Symmetry resolved entanglement: exact results in 1D and beyond

Assuming $\left|h\right|\leqslant2$, we write $\newcommand{\e}{{\rm e}} \mathcal{L}\equiv 2L\left|\sin k_{F}\right|$ and define a natural number $\newcommand{\e}{{\rm e}} m_{c}=m_{c}(n)\equiv \lceil\frac{n}{4}\rceil+1$. We will show that for $\mathcal{L}\gg1$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\exp\left[{\rm i}\frac{k_{F}}{\pi}\alpha L+\left[\frac{1}{6}\left(\frac{1}{n}-n\right)-\frac{\alpha^{2}}{2\pi^{2}n}\right]\ln\mathcal{L}+\Upsilon_{0}(n,\alpha)+\Upsilon_{1}\left(n,\alpha,L,k_{F}\right)+o\left(\mathcal{L}^{-1}\right)\right], \nonumber \end{align} \tag{ 20 }$$

where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0}(n,\alpha)\equiv -\frac{1}{\pi^{2}}\underset{0}{\overset{\infty}{\int}}\ln\left[\frac{2\cos\alpha+2\cosh(nu)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2n}}\right]{\rm d}u\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{\cos\left(\frac{ut}{2\pi}\right)}{2\sinh\left(\frac{t}{2}\right)}\right]{\rm d}t, \nonumber \end{align} \tag{ 21 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{1}\left(n,\alpha,L,k_{F}\right) & \equiv \underset{m=1}{\overset{m_{c}}{\sum}}\ln\left[1+\mathcal{L}^{-\frac{2}{n}\left(2m-1-\frac{\alpha}{\pi}\right)}{\rm e}^{-2{\rm i}k_{F}L}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2}}\right]\nonumber \\ & \quad +\underset{m=1}{\overset{m_{c}}{\sum}}\ln\left[1+\mathcal{L}^{-\frac{2}{n}\left(2m-1+\frac{\alpha}{\pi}\right)}{\rm e}^{2{\rm i}k_{F}L}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2}}\right]. \nonumber \end{align} \tag{ 22 }$$

The term $\left[\frac{1}{6}\left(\frac{1}{n}-n\right)-\frac{\alpha^{2}}{2\pi^{2}n}\right]\ln\mathcal{L}$ in the exponent has been already found before, using CFT techniques [20, 21], and our calculation not only derives it rigorously, but also completes the picture up to corrections of order $o\left(\mathcal{L}^{-1}\right)$.

Furthermore, in 4.3 we will see that this result can be written as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\underset{j=-m_{c}}{\overset{m_{c}}{\sum}}\tilde{S}_{n}(\alpha+2\pi j)+o\left(\mathcal{L}^{-1}\right), \nonumber \end{align} \tag{ 23 }$$

where $\tilde{S}_{n}$ is an analytic function that is defined on the entire real line. This shows that $S_{n}(\alpha)$ has a structure that is natural in the context of CFT, as we explain below.

We will use the notations $\newcommand{\e}{{\rm e}} k\equiv \gamma\slash\sqrt{(h/2){}^{2}+\gamma^{2}-1}$ and $\newcommand{\e}{{\rm e}} k'\equiv \sqrt{1-k^{2}}$, and denote by k_n the positive solution to the equation $\newcommand{\e}{{\rm e}} q^{n}=\exp\left[-\pi I\left(\sqrt{1-k_{n}^{2}}\right)\slash I\left(k_{n}\right)\right]$, where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I(k)=\underset{0}{\overset{1}{\int}}\frac{{\rm d}x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}} \nonumber \end{align} \tag{ 24 }$$

is the complete elliptic integral of the first kind and $\newcommand{\e}{{\rm e}} q\equiv \exp\left[-\pi I\left(k'\right)\slash I(k)\right]$ is the nome [43]. Assuming that $0\leqslant h\neq2$, we will find that as $L\rightarrow\infty$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}(-1)^{L}S_{n}^{(-)}=\left\{\begin{array}{@{}ll@{}} 0, & h<2 \nonumber \\ \left[\frac{\left(k\cdot k'\right)^{2n}\left(1-k_{n}^{2}\right)^{2}}{16^{n-1}k_{n}^{2}}\right]^{\frac{1}{12}}, & h>2 \end{array}\right.,\label{eq: XY REE summary} \nonumber \end{align} \tag{ 25 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}(-1)^{L}\mathcal{S}^{(-)}=\left\{\begin{array}{@{}ll@{}} 0, & h<2 \nonumber \\ \frac{\sqrt{k'}}{3}\left[\ln2-\frac{1}{2}\ln\left(k\cdot k'\right)-\frac{I(k)I\left(k'\right)}{\pi}\left(1+k^{2}\right)\right], & h>2 \end{array}\right..\label{eq: XY vNEE summary} \nonumber \end{align} \tag{ 26 }$$

For finite L, the corrections to these expressions are exponentially small in L. We are not aware of extensions of the Fisher–Hartwig conjecture which allow to calculate these corrections, but we verify numerically that they are negligible even for relatively small values of L. For h  <  2 we get in particular that $\underset{L\rightarrow\infty}{\lim}\left[\mathcal{S}^{\left(\mathrm{even}\right)}-\mathcal{S}^{\left(\mathrm{odd}\right)}\right]=0$, due to a degeneracy between the spectra of the even charge sector and the odd charge sector. This degeneracy stems from the appearance of Majorana zero-modes at the virtual edges of the entanglement Hamiltonian.

For the critical field h  =  2, $S_{n}^{(-)}$ and $\mathcal{S}^{(-)}$ still vanish as $L\rightarrow\infty$, but only as a power law, rather than exponentially. We will find that in this case there is a positive factor $A(n,\gamma)$ such that we can write the following leading order approximation for large L:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}S_{n}^{(-)}\approx A(n,\gamma)L^{-\frac{1}{6n}-\frac{n}{12}}, \nonumber \end{align} \tag{ 27 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}\mathcal{S}^{(-)}\approx-\frac{A(1,\gamma)}{12}L^{-\frac{1}{4}}\ln L,\label{eq: Gapless XY vNEE summary} \nonumber \end{align} \tag{ 28 }$$

in accordance with the CFT results of [20].

These results can be extended to h  <  0 by plugging in the corresponding result for $\left|h\right|$, only in this case the $(-1){}^{L}$ factor that appears in (25)–(28) is absent.

The Jordan–Wigner transformation constitutes the basis for the calculation of the EE for a subsystem A of L sites [44, 45]. One can show that the Majorana operators c_n that belong to subsystem A obey

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \langle GS|c_{n}|GS\rangle=0,\,\,\langle GS|c_{m}c_{n}|GS\rangle=\delta_{mn}+{\rm i}\left(B_{L}\right)_{mn};\,\,m,n=1,\ldots,2L. \nonumber \end{align} \tag{ 29 }$$

Here B_L is a $2L\times2L$ matrix defined as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle B_{L}=\left(\begin{array}{@{}cccc@{}} \Pi_{0} & \Pi_{-1} & \cdots & \Pi_{1-L} \nonumber \\ \Pi_{1} & \Pi_{0} & & \vdots \nonumber \\ \vdots & & \ddots & \vdots \nonumber \\ \Pi_{L-1} & \cdots & \cdots & \Pi_{0} \end{array}\right),\,\,\,\Pi_{m}\equiv \frac{1}{2\pi}\underset{0}{\overset{2\pi}{\int}}{\rm d}\theta {\rm e}^{-{\rm i}m\theta}\mathcal{G}(\theta),\label{eq: subsystem correlation matrix} \nonumber \end{align} \tag{ 30 }$$

where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{G}(\theta)\equiv \left(\begin{array}{@{}cc@{}} 0 & g(\theta) \nonumber \\ -g^{-1}(\theta) & 0 \end{array}\right),\,\,\,g(\theta)\equiv \frac{\cos\theta-{\rm i}\gamma\sin\theta-\frac{h}{2}}{\left|\cos\theta-{\rm i}\gamma\sin\theta-\frac{h}{2}\right|}.\label{eq: Toeplitz generating function} \nonumber \end{align} \tag{ 31 }$$

Using an orthogonal matrix $V$ we can transform B_L into the form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle VB_{L}V^{T}=\oplus_{m=1}^{L}\nu_{m}\left(\begin{array}{@{}cc@{}} 0 & 1 \nonumber \\ -1 & 0 \end{array}\right), \nonumber \end{align} \tag{ 32 }$$

where $\nu_{m}$ are real numbers which satisfy $-1<\nu_{m}<1$. We use $V$ to transform the Majorana operators as well by defining

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle d_{m}=\underset{n=1}{\overset{2L}{\sum}}V_{mn}c_{n},\,\,m=1,\ldots,2L. \nonumber \end{align} \tag{ 33 }$$

Similarly to the transformation of c_n into fermionic operators in (13), one can obtain a set of L fermionic operators by introducing $\newcommand{\e}{{\rm e}} b_{m}\equiv \frac{1}{2}\left(d_{2m}+{\rm i}d_{2m-1}\right)$. In [44, 48] it was shown that the reduced density matrix of subsystem A in the ground state of the entire system can be represented by a quite simple expression involving the fermionic operators b_m:

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \rho_{A}={{\rm Tr}}_{B}(|GS\rangle\langle GS|)=\underset{m=1}{\overset{L}{\Pi}}\left[\left(\frac{1+\nu_{m}}{2}\right)b_{m}^{\dagger}b_{m}+\left(\frac{1-\nu_{m}}{2}\right)b_{m}b_{m}^{\dagger}\right].\label{eq: RDM using fermionic operators} \nonumber \end{align} \tag{ 34 }$$

Since the values $\nu_{m}$ in (34) determine the spectrum of the RDM $\rho_{A}$, considerable efforts were invested in estimating them under certain conditions. The general assumption upon which we will rely is that $L\gg1$. This allows us to use special cases of the Fisher–Hartwig conjecture [39] in order to obtain asymptotic expressions for the EE.

3.2.1. XX model.

We first consider the isotropic case $\gamma=0$, assuming that $\left|h\right|\leqslant2$ (ungapped chain). In this case, further simplification of the expression for the correlation matrix B_L in (30) can be achieved by noticing that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle B_{L}=G_{L}\otimes\left(\begin{array}{@{}cc@{}} 0 & 1 \nonumber \\ -1 & 0 \end{array}\right), \nonumber \end{align} \tag{ 35 }$$

with

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_{L}=\left(\begin{array}{@{}cccc@{}} \phi_{0} & \phi_{-1} & \cdots & \phi_{1-L} \nonumber \\ \phi_{1} & \phi_{0} & & \vdots \nonumber \\ \vdots & & \ddots & \vdots \nonumber \\ \phi_{L-1} & \cdots & \cdots & \phi_{0} \end{array}\right),\,\,\,\phi_{m}\equiv \frac{1}{2\pi}\underset{0}{\overset{2\pi}{\int}}{\rm d}\theta {\rm e}^{-{\rm i}m\theta}\phi(\theta),\label{eq: Simplified correlation matrix} \nonumber \end{align} \tag{ 36 }$$

where we have defined

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \phi(\theta)\equiv \left\{\begin{array}{@{}ll@{}} 1 & -k_{F}<\theta<k_{F} \nonumber \\ -1 & k_{F}<\theta<2\pi-k_{F} \end{array}\right.\,{{\rm and}}\,k_{F}\equiv \arccos\left(\frac{h}{2}\right). \nonumber \end{align} \tag{ 37 }$$

The required values $\nu_{m}$ are therefore just the eigenvalues of the matrix G_L, or equivalently the zeros of the determinant $\newcommand{\e}{{\rm e}} D_{L}(\lambda)\equiv \det\left(\lambda I_{L}-G_{L}\right)$. In [44] it was shown that for large L, $D_{L}(\lambda)$ can be written asymptotically as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle D_{L}(\lambda)\sim D_{L}^{(0)}(\lambda)\equiv \mathcal{L}^{-2\beta^{2}(\lambda)}\left[(\lambda+1)\left(\frac{\lambda+1}{\lambda-1}\right)^{-k_{F}/\pi}\right]^{L}\left[G\left(1+\beta(\lambda)\right)G\left(1-\beta(\lambda)\right)\right]^{2}.\label{eq: FH leading} \nonumber \end{align} \tag{ 38 }$$

