\documentclass[11pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{amsmath,amssymb} \usepackage{physics} \usepackage[margin=1in]{geometry} \usepackage{hyperref} \usepackage{enumitem} \usepackage{braket} \title{Strong Mixing QSF} \begin{document} \maketitle \section{Introduction} We now turn on the coupling between curvature and entropy perturbations. The canonically normalized curvature perturbation is $\zeta_c \equiv \sqrt{2\epsilon}\,\zeta$ and the isocurvature perturbation is denoted by $\sigma$. In a quasi‑de Sitter background with constant turn rate $\Omega_{\rm turn}$, the quadratic action reads \begin{equation}\label{eq:action} S = \int d^4 x \, a^3 \Bigg[ \frac{1}{2} \dot\zeta_c^2 - \lambda H \dot\zeta_c \sigma + \frac{1}{2} \lambda^2 H^2 \sigma^2 - \frac{1}{2a^2} (\nabla \zeta_c)^2 + \frac{1}{2} \dot\sigma^2 - \frac{1}{2a^2} (\nabla\sigma)^2 - \frac{1}{2} \mu^2 \sigma^2 \Bigg] . \end{equation} Here $a(t)$ is the scale factor, $H = \dot a / a$ the Hubble rate, $\lambda = -2\Omega_{\rm turn}/H$ the dimensionless coupling, and $\mu$ the entropy mass. Introduce the two‑component field \begin{equation} \Phi(x) = \begin{pmatrix} \zeta_c(x) \\ \sigma(x) \end{pmatrix} . \end{equation} Defining the connection matrix \begin{equation} \Gamma = \begin{pmatrix} 0 & -\lambda H \\ 0 & 0 \end{pmatrix} , \end{equation} the covariant time derivative is $D_t = \partial_t + \Gamma$. (We use $\Gamma$ instead of $\Omega$ to avoid confusion with the turn rate $\Omega_{\rm turn}$ that enters through $\lambda = -2\Omega_{\rm turn}/H$.) The kinetic part becomes $\frac{1}{2}|D_t \Phi|^2$, and the full action takes the compact form \begin{equation}\label{eq:compact-action} S = \int d^4 x \, a^3 \Bigg[ \frac{1}{2} |D_t \Phi|^2 - \frac{1}{2a^2} (\nabla\Phi)^T (\nabla\Phi) - \frac{1}{2} \Phi^T \tilde M^2 \Phi \Bigg] , \end{equation} where the mass matrix $\tilde M^2$ contains only the entropy mass, \begin{equation} \tilde M^2 = \begin{pmatrix} 0 & 0 \\ 0 & \mu^2 \end{pmatrix} . \end{equation} The $\lambda^2 H^2$ term has been absorbed into $|\Gamma\Phi|^2$ inside the covariant derivative. In Fourier space we have \begin{equation} S = \int dt \, d^3 k \, a^3 \Bigg[ \frac{1}{2} |D_t \Phi_k|^2 - \frac{1}{2} \Phi_k^\dagger \left( \frac{k^2}{a^2} + \tilde M^2 \right) \Phi_k \Bigg] , \end{equation} where $|\Phi_k|^2 \equiv \Phi_k^\dagger \Phi_k$. The Euler–Lagrange equation yields the equation of motion \begin{equation} \frac{1}{a^3} \frac{d}{dt}\bigl[ a^3 D_t \Phi_k \bigr] - \Gamma^\dagger D_t \Phi_k + \left( \frac{k^2}{a^2} + \tilde M^2 \right) \Phi_k = 0 , \end{equation} which, upon expanding $D_t = \partial_t + \Gamma$, becomes \begin{equation} \ddot\Phi_k + 3H \dot\Phi_k + (\Gamma - \Gamma^\dagger) \dot\Phi_k + \left( 3H\Gamma - \Gamma^\dagger\Gamma + \frac{k^2}{a^2} + \tilde M^2 \right) \Phi_k = 0 . \end{equation} We write the on‑shell solution as \begin{equation} \Phi_k(\tau) = K_k(\tau) \, \varphi_k , \end{equation} where $\varphi_k$ is the boundary value at $\tau\to0$, and $K_k(\tau)$ is now a $2\times2$ bulk‑to‑boundary propagator matrix with the boundary condition $K_k(\tau \to 0) = \mathbb{I}$ (the $2\times2$ identity). We now construct the $2\times 2$ bulk‑to‑boundary propagator $K_k(\tau)$ explicitly using the exact Bogoliubov decomposition of the coupled curvature–entropy system. The field $\Phi_k = (\zeta_k,\; \sigma_k)^T$ obeys a linear $2\times2$ system of second‑order ODEs. Two independent Bunch‑Davies solutions exist: $\Phi_k^{(1)} = (q_1,\; \sigma_1)^T$, $\Phi_k^{(2)} = (q_2,\; \sigma_2)^T$. Each is decomposed in the free‑mode basis: \begin{align} \zeta_b(k,z) &= \zeta_b^+(z)\, u_0(k,z) + \zeta_b^-(z)\, u_0^*(k,z), \\ \sigma_b(k,z) &= \sigma_b^+(z)\, u_\mu(k,z) + \sigma_b^-(z)\, u_\mu^*(k,z), \end{align} where $i = 1,2$, $z = k/(aH) = -k\tau$, and the free modes are \begin{align} u_0(k,z) & = i\frac{H}{\sqrt{2k^3}} \, (1 - iz) e^{iz}, \\ u_\mu(k,z) &= \frac{H}{\sqrt{2k^3}} \, z^{3/2} H_{\nu}^{(1)}(z), \qquad \nu = \sqrt{\frac{9}{4} - \frac{\mu^2}{H^2}} . \end{align} For heavy entropy fields ($\mu > \frac{3}{2}H$), $\nu$ is imaginary: $\nu = i\rho$ with $\rho \equiv \sqrt{\mu^2/H^2 - 9/4}$. The mode matrix $U_k(z)$ is the $2\times2$ matrix whose columns are the two BD solutions: \begin{equation} U_k(z) \equiv \begin{pmatrix} \zeta_1(k,z) & \zeta_2(k,z) \\ \sigma_1(k,z) & \sigma_2(k,z) \end{pmatrix}, \end{equation} and the normalization of $U$ is given by \begin{equation} a^3\left[U_k^\dagger D_tU_k-\left(D_tU_k\right)^\dagger U_k\right]=i\,. \end{equation} At the late‑time boundary $z\to0$, $U_k(0)$ is non‑singular (the Bogoliubov coefficients freeze out), so $U_k(0)^{-1}$ is well‑defined. The bulk‑to‑boundary propagator $K_k(\tau)$ connects the boundary value $\varphi_k$ to the bulk field: \begin{equation} \Phi_k(\tau) = K_k(\tau) \, \varphi_k, \end{equation} and \begin{equation} K_k(z) = U_k(z)\, U^{-1}_k(0) . \end{equation} This satisfies $K_k(0) = \mathbb{I}$ by construction. Of particular importance for the $\sigma^3$ bispectrum are the mixed kernels $K_{\sigma q}(k;z)$ and $K_{qq}(k;z)$. From the Bogoliubov decomposition the explicit construction gives the mixed kernels as sums over the two BD branches $b = 1,2$: \begin{align} K_{\sigma \zeta}(k;\tau) &= \sum_{b=1}^2 \sigma_b(k,\tau)\, \zeta_b^*(k,\tau_*), \\ K_{\zeta\zeta}(k;\tau) &= \sum_{b=1}^2 \zeta_b(k,\tau)\, \zeta_b^*(k,\tau_*), \end{align} where $\tau_*$ is the late‑time boundary with $z_* = k/(a_* H) \to 0$. \section{The Bulk--to--Boundary Propagators revisited} The first line of the first order differential equation systems for the Bogoliubov coefficients give \begin{equation} \begin{split} z\frac{\mathrm{d}\zeta_b^+}{\mathrm{d}z}=&-i\frac{\lambda k^3}{H^2z^2}u_0^{*}{'}[u_\mu(k,z)\sigma_b^+(z)+u_\mu^*(k,z)\sigma_b^-(z)]\\=&-i\frac{\lambda k^3}{H^2z^2}u_0^*\sigma_b(k,z). \end{split} \end{equation} Accordingly, it is possible to reconstruct the $\sigma$ field mode function from the solution to the Bogoliubov equations via \begin{equation} \sigma_b(k,z)=i\frac{\sqrt{2}H}{\lambda k^{\frac32}}z^2e^{iz}\frac{\mathrm{d}\zeta_b^+}{\mathrm{d}z}. \end{equation} The bulk--to--boudanry propagator is then \begin{equation} K_{\sigma\zeta}(k,z)=i\frac{\sqrt{2}H}{\lambda k^{\frac32}}z^2e^{iz}\sum_{b=1}^2\frac{\mathrm{d}\zeta_b^+}{\mathrm{d}z}\zeta_b^*(k,0). \end{equation} For general $\nu$, the paper defines \begin{equation} \widehat D_{\nu,\pm} \equiv {}_2F_1\left(\frac12-\nu,\frac12+\nu;1\pm i\lambda;\frac{i}{2}\frac{\rm d}{\mathrm{d} z}\right). \end{equation} This means the formal power series \begin{equation} \widehat D_{\nu,\pm} = \sum_{n=0}^{\infty} \frac{\left(\frac12-\nu\right)_n \left(\frac12+\nu\right)_n} {(1\pm i\lambda)_n n!} \left(\frac{i}{2}\frac{\rm d}{\mathrm{d} z}\right)^n. \end{equation} The exact solution basis is \begin{align} f_1(z)&=\widehat D_{\nu,+} U\left(\frac{i\lambda}{2},1,-2i z\right),\\ f_2(z)&=\widehat D_{\nu,-} U\left(-\frac{i\lambda}{2},1,-2i z\right),\\ f_3(z)&=\widehat D_{\nu,+} M\left(\frac{i\lambda}{2},1,-2i z\right),\\ f_4(z)&=\widehat D_{\nu,-} M\left(-\frac{i\lambda}{2},1,-2i z\right). \end{align} \begin{align} A^{(1)}_1&=\frac{e^{\pi\lambda/4}}{2}\exp\left[\frac{i\lambda}{2}\ln (2z_0)\right]+\mathcal{O}\left(z_0^{-1}\right)\,,\\ A^{(1)}_2&=\frac{e^{-\pi\lambda/4}}{2}\exp\left[\frac{-i\lambda}{2}\ln (2z_0)\right]+\mathcal{O}\left(z_0^{-1}\right)\,,\\ A^{(2)}_1&=-i\frac{e^{\pi\lambda/4}}{2}\exp\left[\frac{i\lambda}{2}\ln (2z_0)\right]+\mathcal{O}\left(z_0^{-1}\right)\,,\\ A^{(2)}_2&=i\frac{e^{-\pi\lambda/4}}{2}\exp\left[-\frac{i\lambda}{2}\ln (2z_0)\right]+\mathcal{O}\left(z_0^{-1}\right)\,, \end{align} \begin{equation} \zeta_b^+(z)=\sum_{i=1}^4A_i^{(b)}f_i(z)\,, \end{equation} \begin{align} \zeta_b^*(k,0)&=\frac{iH}{\sqrt{2k^3}}\left(\zeta_b^+(k,0)-\zeta_b^-(k,0)\right)^*\\ &=\frac{iH}{\sqrt{2k^3}}\left(A^{(b)}_1\mathcal{U}_++A_2^{(b)}\mathcal{U}_-\right)^* \end{align} \begin{equation} \mathcal{U}_s=\frac{1}{\sqrt{\pi}}\frac{\Gamma\left(\frac{1}{2}+a_s\right)\Gamma\left(\frac{3}{4}-\frac{\nu}{2}\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}\right)}{\Gamma\left(\frac{3}{4}-\frac{\nu}{2}+a_s\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}+a_s\right)} \end{equation} \begin{equation} K_{\sigma\zeta}(k,z)=\frac{i}{2\lambda}\frac{H^2}{k^3}z^2e^{iz}\sum_{s={\pm}}e^{s\pi\lambda/2}\mathcal{U}_s^*f'_s(z)\,, \end{equation} \begin{equation} T_{\sigma\zeta}(z)=\frac{i}{2\lambda}z^2e^{iz}\sum_{s={\pm}}e^{s\pi\lambda/2}\mathcal{U}_s^*f'_s(z)\,, \end{equation} with $f_+(z)\equiv f_1(z)$ and $f_-(z)\equiv f_2(z)$. With the integral representation of Tricomi function \begin{equation} U(a,b,z)=\frac{1}{\Gamma(a)}\int_{0}^{\infty}e^{-zt}t^{a-1}(1+t)^{b-a-1}\,\mathrm{d}t, \qquad \Re a>0,\; |\operatorname{ph} z|<\frac12\pi\,, \end{equation} the basis functions can be written as \begin{align} f_s(z)&=\frac{1}{\Gamma\left(a_s\right)}\int_0^{+\infty}\mathrm{d}t\,t^{a_s-1}(1+t)^{-a_s}\sum_{n=0}^{\infty} \frac{\left(\frac12-\nu\right)_n \left(\frac12+\nu\right)_n} {(1+ is\lambda)_n n!} \left(\frac{i}{2}\frac{\rm d}{\mathrm{d} z}\right)^n e^{2izt}\\ &=\frac{1}{\Gamma\left(a_s\right)}\int_0^{+\infty}\mathrm{d}t\,e^{2izt}t^{a_s-1}(1+t)^{-a_s}\sum_{n=0}^{\infty} \frac{\left(\frac12-\nu\right)_n \left(\frac12+\nu\right)_n} {(1+ is\lambda)_n n!}\left(-t\right)^n\\ &=\frac{1}{\Gamma\left(a_s\right)}\int_0^{+\infty}\mathrm{d}t\,e^{2izt}t^{a_s-1}(1+t)^{-a_s}{}_2F_1\left(\frac12-\nu,\frac12+\nu;1+2a_s;-t\right)\,. \end{align} where $a_s=is\lambda/2$ and $s=\pm$. \begin{equation}\label{fperim_int} \begin{split} f_s'(z)&=\frac{2i}{\Gamma\left(a_s\right)}\int_0^{+\infty}\mathrm{d}t\,e^{2izt}t^{a_s}(1+t)^{-a_s}{}_2F_1\left(\frac12-\nu,\frac12+\nu;1+2a_s;-t\right)\,.\\ &=\frac{2i}{\Gamma\left(a_s\right)}\int_0^{+\infty}\mathrm{d}t~e^{2izt}P_{i\rho-1/2}^{-2a_s}(1+2t)\\ &=\frac{i}{\Gamma\left(a_s\right)}\int_1^{+\infty}\mathrm{d}t~e^{iz(t-1)}P_{i\rho-1/2}^{-2a_s}(t)\\ \end{split} \end{equation} We transfer it into the Mellin space \begin{equation} \left(1+t\right)^{-a_s}=\frac{1}{2\pi i}\frac{1}{\Gamma(a_s)}\int_C{\Gamma\left(a_s+u\right)\Gamma\left(-u\right)}t^u\,\mathrm{d}u\,, \end{equation} \begin{equation} \begin{split} {}_2F_1&\left(\frac12-\nu,\frac12+\nu;1+2a_s;-t\right)\\ &=\frac{1}{2\pi i}\frac{\Gamma(1+2a_s)}{\Gamma\left(\frac{1}{2}-\nu\right)\Gamma\left(\frac{1}{2}+\nu\right)}\int_{C}\frac{\Gamma\left(\frac{1}{2}-\nu+v\right)\Gamma\left(\frac{1}{2}+\nu+v\right)\Gamma\left(-v\right)}{\Gamma\left(1+2a_s+v\right)}t^v~\mathrm{d}v\,, \end{split} \end{equation} Inserting the Mellin-Barnes integral into (\ref{fperim_int}), \begin{equation} f'_s(z)=\frac{1}{2i\pi^2}\frac{\Gamma(1+2a_s)}{\Gamma\left(\frac{1}{2}-\nu\right)\Gamma\left(\frac{1}{2}+\nu\right)\Gamma(a_s)^2}I \end{equation} \begin{equation} \begin{split} I&=\int_{C_1,C_2}\mathrm{d}u\mathrm{d}v~\Phi_L\left(\nu,a_s;u,v\right)\frac{\Phi_R(u,v)}{\Gamma(u+v+a_s+1)}\int_0^{+\infty}\mathrm{d}t~e^{2izt}t^{u+v+a_s}\\ &=\int_{C_1,C_2}\mathrm{d}u\mathrm{d}v~\Phi\left(\nu,a_s;u,v\right)\Phi_R(u,v)\left(-2iz\right)^{-(u+v+a_s+1)} \end{split} \end{equation} where \begin{equation} \Phi_L\left(\nu,a_s;u,v\right)=\frac{\Gamma\left(a_s+u\right)\Gamma\left(\frac{1}{2}-\nu+v\right)\Gamma\left(\frac{1}{2}+\nu+v\right)\Gamma(u+v+a_s+1)}{\Gamma\left(1+2a_s+v\right)}\,. \end{equation} \begin{equation} \Phi_R(u,v)=\Gamma(-u)\Gamma(-v) \end{equation} With using the Cauchy theorem, if contour closed in the right half‑plane \begin{equation} \begin{split} I&=-2\pi i\sum_{n\in \mathbf{N}^+}\int_{C_1}\mathrm{d}u\,\Phi_L(\nu,a_s;u,n)\Gamma(-u)\frac{(-1)^n}{n!