two-point correlation functions170 ).
$\begin{eqnarray}{J}_{ba}=\left\langle \displaystyle \frac{\delta z}{\delta {H}_{ab}}\right\rangle =\displaystyle \int {\rm{d}}H{{\rm{e}}}^{-\displaystyle \frac{N}{2}{\rm{Tr}}{H}^{2}}\displaystyle \frac{\delta z}{\delta {H}_{ab}},\end{eqnarray}$
$\begin{eqnarray}=N\displaystyle \int {\rm{d}}H{{\rm{e}}}^{-\displaystyle \frac{N}{2}{\rm{Tr}}{H}^{2}}{H}_{ba}{\rm{Tr}}\left({{\rm{e}}}^{{\rm{i}}TH}\right)=N\left\langle {H}_{ba}z\right\rangle \end{eqnarray}$
then we can rewrite Θ1 as $\begin{eqnarray}{{\rm{\Theta }}}_{1}=\displaystyle \displaystyle \sum _{i,j}{\theta }_{ij}^{\left(1\right)}{J}_{ji}\equiv \left\langle {\rm{Tr}}\left({\theta }^{\left(1\right)}\delta z\right)\right\rangle \mathrm{}.\end{eqnarray}$
By considering that in ( $\begin{eqnarray}{J}_{ba}={\delta }_{ab}{J}_{aa}\end{eqnarray}$
thus $\begin{eqnarray}{J}_{ba}={\delta }_{ab}N\left\langle {H}_{aa}z\right\rangle ={\delta }_{ab}\left\langle {\rm{Tr}}Hz\right\rangle ,\,{{\rm{\Theta }}}_{1}=\displaystyle \displaystyle \sum _{i}{\theta }_{ii}^{\left(1\right)}{J}_{ii}={\rm{Tr}}{\theta }^{\left(1\right)}\left\langle {\rm{Tr}}Hz\right\rangle ,\end{eqnarray}$
where we have used the fact there is a permutation symmetry in the diagonal elements {Hii}. Similarly the second non-self-averaged sector Θ2 can be written as $\begin{eqnarray}{{\rm{\Theta }}}_{2}=\displaystyle \frac{1}{2}{\rm{Tr}}{\theta }^{\left(2\right)}\left\langle \left({\rm{Tr}}{H}^{2}-N\right)z\right\rangle \mathrm{}.\end{eqnarray}$
In general, it is $\begin{eqnarray}{{\rm{\Theta }}}_{k}=\displaystyle \frac{1}{k}{\rm{Tr}}{\theta }^{\left(k\right)}\left\langle \left[{\rm{Tr}}{H}^{k}\right]z\right\rangle ,\end{eqnarray}$
which simply means that {[TrHk]} is an orthogonal basis in the sense $\begin{eqnarray}\left\langle \left[{\rm{Tr}}{H}^{k}\right]\left[{\rm{Tr}}{H}^{l}\right]\right\rangle =k{\delta }_{kl}\mathrm{}.\end{eqnarray}$
Recall ( $\begin{eqnarray}\left[G\left(u\right)\right]=\displaystyle \int {\rm{d}}\tilde{H}{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}{\rm{Tr}}{\tilde{H}}^{2}}{{\rm{e}}}^{-\displaystyle \frac{1}{{t}^{2}}{\rm{Tr}}\left(H\tilde{H}\right)}{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}{\rm{Tr}}{H}^{2}}{{\rm{Tre}}}^{u\tilde{H}}\mathrm{}.\end{eqnarray}$