Here $\newcommand{\e}{{\rm e}} \beta(\lambda)\equiv \frac{1}{2\pi {\rm i}}\ln\left(\frac{\lambda+1}{\lambda-1}\right)$, $\newcommand{\e}{{\rm e}} \mathcal{L}\equiv 2L\left|\sin k_{F}\right|$ and G is the Barnes G-function [43]. Subleading corrections may be obtained from the generalized Fisher–Hartwig conjecture [46]:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle D_{L}(\lambda)=\left[(\lambda+1)\left(\frac{\lambda+1}{\lambda-1}\right)^{-k_{F}/\pi}\right]^{L}\underset{m\in\mathbb{Z}}{\sum}{\rm e}^{-2{\rm i}mk_{F}L}\mathcal{L}^{-2\left(\beta(\lambda)+m\right)^{2}}\left[G\left(1+\beta(\lambda)+m\right)G\left(1-\beta(\lambda)-m\right)\right]^{2}.\label{eq: FH subleading} \nonumber \end{align} \tag{ 39 }$$

3.2.2. XY model.

We now consider the more general case $\gamma\neq0$, i.e. the anisotropic spin chain. We assume that $h\geqslant0$ and focus first on the gapped case $h\neq2$. The system exhibits a quantum phase transition at h  =  2, and therefore we must separate the cases h  <  2 and h  >  2. We define a number $\sigma$ such that $\sigma=1$ for h  <  2 and $\sigma=0$ for h  >  2. Following [42], we also define

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle k\equiv \left\{\begin{array}{@{}ll@{}} \sqrt{\left(1-(h/2)^{2}-\gamma^{2}\right)\slash\left(1-(h/2)^{2}\right)}, & h^{2}<4\left(1-\gamma^{2}\right) \nonumber \\ \sqrt{(h/2)^{2}+\gamma^{2}-1}\slash\gamma, & 4\left(1-\gamma^{2}\right)<h^{2}<4 \nonumber \\ \gamma\slash\sqrt{(h/2)^{2}+\gamma^{2}-1}, & h>2 \end{array}\right.,\label{eq: Definition of k} \nonumber \end{align} \tag{ 40 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tau_{0}\equiv I\left(\sqrt{1-k^{2}}\right)\slash I(k),\label{eq: definition of tau_0} \nonumber \end{align} \tag{ 41 }$$

where $I(k)$ is the complete elliptic integral of the first kind,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I(k)=\underset{0}{\overset{1}{\int}}\frac{{\rm d}x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}.\label{eq: First elliptic integral} \nonumber \end{align} \tag{ 42 }$$

In the XY model, a calculation of a different determinant than that of the XX model is required. Let us define the determinant

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{D}_{L}(\lambda)\equiv \det\left({\rm i}\lambda I_{2L}-B_{L}\right)=\underset{m=1}{\overset{L}{\Pi}}\left(\nu_{m}^{2}-\lambda^{2}\right),\label{eq: definition of XY determinant} \nonumber \end{align} \tag{ 43 }$$

the zeros of which are simply $\pm\nu_{m}$. It was shown in [45] that in the large L limit, the following asymptotic expression for $\tilde{D}_{L}(\lambda)$ is obtained:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{D}_{L}(\lambda)\sim\frac{\left(1-\lambda^{2}\right)^{L}}{\Theta_{3}^{2}\left(\frac{{\rm i}\sigma\tau_{0}}{2}\right)}\Theta_{3}\left(\beta(\lambda)+\frac{{\rm i}\sigma\tau_{0}}{2}\right)\Theta_{3}\left(\beta(\lambda)-\frac{{\rm i}\sigma\tau_{0}}{2}\right).\label{eq: Asymptotic det XY} \nonumber \end{align} \tag{ 44 }$$

Here we have defined $\newcommand{\e}{{\rm e}} \Theta_{3}(s)\equiv \vartheta_{3}\left(\pi s,{\rm e}^{-\pi\tau_{0}}\right)$, where $\vartheta_{3}(z,q)=\underset{m=-\infty}{\overset{\infty}{\sum}}q^{m^{2}}{\rm e}^{2{\rm i}zm}$ is the third Jacobi theta function [49]. The asymptotic expression for $\tilde{D}_{L}(\lambda)$ in (44) has a double zero at each of the points

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda_{l}=\tanh\left[\left(l+\frac{1-\sigma}{2}\right)\pi\tau_{0}\right],\,\,l\in\mathbb{Z}.\label{eq: Limit of nu_m} \nonumber \end{align} \tag{ 45 }$$

This shows that as $L\rightarrow\infty$, the values $\pm\nu_{m}$ are divided into pairs $\tilde{\nu}_{2l-1},\tilde{\nu}_{2l}$ such that for every $l\in\mathbb{Z}$, $\tilde{\nu}_{2l-1},\tilde{\nu}_{2l}\rightarrow\lambda_{l}$. Corrections to the asymptotic expression (44) vanish exponentially as $L\rightarrow\infty$ [47].

The asymptotics of $\tilde{D}_{L}(\lambda)$ in the gapless case h  =  2 differs considerably, due to a discontinuity of the symbol $\mathcal{G}(\theta)$ that was defined in (31). Based on a general conjecture presented in [50, 51] and verified there numerically for several cases, we can predict the two leading terms in the large L approximation of $\ln\tilde{D}_{L}(\lambda)$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln\tilde{D}_{L}(\lambda)\sim\ln\left(1-\lambda^{2}\right)L-2\beta^{2}(\lambda)\ln L.\label{eq: Asymptotic det gapless XY} \nonumber \end{align} \tag{ 46 }$$

We present the derivation of the above expression in appendix A.1.

From the expression for $\rho_{A}$ in (34) we can deduce that the flux-resolved REE may be written as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\underset{m=1}{\overset{L}{\Pi}}\left[\left(\frac{1+\nu_{m}}{2}\right)^{n}{\rm e}^{{\rm i}\alpha}+\left(\frac{1-\nu_{m}}{2}\right)^{n}\right],\label{eq: flux-resolved REE explicit} \nonumber \end{align} \tag{ 47 }$$

where $\nu_{m}$ are the eigenvalues of the matrix G_L defined in (36) [20].

Following [44], we calculate $\ln S_{n}(\alpha)$ for $-\pi<\alpha<\pi$ using integration in the complex plane. We write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}(\alpha)={\rm i}\frac{\alpha}{2}L+\underset{m=1}{\overset{L}{\sum}}e_{n}^{(\alpha)}\left(1,\nu_{m}\right)={\rm i}\frac{\alpha}{2}L+\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\frac{\rm d}{{\rm d}\lambda}\ln D_{L}(\lambda){\rm d}\lambda,\label{eq: Contour Integral} \nonumber \end{align} \tag{ 48 }$$

where $\newcommand{\e}{{\rm e}} D_{L}(\lambda)\equiv \det\left(\lambda I_{L}-G_{L}\right)$ as before, $c(\varepsilon,\delta)$ is the contour presented in figure 1(a), and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle e_{n}^{(\alpha)}(x,\nu)\equiv \ln\left[\left(\frac{x+\nu}{2}\right)^{n}{\rm e}^{{\rm i}\frac{\alpha}{2}}+\left(\frac{x-\nu}{2}\right)^{n}{\rm e}^{-{\rm i}\frac{\alpha}{2}}\right]. \nonumber \end{align} \tag{ 49 }$$

Zoom In Zoom Out Reset image size

We begin by omitting subleading contributions to the asymptotic expression for $D_{L}(\lambda)$, substituting for it the leading order approximation (38). We will accordingly obtain a leading order approximation for $\ln S_{n}(\alpha)$ at large L; this approximation will be denoted by $\ln S_{n}^{(0)}(\alpha)$. One can show that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{\rm d}{{\rm d}\lambda}\ln D_{L}^{(0)}(\lambda)=\left(\frac{k_{F}/\pi}{\lambda-1}+\frac{1-k_{F}/\pi}{\lambda+1}\right)L-\frac{4{\rm i}}{\pi}\cdot\frac{\beta(\lambda)}{(\lambda+1)(\lambda-1)}\left[\ln\mathcal{L}+\left(1+\gamma_{E}\right)+\Upsilon(\lambda)\right],\label{eq: Fisher Hartwig dlog} \nonumber \end{align} \tag{ 50 }$$

where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon(\lambda)\equiv \underset{k=1}{\overset{\infty}{\sum}}\frac{k^{-1}\beta^{2}(\lambda)}{k^{2}-\beta^{2}(\lambda)}. \nonumber \end{align} \tag{ 51 }$$

Substituting (50) into (48) we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}^{(0)}(\alpha) & ={\rm i}\frac{\alpha}{2}L+\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\left(\frac{k_{F}/\pi}{\lambda-1}+\frac{1-k_{F}/\pi}{\lambda+1}\right)L{\rm d}\lambda\nonumber \\ & +\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\left(-\frac{4{\rm i}}{\pi}\cdot\frac{\beta(\lambda)}{(\lambda+1)(\lambda-1)}\left[\ln\mathcal{L}+\left(1+\gamma_{E}\right)+\Upsilon(\lambda)\right]\right){\rm d}\lambda.\label{eq: Contour int. with FH dlog} \nonumber \end{align} \tag{ 52 }$$

Calculating the integrals, we obtain

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}^{(0)}(\alpha)={\rm i}\frac{k_{F}}{\pi}\alpha L+\left[\frac{1}{6}\left(\frac{1}{n}-n\right)-\frac{\alpha^{2}}{2\pi^{2}n}\right]\ln\mathcal{L}+\Upsilon_{0}(n,\alpha)\,\,\,\left(-\pi<\alpha<\pi\right),\label{eq: 0th order expression} \nonumber \end{align} \tag{ 53 }$$

where we have defined

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0}(n,\alpha)\equiv -\frac{1}{\pi^{2}}\underset{0}{\overset{\infty}{\int}}\ln\left[\frac{2\cos\alpha+2\cosh(nu)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2n}}\right]{\rm d}u\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{\cos\left(\frac{ut}{2\pi}\right)}{2\sinh\left(\frac{t}{2}\right)}\right]{\rm d}t. \nonumber \end{align} \tag{ 54 }$$

An equivalent expression for $\Upsilon_{0}(n,\alpha)$, which will be of use later on, is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0}(n,\alpha)=\frac{{\rm i}n}{2\pi}\underset{-\infty}{\overset{\infty}{\int}}\left[\tanh\left(\frac{nu}{2}+{\rm i}\frac{\alpha}{2}\right)-\tanh\left(\frac{u}{2}\right)\right]\ln\frac{\Gamma\left(\frac{1}{2}+\frac{u}{2\pi {\rm i}}\right)}{\Gamma\left(\frac{1}{2}-\frac{u}{2\pi {\rm i}}\right)}{\rm d}u.\label{eq: Upsilon ver2} \nonumber \end{align} \tag{ 55 }$$

It is important to note that the $\alpha^{2}$ term in (53) arises from a Fourier series, $\alpha^{2}= \frac{\pi^{2}}{3}+4\underset{k=1}{\overset{\infty}{\sum}}\frac{(-1){}^{k}}{k^{2}}\cos(k\alpha)$, and therefore it should actually be continued periodically outside the interval $\left[-\pi,\pi\right]$. The calculation of (53) is detailed in appendix A.2.