}\left(-2iz\right)^{-(u+n+a_s+1)}\\ &=(2\pi i)^2\sum_{m\in \mathbf{N}^+}\sum_{n\in \mathbf{N}^+}\Phi_L(\nu,a_s;m,n)\frac{(-1)^m}{m!}\frac{(-1)^n}{n!}\left(-2iz\right)^{-(m+n+a_s+1)}, \end{split} \end{equation} if contour closed in the left half‑plane \begin{equation} \begin{split} I(z)=&(2\pi i)^2\sum_{m\in \mathbf{N}^+}\sum_{n\in \mathbf{N}^+}\frac{(-1)^m}{m!}\frac{(-1)^n}{n!}\\ &\times\left[C^1_{m,n}(-2iz)^{m+n+\nu-1/2}+C^2_{m,n}(-2iz)^{m+n-\nu-1/2}+C^3_{m,n}(-2iz)^m+C^4_{m,n} (-2iz)^n\right] \end{split} \end{equation} \begin{align} C^1_{m,n}&=\frac{\Gamma\left(-n-2\nu\right)\Gamma\left(-m-n-\nu+\frac{1}{2}\right)}{\Gamma\left(-n+2a_s-\nu+\frac{1}{2}\right)}\Gamma\left(n+\nu+\frac{1}{2}\right)\Gamma(m+a_s)\,,\\ C^2_{m,n}&=\frac{\Gamma\left(-n+2\nu\right)\Gamma\left(-m-n+\nu+\frac{1}{2}\right)}{\Gamma\left(-n+2a_s+\nu+\frac{1}{2}\right)}\Gamma\left(n-\nu+\frac{1}{2}\right)\Gamma(m+a_s)\,,\\ C^3_{m,n}&=\sum_{s=\pm}\frac{\Gamma\left(-m+n-\frac{1}{2}-s\nu\right)\Gamma\left(2s\nu-n\right)}{\Gamma\left(n+2a_s+\frac{1}{2}+s\nu\right)}\\ &~~~~~~~~~~~~~~~~~~~~~~~\times\Gamma\left(n+\frac{1}{2}-s\nu\right)\Gamma\left(m-n+a_s+\frac{1}{2}+s\nu\right)\,,\\ C^4_{m,n}&=\frac{\Gamma\left(m-n-\nu-\frac{1}{2}\right)\Gamma\left(m-n+\nu-\frac{1}{2}\right)}{\Gamma\left(m-n+2a_s\right)}\Gamma(m+a_s)\Gamma(-m+n+1)\,. \end{align} \begin{equation} I^{(\sigma^3)}=\int_0^{+\infty}\mathrm{d}x\,x^{-4}\prod_{i=1}^3 (r_ix)^2e^{ir_ix}f'_s(r_ix) \end{equation} For small $z$, the leading contribution is given by setting $m=n=0$, \begin{equation} I_s(z)=(2\pi i)^2\left[C^1_{(0,0)}(-2iz)^{\nu-1/2}+C^2_{(0,0)}(-2iz)^{-\nu-1/2}+\left(C^3_{(0,0)}+C^4_{(0,0)}\right)\right]+\mathcal{O}(z^{3/2})\,, \end{equation} \begin{equation}\label{eqn:transf_smallz} T_{\sigma\zeta}(z)=-\lambda\sum_{s=\pm}\left[e^{i\phi_s\pi/2}\,2^{\nu-1/2}\frac{\Gamma(-2\nu)\Gamma(1+2a_s)}{\Gamma(a_s)\Gamma\left(2a_s-\nu+\frac{1}{2}\right)}\mathcal{U}^*_s\right]z^{\nu+3/2}+(\nu\to-\nu)+\mathcal{O}\left(z^2\right)\,, \end{equation} \begin{equation} \phi_s=\left(-\nu+\frac{1}{2}\right)-is\lambda\,. \end{equation} \begin{equation} \phi^*_s=\left(\nu+\frac{1}{2}\right)+is\lambda\,. \end{equation} We denote \begin{equation} T_{\sigma\zeta}(z)=C^+z^{\nu+3/2}+C^-z^{-\nu+3/2}\,. \end{equation} \begin{equation} \begin{split} \bar C^+&=-\lambda\sum_{s=\pm}\left[e^{-i\phi_s^*\pi/2}\,2^{-\nu-1/2}\frac{\Gamma(2\nu)\Gamma(1+2a_{-s})}{\Gamma(a_{-s})\Gamma\left(2a_{-s}+\nu+\frac{1}{2}\right)}\mathcal{U}^*_{-s}\right]\\ &=-\lambda\sum_{s=\pm}\left[e^{-i\phi_{-s}^*\pi/2}\,2^{-\nu-1/2}\frac{\Gamma(2\nu)\Gamma(1+2a_{s})}{\Gamma(a_{s})\Gamma\left(2a_{s}+\nu+\frac{1}{2}\right)}\mathcal{U}^*_{s}\right]\\ &=e^{-i\phi_-^*\pi/2}\mathcal{W}_++e^{-i\phi_+^*\pi/2}\mathcal{W}_- \end{split} \end{equation} \begin{equation} \mathcal{W}_s=-\lambda 2^{-\nu-1/2}\frac{\Gamma(2\nu)\Gamma(1+2a_{s})}{\Gamma(a_{s})\Gamma\left(2a_{s}+\nu+\frac{1}{2}\right)}\mathcal{U}^*_{s} \end{equation} while \begin{equation} \begin{split} C^-&=-\lambda\sum_{s=\pm}\left[e^{i\varphi_{s}\pi/2}\,2^{-\nu-1/2}\frac{\Gamma(2\nu)\Gamma(1+2a_{s})}{\Gamma(a_{s})\Gamma\left(2a_{s}+\nu+\frac{1}{2}\right)}\mathcal{U}^*_{s}\right]\\ &=e^{i\varphi_+\pi/2}\mathcal{W}_++e^{i\varphi_-\pi/2}\mathcal{W}_-\\ &=e^{i\phi_-^*\pi/2}\mathcal{W}_++e^{i\phi_+^*\pi/2}\mathcal{W}_- \end{split} \end{equation} \begin{equation} \varphi_s=\left(\nu+\frac{1}{2}\right)-is\lambda=\phi_{-s}^* \end{equation} \begin{equation} \begin{split} \bar C^+-C^-=\left(e^{-i\phi_-^*\pi/2}-e^{i\phi_-^*\pi/2}\right)\mathcal{W}_++\left(e^{-i\phi_+^*\pi/2}-e^{i\phi_+^*\pi/2}\right)\mathcal{W}_- \end{split} \end{equation} If $C^+$ is the complex conjugate of $C^-$, \begin{equation} \frac{\mathcal{W}_+}{\mathcal{W}_-}=-\frac{\sin\left(\frac{\pi}{2}\phi_+^*\right)}{\sin\left(\frac{\pi}{2}\phi_-^*\right)} \end{equation} \begin{equation} \begin{split} \frac{\mathcal{W}_+}{\mathcal{W}_-}&=\frac{\Gamma\left(1+2a_+\right)\Gamma\left(\frac{1}{2}-a_+\right)\Gamma\left(a_-\right)\Gamma\left(2a_--\nu+\frac{1}{2}\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}-a_-\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}-a_-\right)}{\Gamma\left(1+2a_-\right)\Gamma\left(\frac{1}{2}-a_-\right)\Gamma\left(a_+\right)\Gamma\left(2a_+-\nu+\frac{1}{2}\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}-a_+\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}-a_+\right)}\\ &=(-2^{2i\lambda})\frac{\Gamma\left(-i\lambda+\nu+\frac{1}{2}\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}+\frac{i\lambda}{2}\right)\Gamma\left(\frac{3}{4}-\frac{\nu}{2}+\frac{i\lambda}{2}\right)}{\Gamma\left(i\lambda+\nu+\frac{1}{2}\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}-\frac{i\lambda}{2}\right)\Gamma\left(\frac{3}{4}-\frac{\nu}{2}-\frac{i\lambda}{2}\right)}\\ &=-\frac{\Gamma\left(\frac{1}{4}+\frac{\nu}{2}-\frac{i\lambda}{2}\right)\Gamma\left(\frac{3}{4}-\frac{\nu}{2}+\frac{i\lambda}{2}\right)}{\Gamma\left(\frac{1}{4}+\frac{\nu}{2}+\frac{i\lambda}{2}\right)\Gamma\left(\frac{3}{4}-\frac{\nu}{2}-\frac{i\lambda}{2}\right)}\\ &=-\frac{\sin\left(\frac{1}{4}+\frac{\nu}{2}+\frac{i\lambda}{2}\right)}{\sin\left(\frac{1}{4}+\frac{\nu}{2}-\frac{i\lambda}{2}\right)}\\ &=-\frac{\sin\left(\frac{\pi}{2}\phi_+^*\right)}{\sin\left(\frac{\pi}{2}\phi_-^*\right)}\,. \end{split} \end{equation} The Schwinger-Keldysh bulk-to-bulk kernels is given by \begin{align} D_{\sigma\sigma}^{>}(k;\eta_1\eta_2)=\sum_{b=1}^2\sigma_b(k,\eta_1)\sigma^*_b(k,\eta_2) \end{align} \begin{equation} \sigma_b(k,z)=i\frac{\sqrt{2}H}{\lambda k^{\frac32}}z^2e^{iz}\frac{\mathrm{d}\zeta_b^+}{\mathrm{d}z}. \end{equation} \begin{align} A_1^{(2)}A_1^{(2)*}=A_1^{(1)}A_1^{(1)*}=\frac{e^{\pi\lambda/2}}{4}\\ A_2^{(2)}A_2^{(2)*}=A_2^{(1)}A_2^{(1)*}=\frac{e^{-\pi\lambda/2}}{4}\\ A_1^{(1)}A_2^{(1)*}=\frac{e^{i\lambda\ln(2z_0)}}{4}\\ A_1^{(2)}A_2^{(2)*}=-\frac{e^{i\lambda\ln(2z_0)}}{4} \end{align} \begin{equation} \sum_{b=1}^2\frac{\mathrm{d}\zeta^+_b(z_1)}{\mathrm{d}z}\frac{\mathrm{d}\zeta^{+*}_b(z_2)}{\mathrm{d}z}=\frac{e^{\pi\lambda/2}}{2}f'_1(z_1)f'^*_1(z_2)+\frac{e^{-\pi\lambda/2}}{2}f'_2(z_1)f'^*_2(z_2) \end{equation} \begin{equation} D_{\sigma\sigma}^>(k;z_1,z_2)=\frac{2H^2}{\lambda^2k^3}z_1^2z_2^2e^{i(z_1-z_2)}\left(\frac{e^{\pi\lambda/2}}{2}f'_1(z_1)f'^*_1(z_2)+\frac{e^{-\pi\lambda/2}}{2}f'_2(z_1)f'^*_2(z_2)\right) \end{equation} \begin{equation} D_{\sigma\sigma}^<(k;z_1,z_2)=\frac{2H^2}{\lambda^2k^3}z_1^2z_2^2e^{-i(z_1-z_2)}\left(\frac{e^{\pi\lambda/2}}{2}f'^*_1(z_1)f'_1(z_2)+\frac{e^{-\pi\lambda/2}}{2}f'^*_2(z_1)f'_2(z_2)\right) \end{equation} \begin{equation} f'_s(z_1)f'^*_s(z_2)=\frac{1}{4\pi^4}\frac{\Gamma(1+2a_s)\Gamma(1+2a_{-s})}{\Gamma\left(\frac{1}{2}-\nu\right)^2\Gamma\left(\frac{1}{2}+\nu\right)^2\Gamma(a_s)^2\Gamma(a_{-s})^2}I_s(z_1)I_s^*(z_2) \end{equation} Therefore, \begin{equation} D_{\sigma\sigma}\sim C_1c^{2\nu+3}+C_2c^{-2\nu+3}+C_3c^{3}\,, \end{equation} \begin{equation} \sum_{b=1}^2\frac{\mathrm{d}\zeta^+_b(z_1)}{\mathrm{d}z}\frac{\mathrm{d}\zeta^{+*}_b(z_2)}{\mathrm{d}z}=\frac{\lambda^2}{c}\left(\eta_1\eta_2\right)^{-1/2}\left[\left(4\eta_1\eta_2\right)^{i\rho}Q\exp\left(2i\rho\ln c\right)+c.c+real~const\right] \end{equation} \begin{equation} Q=\frac{1}{2}\frac{\Gamma\left(-2i\rho\right)\Gamma\left(2i\rho\right)}{\Gamma\left(\frac{1}{2}+i\lambda-i\rho\right)\Gamma\left(\frac{1}{2}-i\lambda-i\rho\right)} \end{equation} The leading order \begin{equation} D_{\sigma\sigma}^<(k;z_1,z_2)=D_{\sigma\sigma}^>(k;z_1,z_2)=2H^2\left(\eta_1\eta_2\right)^{3/2}\left[\left(4\eta_1\eta_2\right)^{i\rho}Q\exp\left(2i\rho\ln c\right)+c.c+C_0\right] \end{equation} \section{Power Spectrum} Inserting $\delta\Phi_k(\tau) = K_k(\tau)\, \delta\varphi_k$ into the action, \begin{equation} S_{\rm on-shell} = \frac{1}{2} \int d^3 k \, \varphi_k^\dagger\, \Bigl[ a(\tau)^3 K_k(\tau)^\dagger D_t K_k(\tau) \Bigr]_{\tau \to 0}\, \varphi_k . \end{equation} The second‑order wavefunction coefficient $\psi_2(k)$ is therefore \begin{equation} \psi_2(k) = -i \lim_{\tau \to 0} a(\tau)^3 \, K_k(\tau)^\dagger D_t K_k(\tau) , \end{equation} and \begin{equation} \begin{split} \Re \psi_2&=\psi_2+\psi_2^\dagger\\ &=(U_k(0)U_k(0)^\dagger)^{-1}\lim_{\tau \to 0}\left(-ia^3U_k^\dagger D_tU_k+ia^3U_k\left(D_tU_k\right)^\dagger\right)\\ &=(U_k(0)U_k(0)^\dagger)^{-1}\,, \end{split} \end{equation} the last equality is given by the normalization condition of $U_k$. Therefore, \begin{equation} \braket{\varphi_k\varphi_k^\dagger}^\prime=\frac{1}{2\Re\psi_2}=U_k(0)U_k(0)^\dagger\,, \end{equation} and the dimensionless power spectrum of curvature perturbation \begin{equation} \Delta_\zeta(k,z)=\frac{k^3}{4\pi^2\epsilon}\sum_b|\zeta_b(k,z)|^2\,, \end{equation} \section{Non-Gaussianities} Consider the interaction \begin{equation} -\frac{a^4}{3!}g_{\sigma^3}\sigma^3\subset \mathcal{L}_{\rm int}\,. \end{equation} We insert $\sigma(z)=K_{\sigma\zeta}(k;z)\zeta(0)$ into the action, \begin{equation} \psi_{\zeta\zeta\zeta}^{(\sigma^3)}(k_1,k_2,k_3)=-\frac{g_{\sigma^3}}{3!}\int_{-\infty}^0\mathrm{d}\eta~a(\eta)^4\prod_{i=1}^3 K_{\sigma\zeta}(k_i;\eta)\,. \end{equation} Define the dimensionless transfer function \begin{equation} T_{\sigma\zeta}(z)=\frac{k^3}{H^2}K_{\sigma\zeta}(k,\eta)\,, \end{equation} and the squeezed limit 3-point wave function coefficient, \begin{equation} \psi_{\zeta\zeta\zeta}^{(\sigma^3)}(k,k,ck)= \end{equation} The exact mode-function expression is \begin{equation} B_q^{(\sigma^3)}(k_1,k_2,k_3) = -2g_{\sigma^3}\,\mathrm{Im} \int_{-\infty(1-i\epsilon)}^{0} \mathrm{d}\tau\,a^4(\tau) \prod_{i=1}^3 K_{\sigma q}(k_i;\tau) \,. \end{equation} Using $a=-1/(H\tau)$ and choosing a reference scale $k_*$ with $r_i\equiv k_i/k_*$ and $x=-k_*\tau$, this becomes \begin{equation} B_q^{(\sigma^3)} = -2g_{\sigma^3}\frac{H^2}{k_*^6} \frac{1}{r_1^3r_2^3r_3^3} \mathrm{Im}\int_0^\infty\mathrm{d} x\,x^{-4} \prod_{i=1}^3 T_\sigma(r_ix)\,, \end{equation} where the $i\epsilon$ prescription is implicit. \subsection{Squeezed Limit} We denote $k$ be the momentum of the hard mode and $ck$ be the momentum of the soft mode, where $c\ll1$. For convenience, we take $x=-k\eta$, \begin{equation} B_q^{(\sigma^3)}(k,k,ck) = -2g_{\sigma^3}\frac{H^2}{k^6} \frac{1}{c^3} \mathrm{Im}\int_0^\infty\mathrm{d} x\,x^{-4} T_\sigma(x)^2T_\sigma(cx)\,. \end{equation} The transfer function of the soft mode could be written in the form \begin{equation} T(cx)=C_{\lambda,\rho}^+(cx)^{3/2+i\rho}+C_{\lambda,\rho}^-(cx)^{3/2-i\rho}\,. \end{equation} We denote \begin{equation} I^{(\sigma^3)}(k_1,k_2,k_3)=\int_0^\infty\mathrm{d} x\,x^{-4} \prod_{i=1}^3 T_\sigma(r_ix)\, \end{equation} \begin{equation} I^{(\sigma^3)}(k,k,ck)=C_{\lambda,\rho}^+J_{\lambda,\rho}^+(k)c^{3/2+i\rho}+C_{\lambda,\rho}^-J_{\lambda,\rho}^-(k)c^{3/2-i\rho}\,, \end{equation} where \begin{align} J_{\lambda,\rho}^+(k)&=\int_0^{\infty}\mathrm{d}x\,x^{-5/2+i\rho}T_\sigma(x)^2\,,\\ J_{\lambda,\rho}^-(k)&=\int_0^{\infty}\mathrm{d}x\,x^{-5/2-i\rho}T_\sigma(x)^2\,. \end{align} \begin{equation} \begin{split} &B_q^{(\sigma^3)}(k,k,ck) = -2g_{\sigma^3}\frac{H^2}{k^6} \frac{1}{c^{3/2}}\\ &\times\left[\Im\left(C_{\lambda,\rho}^+J_{\lambda,\rho}^+(k)+C_{\lambda,\rho}^-J_{\lambda,\rho}^-(k)\right)\cos\left(\rho\ln c\right)+\Re\left(C_{\lambda,\rho}^+J_{\lambda,\rho}^+(k)-C_{\lambda,\rho}^-J_{\lambda,\rho}^-(k)\right)\sin\left(\rho\ln c\right)\right] \end{split} \end{equation} \begin{equation} \Im\left(CJ\right)\cos\left(\rho\ln c\right)+\Re\left(CJ\right)\sin\left(\rho\ln c\right) \end{equation} \begin{equation} J\equiv J^+-J^{-*}=-2i\int_0^{\infty}\mathrm{d}x\,x^{-5/2+i\rho}\,\Im\left(T_\sigma(x)^2\right) \end{equation} \begin{equation} B_q^{(\sigma^3)}(k,k,ck) = -2g_{\sigma^3}\frac{H^2}{k^6} \frac{|CJ|}{c^{3/2}}\sin\left(\rho\ln