Therefore similar to the computation of ( $\begin{eqnarray}\begin{array}{l}\left\langle \left[{\mathrm{Tre}}^{{u}_{L}\tilde{H}}\right]\left[{\mathrm{Tre}}^{{u}_{R}\tilde{H}}\right]\right\rangle =\int \left[{\rm{d}}H{\rm{d}}{H}_{L}{\rm{d}}{H}_{R}\right]{{\rm{e}}}^{\frac{N}{2}\left(\mathrm{Tr}{H}_{L}^{2}+\mathrm{Tr}{H}_{R}^{2}+\mathrm{Tr}{H}^{2}\right)}{{\rm{e}}}^{-N\mathrm{Tr}H\left({H}_{L}+{H}_{R}\right)}{\mathrm{Tre}}^{{u}_{L}{H}_{L}}{\mathrm{Tre}}^{{u}_{R}{H}_{R}}\\ =\int \left[{\rm{d}}{H}_{L}{\rm{d}}{H}_{R}\right]{{\rm{e}}}^{-N\mathrm{Tr}{H}_{L}{H}_{R}}{\mathrm{Tre}}^{{u}_{L}{H}_{L}}{\mathrm{Tre}}^{{u}_{R}{H}_{R}}=\displaystyle \sum _{k}\frac{{u}_{L}^{k}}{k!}\frac{{u}_{R}^{k}}{k!}k,\end{array}\end{eqnarray}$
where we have used the formal integral $\begin{eqnarray}\displaystyle \int \left[{\rm{d}}{H}_{L}{\rm{d}}{H}_{R}\right]{{\rm{e}}}^{-N{\rm{Tr}}{H}_{L}{H}_{R}}{\left[{H}_{L}\right]}_{ij}{\left[{H}_{R}\right]}_{ji}=\displaystyle \frac{1}{N}\mathrm{}.\end{eqnarray}$
By expanding both sides of ( $\begin{eqnarray}\left\langle \left[{\rm{Tr}}{H}^{n}\right]z\right\rangle ,\,{\rm{or}}\,\left\langle {\rm{Tr}}{H}^{n}z\right\rangle \end{eqnarray}$
or more conveniently the generating function $\begin{eqnarray}G\left(u\right)=\left\langle {\rm{Tr}}\left({{\rm{e}}}^{uH}\right)z\right\rangle =\left\langle z\left(u\right)\right\rangle \left\langle z\left({\rm{i}}T\right)\right\rangle +\displaystyle \sum _{l\mathrm{=0}}^{\infty }\left(l+1\right){\left(-1\right)}^{l+1}{J}_{l+1}\left(-2{\rm{i}}u\right){J}_{l+1}\left(2T\right),\end{eqnarray}$
$\begin{eqnarray}=N\displaystyle \frac{{J}_{1}\left(-2{\rm{i}}u\right)}{-{\rm{i}}u}\left\langle z\left({\rm{i}}T\right)\right\rangle +\displaystyle \sum _{l\mathrm{=0}}^{\infty }\left(l+1\right){\left(-1\right)}^{l+1}{J}_{l+1}\left(-2{\rm{i}}u\right){J}_{l+1}\left(2T\right)\mathrm{}.\end{eqnarray}$
Expanding the generating function gives $\begin{eqnarray}\left\langle {\rm{Tr}}\left(H\right)z\right\rangle ={\rm{i}}{J}_{1}\left(2T\right),\,\left\langle {\rm{Tr}}\left({H}^{2}\right)z\right\rangle =N\left\langle z\right\rangle -2{J}_{2}\left(2T\right),\end{eqnarray}$
$\begin{eqnarray}\left\langle {\rm{Tr}}\left({H}^{3}\right)z\right\rangle =-3{\rm{i}}{J}_{3}\left(2T\right)+3{\rm{i}}{J}_{1}\left(2T\right),\end{eqnarray}$
$\begin{eqnarray}\left\langle {\rm{Tr}}\left({H}^{4}\right)z\right\rangle \mathrm{=2}N\left\langle z\right\rangle -8{J}_{2}\left(2T\right)+4{J}_{4}\left(2T\right),\ldots \end{eqnarray}$
which indeed lead to (It would be desired to derive a generating function of the normal ordered operators [TrHn] which has the integral form234 ) describes a GUE model coupled with an external source. As shown in [65] it can be rewritten as230 ).
