More on symmetry resolved operator entanglement

The entanglement properties between a region A and its complementary B are usually defined starting from a state $\left| \psi \right|$. Nevertheless, one can also study the entanglement properties of an operator O by vectorizing it. Namely, if the operator O lives in $\mathrm{End}(\mathcal{H}_A)\otimes \mathrm{End}(\mathcal{H}_B)$, we can view it equivalently as a vector in $(\mathcal{H}_A\otimes\mathcal{H}_B) \otimes (\mathcal{\bar{H}}_A\otimes\mathcal{\bar{H}}_B)$, where $\mathcal{H}_{A,B}$ denotes the Hilbert space in part A or B, and $\mathcal{\bar{H}}_{A,B}$ its dual. To work with a properly normalised state after the vectorization, we divide the operator O by a normalisation factor $\sqrt{\mathrm{Tr}(O^{\dagger}O)}$. Operationally, one can pick an orthonormal basis $\{\vert i\rangle \}$ for $(\mathcal{H}_A\otimes\mathcal{H}_B)$ and the corresponding basis $\{\vert j\rangle \}$ for its dual $(\mathcal{\bar{H}}_A\otimes\mathcal{\bar{H}}_B)$, and write $O = \sum_{ij}O_{ij}\vert i\rangle \langle j\vert$, where $O_{ij} = \langle i\vert O\vert j\rangle$. Then the normalised operator-state is obtained as [64, 65]

$$\begin{align} \vert O\rangle = \frac{1}{\sqrt{\mathrm{Tr}\left(O^{\dagger}O\right)}}\sum_{ij}O_{ij}\vert i\rangle \vert j\rangle. \end{align} \tag{ 1 }$$

Importantly, O admits a Schmidt decomposition,

$$\begin{align} \frac{O}{\sqrt{\mathrm{Tr}\left(O^{\dagger}O\right)}} = \sum_{i = 1}^r \lambda_i O_{A,i} \otimes O_{B,i} , \end{align} \tag{ 2 }$$

where r is the operator Schmidt rank, and the λ_i are real positive coefficients that satisfy $\sum_{i = 1}^r \lambda_i^2 = 1$. The operators $O_{A,i} \in \mathrm{End}(\mathcal{H}_A)$ (same for $O_{B,i}$) obey the orthonormality condition $\mathrm{Tr}[O_{A,i}^{\dagger},O_{A,j}] = \delta_{ij}$. This can be seen by performing the ordinary Schmidt decomposition of the pure state $\vert O\rangle$ and eventually reverting the vectorization to get back to the space of operators.

The OE is defined as follows. From the vectorization $\vert O\rangle$, we can build the super-reduced-density-matrix $\mathrm{Tr}_{B\otimes B}(\vert O\rangle\langle O\vert)$—where 'super' refers to the operators in the space of operators—which is a super-operator acting on operators on $\mathcal{H}_A$. Then the OE is the Rényi entropy of that super-reduced-density-matrix,

$$\begin{align} S^{\left(n\right)}\left(O\right) = \frac{1}{1-n}\log\mathrm{Tr}\left[ \left(\mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right) \right)^n\right]. \end{align} \tag{ 3 }$$

Alternatively, the OE is given in terms of the Schmidt values λ_i in equation (2) as

$$\begin{align} S^{\left(n\right)}\left(O\right) = \frac{1}{1-n}\log \sum_{i = 1}^r\left(\lambda_i^2\right)^n. \end{align} \tag{ 4 }$$

As usual, the limit n → 1 produces the von Neumann OE

$$\begin{align} S\left(O\right) = -\sum_{i = 1}^r\lambda^2_i\log \lambda^2_i. \end{align} \tag{ 5 }$$

We now assume that there is a charge operator Q acting on the full Hilbert space $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$ which generates a U(1) symmetry, and which is a sum of the two charge operators acting on subsystems A and B, i.e. $Q = Q_{A}\otimes \unicode{x1D7D9}_B+\unicode{x1D7D9}_A\otimes Q_{B}$. A natural question is how to define a symmetry resolution of the OE of an operator O that possesses a fixed charge q_O , i.e. $[Q,O] = q_O Q$. The answer is based on the symmetry resolution for $\vert O\rangle$, as we are going to review. This problem has been already addressed in [26, 27, 41] and we report here the main definitions we will use in this manuscript.