c+\theta\right) \end{equation} \begin{equation} \theta=\arg\left(CJ\right) \end{equation} \paragraph{Large Mass Limit} \begin{equation} C=-\lambda\sum_{s=\pm}\left[e^{i\phi_s\pi/2}\,2^{\nu-1/2}\frac{\Gamma(-2\nu)\Gamma(1+2a_s)}{\Gamma(a_s)\Gamma\left(2a_s-\nu+\frac{1}{2}\right)}\mathcal{U}^*_s\right] \end{equation} \begin{equation} \mathcal{U}_s=\frac{1}{\sqrt{\pi}}\frac{\Gamma\left(\frac{1}{2}+a_s\right)\Gamma\left(\frac{3}{4}-\frac{\nu}{2}\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}\right)}{\Gamma\left(\frac{3}{4}-\frac{\nu}{2}+a_s\right)\Gamma\left(\frac{3}{4}+\frac{\nu}{2}+a_s\right)} \end{equation} For large $\rho$ \begin{equation} \mathcal{U}_s=\frac{1}{\sqrt{\pi}}\Gamma\left(\frac{1}{2}+a_s\right)\left(\frac{\rho}{2}\right)^{-2a_s} \end{equation} \begin{equation} \begin{split} C&=-\lambda\frac{i}{\cosh\left(\frac{\pi\lambda}{2}\right)}\sum_{s=\pm}a_s\exp\left[-i\rho\left(\ln\rho+\ln 2-1\right)+4a_s\left(\ln2-\ln\rho\right)-\ln2-\frac{1}{2}\ln\rho\right]\\ &=\frac{\lambda^2\sin\left(2\lambda\ln\frac{\rho}{2}\right)}{2\sqrt{\rho}\cosh\left(\frac{\pi\lambda}{2}\right)}\exp\left[-i\left(\frac{\pi}{2}+\rho\left(\ln\rho+\ln 2-1\right)\right)\right] \end{split} \end{equation} \begin{equation} J\equiv J^+-J^{-*}=-2i\int_0^{\infty}\mathrm{d}x\,x^{-5/2+i\rho}\,\Im\left(T_\sigma(x)^2\right) \end{equation} \begin{equation} T(x)=x^2e^{ix}\sum_{s=\pm}\mathcal{A}_s\int_0^{\infty}\mathrm{d}t\mathrm{d}u~{e}^{2ixt}t^{a_s}(1+t)^{-a_s}u^{-1/2+i\rho}(u+1)^{-1/2-i\rho-2a_s}(u+tu+1)^{i\rho-1/2} \end{equation} \begin{equation} \mathcal{A}_s=\frac{i}{2\lambda}e^{s\pi\lambda/2}\mathcal{U}_s^*\frac{\Gamma(1+2a_s)}{\Gamma\left(\frac{1}{2}+i\rho\right)\Gamma\left(\frac{1}{2}+2a_s-i\rho\right)}\frac{2i}{\Gamma(a_s)} \end{equation} \begin{equation} \mathcal{A}_s=-\frac{is}{4\pi}\frac{e^{\pi\rho}}{\cosh\left(\frac{\pi\lambda}{2}\right)} \end{equation} \begin{equation} \begin{split} J^+&=\int_0^{\infty}\mathrm{d}x~x^{-5/2+i\rho}T(x)^2\\ &=\sum_{s,s'=\pm}\mathcal{A}_s\mathcal{A}_{s'}\\ &\times\int_0^\infty\prod_{i=1}^2\mathrm{d}t_i\mathrm{d}u_i~\left(\frac{t_i}{(1+t_i)(1+u_i)}\right)^{a_s}\left(\frac{u_i(u_i+t_iu_i+1)}{u_i+1}\right)^{i\rho}\left(u(u+1)(u+t_iu_i+1)\right)^{-1/2}\\ &\times\int_0^\infty \mathrm{d}x~x^{-1/2+i\rho+ix}{e}^{2ixt_i}\\ &=\Gamma\left(i\rho+\frac{1}{2}\right)(i)^{-i\rho-1/2}\sum_{s,s'=\pm}\mathcal{A}_s\mathcal{A}_{s'}\int_0^\infty\mathrm{d}t_i\mathrm{d}u_i~F_s(t_a,u_a)\exp\left(i\rho\Phi^+(t_a,u_a)\right) \end{split} \end{equation} \begin{equation} F_s(t_a,u_a)=\prod_{i=1}^2\left(\frac{t_i}{(1+t_i)(1+u_i)}\right)^{a_s}\left(u_i(u_i+1)(u_i+t_iu_i+1)\right)^{-1/2}\left(2(t_1+t_2)+1\right)^{-1/2} \end{equation} \begin{equation} \Phi^+(t_a,u_a)=\sum_{i=1}^2\left[\ln\frac{u_i(u_i+t_iu_i+1)}{u_i+1}\right]-\ln\left(2(t_1+t_2)+1\right) \end{equation} \begin{equation} J^-=\Gamma\left(-i\rho+\frac{1}{2}\right)(i)^{i\rho-1/2}\sum_{s,s'=\pm}\mathcal{A}_s\mathcal{A}_{s'}\int_0^\infty\mathrm{d}t_i\mathrm{d}u_i~F_s(t_a,u_a)\exp\left(i\rho\Phi^-(t_a,u_a)\right) \end{equation} \begin{equation} \Phi^-(t_a,u_a)=\sum_{i=1}^2\left[\ln\frac{u_i(u_i+t_iu_i+1)}{u_i+1}\right]+\ln\left(2(t_1+t_2)+1\right) \end{equation} \begin{equation} \Gamma\left(i\rho+\frac{1}{2}\right)(i)^{-i\rho-1/2}=\sqrt{2\pi}\exp\left[i\left(\rho\ln\rho-\rho-\frac{1}{4}\right)\right] \end{equation} \begin{equation} \Gamma\left(-i\rho+\frac{1}{2}\right)(i)^{i\rho-1/2}=\sqrt{2\pi}\exp\left[i\left(-\rho\ln\rho+\rho-\frac{1}{4}\right)\right] \end{equation} Next, we are going to find the branch cut and singularity of the integrand. Notice that $t_1,u_1$ and $t_2,u_2$ are symmetric in the integrand, we could choice the contour such that $t_1=t_2$ and $u_1=u_2$. Therefore, the branch cuts are given by $t_1,t_2,u_1,u_2\in\mathbb{R}^-;u_i+t_iu_i+1\in\mathbb{R}^-$ and