It is noteworthy that the term $\Upsilon_{0}(n,\alpha)$ is independent of L and k_F, and that it is real and even with respect to $\alpha$. We can therefore write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0}(n,\alpha)=c_{0}(n)+c_{2}(n)\alpha^{2}+\mathcal{O}\left(\alpha^{4}\right). \nonumber \end{align} \tag{ 56 }$$

Knowing the values $c_{0}(n)$ and $c_{2}(n)$ lets us write $\ln S_{n}^{(0)}(\alpha)$ as a quadratic polynomial in $\alpha$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}^{(0)}(\alpha)\approx c_{0}(n)+\frac{1}{6}\left(\frac{1}{n}-n\right)\ln\mathcal{L}+{\rm i}\frac{k_{F}}{\pi}L\alpha-\frac{1}{2}\left(\frac{\ln\mathcal{L}}{\pi^{2}n}-2c_{2}(n)\right)\alpha^{2}\equiv \ln S_{n}^{(G)}(\alpha).\label{eq: flux-resolved XX gaussian} \nonumber \end{align} \tag{ 57 }$$

In such a way the flux-resolved REE is approximated (up to a phase and a normalization constant) as a density function of a Gaussian distribution $S_{n}^{(G)}(\alpha)$, which implies that under this approximation its Fourier transform — the charge-resolved REE — represents a Gaussian distribution as well:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}\left(Q_{A}\right)\approx {\rm e}^{c_{0}(n)}{{\mathcal L}}^{\frac{1}{6}\left(\frac{1}{n}-n\right)}\sqrt{\frac{1}{\frac{2\ln\mathcal{L}}{\pi n}-4\pi c_{2}(n)}}\exp\left[-\frac{\pi\left(Q_{A}-\frac{k_{F}}{\pi}L\right)^{2}}{\frac{2\ln\mathcal{L}}{\pi n}-4\pi c_{2}(n)}\right].\label{eq: charge-resolved XX gaussian} \nonumber \end{align} \tag{ 58 }$$

The deviation of $S_{n}^{(G)}(\alpha)$ from $S_{n}^{(0)}(\alpha)$ is obviously small as long as $\left|\alpha\right|\ll\pi$. If we demand that $\ln\mathcal{L}\slash n\gg1$, subleading corrections to $S_{n}^{(0)}(\alpha)$ do not spoil this (for $\left|\alpha\right|\ll\pi$ these subleading corrections, which we obtain below, vanish exponentially as $\ln\mathcal{L}\slash n\rightarrow\infty$), meaning that $S_{n}(\alpha)\approx S_{n}^{(G)}(\alpha)$ constitutes a decent approximation in the $\left|\alpha\right|\ll\pi$ regime. Furthermore, the condition $\ln\mathcal{L}\slash n\gg1$ guarantees that the main contribution to the integral in (8) will come from the $\left|\alpha\right|\ll\pi$ regime, due to the fast decay of the $\newcommand{\e}{{\rm e}} \exp\left[-\frac{1}{2}\left(\frac{\ln\mathcal{L}}{\pi^{2}n}-2c_{2}(n)\right)\alpha^{2}\right]$ term away from $\alpha=0$. We can therefore deduce that the Gaussian approximation (58) is valid as long as $\ln\mathcal{L}\slash n\gg1$. We will test the quality of this approximation in the next subsection.

The value of $c_{2}(n)$ for the case n  =  1 is of special interest: since $S_{1}\left(Q_{A}\right)$ is the charge distribution in subsystem A, the expression $\left(\frac{\ln\mathcal{L}}{\pi^{2}}-2c_{2}(1)\right)$ corresponds to the charge variance. Substituting n  =  1, the value $c_{2}(1)=-\frac{1+\gamma_{E}}{2\pi^{2}}$ is obtained (a detailed proof is presented in appendix A.3). This agrees with [52], where it was proven that for a half-filled chain ($k_{F}=\frac{\pi}{2}$, and accordingly $\mathcal{L}=2L$) the charge variance is $\frac{\ln2L+1+\gamma_{E}}{\pi^{2}}$.

Corrections to the leading order approximation (53) can be calculated by taking into account subleading contributions that appear in (39). Following [46], we use the fact that $G(1+x)/G(x)=\Gamma(x)$ and, omitting terms which will contribute corrections of order $\mathcal{O}\left(\mathcal{L}^{-4}\right)$, we rewrite (39) as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle D_{L}(\lambda) & =D_{L}^{(0)}(\lambda)\left[1+{\rm e}^{2{\rm i}k_{F}L}\mathcal{L}^{-2+4\beta(\lambda)}\frac{\Gamma\left(1-\beta(\lambda)\right)^{2}}{\Gamma\left(\beta(\lambda)\right)^{2}}+{\rm e}^{-2{\rm i}k_{F}L}\mathcal{L}^{-2-4\beta(\lambda)}\frac{\Gamma\left(1+\beta(\lambda)\right)^{2}}{\Gamma\left(-\beta(\lambda)\right)^{2}}\right]\nonumber \\ & \equiv D_{L}^{(0)}(\lambda)\left[1+H(\lambda)\right]. \nonumber \end{align} \tag{ 59 }$$

Substituting this into the integral expression for $\ln S_{n}(\alpha)$ (48), we obtain

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}(\alpha) & =\ln S_{n}^{(0)}(\alpha)+\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\frac{\rm d}{{\rm d}\lambda}\ln\left[1+H(\lambda)\right]{\rm d}\lambda+\mathcal{O}\left(\mathcal{L}^{-4}\right)\nonumber \\ & =\ln S_{n}^{(0)}(\alpha)-\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}\frac{de_{n}^{(\alpha)}(1+\varepsilon,\lambda)}{{\rm d}\lambda}\ln\left[1+H(\lambda)\right]{\rm d}\lambda+\mathcal{O}\left(\mathcal{L}^{-4}\right). \nonumber \end{align} \tag{ 60 }$$

Using the fact that for every  −1  <  x  <  1,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \beta\left(x+{\rm i}0^{\pm}\right)=-{\rm i}W(x)\mp\frac{1}{2}, \nonumber \end{align} \tag{ 61 }$$

where $\newcommand{\e}{{\rm e}} W(x)\equiv \frac{1}{2\pi}\ln\frac{1+x}{1-x}$, we obtain that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H\left(x+{\rm i}0^{\pm}\right)={\rm e}^{\pm2{\rm i}\left(2\ln\mathcal{L}W(x)-k_{F}L\right)}\frac{\Gamma\left(\frac{1}{2}\mp {\rm i}W(x)\right)^{2}}{\Gamma\left(\frac{1}{2}\pm {\rm i}W(x)\right)^{2}}+\mathcal{O}\left(\mathcal{L}^{-4}\right). \nonumber \end{align} \tag{ 62 }$$

Now we write $\ln\left[1+H(\lambda)\right]=\underset{k=1}{\overset{\infty}{\sum}}\frac{(-1){}^{k+1}}{k}H(\lambda){}^{k}$ and take the limit $\varepsilon,\delta\rightarrow0^{+}$, omitting terms of order $\mathcal{O}\left(\mathcal{L}^{-4}\right)$, so that we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}(\alpha)-\ln S_{n}^{(0)}(\alpha) & =\frac{1}{2\pi {\rm i}}\underset{-\infty}{\overset{\infty}{\int}}{\rm d}u\underset{k=1}{\overset{\infty}{\sum}}\frac{(-1)^{k+1}n}{2k}\left[\tanh\left(\frac{nu}{2}+{\rm i}\frac{\alpha}{2}\right)-\tanh\left(\frac{u}{2}\right)\right]\nonumber \\ & \times\left\{{\rm e}^{2{\rm i}k\left(\frac{\ln\mathcal{L}}{\pi}u-k_{F}L\right)}\frac{\Gamma\left(\frac{1}{2}+\frac{u}{2\pi {\rm i}}\right)^{2k}}{\Gamma\left(\frac{1}{2}-\frac{u}{2\pi {\rm i}}\right)^{2k}}-{\rm e}^{-2{\rm i}k\left(\frac{\ln\mathcal{L}}{\pi}u-k_{F}L\right)}\frac{\Gamma\left(\frac{1}{2}-\frac{u}{2\pi {\rm i}}\right)^{2k}}{\Gamma\left(\frac{1}{2}+\frac{u}{2\pi {\rm i}}\right)^{2k}}\right\} . \nonumber \end{align} \tag{ 63 }$$

Let us define a natural number $\newcommand{\e}{{\rm e}} m_{c}=m_{c}(n)\equiv \lceil\frac{n}{4}\rceil+1$, so that $m_{c}\geqslant\frac{n}{4}+1$, and thus $\frac{2}{n}\left(2m_{c}-1\pm\frac{\alpha}{\pi}\right)\geqslant1$ for every $-\pi<\alpha<\pi$. For each $k\geqslant1$ and every $-\pi<\alpha<\pi$, we can estimate the integral

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I_{k}^{+}\equiv \frac{1}{2\pi {\rm i}}\underset{-\infty}{\overset{\infty}{\int}}{\rm d}u\frac{(-1)^{k+1}n}{2k}\left[\tanh\left(\frac{nu}{2}+{\rm i}\frac{\alpha}{2}\right)-\tanh\left(\frac{u}{2}\right)\right]{\rm e}^{2{\rm i}k\left(\frac{\ln\mathcal{L}}{\pi}u-k_{F}L\right)}\frac{\Gamma\left(\frac{1}{2}+\frac{u}{2\pi {\rm i}}\right)^{2k}}{\Gamma\left(\frac{1}{2}-\frac{u}{2\pi {\rm i}}\right)^{2k}}\label{eq: definition of I_k^+} \nonumber \end{align} \tag{ 64 }$$

by enclosing the m_c poles of $\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)$ that are in the upper half-plane (at $z=\frac{{\rm i}\pi}{n}\left(2m-1-\frac{\alpha}{\pi}\right)$ for $m\in\mathbb{N}$) and are closest to the real line using a rectangular contour, the vertical sides of which are infinitely far from the imaginary line, and whose upper horizontal side crosses the imaginary line through the segment between the m_cth and the $\left(m_{c}+1\right)$th pole of $\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)$ (see figure 1(b)). Thus we make sure that the integral over the upper horizontal side of the contour is of order $\mathcal{O}\left(\mathcal{L}^{-1-\frac{4}{n}}\right)$ at most. Ignoring the poles of $\tanh\left(\frac{z}{2}\right)$, considering that the contribution of their residues is only of order $\mathcal{O}\left(\mathcal{L}^{-2}\right)$, we can write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I_{k}^{+}=\underset{m=1}{\overset{m_{c}}{\sum}}\frac{(-1)^{k+1}}{k}\mathcal{L}^{-\frac{2k}{n}\left(2m-1-\frac{\alpha}{\pi}\right)}{\rm e}^{-2{\rm i}k_{F}Lk}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2k}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2k}}+\mathcal{O}\left(\mathcal{L}^{-1-\frac{4}{n}}+\mathcal{L}^{-2}\right). \nonumber \end{align} \tag{ 65 }$$

In a similar way, we define

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I_{k}^{-}\equiv -\frac{1}{2\pi {\rm i}}\underset{-\infty}{\overset{\infty}{\int}}{\rm d}u\frac{(-1)^{k+1}n}{2k}\left[\tanh\left(\frac{nu}{2}+{\rm i}\frac{\alpha}{2}\right)-\tanh\left(\frac{u}{2}\right)\right]{\rm e}^{-2{\rm i}k\left(\frac{\ln\mathcal{L}}{\pi}u-k_{F}L\right)}\frac{\Gamma\left(\frac{1}{2}-\frac{u}{2\pi {\rm i}}\right)^{2k}}{\Gamma\left(\frac{1}{2}+\frac{u}{2\pi {\rm i}}\right)^{2k}}, \nonumber \end{align} \tag{ 66 }$$

and sum over the residues of $\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)$ at its poles in the lower half-plane up to m  =  m_c, so that we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I_{k}^{-}=\underset{m=1}{\overset{m_{c}}{\sum}}\frac{(-1)^{k+1}}{k}\mathcal{L}^{-\frac{2k}{n}\left(2m-1+\frac{\alpha}{\pi}\right)}{\rm e}^{2{\rm i}k_{F}Lk}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2k}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2k}}+\mathcal{O}\left(\mathcal{L}^{-1-\frac{4}{n}}+\mathcal{L}^{-2}\right). \nonumber \end{align} \tag{ 67 }$$

The summation over m is truncated due the fact that infinite summation will not converge. Further corrections of order $o\left(\mathcal{L}^{-1}\right)$ that are not captured by the generalized Fisher–Hartwig conjecture, and stem from a calculation related to random matrix theory, were found in the calculation of the total REE in [46].