From the commutation relation between O and Q, the terms in the Schmidt decomposition (2) can be reorganised according to their charge q as [41, 61]

$$\begin{align} \frac{O}{\sqrt{\mathrm{Tr}\left(O^{\dagger}O\right)}} = \sum_q \sum_j \lambda^{\left(q \right)}_j O_{A,j}^{\left(q\right)} \otimes O_{B,j}^{\left(q_O-q\right)}, \end{align} \tag{ 6 }$$

where

$$\begin{align} \left[Q_A , O_{A,\,\,j}^{\left(q\right)}\right] = q \, O_{A,\,\,j}^{\left(q\right)}, \quad \left[Q_B , O_{B,\,\,j}^{\left(q_O-q\right)}\right] = \left(q_O-q\right) \, O_{B,\,\,j}^{\left(q_O-q\right)}, \end{align} \tag{ 7 }$$

such that $[Q, O_{A,j}^{(q)} \otimes O_{B,j}^{(q_O-q)} ] = q_O O_{A,j}^{(q)} \otimes O_{B,j}^{(q_O-q)}$. The charge q that appears in these equations can be introduced as the eigenvalue of a charge 'superoperator' $\mathcal{Q}$ living in the Hilbert space $\mathrm{End}(\mathcal{H}) \otimes \mathrm{End}(\bar{\mathcal{H}})$

$$\begin{align} \mathcal{Q} = Q\otimes \unicode{x1D7D9}-\unicode{x1D7D9}\otimes Q^T. \end{align} \tag{ 8 }$$

Such superoperator satisfies the commutation relation

$$\begin{align} \left[\vert O\rangle\langle O\vert,\mathcal{Q}\right] = 0, \end{align} \tag{ 9 }$$

where we have used the vectorization introduced in equation (1). Using the local structure of Q in $A\cup B$, we can write

$$\begin{align} \mathcal{Q} = \mathcal{Q}_A \otimes \unicode{x1D7D9}_{B\otimes B}+ \unicode{x1D7D9}_{A\otimes A}\otimes\mathcal{Q}_B,\quad \mathcal{Q}_A = Q_A\otimes \unicode{x1D7D9}_A-\unicode{x1D7D9}_A\otimes Q_A^T, \end{align} \tag{ 10 }$$

and the following commutation relation holds

$$\begin{align} \left[\mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right),\mathcal{Q}_A\right] = 0. \end{align} \tag{ 11 }$$

We can exploit the result above such that $\mathrm{Tr}_{B\otimes B}(\vert O\rangle\langle O\vert)$ can be decomposed as

$$\begin{align} \mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right) = \bigoplus_q p\left(q\right) \mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right)\left(q\right), \quad p\left(q\right)\equiv\mathrm{Tr}\left[\Pi_q \mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right)\right], \end{align} \tag{ 12 }$$

where $\Pi_q$ is a projector onto the eigenspace of $\mathcal{Q}_A$ with fixed q and $\mathrm{Tr}_{B\otimes B}(\vert O\rangle\langle O\vert)(q)$ denotes the (normalised) reduced density matrix built from the vector $\vert O\rangle$ and restricted to the charge q. The partition function projected in a given charge sector reads

$$\begin{align} \mathcal{Z}^{\left(n\right)}_{q}\left(O\right)\equiv\mathrm{Tr}\left[\Pi_q \left(\mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right)\right)^{n}\right], \end{align} \tag{ 13 }$$

and the SROE is given by

$$\begin{align} S_{q}^{\left(n\right)}\left(O\right)& = \frac{1}{1-n}\log \frac{\mathcal{Z}^{\left(n\right)}_q\left(O\right)}{\left[\mathcal{Z}^{\left(1\right)}_{q}\left(O\right)\right]^n},\qquad S_{q}\left(O\right) = \displaystyle \lim_{n\to 1} S_{q}^{\left(n\right)}\left(O\right). \end{align} \tag{ 14 }$$

According to equation (12), the total von Neumann OE associated to O splits into

$$\begin{align} S\left(O\right) = \displaystyle \sum_q p\left(q\right) S_q\left(O\right)- \displaystyle \sum_q p\left(q\right) \log p\left(q\right) . \end{align} \tag{ 15 }$$

As it was shown in detail in [41], using the uniqueness of the Schmidt coefficients, the set of all (non-zero) values $\{\lambda_j^{(q)}\}$ altogether must be the same as the set of values $\{\lambda_i\}$ from equation (2). Therefore, all quantities defined above can be defined in terms of $\{\lambda_j^{(q)}\}$ as follows

$$\begin{align} p\left(q\right) = \sum_{j} \left(\lambda_j^{\left(q\right)}\right)^2,\qquad S_q^{\left(n\right)} \left(O\right) \, = \, \frac{1}{1-n} \log \left( \sum_j \left(\frac{\left(\lambda_j^{\left(q\right)}\right)^2}{p\left(q\right)} \right)^{\!\!n\,} \right). \end{align} \tag{ 16 }$$

Our goal is to extend our previous study on SROE [41]. The manuscript is organised as follows. In section 2 we review the known analytical techniques for computing the (non-resolved) OE in CFT and in free fermion spin chains, and we extend these to include the symmetry resolution. In section 3 we apply these techniques to the case of the density matrix of a critical 1D system at thermal equilibrium. In section 4 we study the SROE of local operators evolving in Heisenberg picture for free fermion chains. We draw our conclusions in section 5.

In this section, we explain how to compute the symmetry resolved entanglement entropy in a given charge sector for a state $\vert \psi\rangle$, denoted by $S_{q}^{(n)}(\vert \psi\rangle)$.