$\begin{eqnarray}\left[G\left(u\right)\right]=\displaystyle \int {\rm{d}}\tilde{H}\,{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}{\rm{Tr}}{\tilde{H}}^{2}}{{\rm{e}}}^{-\displaystyle \frac{1}{{t}^{2}}{\rm{Tr}}\left(H\tilde{H}\right)}{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}{\rm{Tr}}{H}^{2}}{{\rm{Tre}}}^{u\tilde{H}}\mathrm{}.\end{eqnarray}$
Note that ( $\begin{eqnarray}\left[G\left(u\right)\right]=\displaystyle \int \displaystyle \prod _{i}{\rm{d}}{\tilde{\lambda }}_{i}{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}\displaystyle \sum {\tilde{\lambda }}_{i}-\displaystyle \frac{1}{{t}^{2}}\displaystyle \sum {\tilde{\lambda }}_{i}{\lambda }_{i}+\displaystyle \frac{1}{2{t}^{2}}\displaystyle \sum {\lambda }_{i}^{2}}\displaystyle \frac{{\rm{\Delta }}\left(\tilde{\lambda }\right)}{{\rm{\Delta }}\left(\lambda \right)}\displaystyle \displaystyle \sum _{k}{{\rm{e}}}^{u{\tilde{\lambda }}_{k}}\end{eqnarray}$
$\begin{eqnarray}=\,\displaystyle \displaystyle \sum _{j}{{\rm{e}}}^{\displaystyle \frac{1}{2{t}^{2}}\displaystyle \sum {\lambda }_{i}^{2}-\displaystyle \frac{1}{2{t}^{2}}\displaystyle \displaystyle \sum _{i}{\left({\lambda }_{i}-{\delta }_{ji}{t}^{2}u\right)}^{2}}\displaystyle \prod _{i\ne j}^{N}\left(1+\displaystyle \frac{-u{t}^{2}}{{\lambda }_{i}-{\lambda }_{j}}\right)\end{eqnarray}$
$\begin{eqnarray}=\,{{\rm{e}}}^{-\displaystyle \frac{{t}^{2}{u}^{2}}{2}}\displaystyle \frac{1}{-u{t}^{2}}\displaystyle {\oint }_{H}\displaystyle \frac{{\rm{d}}w}{2\pi {\rm{i}}}\displaystyle \prod _{i\mathrm{=1}}^{N}\left(1+\displaystyle \frac{-u{t}^{2}}{w-{\lambda }_{i}}\right){{\rm{e}}}^{wu}\mathrm{}.\end{eqnarray}$
Notice that in the large N limit [TrHn] is a linear combination of single trace operator so we should expand each 1/(w - λi) into Taylor series and only keep terms with $\displaystyle {\sum }_{i}{\lambda }_{i}^{k}$ $\begin{eqnarray*}\begin{array}{lll}\left[G\left(u\right)\right] & = & {{\rm{e}}}^{-\displaystyle \frac{{u}^{2}}{2N}}\left(-\displaystyle \frac{N}{u}\right)\displaystyle {\oint }_{H}\displaystyle \frac{{\rm{d}}w}{2\pi {\rm{i}}}\left({\left(1+\displaystyle \frac{-u/N}{w}\right)}^{N}\right.\\ & & \left.+\,\displaystyle \frac{-u}{N}{\left(1+\displaystyle \frac{-u/N}{w}\right)}^{N-1}\displaystyle \displaystyle \sum _{k}\displaystyle \frac{\displaystyle {\sum }_{i}{\lambda }_{i}^{k}}{{w}^{k+1}}\right){{\rm{e}}}^{uw},\end{array}\end{eqnarray*}$