Summing over k, we finally get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}(\alpha)=\ln S_{n}^{(0)}(\alpha)+\Upsilon_{1}\left(n,\alpha,L,k_{F}\right)+o\left(\mathcal{L}^{-1}\right)\,\,\,\left(-\pi<\alpha<\pi\right),\label{eq: 1st order expression} \nonumber \end{align} \tag{ 68 }$$

where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{1}\left(n,\alpha,L,k_{F}\right) & \equiv \underset{m=1}{\overset{m_{c}}{\sum}}\ln\left[1+\mathcal{L}^{-\frac{2}{n}\left(2m-1-\frac{\alpha}{\pi}\right)}{\rm e}^{-2{\rm i}k_{F}L}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2}}\right]\nonumber \\ & \quad +\underset{m=1}{\overset{m_{c}}{\sum}}\ln\left[1+\mathcal{L}^{-\frac{2}{n}\left(2m-1+\frac{\alpha}{\pi}\right)}{\rm e}^{2{\rm i}k_{F}L}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2}}\right]. \nonumber \end{align} \tag{ 69 }$$

Figure 2(a) shows the dependence of the flux-resolved REE on $\alpha$ in a half-filled system ($k_{F}=\frac{\pi}{2}$), for different values of n. The numerical evaluation of (47) is compared to the analytical results, and it can be seen that while the leading order approximation $S_{n}^{(0)}(\alpha)$ exhibits an $\mathcal{O}(1)$ deviation from the numerical values as $\alpha\rightarrow\pm\pi$, this deviation practically vanishes after we include corrections up to order ${{\mathcal O}}\left({{\mathcal L}}^{-1}\right)$. Figure 2(b) shows a more detailed comparison between the analytical result up to order $\mathcal{O}\left(\mathcal{L}^{-1}\right)$ and the numerical result, for the case of half-filling. In this figure we denote the analytical result by $\newcommand{\e}{{\rm e}} \ln S_{n}^{(1)}(\alpha)\equiv \ln S_{n}^{(0)}(\alpha)+\Upsilon_{1}\left(n,\alpha,L,k_{F}\right)$, while the numerical result is denoted by $\ln S_{n}(\alpha)$. The negligible difference between the two calculations indicates that $\ln S_{n}^{(1)}(\alpha)$ provides a very good approximation even for a subsystem of relatively moderate length.

Zoom In Zoom Out Reset image size

We can now use the analytical results for $S_{n}(\alpha)$ in order to calculate the charge-resolved REE through (8), and then the charge-resolved vNEE. Figure 3 shows that when we use the analytical approximation $S_{n}^{(1)}(\alpha)$, these calculations are in good agreement with numerical results. On the other hand, the Gaussian approximation derived from (58) exhibits a discernible deviation from numerical results for both $S_{1}\left(Q_{A}\right)$ and ${{\mathcal S}}\left(Q_{A}\right)$, since $\ln\mathcal{L}$ is not large enough.

Zoom In Zoom Out Reset image size

$\ln S_{n}^{(0)}(\alpha)$ in (53) was defined for $-\pi<\alpha<\pi$, and its real part is $2\pi$-periodic in $\alpha$ (remember that the $\alpha^{2}$ term originated from a Fourier series). Nevertheless, its periodic continuation is not analytic (nor is it even differentiable), so we would like to define the analytic continuation of $S_{n}^{(0)}(\alpha)$ for $\alpha\in\mathbb{R}$. For this purpose, we will construct a natural continuation of $\ln S_{n}^{(0)}(\alpha)$ so that the corresponding continuation of $S_{n}^{(0)}(\alpha)$, which will be denoted by $S_{a,n}^{(0)}(\alpha)$, would turn out analytic. The linear and quadratic terms in (53) naturally remain as before, so we need only to construct an appropriate continuation $\Upsilon_{0,a}(n,\alpha)$ of the term $\Upsilon_{0}(n,\alpha)$, and then obtain for $\alpha\in\mathbb{R}$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{a,n}^{(0)}(\alpha)=\exp\left\{{\rm i}\frac{k_{F}}{\pi}\alpha L+\left[\frac{1}{6}\left(\frac{1}{n}-n\right)-\frac{\alpha^{2}}{2\pi^{2}n}\right]\ln\mathcal{L}+\Upsilon_{0,a}(n,\alpha)\right\} .\label{eq: Renyi Analytic Continuation} \nonumber \end{align} \tag{ 70 }$$

Regarding the term $\Upsilon_{0}(n,\alpha)$ as it is written in (55), note that as $\alpha$ approaches $\pi^{-}$ or $-\pi^{+}$, a pole of the function $\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)$ approaches the real line. A shift of $\alpha\rightarrow\alpha+2\pi$ maintains the positions of all poles of $\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)$ in the upper half-plane (at $z=\frac{{\rm i}\pi}{n}\left(2m-1-\frac{\alpha}{\pi}\right),\,m\in\mathbb{N}$), but during a continuous shift of such kind the pole that was originally at $z=\frac{{\rm i}\pi}{n}\left(1-\frac{\alpha}{\pi}\right)$ crosses the real line, and ends up at $z=\frac{{\rm i}\pi}{n}\left(-1-\frac{\alpha}{\pi}\right)$. We can now think of $\Upsilon_{0,a}(n,\alpha+2\pi)$ as the value obtained by calculating the integral in (55) while deforming the contour of integration (originally just the real line) so that it also encircles the pole that crossed the real line, thus counting the residue of the integrand at $z=\frac{{\rm i}\pi}{n}\left(-1-\frac{\alpha}{\pi}\right)$ (see figure 1(c)). In such a way we get for every $-\pi<\alpha<\pi$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0,a}(n,\alpha+2\pi)-\Upsilon_{0}(n,\alpha) & ={{\rm Res}}\left\{-n\left[\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)-\tanh\left(\frac{z}{2}\right)\right]\ln\frac{\Gamma\left(\frac{1}{2}+\frac{z}{2\pi {\rm i}}\right)}{\Gamma\left(\frac{1}{2}-\frac{z}{2\pi {\rm i}}\right)},z=\frac{{\rm i}\pi}{n}\left(-1-\frac{\alpha}{\pi}\right)\right\} \nonumber \\ & =2\ln\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(1+\frac{\alpha}{\pi}\right)\right)}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(1+\frac{\alpha}{\pi}\right)\right)}.\label{eq: 2pi shift of upsilon_0} \nonumber \end{align} \tag{ 71 }$$

By the same logic, for every natural number $m\geqslant1$ we can deform the integration contour so that it encircles the m poles which cross the real line from the upper half-plane to the lower half-plane during the shift $\alpha\rightarrow\alpha+2\pi m$, so that for every $-\pi<\alpha<\pi$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle & \Upsilon_{0,a}(n,\alpha+2\pi m)-\Upsilon_{0}(n,\alpha) \nonumber \\ & =\underset{j=1}{\overset{m}{\sum}}{{\rm Res}}\left\{-n\left[\tanh\left(\frac{nz}{2}+{\rm i}\frac{\alpha}{2}\right)-\tanh\left(\frac{z}{2}\right)\right]\ln\frac{\Gamma\left(\frac{1}{2}+\frac{z}{2\pi {\rm i}}\right)}{\Gamma\left(\frac{1}{2}-\frac{z}{2\pi {\rm i}}\right)},z=\frac{{\rm i}\pi}{n}\left(-2j+1-\frac{\alpha}{\pi}\right)\right\} \nonumber \\ & =\underset{j=1}{\overset{m}{\sum}}2\ln\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2j-1+\frac{\alpha}{\pi}\right)\right)}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2j-1+\frac{\alpha}{\pi}\right)\right)}.\label{eq: Analytic upsilon positive shift} \nonumber \end{align} \tag{ 72 }$$

For a shift of $\alpha\rightarrow\alpha-2\pi m$ (this time encircling poles that cross the real line from the lower half-plane to the upper half-plane), we get for every $-\pi<\alpha<\pi$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0,a}(n,\alpha-2\pi m)-\Upsilon_{0}(n,\alpha)=\underset{j=1}{\overset{m}{\sum}}2\ln\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2j-1-\frac{\alpha}{\pi}\right)\right)}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2j-1-\frac{\alpha}{\pi}\right)\right)}.\label{eq: Analytic upsilon negative shift} \nonumber \end{align} \tag{ 73 }$$

For fixed n, the terms $\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2j-1\pm\frac{\alpha}{\pi}\right)\right)$ might diverge for certain values of j  and $\alpha$, in which case (72) or (73) diverge, respectively. This however does not pose a problem, since we are eventually interested in the exponents of (72) and (73), and when $\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2j-1\pm\frac{\alpha}{\pi}\right)\right)$ diverge for some $1\leqslant j\leqslant m$ it just means that $\newcommand{\e}{{\rm e}} \exp\left(\Upsilon_{0,a}(n,\alpha\pm2\pi m)\right)=0$, respectively. Both the periodic and the analytic contintuations of $\newcommand{\e}{{\rm e}} \exp\Upsilon_{0}$ are presented in figure 4.

Zoom In Zoom Out Reset image size

Defining $\Upsilon_{0,a}(n,\alpha)$ this way and substituting it into the analytic continuation of $S_{n}^{(0)}(\alpha)$ in (70), we obtain for each $m\in\mathbb{N}$ and every $-\pi<\alpha<\pi$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{S_{a,n}^{(0)}(\alpha\pm2\pi m)}{S_{n}^{(0)}(\alpha)}=\mathcal{L}^{-\frac{2}{n}\left(m^{2}\pm\frac{\alpha}{\pi}m\right)}{\rm e}^{\pm2{\rm i}mk_{F}L}\underset{j=1}{\overset{m}{\Pi}}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2j-1\pm\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}(2j-1\pm\alpha)\right)^{2}}, \nonumber \end{align} \tag{ 74 }$$

and in particular

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{S_{a,n}^{(0)}(\alpha+2\pi m)}{S_{a,n}^{(0)}\left(\alpha+2\pi(m-1)\right)} & =\mathcal{L}^{-\frac{2}{n}\left(2m-1+\frac{\alpha}{\pi}\right)}{\rm e}^{2{\rm i}k_{F}L}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1+\frac{\alpha}{\pi}\right)\right)^{2}},\nonumber \\ \frac{S_{a,n}^{(0)}(\alpha-2\pi m)}{S_{a,n}^{(0)}\left(\alpha-2\pi(m-1)\right)} & =\mathcal{L}^{-\frac{2}{n}\left(2m-1-\frac{\alpha}{\pi}\right)}{\rm e}^{-2{\rm i}k_{F}L}\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2}}{\Gamma\left(\frac{1}{2}-\frac{1}{2n}\left(2m-1-\frac{\alpha}{\pi}\right)\right)^{2}}.\label{eq: 2 pi shift relations} \nonumber \end{align} \tag{ 75 }$$

Let us now define $\newcommand{\e}{{\rm e}} \sigma_{m}(\alpha)\equiv S_{a,n}^{(0)}(\alpha+2\pi m)$ for every $m\in\mathbb{Z}$ and $-\pi<\alpha<\pi$. We can rewrite (68) as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}(\alpha)=\ln\sigma_{0}(\alpha)+\underset{m=1}{\overset{m_{c}}{\sum}}\left\{\ln\left[1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right]+\ln\left[1+\frac{\sigma_{-m}(\alpha)}{\sigma_{-m+1}(\alpha)}\right]\right\} +o\left(\mathcal{L}^{-1}\right), \nonumber \end{align} \tag{ 76 }$$

and therefore, up to $o\left(\mathcal{L}^{-1}\right)$ corrections,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\left[\underset{m=1}{\overset{m_{c}}{\Pi}}\left(\frac{\sigma_{-m}(\alpha)}{\sigma_{-m+1}(\alpha)}+1\right)\right]\sigma_{0}(\alpha)\left[\underset{m=1}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right)\right].\label{eq: Renyi truncated multiple} \nonumber \end{align} \tag{ 77 }$$

We could have formally represented the result in (68) as an asymptotic (divergent) series had we not defined the cutoff index m_c. Such a representation would have brought us to the asymptotic (divergent) product

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\left[\underset{m=1}{\overset{\infty}{\Pi}}\left(\frac{\sigma_{-m}(\alpha)}{\sigma_{-m+1}(\alpha)}+1\right)\right]\sigma_{0}(\alpha)\left[\underset{m=1}{\overset{\infty}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right)\right], \nonumber \end{align} \tag{ 78 }$$

which for any arbitrary $j\in\mathbb{Z}$ can be written as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\left[\underset{m=1}{\overset{\infty}{\Pi}}\left(\frac{\sigma_{j-m}(\alpha)}{\sigma_{j-m+1}(\alpha)}+1\right)\right]\sigma_{j}(\alpha)\left[\underset{m=1}{\overset{\infty}{\Pi}}\left(1+\frac{\sigma_{j+m}(\alpha)}{\sigma_{j+m-1}(\alpha)}\right)\right]. \nonumber \end{align} \tag{ 79 }$$

This result is just $S_{n}(\alpha+2\pi j)=S_{n}(\alpha)$, as long as we ignore $o\left(\mathcal{L}^{-1}\right)$ corrections and treat it as an asymptotic product.