Let us consider a system with an internal U(1) symmetry and its bipartition into two subsystems, A and B. Tracing out the degrees of freedom of B, we obtain the reduced density matrix (RDM) of A, $\rho_A = \mathrm{Tr}_B \vert \psi\rangle\langle \psi\vert$. A measure of the entanglement between A and its complementary part is provided by the Rényi entropies, defined as

$$\begin{align} S^{\left(n\right)}\left(\vert \psi\rangle\right) = \frac{1}{1-n} \ln\mathrm{Tr} \rho_A^{n}, \end{align} \tag{ 17 }$$

and the limit n → 1 gives the von Neumann entropy as usual. When $\vert \psi\rangle$ is an eigenstate of the Hermitian charge operator Q, by taking the partial trace of the commutator $[ \vert \psi\rangle\langle \psi\vert,Q] = 0$ over B one finds that $[\rho_A,Q_A] = 0$, using $Q = Q_{A}\otimes \unicode{x1D7D9}_B+\unicode{x1D7D9}_A\otimes Q_{B}$. This means that the reduced density matrix ρ_A has a block-diagonal structure where each block corresponds to an eigenvalue q' of Q_A ,

$$\begin{align} \rho_A = \oplus_{q^{^{\prime}}} p_{A}\left(q^{^{\prime}}\right)\rho_{A}\left(q^{^{\prime}}\right), \end{align} \tag{ 18 }$$

where $p_{A}(q^{^{\prime}})$ is the probability of finding q' in a measurement of Q_A in the RDM ρ_A , i.e. $p_A(q^{^{\prime}}) = \mathrm{Tr} \left(\Pi_{q^{^{\prime}}} \rho_A\right)$. Within this convention, the density matrices $\rho_A (q^{^{\prime}})$ of different blocks are normalised as $\textrm{tr}\rho_A (q^{^{\prime}}) = 1$. Thus, from the normalised $\rho_A(q^{^{\prime}})$, we can define the symmetry resolved Rényi entropies as

$$\begin{align} S_{q^{^{\prime}}}^{\left(n\right)}\left(\vert \psi\rangle\right) \equiv \dfrac{1}{1-n}\ln \mathrm{Tr} \rho^n_A\left(q^{^{\prime}}\right)\quad S_{q^{^{\prime}}}\left(\vert \psi\rangle\right) = \lim_{n\to 1} S_{q^{^{\prime}}}^{\left(n\right)}\left(\vert \psi\rangle\right). \end{align} \tag{ 19 }$$

The total von Neumann entanglement entropy associated to ρ_A in equation (18) admits a decomposition as in equation (15),

$$\begin{align} S\left(\vert \psi\rangle\right) = \displaystyle \sum_{q^{^{\prime}}} p\left(q^{^{\prime}}\right) S_{q^{^{\prime}}}\left(\vert \psi\rangle\right)- \displaystyle \sum_{q^{^{\prime}}} p\left(q^{^{\prime}}\right) \ln p\left(q^{^{\prime}}\right). \end{align} \tag{ 20 }$$

The two terms are known as 'configurational entanglement entropy' and 'fluctuation entanglement entropy' (or 'number entanglement entropy') respectively [54]. The configurational entropy is also related to the operationally accessible entanglement entropy of [66–68], while the number entropy is the subject of a substantial recent activity [54, 69–72].

The calculation of the symmetry resolved entropies by the definition (19) is a difficult task, especially for an analytic derivation. As proposed in [42], it is convenient to focus on the charged moments of ρ_A , $\mathrm{Tr}\left(\rho_A^ne^{iQ_A \alpha}\right)$ [73–76]. Their Fourier transforms with respect to α are the moments of the RDM restricted to the sector of fixed charge q' [42], i.e.

$$\begin{align} \mathcal{Z}^{\left(n\right)}_{q^{^{\prime}}}\left(\vert \psi\rangle\right)\equiv \mathrm{Tr} \left(\Pi_{q^{^{\prime}}}\,\rho^n_A\right) = \displaystyle \int_{-\pi}^{\pi}\dfrac{d\alpha}{2\pi}e^{-iq^{^{\prime}}\alpha}\mathrm{Tr}\left(\rho_A^ne^{iQ_A \alpha}\right). \end{align} \tag{ 21 }$$

Finally the symmetry resolved entropies are obtained as

$$\begin{align} S^{\left(n\right)}_{q^{^{\prime}}}\left(\vert \psi\rangle\right) = \dfrac{1}{1-n}\ln \left[ \dfrac{\mathcal{Z}^{\left(n\right)}_{q^{^{\prime}}}\left(\vert \psi\rangle\right)}{\left(\mathcal{Z}^{\left(1\right)}_{q^{^{\prime}}}\left(\vert \psi\rangle\right)\right)^n}\right]. \end{align} \tag{ 22 }$$