the branch points are $t_1,t_2=0,-1/4,-1$, $u_1,u_2=0,-1$ and $u_i+t_iu_i+1=0$. The saddle points is given by \begin{equation} \nabla\Phi^\pm(t_a,u_a)=0 \end{equation} \begin{align} &0=\nabla_{u_i}\Phi^\pm(t_a,u_a)=\frac{1}{u_i}+\frac{1+t_i}{u+t_iu_i+1}-\frac{1}{u_i+1}\\ &0=\nabla_{v_i}\Phi^\pm(t_a,u_a)=\frac{u_i}{u_i+t_iu_i+1}\mp\frac{2}{2(t_1+t_2)+1} \end{align} \[ (u_i,t_i)^+ = \left(-\frac{4}{3} - \frac{\sqrt{2}}{3}i,\; -\frac{1}{6} + \frac{\sqrt{2}}{6}i\right),\; \left(-\frac{4}{3} + \frac{\sqrt{2}}{3}i,\; -\frac{1}{6} - \frac{\sqrt{2}}{6}i\right). \] \[ (u_i,t_i)^- = \left(-\frac{2}{3} - \frac{\sqrt{2}}{3}i,\; -\frac{1}{6} - \frac{\sqrt{2}}{6}i\right),\; \left(-\frac{2}{3} + \frac{\sqrt{2}}{3}i,\; -\frac{1}{6} + \frac{\sqrt{2}}{6}i\right). \] \begin{equation} \begin{split} I^-_{(1)}&=-i 3^{-i A-i \rho +\frac{3}{2}} \exp \left(2 \tan ^{-1}\left(\sqrt{2}\right) (A-\rho )+2 \pi \rho +\rho \tan ^{-1}\left(2 \sqrt{2}\right)\right)\\ &=-i3^{-i\rho+3/2}\exp\left[\rho\left(-2\arctan\sqrt{2}+\arctan 2\sqrt{2}+2\pi\right)\right]\exp\left[\frac{\lambda(s+s')}{2}(2\arctan\sqrt{2}-i)\right] \end{split} \end{equation} \begin{equation} \begin{split} I^-_{(2)}&=i 3^{-i A-i \rho +\frac{3}{2}} \exp \left(2 \tan ^{-1}\left(\sqrt{2}\right) (\rho -A)-2 \pi \rho +\rho \left(-\tan ^{-1}\left(2 \sqrt{2}\right)\right)\right)\\ &=i3^{-i\rho+3/2}\exp\left[\rho\left(2\arctan\sqrt{2}-\arctan 2\sqrt{2}-2\pi\right)\right]\exp\left[\frac{\lambda(s+s')}{2}(-2\arctan\sqrt{2}-i)\right] \end{split} \end{equation} \begin{equation} \begin{split} I^+_{(2)}&=\left(2 \sqrt{2}+i\right) 3^{-i A+i \rho -\frac{1}{2}} \exp \left(\rho \left(-\tan ^{-1}\left(2 \sqrt{2}\right)\right)-2 \left(\pi -\tan ^{-1}\left(\sqrt{2}\right)\right) (A+\rho )\right)\\ &=\left(2 \sqrt{2}+i\right) 3^{i \rho -\frac{1}{2}}\exp\left[\rho\left(2\arctan\sqrt{2}-\arctan2\sqrt{2}-2\pi\right)\right]\exp\left[\frac{\lambda(s+s')}{2}\left(2\arctan\sqrt{2}+i-2\pi\right)\right] \end{split} \end{equation} \begin{equation} \begin{split} I^+_{(1)}&=\left(2 \sqrt{2}-i\right) 3^{-i A+i \rho -\frac{1}{2}} \exp \left(2 \left(\pi -\tan ^{-1}\left(\sqrt{2}\right)\right) (A+\rho )+\rho \tan ^{-1}\left(2 \sqrt{2}\right)\right)\\ &=\left(2 \sqrt{2}-i\right) 3^{i \rho -\frac{1}{2}}\exp\left[\rho\left(-2\arctan\sqrt{2}+\arctan2\sqrt{2}+2\pi\right)\right]\exp\left[\frac{\lambda(s+s')}{2}\left(-2\arctan\sqrt{2}-i+2\pi\right)\right] \end{split} \end{equation} \begin{equation} H^\pm(t,u)=\begin{pmatrix} -\frac{1}{u^2}-\frac{(1+t)^2}{(u+tu+1)^2}-\frac{1}{(u+1)^2}&\frac{1}{(u+tu+1)^2}\\ \frac{1}{(u+tu+1)^2}&-\frac{u^2}{(u+tu+1)^2}\pm\frac{4}{(4t+1)^2} \end{pmatrix} \end{equation} \begin{equation} H^\pm_\Phi(t,u)=\begin{pmatrix} H^\pm(t,u)&\begin{matrix} 0&0\\0&\pm\frac{4}{(4t+1)^2} \end{matrix}\\\begin{matrix} 0&0\\0&\pm\frac{4}{(4t+1)^2} \end{matrix}& H^\pm(t,u) \end{pmatrix} \end{equation} \begin{equation} H^{+}_{(1)} = \begin{pmatrix} \frac{17}{9} + \frac{25\sqrt{2}}{9}\mathrm{i} & -2 \\ -2 & 0 \end{pmatrix}, \quad H^{+}_{(2)} = \begin{pmatrix} \frac{17}{9} - \frac{25\sqrt{2}}{9}\mathrm{i} & -2 \\ -2 & 0 \end{pmatrix} \end{equation} \begin{equation} H^{-}_{(1)} = \begin{pmatrix} -3 - 3\sqrt{2}\mathrm{i} & 2 + 4\sqrt{2}\mathrm{i} \\ 2 + 4\sqrt{2}\mathrm{i} & \frac{56}{9} - \frac{32\sqrt{2}}{9}\mathrm{i} \end{pmatrix}, \quad H^{-}_{(2)} = \begin{pmatrix} -3 + 3\sqrt{2}\mathrm{i} & 2 - 4\sqrt{2}\mathrm{i} \\ 2 - 4\sqrt{2}\mathrm{i} & \frac{56}{9} + \frac{32\sqrt{2}}{9}\mathrm{i} \end{pmatrix} \end{equation} \begin{equation} \det H^{+}_{(1)} = -4, \quad \det H^{+}_{(2)} = -4, \quad \det H^{-}_{(1)} = -12 - 24\sqrt{2}\,\mathrm{i}, \quad \det H^{-}_{(2)} = -12 + 24\sqrt{2}\,\mathrm{i} \end{equation} \begin{equation} \int_0^\infty\mathrm{d}t_i\mathrm{d}u_i~F_s(t_a,u_a)\exp\left(i\rho\Phi^\pm(t_a,u_a)\right)=F_s(y^A_0)\exp\left(i\rho\Phi^{\pm}(y_0)\right)\int \end{equation} \end{document}