Note that the result in (77) can also be written as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\underset{j=-m_{c}}{\overset{m_{c}}{\sum}}S_{a,n}^{(0)}(\alpha+2\pi j)+o(1),\label{eq: periodic entropy structure} \nonumber \end{align} \tag{ 80 }$$

a structure which is natural from the CFT perspective. Indeed, there one writes the flux-resolved entropy $S_{n}(\alpha)$ as a correlation function over n copies of space-time of $\mathcal{T}_{\mathcal{V}}=\mathcal{T}\times\mathcal{V}$, twist fields (appearing in the calculation of the total entropies) $\mathcal{T}$ modified by fusion of vertex operators $\mathcal{V}$, which assign a phase $\alpha$ to every particle encircling them [20]. In a bosonized language it can be written in terms of the appropriate boson field $\phi$ as $\mathcal{V}_{0}(\alpha)={\rm e}^{{\rm i}\frac{\alpha}{2\pi}\phi}$. However, the periodicity in $\alpha$ implies that $\mathcal{V}$ could actually be taken as a sum over all possible shifts of $\alpha$ by integer multiples of $2\pi$, that is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{T}_{\mathcal{V}}=\sum_{j}a_{j}(n,\alpha)\mathcal{T}\times\mathcal{V}_{0}(\alpha+2\pi j), \nonumber \end{align} \tag{ 81 }$$

with some coefficients $a_{j}(n,\alpha)$. Computing the entropies as in [20] would then lead to the form of (80). Our exact results allow one to go beyond CFT and find the coefficients for the XX system, which take the values $\newcommand{\e}{{\rm e}} a_{j}(n,\alpha)=\exp\Upsilon_{0,a}(n,\alpha+2\pi j)$.

Interestingly, this structure is maintained even when we include all terms up to an order of $\mathcal{O}\left(\mathcal{L}^{-1}\right)$. Let us define for every $-\pi<\alpha<\pi$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sigma_{\mathrm{right}}(\alpha)\equiv \sigma_{0}(\alpha)\left[\underset{m=2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right)\right]. \nonumber \end{align} \tag{ 82 }$$

First, note that (77) can be also written as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\sigma_{-m_{c}}(\alpha)\left[\underset{m=-m_{c}+1}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right)\right]=\underset{j=-m_{c}}{\overset{m_{c}}{\sum}}\sigma_{j}(\alpha)\underset{m=j+2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right).\label{eq: Renyi rewritten} \nonumber \end{align} \tag{ 83 }$$

By definition of m_c, for any m  >  m_c and every $-\pi<\alpha<\pi$ it is true that $\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}=o\left(\mathcal{L}^{-1}\right)$, and therefore for every $0\leqslant j\leqslant m_{c}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sigma_{j}(\alpha)\underset{m=j+2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right) & =\sigma_{j}(\alpha)\underset{m=2}{\overset{m_{c}-j}{\Pi}}\left(1+\frac{\sigma_{j+m}(\alpha)}{\sigma_{j+m-1}(\alpha)}\right)\nonumber \\ & =\sigma_{j}(\alpha)\underset{m=2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{j+m}(\alpha)}{\sigma_{j+m-1}(\alpha)}\right)+o\left(\mathcal{L}^{-1}\right)\nonumber \\ & =\sigma_{\mathrm{right}}(\alpha+2\pi j)+o\left(\mathcal{L}^{-1}\right).\label{eq: non-negative j's} \nonumber \end{align} \tag{ 84 }$$

It is also evident from the relations in (75) that for $m_{1},m_{2}\geqslant1$ such that $m_{1}+m_{2}>m_{c}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{\sigma_{-m_{1}}(\alpha)}{\sigma_{-m_{1}-1}(\alpha)}\cdot\frac{\sigma_{m_{2}}(\alpha)}{\sigma_{m_{2}-1}(\alpha)}=o\left(\mathcal{L}^{-1}\right). \nonumber \end{align} \tag{ 85 }$$

We can thus conclude that for every $-m_{c}\leqslant j<0$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sigma_{j}(\alpha)\underset{m=j+2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right) & =\sigma_{j}(\alpha)\underset{m=j+2}{\overset{m_{c}+j}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right)+o\left(\mathcal{L}^{-1}\right)\nonumber \\ & =\sigma_{\mathrm{right}}(\alpha+2\pi j)+o\left(\mathcal{L}^{-1}\right).\label{eq: negative j's} \nonumber \end{align} \tag{ 86 }$$

From (83), (84) and (86) we can now derive that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\underset{j=-m_{c}}{\overset{m_{c}}{\sum}}\sigma_{\mathrm{right}}(\alpha+2\pi j)+o\left(\mathcal{L}^{-1}\right). \nonumber \end{align} \tag{ 87 }$$

The symmetry of the expression for $S_{n}(\alpha)$ in (77) obviously enables us to equivalently write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\underset{j=-m_{c}}{\overset{m_{c}}{\sum}}\sigma_{\mathrm{left}}(\alpha+2\pi j)+o\left(\mathcal{L}^{-1}\right), \nonumber \end{align} \tag{ 88 }$$

where we have defined

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sigma_{\mathrm{left}}(\alpha)\equiv \sigma_{0}(\alpha)\left[\underset{m=2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{-m}(\alpha)}{\sigma_{-m+1}(\alpha)}\right)\right]. \nonumber \end{align} \tag{ 89 }$$

This means that we can define

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{S}_{n}(\alpha)\equiv \frac{\sigma_{\mathrm{left}}(\alpha)+\sigma_{\mathrm{right}}(\alpha)}{2}=\frac{\sigma_{0}(\alpha)}{2}\left[\underset{m=2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{-m}(\alpha)}{\sigma_{-m+1}(\alpha)}\right)+\underset{m=2}{\overset{m_{c}}{\Pi}}\left(1+\frac{\sigma_{m}(\alpha)}{\sigma_{m-1}(\alpha)}\right)\right], \nonumber \end{align} \tag{ 90 }$$

and obtain the desired structure, namely

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{n}(\alpha)=\underset{j=-m_{c}}{\overset{m_{c}}{\sum}}\tilde{S}_{n}(\alpha+2\pi j)+o\left(\mathcal{L}^{-1}\right). \nonumber \end{align} \tag{ 91 }$$

We first estimate $S_{n}^{(-)}$ at the limit $L\rightarrow\infty$ assuming that the system is gapped, i.e. $h\neq2$. As was explained in 3.2.2, as $L\rightarrow\infty$ the values $\pm\nu_{m}$ converge in pairs to the values $\lambda_{l}$ defined in (45), which in turn depend on h.

The case h  <  2 is simple: since $\lambda_{0}=0$, we obtain $S_{n}^{(-)}\rightarrow0$. For h  >  2, on the other hand, the asymptotic expression for $S_{n}^{(-)}$ does not vanish. Indeed, we can write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}\underset{m=1}{\overset{L}{\Pi}}\left|\left(\frac{1-\nu_{m}}{2}\right)^{n}-\left(\frac{1+\nu_{m}}{2}\right)^{n}\right|= & \underset{m=-\infty}{\overset{\infty}{\Pi}}\left|\left(\frac{1-\lambda_{m}}{2}\right)^{n}-\left(\frac{1+\lambda_{m}}{2}\right)^{n}\right|,\nonumber \\ \nonumber \end{align} \tag{ 93 }$$

and writing $\newcommand{\e}{{\rm e}} q\equiv {\rm e}^{-\pi\tau_{0}}$ ($\tau_{0}$ was defined in (41)) we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{m=-\infty}{\overset{\infty}{\Pi}}\left|\left(\frac{1-\lambda_{m}}{2}\right)^{n}-\left(\frac{1+\lambda_{m}}{2}\right)^{n}\right|=\underset{m=0}{\overset{\infty}{\Pi}}\left[\left(\frac{1}{1+q^{2m+1}}\right)^{n}-\left(\frac{q^{2m+1}}{1+q^{2m+1}}\right)^{n}\right]^{2}, \nonumber \end{align} \tag{ 94 }$$

so that eventually we obtain

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}\left|S_{n}^{(-)}\right|=\underset{m=0}{\overset{\infty}{\Pi}}\left[\left(\frac{1}{1+q^{2m+1}}\right)^{n}-\left(\frac{q^{2m+1}}{1+q^{2m+1}}\right)^{n}\right]^{2}=\left(\frac{\underset{m=0}{\overset{\infty}{\Pi}}\left[1-q^{n(2m+1)}\right]}{\underset{m=0}{\overset{\infty}{\Pi}}\left[1+q^{(2m+1)}\right]^{n}}\right)^{2}. \nonumber \end{align} \tag{ 95 }$$

In order to further simplify this result for $\underset{L\rightarrow\infty}{\lim}\left|S_{n}^{(-)}\right|$, we remind the reader of the definition of the Jacobi theta functions [49]:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \vartheta_{2}(z,q) & =\underset{m=-\infty}{\overset{\infty}{\sum}}q^{\left(m+\frac{1}{2}\right)^{2}}{\rm e}^{2{\rm i}z\left(m+\frac{1}{2}\right)},\nonumber \\ \vartheta_{3}(z,q) & =\underset{m=-\infty}{\overset{\infty}{\sum}}q^{m^{2}}{\rm e}^{2{\rm i}zm},\nonumber \\ \vartheta_{4}(z,q) & =\underset{m=-\infty}{\overset{\infty}{\sum}}(-1)^{m}q^{m^{2}}{\rm e}^{2{\rm i}zm}. \nonumber \end{align} \tag{ 96 }$$

We write $\newcommand{\e}{{\rm e}} \theta_{j}(q)\equiv \vartheta_{j}(0,q)$ and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle k(q)\equiv \frac{\theta_{2}^{2}(q)}{\theta_{3}^{2}(q)},\,\,\,k'(q)\equiv \sqrt{1-k^{2}(q)}=\frac{\theta_{4}^{2}(q)}{\theta_{3}^{2}(q)}. \nonumber \end{align} \tag{ 97 }$$