We can apply the same machinery to compute the symmetry resolution of the OE in the charge sectors of the operator in equation (10) starting from the charged moments of the super-density-matrix built from $\vert O\rangle$. They are defined as

$$\begin{align} Z_n\left(\alpha\right)\equiv \mathrm{Tr} \left[\left(\mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right)\right)^ne^{i\alpha\mathcal{Q}_A}\right], \end{align} \tag{ 23 }$$

and their Fourier transform gives

$$\begin{align} \mathcal{Z}^{\left(n\right)}_{q}\left(O\right) = \mathrm{Tr}\left(\Pi_q\left(\mathrm{Tr}_{B\otimes B}\left(\vert O\rangle\langle O\vert\right)\right)^n\right) = \int_{-\pi}^{\pi}\frac{d\alpha}{2\pi}e^{-iq\alpha}Z_n\left(\alpha\right), \end{align} \tag{ 24 }$$

where q is the eigenvalue of $\mathcal{Q}_A$. Thus, we have described a procedure to compute the SROE without the explicit knowledge of how the spectrum of the operator is resolved into the different charge sectors.

2.2.1. Brief review of the replica trick for the OE in CFT.

In field theory, the reason for looking at Rényi entropies—instead of focusing directly on the von Neumann entropy—is that, for integer n, $\mathrm{Tr} \rho_A^{n}$ can be expressed in path-integral formalism as a partition function on an n-sheeted Riemann surface $\mathcal{R}_n$ obtained by joining cyclically the n sheets along the region A [11, 13]. One approach to compute these Rényi entropies is based on a particular type of twist fields in quantum field theory that are associated with the branch points of the Riemann surface $\mathcal{R}_n$. We denote them by $\mathcal{T}_n$. Their action, in operator formalism, is defined by [14, 77, 78]

$$\begin{align} \mathcal{T}_n\left(x_1\right)\ \phi_i\left(x^{^{\prime}}\right) = \phi_{i+1}\left(x^{^{\prime}}\right) \mathcal{T}_n\left(x_1\right),\quad \mathcal{T}_{-n}\left(x_2\right)\ \phi_i\left(x^{^{\prime}}\right)& = \phi_{i-1}\left(x^{^{\prime}}\right) \mathcal{T}_{-n}\left(x_2\right), \end{align} \tag{ 25 }$$

where x₁ and x₂ are the endpoints of the interval $A = [x_1,x_2]$ on the real axis, x' is a point in A and $i = 1,\dots,n$ indexes the n sheets modulo n. Here $\phi_i(z^{^{\prime}}) = \unicode{x1D7D9} \otimes \dots \otimes \phi(z^{^{\prime}})\dots \otimes \unicode{x1D7D9}$ denotes any operator acting in the single i-th copy. By definition, the two-point function of the twist fields is the partition function on $\mathcal{R}_n$ which enters the definition of the Rényi entropies [14]. In conformal invariant theories the two-point function is fixed by the scaling dimension of the fields, which here is

$$\begin{align} \Delta_{\mathcal{T}_{\sigma}} = \frac{c}{12}\left(n-\frac{1}{n}\right). \end{align} \tag{ 26 }$$

We can generalise the action of the twist operators to arbitrary permutations $\sigma \in S_{n}$ of the permutation group of the n copies of the field theory, namely we can define a twist operator $\mathcal{T}_{\sigma}$ as the operator with smallest possible scaling dimension such that [22]

$$\begin{align} \mathcal{T}_{\sigma}\left(x_1\right)\ \phi_i\left(x^{^{\prime}}\right)& = \phi_{\sigma\left(i\right)}\left(x^{^{\prime}}\right) \mathcal{T}_{\sigma}\left(x_1\right). \end{align} \tag{ 27 }$$

If σ has cycles of lengths $n_1+\dots + n_c = n$, then the scaling dimension of $\mathcal{T}_{\sigma}$ is

$$\begin{align} \Delta_{\mathcal{T}_{\sigma}} = \Delta_{n_1}+\dots+\Delta_{n_c},\quad \Delta_{n_i} = \frac{c}{12}\left(n_i-\frac{1}{n_i} \right), \end{align} \tag{ 28 }$$

where c is the central charge of the theory. The Rényi entropy is given in terms of the correlator of the twist fields located at the two endpoints of A [14, 22]

$$\begin{align} S^{\left(n\right)}\left(\vert \psi\rangle\right) = \frac{1}{1-n}\log\left(\langle \psi\vert^{\otimes n} \mathcal{T}_{\sigma}\left(x_1\right) \mathcal{T}_{\sigma^{-1}}\left(x_2\right)\vert \psi\rangle^{\otimes n}\right), \end{align} \tag{ 29 }$$

where $\vert \psi\rangle^{\otimes n} = \vert \psi\rangle \otimes \vert \psi\rangle \dots \otimes\vert \psi\rangle \in \mathcal{H}^n$ and $\sigma = (1,2,\dots n)$. Here σ has a single cycle of length n, so it has the scaling dimension (26). The OE of an operator O can be computed in a similar fashion. One has to consider n replicas of O, $O^{\otimes n} = O \otimes O \dots \otimes O$. Then the analogous expression of equation (29) reads [22]

$$\begin{align} S^{\left(n\right)}\left(O\right) = \frac{1}{1-n}\log \left( \frac{\mathrm{Tr}\left[ \left(O^{\dagger}\right)^{\otimes n} \mathcal{T}_{\sigma}\left(x_1\right) \mathcal{T}_{\sigma^{-1}}\left(x_2\right)O^{\otimes n}\mathcal{T}_{\sigma^{-1}}\left(x_1\right) \mathcal{T}_{\sigma}\left(x_2\right)\right]}{\mathrm{Tr}\left(\left(O^{\dagger}\right)^{\otimes n}O^{\otimes n}\right)}\right) . \end{align} \tag{ 30 }$$

We now generalise this formula to include the charge operators needed for the charged moments.