where we have substituted t2 = 1/N. Sending N to infinity gives $\begin{eqnarray}{{\rm{e}}}^{-\displaystyle \frac{{u}^{2}}{2N}}\sim \mathrm{1,}\,{\left(1+\displaystyle \frac{-u/N}{w}\right)}^{N}\sim {\left(1+\displaystyle \frac{-u/N}{w}\right)}^{N-1}\sim {{\rm{e}}}^{-u/w},\end{eqnarray}$
thus we arrive at the final result $\begin{eqnarray}\left[G\left(u\right)\right]=\left(-\displaystyle \frac{N}{u}\right)\displaystyle {\oint }_{H}\displaystyle \frac{{\rm{d}}w}{2\pi {\rm{i}}}{{\rm{e}}}^{u\left(w-\displaystyle \frac{1}{w}\right)}+\displaystyle \displaystyle \sum _{k\mathrm{=1}}{\rm{Tr}}{H}^{k}\displaystyle {\oint }_{H}\displaystyle \frac{{\rm{d}}w}{2\pi {\rm{i}}}{{\rm{e}}}^{u\left(w-\displaystyle \frac{1}{w}\right)}\displaystyle \frac{1}{{w}^{k+1}}\mathrm{}.\end{eqnarray}$
These contour integral can be evaluated exactly by using the expansion $\begin{eqnarray}{{\rm{e}}}^{u\left(w-\displaystyle \frac{1}{w}\right)}=\displaystyle \sum _{i=-\infty }^{\infty }{w}^{k}{J}_{k}\left(2u\right),\end{eqnarray}$
which leads to $\begin{eqnarray}\left[G\left(u\right)\right]=\displaystyle \frac{N}{u}{J}_{1}\left(2u\right)+\displaystyle \displaystyle \sum _{k}{\rm{Tr}}{H}^{k}{J}_{k}\left(2u\right)\mathrm{}.\end{eqnarray}$
By expanding with respect to u, indeed we get the correct normal-ordered operators $\begin{eqnarray}\begin{array}{l}\left[G\left(u\right)\right]=N+u{\rm{Tr}}H+\displaystyle \frac{{u}^{2}}{\mathrm{2!}}\left({\rm{Tr}}{H}^{2}-N\right)+\displaystyle \frac{{u}^{3}}{\mathrm{3!}}\left({\rm{Tr}}{H}^{3}-3{\rm{Tr}}H\right)\\ \,+\,\displaystyle \frac{{u}^{4}}{\mathrm{4!}}\left({\rm{Tr}}{H}^{4}-4{\rm{Tr}}{H}^{2}+2N\right)+\displaystyle \frac{{u}^{5}}{5!}\left({\rm{Tr}}{H}^{5}-5{\rm{Tr}}{H}^{3}+10{\rm{Tr}}H\right)\\ \,+\,\displaystyle \frac{1}{6!}\left({\rm{Tr}}{H}^{6}-6{\rm{Tr}}{H}^{4}+15{\rm{Tr}}{H}^{2}-5N\right)\ldots .\end{array}\end{eqnarray}$
We can also obtain a generating function of Θk $\begin{eqnarray}\left\langle \left[G\left(u\right)\right]z\right\rangle =\displaystyle \int {\rm{d}}H\displaystyle \int {\rm{d}}\tilde{H}{{\rm{e}}}^{\displaystyle \frac{N}{2}{\rm{Tr}}{\tilde{H}}^{2}}{{\rm{e}}}^{-N{\rm{Tr}}\left(H\tilde{H}\right)}{{\rm{Tre}}}^{u\tilde{H}}{{\rm{Tre}}}^{{\rm{i}}TH}\end{eqnarray}$
$\begin{eqnarray}=\displaystyle \frac{N}{u}{J}_{1}\left(2u\right)\left\langle z\right\rangle +\displaystyle \displaystyle \sum _{k}{J}_{k}\left(2u\right)\left\langle {\rm{Tr}}{H}^{k}{{\rm{Tre}}}^{{\rm{i}}TH}\right\rangle \end{eqnarray}$
which, unfortunately, does not have a simple closed form but the ensemble average ⟨TrHkTreiTH⟩ can be computed with the generating function (