This definition of k implies that $\newcommand{\e}{{\rm e}} q=\exp\left[-\pi\frac{I\left(k'\right)}{I(k)}\right]$ (I was defined in (42)) [49], and thus it agrees with the definition of k previously presented in (40). We also write $\newcommand{\e}{{\rm e}} k_{n}(q)\equiv k\left(q^{n}\right)$ and $\newcommand{\e}{{\rm e}} k_{n}'(q)\equiv k'\left(q^{n}\right)$, and rely on the following identities from [49] that hold for every 0  <  q  <  1:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{m=0}{\overset{\infty}{\Pi}}\left[1+q^{(2m+1)}\right] & =\left(\frac{16q}{k^{2}k'^{2}}\right)^{\frac{1}{24}},\nonumber \\ \underset{m=0}{\overset{\infty}{\Pi}}\left[1-q^{(2m+1)}\right] & =k'^{\frac{1}{4}}\underset{m=0}{\overset{\infty}{\Pi}}\left[1+q^{(2m+1)}\right]=\left(\frac{16qk'^{4}}{k^{2}}\right)^{\frac{1}{24}}. \nonumber \end{align} \tag{ 98 }$$

We then obtain that for h  >  2,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}\left|S_{n}^{(-)}\right|=\left[\frac{\left(kk'\right)^{2n}k_{n}'^{4}}{16^{n-1}k_{n}^{2}}\right]^{\frac{1}{12}}. \nonumber \end{align} \tag{ 99 }$$

Since $S_{n}^{(-)}$ is real by definition we can only have $S_{n}^{(-)}=\pm\left|S_{n}^{(-)}\right|$, but this still leaves us with an ambiguity regarding the sign of $S_{n}^{(-)}$. To resolve this ambiguity we turn to the large h limit of the above expression. The definition of k in (40) implies that as $h\rightarrow\infty$, $k\rightarrow0$ and therefore $k'\rightarrow1$ and $q\rightarrow0$. Furthermore, one can show that as $q\rightarrow0$, $k\sim4q^{\frac{1}{2}}$ [49] and consequently

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{h\rightarrow\infty}{\lim}\left[\frac{\left(kk'\right)^{2n}k_{n}'^{4}}{16^{n-1}k_{n}^{2}}\right]^{\frac{1}{12}}=1. \nonumber \end{align} \tag{ 100 }$$

On the other hand, as can be easily seen from the Hamiltonian in (14), in the large h limit the system in question is ferromagnetic, and we therefore expect that as $h\rightarrow\infty$ all L fermion sites of subsystem A will be occupied in the ground state (i.e. $\rho_{A}$ has a non-vanishing eigenvalue only for the state that corresponds to Q_A  =  L). This, in turn, suggests that for every finite L, as $h\rightarrow\infty$ we obtain $S_{n}^{(-)}\rightarrow1$ for even L and $S_{n}^{(-)}\rightarrow-1$ for odd L. By continuity, the sign should remain the same for finite h  >  2.

This finally brings us to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}(-1)^{L}S_{n}^{(-)}=\left\{\begin{array}{@{}ll@{}} 0, & h<2 \nonumber \\ \left[\frac{\left(kk'\right)^{2n}k_{n}'^{4}}{16^{n-1}k_{n}^{2}}\right]^{\frac{1}{12}}, & h>2 \end{array}\right..\label{eq: XY Renyi symmetry block} \nonumber \end{align} \tag{ 101 }$$

Figure 5(a) shows a comparison between the asymptotic analytical result for $S_{n}^{(-)}$ and the numerical result. It indicates a very good agreement between the two calculations, and in particular confirms two conspicuous properties of the analytical result in the large L limit: that $S_{n}^{(-)}\rightarrow0$ in the h  <  2 regime, and that $\left|S_{n}^{(-)}\right|\rightarrow1$ as $h\rightarrow\infty$. A numerical calculation of $S_{2}^{(-)}$ for several values of L has previously appeared in [37].

Zoom In Zoom Out Reset image size

We use our calculation of $S_{n}^{(-)}$ in order to calculate $\mathcal{S}^{(-)}=-\underset{n\rightarrow1}{\lim}\partial_{n}S_{n}^{(-)}$ at the large L limit. Relying on (92), we can obtain an explicit expression for $\mathcal{S}^{(-)}$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{S}^{(-)}=(-1)^{L}\underset{m=1}{\overset{L}{\sum}}\left(\underset{j\neq m}{\Pi}\nu_{j}\right)\cdot\left[\frac{1-\nu_{m}}{2}\ln\left(\frac{1-\nu_{m}}{2}\right)-\frac{1+\nu_{m}}{2}\ln\left(\frac{1+\nu_{m}}{2}\right)\right].\label{eq: XY vNEE explicit} \nonumber \end{align} \tag{ 102 }$$

This expression can be used for numerical estimates of ${{\mathcal S}}^{(-)}$.

From (101) we can now calculate $\mathcal{S}^{(-)}$ as $L\rightarrow\infty$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}(-1)^{L}\mathcal{S}^{(-)}=\left\{\begin{array}{@{}ll@{}} 0, & h<2 \nonumber \\ \frac{\sqrt{k'}}{3}\left[\ln2-\frac{1}{2}\ln\left(k\cdot k'\right)-\frac{I(k)I\left(k'\right)}{\pi}\left(1+k^{2}\right)\right], & h>2 \end{array}\right..\label{eq: XY vNEE} \nonumber \end{align} \tag{ 103 }$$

The details of this calculation appear in appendix A.4. ${{\mathcal S}}^{(-)}$ is plotted in figure 5(b), where again good agreement between the analytical estimate and the numerical result is evident. Figures 6(a) and (b) show the difference between the analytical limit for $L\rightarrow\infty$ and the numerical results for finite L. They demonstrate that away from the vicinity of h  =  2, where the phase transition occurs, corrections to the asymptotic result vanish rapidly as L grows, and it is apparent that e.g. for $\gamma=0.5$ these corrections turn negligible even for a relatively short subsystem. As h nears h  =  2, we need a larger value of L in order for the deviation to be small.

Zoom In Zoom Out Reset image size

Both $S_{n}^{(-)}$ and ${{\mathcal S}}^{(-)}$ illustrate a striking property of the phase in which the system is found for h  <  2: since $S_{n}^{(-)}={{\mathcal S}}^{(-)}=0$, we obtain that for h  <  2, the system satisfies $S_{n}^{\left(\mathrm{even}\right)}=S_{n}^{\left(\mathrm{odd}\right)}$ and ${{\mathcal S}}^{\left(\mathrm{even}\right)}={{\mathcal S}}^{\left(\mathrm{odd}\right)}$. This property stems from the fact that we can write the RDM as $\newcommand{\e}{{\rm e}} \rho_{A}=\exp\left(-H_{A}\right)$ where the entanglement Hamiltonian H_A is quadratic [38, 48], and treat H_A as the Hamiltonian of an effective system of a 1D open fermionic chain with L sites. H_A is expected to have the same modes at the virtual edges of the subsystem as the original system (the Kitaev chain) would host at a physical edge [29]. Thus the phase h  <  2 corresponds to a topologically non-trivial phase of H_A where two Majorana zero-modes — one at each end of the system — remain decoupled, provided that the virtual chain is long enough [53]. Combining these two Majorana operators yields a fermionic operator whose occupancy does not change the eigenvalues of H_A, and thus induces a two-fold degeneracy in the system: every eigenstate of H_A with an even total fermionic number has a corresponding eigenstate with the same eignevalue but with an odd total fermionic number, and vice versa. This degeneracy persists as long as h  <  2. This explains why in the large L limit, the contributions to the entropy from the block that corresponds to an even Q_A and the block that corresponds to an odd Q_A are exactly the same. Our work provides a rigorous proof of this behavior for the system considered. These observations allow us to explain the finite L corrections to our results, which become noticeable for $L\lesssim 50$, as depicted in figures 6(a) and (b).

Since for $h\neq2$ the system is gapped, the correlations vanish exponentially as $L\rightarrow\infty$ [54], and therefore so do the corrections to the limiting values of $S_{n}^{(-)}$ and $\mathcal{S}^{(-)}$. For h  <  2 the corrections are dominated by the hybridization of the entanglement Majorana edge-modes: though they are localized exponentially at the ends of the virtual chain [53], for finite L the virtual edge Majorana fermions exhibit some overlap, and therefore a true degeneracy is not achieved [53] for most values of h  <  2, resulting in a finite nonzero value of the lowest eigenvalue $\newcommand{\e}{{\rm e}} \left|\nu_{1}\right|\equiv \min\left|\nu_{m}\right|$. Yet for certain values of h  <  2 the virtual Majorana wave functions interfere destructively, and this creates the minima apparent in figures 6(a) through (d) in both $\left|\mathcal{S}^{(-)}\right|$ and $\left|\nu_{1}\right|$. This in fact suggests that the finite size corrections to $\mathcal{S}^{(-)}$ are dominated by $\nu_{1}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \left|{{\mathcal S}}^{(-)}\right|\approx\left|\frac{1-\nu_{1}}{2}\ln\left(\frac{1-\nu_{1}}{2}\right)-\frac{1+\nu_{1}}{2}\ln\left(\frac{1+\nu_{1}}{2}\right)\right|. \nonumber \end{align} \tag{ 104 }$$

The accuracy of this relation can serve as a quantitative test for the above arguments. And indeed, calculating the ratio between this approximation and $\left|{{\mathcal S}}^{(-)}\right|$ for the cases that appear in figures 6(a) and (b), we get that for h  <  1.9 (outside the vicinity of the critical point h  =  2), the contribution of $\nu_{1}$ to ${{\mathcal S}}^{(-)}$ is always above $85\%$ for $\gamma=0.5$, and always above $65\%$ for $\gamma=0.1$.

Considerations similar to those detailed above allow the extension of our main results (101) and (103) to $0>h\neq-2$. The limit $\underset{L\rightarrow\infty}{\lim}\left|S_{n}^{(-)}\right|$ is symmetric in h, and therefore, in particular, it tends to 1 as $h\rightarrow-\infty$. However, the sign ambiguity is resolved in a different way than in the h  >  0 case: for finite L we expect that in the $h\rightarrow-\infty$ limit, all sites of A become unoccupied such that Q_A  =  0 with probability 1. We thus obtain that as $h\rightarrow-\infty$, $S_{n}^{(-)}\rightarrow1$ both for even and odd L, and so for $0>h\neq-2$ the limit $\underset{L\rightarrow\infty}{\lim}S_{n}^{(-)}$ exists and is also positive.

The extensions of (101) and (103) to $0>h\neq-2$ are therefore symmetric, apart from the absence of the $(-1){}^{L}$ prefix, namely

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}S_{n}^{(-)}\Big\rvert_{h}=\underset{L\rightarrow\infty}{\lim}(-1)^{L}S_{n}^{(-)}\Big\rvert_{-h}\,\,\,\left(0>h\neq-2\right), \nonumber \end{align} \tag{ 105 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{L\rightarrow\infty}{\lim}{{\mathcal S}}^{(-)}\Big\rvert_{h}=\underset{L\rightarrow\infty}{\lim}(-1)^{L}{{\mathcal S}}^{(-)}\Big\rvert_{-h}\,\,\,\left(0>h\neq-2\right). \nonumber \end{align} \tag{ 106 }$$

Here we calculate the asymptotics of $S_{n}^{(\pm)}$ for the case where the system is gapless, i.e. h  =  2. Following [45] we write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln\left|S_{n}^{(\pm)}\right|=\mathrm{Re}\left\{\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{4\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\pm)}(1+\varepsilon,\lambda)\frac{\rm d}{{\rm d}\lambda}\ln\tilde{D}_{L}(\lambda){\rm d}\lambda\right\} , \nonumber \end{align} \tag{ 107 }$$

where $c(\varepsilon,\delta)$ is the contour shown in figure 1(a), and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle e_{n}^{(\pm)}(x,\nu)\equiv \ln\left[\left(\frac{x-\nu}{2}\right)^{n}\pm\left(\frac{x+\nu}{2}\right)^{n}\right]. \nonumber \end{align} \tag{ 108 }$$

Using the asymptotic approximation for $\tilde{D}_{L}(\lambda)$ in (46), we obtain that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{\rm d}{{\rm d}\lambda}\ln\tilde{D}_{L}(\lambda)\sim\left(\frac{1}{\lambda+1}+\frac{1}{\lambda-1}\right)L-\frac{4{\rm i}}{\pi}\cdot\frac{\beta(\lambda)}{(\lambda+1)(\lambda-1)}\ln L. \nonumber \end{align} \tag{ 109 }$$

This expression is reminiscent of (50) from the calculation of the REE in the gapless XX model, and we can therefore carry out the integration along the same lines of argument, so that eventually we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln\left|S_{n}^{(\pm)}\right|\sim\left[\frac{1}{12}\left(\frac{1}{n}-n\right)-\frac{\eta_{\pm}}{4n}\right]\ln L, \nonumber \end{align} \tag{ 110 }$$

where $\newcommand{\e}{{\rm e}} \eta_{+}=0$ and $\newcommand{\e}{{\rm e}} \eta_{-}=1$. Since $S_{n}^{(+)}=\left|S_{n}^{(+)}\right|$, we have in particular obtained that for the gapless XY model

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}^{(+)}\sim\frac{1}{12}\left(\frac{1}{n}-n\right)\ln L, \nonumber \end{align} \tag{ 111 }$$

a special case of a result which was derived and verified numerically in [50]. The coefficient of the logarithm is halved as compared to (53) with $\alpha=0$, since a Majorana mode rather than a complex fermion is gapless here, in accordance with CFT predictions [4].