2.2.2. Method for computing the charged moments.

The replica trick can be adapted as follows to obtain the charged moments $Z_n(\alpha)$. The trick consists in inserting an Aharonov–Bohm flux through the multi-sheeted Riemann surface $\mathcal{R}_n$ such that the total phase accumulated by the field upon going through the entire surface is α, similarly to what is done for the standard entanglement entropy [42]. In terms of twist fields, this amounts to computing

$$\begin{align} Z_n\left(\alpha\right) = \frac{\mathrm{Tr}\left[ \left(O^{\dagger}\right)^{\otimes n} \mathcal{T}_{\sigma}\left(x_1\right) \mathcal{T}_{\sigma^{-1}}\left(x_2\right) e^{-i\alpha Q_{A,1}} O^{\otimes n}e^{i\alpha Q_{A,1}}\mathcal{T}_{\sigma^{-1}}\left(x_1\right) \mathcal{T}_{\sigma}\left(x_2\right)\right]}{\mathrm{Tr}\left(\left(O^{\dagger}\right)^{\otimes n}O^{\otimes n}\right)}, \end{align} \tag{ 31 }$$

where $Q_{A,1} = Q_1 \otimes \unicode{x1D7D9} \otimes \dots \otimes \unicode{x1D7D9}$ is the charge operator acting in the first replica. Let us stress that the operator $e^{i\alpha Q_{A,1}}$ appears with opposite signs because of the definition of the charge operator in the doubled Hilbert space (see equation (10)).

In this paper, we will apply formula (31) in the context of a spinless Luttinger liquid, which is equivalent to a c = 1 compactified boson CFT parameterised by a coupling constant K called 'Luttinger parameter' see e.g. [79, 80]. Crucially, in the Luttinger liquid, the U(1) symmetry is generated by $Q = \frac{\sqrt{K}}{2\pi}\int_{-\infty}^\infty dx\, \partial \varphi(x)$, where $\varphi(x)$ is the boson field. Then when the charge operator is restricted to the interval $A = [x_1,x_2]$, the operators $e^{iQ_A\alpha}$ appearing in equation (31) can be written as

$$\begin{align} e^{iQ_A\alpha} = e^{i\frac{\alpha \sqrt{K}}{2\pi}\int_{x_1}^{x_2} dx \partial \varphi\left(x\right)} = e^{i\frac{\alpha \sqrt{K}}{2\pi}\left(\varphi\left(x_2\right)-\varphi\left(x_1\right)\right)}, \end{align} \tag{ 32 }$$

which is a product of two vertex operators $V_\alpha(x_2) V_{-\alpha}(x_1)$ with $V_{\alpha}(x) = e^{i\frac{\alpha \sqrt{K}}{2\pi} \varphi(x)}$. The vertex operator $V_{-\alpha}(x_1)$ can be fused with the twist field $\mathcal{T}_{\sigma^{-1}}(x_1)$ into a single 'charged twist field', $\mathcal{T}_{\sigma^{-1}}^{-\alpha}(x_1) \equiv \mathcal{T}_{\sigma^{-1}}(x_1) V_{-\alpha}(x_1)$ [42]. In other words, when one turns around the operator $\mathcal{T}_{\sigma^{-1}}^{-\alpha}(x_1)$ on the Riemann sheet, one goes from replica j to replica $\sigma(j)$ and one also picks up a phase α when going through the first replica. The scaling dimension of the composite (or charged) twist fields reads [42]

$$\begin{align} \Delta_{\mathcal{T}_{\sigma}^{\alpha}} = \frac{1}{12}\left(n-\frac{1}{n} \right)+\frac{K}{n}\left(\frac{\alpha}{2\pi} \right)^2. \end{align} \tag{ 33 }$$

In section 3 we will apply formulas (31)–(33) to compute the SROE of the thermal density matrix.

It is well established that, for the eigenstates of quadratic Hamiltonians, the entanglement entropies can be efficiently computed in terms of the eigenvalues of the correlation matrix of the subsystem [81–83]. As pointed out in [41], this strategy can be adapted to compute the charged moments of the OE, see equation (23). Let us briefly review the resulting formalism here. The operator O that we are interested in is the Gaussian density matrix ρ of a free fermionic chain of length L. The density matrix can be diagonalised and be put in the form $\rho \propto e^{-\sum_k \lambda_kc^{\dagger}_k c_k}$, where $e^{-\lambda_k} = \frac{n_k}{1-n_k}$ with n_k the occupation number of the orbital $k = 1, \dots, L$. Here $c_k^{\dagger}$ (c_k ) creates (annihilates) a fermion in the orbital k; the creation/annihilation operators satisfy $\{c_k,c_{k^{^{\prime}}}^{\dagger}\} = \delta_{kk^{^{\prime}}}$.