$S_{n}^{(-)}$ is again determined only up to a sign,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \left|S_{n}^{(-)}\right|\approx A(n,\gamma)L^{-\frac{1}{6n}-\frac{n}{12}}, \nonumber \end{align} \tag{ 112 }$$

where $A(n,\gamma)$ is some positive factor independent of L, assuming that the largest subleading contribution not appearing in the approximation (46) does not depend on L. We have verified this assumption numerically by fitting to the results a function of L that scales as L^{−1/6n−n/12}, while the proportionality constant remained a free parameter. An example is shown in figure 7(a), where good agreement between numerical and analytical results is evident. Lead by similar considerations as in the case of the gapped XY model, we can determine the sign of $S_{n}^{(-)}$ to be $(-1){}^{L}$, so that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}S_{n}^{(-)}\approx A(n,\gamma)L^{-\frac{1}{6n}-\frac{n}{12}}.\label{eq: Ungapped XY parity-resolved Renyi} \nonumber \end{align} \tag{ 113 }$$

Zoom In Zoom Out Reset image size

Specific attributes of the factor $A(n,\gamma)$ are generally not captured by known theorems or conjectures we are aware of, and its analysis is beyond the scope of this work. We show its typical behavior, as extracted from numerical results, in figure 7(b). The result (113) confirms a previous prediction based on CFT considerations [20].

The parity-resolved vNEE $\mathcal{S}^{(-)}$ for h  =  2 is therefore

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}\mathcal{S}^{(-)}\approx-\frac{A(1,\gamma)}{12}L^{-\frac{1}{4}}\ln L. \nonumber \end{align} \tag{ 114 }$$

As before, we can extend the results for $S_{n}^{(-)}$ and $\mathcal{S}^{(-)}$ to h  =  −2 by simply omitting the $(-1){}^{L}$ prefix.

We derive here the leading order asymptotic approximation for the determinant $\tilde{D}_{L}(\lambda)$ defined in (43), for the case of a gapless XY chain (h  =  2). A generalized version of the Fisher–Hartwig formula was conjectured in [50, 51], regarding the determinant of a block Toeplitz matrix of the form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T_{L}\left[\mathcal{M}\right]=\left(\begin{array}{@{}cccc@{}} \tilde{\Pi}_{0} & \tilde{\Pi}_{-1} & \cdots & \tilde{\Pi}_{1-L} \nonumber \\ \tilde{\Pi}_{1} & \tilde{\Pi}_{0} & & \vdots \nonumber \\ \vdots & & \ddots & \vdots \nonumber \\ \tilde{\Pi}_{L-1} & \cdots & \cdots & \tilde{\Pi}_{0} \end{array}\right),\,\,\,\tilde{\Pi}_{m}\equiv \frac{1}{2\pi}\underset{0}{\overset{2\pi}{\int}}{\rm d}\theta {\rm e}^{-{\rm i}m\theta}\mathcal{M}(\theta), \nonumber \end{align} \tag{ A.1 }$$

where $\mathcal{M}(\theta)$ is a piecewise continuous $d\times d$ matrix with jump discontinuities at the points $\theta_{r}$, $r=0,\ldots,R$. We define for each discontinuity $\newcommand{\e}{{\rm e}} \mathcal{M}_{r}^{\pm}\equiv \underset{\theta\rightarrow\theta_{r}^{\pm}}{\lim}\mathcal{M}(\theta)$, and assume that for each r, $\mathcal{M}_{r}^{+}$ and $\mathcal{M}_{r}^{-}$ commute. This allows us to find a joint diagonalizing basis for $\mathcal{M}_{r}^{\pm}$, and we denote the corresponding eigenvalues by $\mu_{r,j}^{\pm}$, $j=1,\ldots,d$. According to the conjecture [51], for the first two leading terms of the large L approximation of $\ln\det T_{L}\left[\mathcal{M}\right]$ we then have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln\det T_{L}\left[\mathcal{M}\right]=\frac{L}{2\pi}\underset{0}{\overset{2\pi}{\int}}{\rm d}\theta\ln\left(\det\mathcal{M}(\theta)\right)+\frac{\ln L}{4\pi^{2}}\underset{r=0}{\overset{R}{\sum}}\underset{j=1}{\overset{\rm d}{\sum}}\left(\ln\left(\frac{\mu_{r,j}^{-}}{\mu_{r,j}^{+}}\right)\right)^{2}+\cdots. \nonumber \end{align} \tag{ A.2 }$$

For h  =  2, $\tilde{D}_{L}(\lambda)$ is of the form described above, i.e. $\tilde{D}_{L}(\lambda)=\det T_{L}\left[\mathcal{M}\right]$ for $\mathcal{M}(\theta)={\rm i}\lambda I_{2}-\mathcal{G}(\theta)$. Now $\mathcal{M}$ has a single discontinuity at $\theta_{0}=0$, with

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \mathcal{M}_{0}^{\pm}=\left(\begin{array}{@{}cc@{}} {\rm i}\lambda & \pm {\rm i} \nonumber \\ \pm {\rm i} & {\rm i}\lambda \end{array}\right), \nonumber \end{align} \tag{ A.3 }$$

from which we obtain that $\mu_{0,1}^{\pm}={\rm i}\lambda\pm {\rm i}$ and $\mu_{0,2}^{\pm}={\rm i}\lambda\mp {\rm i}$. Since $\det\mathcal{M}(\theta)=1-\lambda^{2}$ is independent of $\theta$, we finally arrive at

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln\tilde{D}_{L}(\lambda)=\ln\left(1-\lambda^{2}\right)L-2\beta^{2}(\lambda)\ln L+\cdots. \nonumber \end{align} \tag{ A.4 }$$

From the Fisher–Hartwig conjecture we have derived the leading order approximation for the asymptotic expression for $S_{n}(\alpha)$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}^{(0)}(\alpha) & ={\rm i}\frac{\alpha}{2}L+\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\left(\frac{k_{F}/\pi}{\lambda-1}+\frac{1-k_{F}/\pi}{\lambda+1}\right)L{\rm d}\lambda\nonumber \\ & +\underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\left(-\frac{4{\rm i}}{\pi}\cdot\frac{\beta(\lambda)}{(\lambda+1)(\lambda-1)}\left[\ln\mathcal{L}+\left(1+\gamma_{E}\right)+\Upsilon(\lambda)\right]\right){\rm d}\lambda. \nonumber \end{align} \tag{ A.5 }$$

Regarding the first integral, it is easily shown that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\left(\frac{k_{F}/\pi}{\lambda-1}+\frac{1-k_{F}/\pi}{\lambda+1}\right)L{\rm d}\lambda={\rm i}\left(-1+\frac{2k_{F}}{\pi}\right)\frac{\alpha}{2}L. \nonumber \end{align} \tag{ A.6 }$$

As for the second integral, we use the fact that for every  −1  <  x  <  1,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \beta\left(x+{\rm i}0^{\pm}\right)=-{\rm i}W(x)\mp\frac{1}{2}, \nonumber \end{align} \tag{ A.7 }$$

where $\newcommand{\e}{{\rm e}} W(x)\equiv \frac{1}{2\pi}\ln\frac{1+x}{1-x}$. It can be shown that the contribution from the circular arcs of the contour $c(\varepsilon,\delta)$ vanishes as $\varepsilon,\delta\rightarrow0^{+},$ and therefore we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle & \underset{\varepsilon,\delta\rightarrow0^{+}}{\lim}\frac{1}{2\pi {\rm i}}\underset{c(\varepsilon,\delta)}{\int}e_{n}^{(\alpha)}(1+\varepsilon,\lambda)\left(-\frac{4{\rm i}}{\pi}\cdot\frac{\beta(\lambda)}{(\lambda+1)(\lambda-1)}\left[\ln\mathcal{L}+\left(1+\gamma_{E}\right)+\Upsilon(\lambda)\right]\right){\rm d}\lambda\nonumber \\ & =\frac{\left(\ln\mathcal{L}+\left(1+\gamma_{E}\right)\right)}{\pi^{2}}\underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\frac{2e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x\nonumber \\ & +\underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{k=1}{\overset{\infty}{\sum}}\frac{1}{\pi^{2}k}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\left[\frac{\left(\frac{1}{2}-{\rm i}W(x)\right)^{3}}{k^{2}-\left(\frac{1}{2}-{\rm i}W(x)\right)^{2}}+\frac{\left(\frac{1}{2}+{\rm i}W(x)\right)^{3}}{k^{2}-\left(\frac{1}{2}+{\rm i}W(x)\right)^{2}}\right]\frac{2e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x\nonumber \\ & =\frac{\ln\mathcal{L}}{\pi^{2}}\underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\frac{2e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x-\frac{1}{\pi^{2}}\underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\left[\psi\left(\frac{1}{2}+{\rm i}W(x)\right)+\psi\left(\frac{1}{2}-{\rm i}W(x)\right)\right]\frac{e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x. \nonumber \end{align} \tag{ A.8 }$$

Here we denoted by $\psi(x)$ the Digamma function, $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$, and used the identity [44]

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{k=1}{\overset{\infty}{\sum}}\frac{1}{k}\left[\frac{\left(\frac{1}{2}+{\rm i}w\right)^{3}}{k^{2}-\left(\frac{1}{2}+{\rm i}w\right)^{2}}+\frac{\left(\frac{1}{2}-{\rm i}w\right)^{3}}{k^{2}-\left(\frac{1}{2}-{\rm i}w\right)^{2}}\right]=-1-\gamma_{E}-\frac{1}{2}\left[\psi\left(\frac{1}{2}+{\rm i}w\right)+\psi\left(\frac{1}{2}-{\rm i}w\right)\right]. \nonumber \end{align} \tag{ A.9 }$$

Using a change of variables $u=\ln\frac{1+x}{1-x}$, and taking the limit $\varepsilon\rightarrow0^{+}$, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\frac{2e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x & =-n\underset{-\infty}{\overset{\infty}{\int}}\frac{u}{{\rm e}^{u}+1}\cdot\frac{{\rm e}^{nu+{\rm i}\frac{\alpha}{2}}-{\rm e}^{u-{\rm i}\frac{\alpha}{2}}}{{\rm e}^{nu+{\rm i}\frac{\alpha}{2}}+{\rm e}^{-{\rm i}\frac{\alpha}{2}}}{\rm d}u\nonumber \\ & =-\frac{\pi^{2}}{6}n+\frac{2}{n}\underset{k=1}{\overset{\infty}{\sum}}\frac{(-1)^{k+1}}{k^{2}}\cos(\alpha k).\nonumber \\ \nonumber \end{align} \tag{ A.10 }$$