For our purposes it is convenient to write the density matrix ρ as

$$\begin{align} \rho = \bigotimes_{k = 1}^{L} \frac{\vert 0\rangle_k\langle 0\vert_k+e^{-\lambda_k}\vert 1\rangle_k\langle 1\vert_k}{1+e^{-\lambda_k}} = \bigotimes_{k = 1}^{L} \left[\left(1-n_k\right)\vert 0\rangle_k\langle 0\vert_k+n_k\vert 1\rangle_k\langle 1\vert_k\right], \end{align} \tag{ 34 }$$

where $\left| 0 \right|_k$ (resp. $\left| 1 \right|_k$) are states where the orbital k is empty (resp. filled), so that by applying the vectorization trick in equation (1) for ρ, we get

$$\begin{align} \frac{\vert \rho\rangle}{\sqrt{\mathrm{Tr}\left[\rho^2\right]}} = \bigotimes_{k = 1}^{L} \frac{\left[\left(1-n_k\right)\vert 0\rangle_k\vert 0\rangle_{\tilde{k}}+n_k\vert 1\rangle_k\vert 1\rangle_{\tilde{k}}\right]}{\sqrt{n^2_k+\left(1-n_k\right)^2}} = \bigotimes_{k = 1}^{L} \frac{\left[1-n_k+n_kc^{\dagger}_k\tilde{c}^{\dagger}_k\right] \vert 0\rangle}{\sqrt{n^2_k+\left(1-n_k\right)^2}}, \end{align} \tag{ 35 }$$

where the $\tilde{c}_k$ operators are copies of the c_k 's introduced in the vectorization process, and $\vert 0\rangle$ is the vacuum annihilated by all the c_k 's and $\tilde{c}_k$'s. From this pure state, we can build the super-reduced-density-matrix $\mathrm{Tr}_{B\otimes B}(\vert \rho\rangle\langle \rho\vert)$. The correlation matrix of the state $\vert \rho\rangle$ reads [41, 84]

$$\begin{align} C_{kk^{^{\prime}}} = \langle \rho\vert \begin{pmatrix} c_k^{\dagger}\\ \tilde{c}_k \end{pmatrix} \begin{pmatrix} c_{k^{^{\prime}}} \, \tilde{c}_{k^{^{\prime}}}^{\dagger} \end{pmatrix}\vert \rho\rangle = \frac{\delta_{kk^{^{\prime}}}}{n^2_k+\left(1-n_k\right)^2}\begin{pmatrix} & n_k^2 &n_k\left(1-n_k\right) \\ \kern-2pc&n_k\left(1-n_k\right) & \left(1-n_k\right)^2 \\ \end{pmatrix}. \end{align} \tag{ 36 }$$

In the basis of $c_k, \tilde{c}_k$'s, the supercharge operator takes the form

$$\begin{align} \mathcal{Q} = \left(\sum_k c^{\dagger}_k c_k\right)\otimes \unicode{x1D7D9}-\unicode{x1D7D9}\otimes \left(\sum_k\tilde{c}^{\dagger}_k \tilde{c}_k\right)^T. \end{align} \tag{ 37 }$$

At this point, we can compute the $2L\times 2L$ correlation matrix as

$$\begin{align} C = \bigoplus_{k = 1}^{L}C_{kk}, \end{align} \tag{ 38 }$$

and by doing a Fourier transform, we can write C in the spatial basis. To evaluate the charged moments in equation (23), we have to restrict the supercharge operator to $\mathcal{Q}_A$ and do the Fourier transform of the correlation matrix in equation (38) to the subspace corresponding to the subsystem A, of size L_A . Diagonalizing the latter matrix, we get $2L_A$ real eigenvalues ξ_i between 0 and 1.

Therefore, one can compute the charged moments of the reduced density matrix built from $\vert \rho\rangle$ in terms of the eigenvalues ξ_i as [41, 83]

$$\begin{align} Z_{n}\left(\alpha\right) = e^{-i\alpha \left(L_A\right)}\prod_{a = 1}^{2L_A}\left(\xi_a^{n}e^{i\alpha}+\left(1-\xi_a\right)^{n}\right). \end{align} \tag{ 39 }$$

Using equation (24), we can compute exactly the SROE for the reduced density matrix of a free fermionic chain. We will use similar techniques to compute the SROE of an operator in Heisenberg picture in section 4.