Recalling the Fourier series of $\alpha^{2}$ in $(-\pi,\pi)$, we can now write for every $-\pi<\alpha<\pi$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{\ln\mathcal{L}}{\pi^{2}}\underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\frac{2e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x=\left[\frac{1}{6}\left(\frac{1}{n}-n\right)-\frac{\alpha^{2}}{2\pi^{2}n}\right]\ln\mathcal{L}. \nonumber \end{align} \tag{ A.11 }$$

Changing variables and taking the limit $\varepsilon\rightarrow0^{+}$ as before, the second part of the integral turns out to be

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle & -\frac{1}{\pi^{2}}\underset{\varepsilon\rightarrow0^{+}}{\lim}\underset{-1+\frac{\varepsilon}{2}}{\overset{1-\frac{\varepsilon}{2}}{\int}}\left[\psi\left(\frac{1}{2}+{\rm i}W(x)\right)+\psi\left(\frac{1}{2}-{\rm i}W(x)\right)\right]\frac{e_{n}^{(\alpha)}(1+\varepsilon,x)}{1-x^{2}}{\rm d}x\nonumber \\ & =-\frac{{\rm i}}{\pi}\underset{-\infty}{\overset{\infty}{\int}}\ln\left[\frac{2\cosh\left(\frac{nu}{2}+{\rm i}\frac{\alpha}{2}\right)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{n}}\right]\frac{\rm d}{{\rm d}u}\ln\frac{\Gamma\left(\frac{1}{2}+\frac{u}{2\pi {\rm i}}\right)}{\Gamma\left(\frac{1}{2}-\frac{u}{2\pi {\rm i}}\right)}{\rm d}u\nonumber \\ & =-\frac{1}{\pi^{2}}\underset{0}{\overset{\infty}{\int}}\ln\left[\frac{2\cos\alpha+2\cosh(nu)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2n}}\right]{\rm d}u\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{\cos\left(\frac{ut}{2\pi}\right)}{2\sinh\left(\frac{t}{2}\right)}\right]{\rm d}t. \nonumber \end{align} \tag{ A.12 }$$

Finally, we arrive at

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle \ln S_{n}^{(0)}(\alpha)={\rm i}\frac{k_{F}}{\pi}\alpha L+\left[\frac{1}{6}\left(\frac{1}{n}-n\right)-\frac{\alpha^{2}}{2\pi^{2}n}\right]\ln\mathcal{L}+\Upsilon_{0}(n,\alpha), \nonumber \end{align*}$$

where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0}(n,\alpha)\equiv -\frac{1}{\pi^{2}}\underset{0}{\overset{\infty}{\int}}\ln\left[\frac{2\cos\alpha+2\cosh(nu)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2n}}\right]{\rm d}u\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{\cos\left(\frac{ut}{2\pi}\right)}{2\sinh\left(\frac{t}{2}\right)}\right]{\rm d}t. \nonumber \end{align} \tag{ A.13 }$$

We write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Upsilon_{0}(n,\alpha)=-\frac{1}{\pi^{2}}\underset{0}{\overset{\infty}{\int}}\ln\left[\frac{2\cos\alpha+2\cosh(nu)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2n}}\right]{\rm d}u\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{\cos\left(\frac{ut}{2\pi}\right)}{2\sinh\left(\frac{t}{2}\right)}\right]{\rm d}t=c_{0}(n)+c_{2}(n)\alpha^{2}+\mathcal{O}\left(\alpha^{4}\right), \nonumber \end{align} \tag{ A.14 }$$

and prove that $c_{2}(1)=-\frac{1+\gamma_{E}}{2\pi^{2}}$. Indeed, substituting n  =  1,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \ln\left[\frac{2\cos\alpha+2\cosh(u)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2}}\right]=\ln\left[\frac{2+2\cosh(u)}{\left(2\cosh\left(\frac{u}{2}\right)\right)^{2}}\right]-\frac{1}{4\cosh^{2}\left(\frac{u}{2}\right)}\alpha^{2}+\mathcal{O}\left(\alpha^{4}\right), \nonumber \end{align} \tag{ A.15 }$$

and so

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle c_{2}(1)=\frac{1}{4\pi^{2}}\underset{0}{\overset{\infty}{\int}}\frac{1}{\cosh^{2}\left(\frac{u}{2}\right)}{\rm d}u\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{\cos\left(\frac{ut}{2\pi}\right)}{2\sinh\left(\frac{t}{2}\right)}\right]{\rm d}t. \nonumber \end{align} \tag{ A.16 }$$

We use

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{0}{\overset{\infty}{\int}}\frac{\cos\left(\frac{ut}{2\pi}\right)}{\cosh^{2}\left(\frac{u}{2}\right)}{\rm d}u=\underset{-\infty}{\overset{\infty}{\int}}\frac{{\rm e}^{{\rm i}\frac{t}{\pi}x}}{\cosh^{2}(x)}{\rm d}x, \nonumber \end{align} \tag{ A.17 }$$

where the complex integral can be calculated using a rectangular contour with infinite horizontal sides at $\Im z=0$ and $\Im z={\rm i}\pi$, so that we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \underset{0}{\overset{\infty}{\int}}\frac{\cos\left(\frac{ut}{2\pi}\right)}{\cosh^{2}\left(\frac{u}{2}\right)}{\rm d}u=\frac{2t{\rm e}^{-\frac{t}{2}}}{1-{\rm e}^{-t}}. \nonumber \end{align} \tag{ A.18 }$$

We can therefore write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle c_{2}(1) & =\frac{1}{2\pi^{2}}\underset{0}{\overset{\infty}{\int}}\left[\frac{{\rm e}^{-t}}{t}-\frac{t{\rm e}^{-\frac{t}{2}}}{\left(1-{\rm e}^{-t}\right)\left({\rm e}^{\frac{t}{2}}-{\rm e}^{-\frac{t}{2}}\right)}\right]{\rm d}t\nonumber \\ & =\frac{1}{2\pi^{2}}\underset{0}{\overset{\infty}{\int}}\left[\frac{1-{\rm e}^{-t}-t}{t\left({\rm e}^{t}-1\right)}+\frac{1}{{\rm e}^{t}-1}-\frac{t{\rm e}^{t}}{\left({\rm e}^{t}-1\right)^{2}}\right]{\rm d}t\nonumber \\ & =\frac{-\gamma_{E}}{2\pi^{2}}+\frac{1}{2\pi^{2}}\underset{0}{\overset{\infty}{\int}}\frac{\rm d}{{\rm d}t}\left(\frac{t}{{\rm e}^{t}-1}\right){\rm d}t\nonumber \\ & =-\frac{\gamma_{E}+1}{2\pi^{2}},\nonumber \\ \nonumber \end{align} \tag{ A.19 }$$

where we have used the identity $\gamma_{E}=\underset{0}{\overset{\infty}{\int}}\frac{{\rm e}^{-t}+t-1}{t\left({\rm e}^{t}-1\right)}{\rm d}t$ [49].

We present here a detailed calculation of $\mathcal{S}^{(-)}=-\underset{n\rightarrow1}{\lim}\partial_{n}S_{n}^{(-)}$ as $L\rightarrow\infty$, based on the result for $S_{n}^{(-)}$ in (101). For h  <  2 we obviously have ${{\mathcal S}}^{(-)}\rightarrow0$. For h  >  2, we can calculate the derivative of the expression for $S_{n}^{(-)}$ by rewriting it in terms of the Jacobi theta functions:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}S_{n}^{(-)}\rightarrow\left[\frac{\left(kk'\right)^{2n}k_{n}'^{4}}{16^{n-1}k_{n}^{2}}\right]^{\frac{1}{12}}=\left[\frac{\theta_{2}^{4n}(q)\theta_{4}^{4n}(q)\theta_{4}^{8}\left(q^{n}\right)}{16^{n-1}\theta_{3}^{8n}(q)\theta_{2}^{4}\left(q^{n}\right)\theta_{3}^{4}\left(q^{n}\right)}\right]^{\frac{1}{12}}. \nonumber \end{align} \tag{ A.20 }$$

After some elementary steps, we arrive at

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}\mathcal{S}^{(-)}\rightarrow\frac{\sqrt{k'}}{3}\left[\ln2-\frac{1}{2}\ln\left(k\cdot k'\right)+q\ln q\cdot\left(\frac{\theta_{3}'(q)}{\theta_{3}(q)}+\frac{\theta_{2}'(q)}{\theta_{2}(q)}-\frac{2\theta_{4}'(q)}{\theta_{4}(q)}\right)\right], \nonumber \end{align} \tag{ A.21 }$$

where $\newcommand{\e}{{\rm e}} \theta_{j}'(q)\equiv \frac{\rm d}{{\rm d}q}\theta_{j}(q)$. For further simplification, we use the fact that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{\theta_{3}'(q)}{\theta_{3}(q)}+\frac{\theta_{2}'(q)}{\theta_{2}(q)}-\frac{2\theta_{4}'(q)}{\theta_{4}(q)}=\frac{\rm d}{{\rm d}q}\ln\left(\frac{\theta_{2}\theta_{3}}{\theta_{4}^{2}}\right)=\frac{\rm d}{{\rm d}q}\ln\left(\frac{k^{\frac{1}{2}}}{k'}\right), \nonumber \end{align} \tag{ A.22 }$$

along with the identity [49]

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{k^{\frac{1}{2}}}{k'}=2q^{\frac{1}{4}}\underset{m=1}{\overset{\infty}{\Pi}}\left(1+q^{m}\right)^{6}, \nonumber \end{align} \tag{ A.23 }$$

in order to obtain that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle q\left(\frac{\theta_{3}'(q)}{\theta_{3}(q)}+\frac{\theta_{2}'(q)}{\theta_{2}(q)}-\frac{2\theta_{4}'(q)}{\theta_{4}(q)}\right)=\frac{1}{4}+6\underset{m=1}{\overset{\infty}{\sum}}\frac{mq^{m}}{1+q^{m}}. \nonumber \end{align} \tag{ A.24 }$$

To calculate the sum of the remaining series, we use [65]

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \theta_{3}^{4}(q)=1+8\underset{m=1}{\overset{\infty}{\sum}}\frac{mq^{m}}{1+(-q)^{m}}, \nonumber \end{align} \tag{ A.25 }$$

and also $\theta_{4}(q)=\theta_{3}(-q)$ and $\theta_{2}^{4}+\theta_{4}^{4}=\theta_{3}^{4}$, in order to arrive at

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{1}{24}\left(\theta_{3}^{4}+\theta_{2}^{4}-1\right) & =\frac{1}{24}\left(2\theta_{3}^{4}-\theta_{4}^{4}-1\right)\nonumber \\ & =\frac{1}{3}\underset{m=1}{\overset{\infty}{\sum}}\left[\frac{2mq^{m}}{1+(-q)^{m}}-\frac{m(-q)^{m}}{1+q^{m}}\right]\nonumber \\ & =\underset{m=1}{\overset{\infty}{\sum}}\frac{mq^{m}}{1+q^{m}}. \nonumber \end{align} \tag{ A.26 }$$

Additionally, we note that the following identity holds [43]:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle I(k)=\frac{\pi}{2}\theta_{3}^{2}(q). \nonumber \end{align} \tag{ A.27 }$$

We can therefore write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \frac{1}{4}+6\underset{m=1}{\overset{\infty}{\sum}}\frac{mq^{m}}{1+q^{m}}=\frac{1}{4}\left(\theta_{3}^{4}(q)+\theta_{2}^{4}(q)\right)=\frac{I^{2}(k)}{\pi^{2}}\left(1+k^{2}\right), \nonumber \end{align} \tag{ A.28 }$$

and consequently we obtain for h  >  2 that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (-1)^{L}\mathcal{S}^{(-)}\rightarrow\frac{\sqrt{k'}}{3}\left[\ln2-\frac{1}{2}\ln\left(k\cdot k'\right)-\frac{I(k)I\left(k'\right)}{\pi}\left(1+k^{2}\right)\right]. \nonumber \end{align} \tag{ A.29 }$$