As reviewed in the previous section, in CFT the charged moments can be computed as

$$\begin{align} Z_n\left(\alpha\right) = \frac{\mathrm{Tr}\left(\rho_{\beta}^{\otimes n}\mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(x_1\right) \mathcal{T}^{\alpha}_{\sigma}\left(x_1\right)\rho_{\beta}^{\otimes n} \mathcal{T}^{\alpha}_{\sigma}\left(x_2\right) \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(x_2\right)\right)}{\left[\mathrm{Tr}\rho_{\beta}^{\otimes 2}\right]^n}, \end{align} \tag{ 41 }$$

where we have used the charged twist fields $\mathcal{T}_\sigma^\alpha(x_1)$ that cyclically permute the replicas of the subsystem A, leaving B untouched as in equation (30), and are simultaneously the end points of the charge operator $Q_A = \int^{x_2}_{x_1} dx \partial\varphi(x)$ as in equation (32). If we view the ratio (41) as a correlation function of twist operators living on an infinitely long cylinder of circumference $2\beta v$, parametrised by the complex coordinate x + iy with $(x, y) \in \mathbb{R} \times [0, 2\beta v]$, then the four charged twist operators are located at the points $0, L_A, i\beta v$ and $L_A + i\beta v$, where L_A is the size of the subsystem A (see figure 1).

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Therefore, the four-point function we need to calculate reads

$$\begin{align} Z_n\left(\alpha\right) = \langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(0\right) \mathcal{T}^{\alpha}_{\sigma}\left(i\beta v\right) \mathcal{T}^{\alpha}_{\sigma} \left(L_A\right)\mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(i\beta v +L_A\right)\rangle, \end{align} \tag{ 42 }$$

where $\sigma = (1, 2, \dots, n)$ is a cyclic permutation. The scaling dimension of the composite twist field is given by equation (33) with K = 1. For free fermions, this object corresponds to the evaluation of the four-point function of charged twist fields on the cylinder of circumference $2\beta v$. The result is given in [22] for α = 0, and it can be generalised straightforwardly to α ≠ 0 by replacing the scaling dimension (26) by (33), leading to

$$\begin{align} \log Z_n\left(\alpha\right) = -\left[ \frac{n^2-1}{3n}+\frac{\alpha^2}{\pi^2 n} \right]\log \left( \frac{2\beta v}{\pi} \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right) + \log c_{n,\alpha}, \end{align} \tag{ 43 }$$

with $c_{n,\alpha}$ a non-universal constant which depends on the microscopic details of the model, and that we will discuss in more detail below. We implicitly assume that the lattice spacing is set equal to 1. We use this result to compute the SROE in section 3.3.

The result (43) for the charged moments can be generalised to a Luttinger liquid—or equivalently a compactified boson CFT—with a Luttinger parameter K ≠ 1, corresponding to interacting fermions. The scaling dimension of the composite twist field in this case changes according to equation (33).

We are interested in computing $\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}(u_1) \mathcal{T}^{\alpha}_{\sigma}(v_1) \mathcal{T}^{\alpha}_{\sigma} (u_2)\mathcal{T}^{-\alpha}_{\sigma^{-1}}(v_2)\rangle$, with $u_1 = 0,v_1 = L_A,u_2 = i\beta v, v_2 = i\beta v+L_A$. The calculation of that four-point function is much more complicated for K ≠ 1 than for K = 1, as it involves more data from the underlying CFT, including operator content and OPE coefficients (see e.g. [85–88]). Here we exploit the result of the calculation for two intervals on the infinite line from [89] (see also [90]), and we use the conformal transformation $w \to z = \beta v/\pi \log w$ that maps each sheet in the w-plane into an infinitely long cylinder of circumference $2\beta v$, to obtain the result that we need. It is convenient to introduce the cross-ratio

$$\begin{align} x = \frac{\sinh \left(\frac{\pi \left(v_1-u_1\right)}{2 \beta v}\right) \sinh \left(\frac{\pi \left(u_2-v_2\right)}{2 \beta v}\right)}{\sinh\left(\frac{\pi \left(v_2-u_1\right)}{2 \beta v}\right) \sinh\left(\frac{\pi \left(u_2-v_1\right)}{2 \beta v}\right)} = \left[\tanh \frac{\pi L_A}{2\beta v}\right]^2 . \end{align} \tag{ 44 }$$

Then the four-point function of the charged twist fields splits into a product of the four-point function of the twist fields in the plane, times the four-point function of vertex operators in the n-sheeted Riemann surface $\mathcal{R}_n$ [89],

$$\begin{align} Z_n\left(\alpha\right) & = \langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(u_1\right) \mathcal{T}^{\alpha}_{\sigma}\left(v_1\right) \mathcal{T}^{\alpha}_{\sigma} \left(u_2\right)\mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(v_2\right)\rangle \nonumber\\ & = \frac{c_{n,\alpha}}{c_{n,0}} \langle \mathcal{T}_{\sigma^{-1}}\left(u_1\right) \mathcal{T}_{\sigma}\left(v_1\right) \mathcal{T}_{\sigma} \left(u_2\right)\mathcal{T}_{\sigma^{-1}}\left(v_2\right)\rangle \, \langle V_{-\alpha} \left(u_1\right) V_\alpha \left(v_1\right) \nonumber\\ & \quad \times V_\alpha \left(u_2\right) V_{-\alpha}\left(v_2\right)\rangle_{\mathcal{R}_n} . \end{align} \tag{ 45 }$$

Here $c_{n,\alpha}$ denotes again ultra-violet non-universal constant. The factor $\langle \mathcal{T}_{\sigma^{-1}}(u_1) \mathcal{T}_{\sigma}(v_1) \mathcal{T}_{\sigma} (u_2)$ $\mathcal{T}_{\sigma^{-1}}(v_2)\rangle$ is the partition function (41) at α = 0, while the product of vertex operators is readily evaluated [89], leading finally to

$$\begin{align} &\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(u_1\right) \mathcal{T}^{\alpha}_{\sigma}\left(v_1\right) \mathcal{T}^{\alpha}_{\sigma} \left(u_2\right)\mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(v_2\right)\rangle \nonumber\\ & \quad = c_{n,\alpha}\frac{Z_n\left(0\right)}{c_{n,0}}\left[\frac{4 \beta ^2 v^2}{\pi ^2} \sinh \left(\frac{\pi \left(v_1-u_1\right)}{2 \beta v}\right) \sinh \left(\frac{\pi \left(v_2-u_2\right)}{2 \beta v}\right) \right.\nonumber\\ & \qquad \left. \times \frac{\sinh \left(\frac{\pi \left(u_2-u_1\right)}{2 \beta v}\right) \sinh \left(\frac{\pi \left(v_2-v_1\right)}{2 \beta v}\right)}{\sinh \left(\frac{\pi \left(v_2-u_1\right)}{2 \beta v}\right) \sinh \left(\frac{\pi \left(u_2-v_1\right)}{2 \beta v}\right)}\right]^{-\frac{\alpha^2 K}{2\pi^2 n}} \nonumber\\ & = c_{n,\alpha}\frac{Z_n\left(0\right)}{c_{n,0}}\left[\frac{2 \beta v}{\pi}\tanh\left(\frac{L_A\pi}{2\beta v}\right)\right]^{-\frac{K\alpha^2}{\pi^2 n}}. \end{align} \tag{ 46 }$$

The partition function $Z_n(0)$ defined in equation (41) with α = 0 can be computed by combining the expression for the OE in [22] with the result for the four-point function of twist fields in [86]. The result is

$$\begin{align} Z_n\left(0\right) = c_{n,0}\left[\frac{2 \beta v}{\pi}\tanh\left(\frac{L_A\pi}{2\beta v}\right)\right]^{-\frac{n^2-1}{3n}}\mathcal{F}_n\left(x\right), \end{align} \tag{ 47 }$$

where $\mathcal{F}_n(x)$ is the conformal block of twist fields [86],

$$\begin{align} \mathcal{F}_n\left(x\right) = \frac{\Theta\left(\boldsymbol{0}|\Gamma\left(x\right)/K\right)\Theta \left(\boldsymbol{0}|\Gamma\left(x\right)K\right)}{\left[\Theta \left(\boldsymbol{0}|\Gamma\left(x\right)\right)\right]^2}. \end{align} \tag{ 48 }$$

Here $\Theta( \textbf{u} | \Omega )$ is the Riemann–Siegel Theta function defined for an $(n-1)\times(n-1)$ complex matrix Ω and an $(n-1)$-dimensional vector u:

$$\begin{align} \Theta \left(\boldsymbol{u}|\Omega\right)\equiv \sum_{\boldsymbol{m}\in\mathbb{Z}^{n-1}} e^{i\pi\boldsymbol{m}^t \cdot \Omega\cdot \boldsymbol{m}+2\pi i \, \boldsymbol{m}^t\cdot \boldsymbol{u}} . \end{align} \tag{ 49 }$$

The matrix $\Gamma(x)$ in equation (48) has entries given by [86]

$$\begin{align} \Gamma_{rs}\left(x\right) = \frac{2i}{n} \sum_{l = 1}^{n-1}\cos\left[\frac{2 \pi l \left(r-s\right)}{n}\right]\sin\left(\frac{\pi l}{n}\right) \beta_{l/n}\left(x\right),\quad r, s = 1, \dots,n-1, \end{align} \tag{ 50 }$$

and

$$\begin{align} \beta_p\left(x\right) = \frac{I_p\left(1-x\right)}{I_p\left(x\right)}, \end{align} \tag{ 51 }$$

with $I_p(x)\equiv \,_2F_1(p,1-p,1,1-x)$. The conformal block $\mathcal{F}_n(x)$ is invariant under $x\mapsto 1-x$ and it is normalised such that $\mathcal{F}_n(0) = \mathcal{F}_n(1) = 1$.

Our final result for the charged moment is then

$$\begin{align} Z_n\left(\alpha\right) = c_{n,\alpha} \left[\frac{2 \beta v}{\pi}\tanh\left(\frac{L_A\pi}{2\beta v}\right)\right]^{-\frac{n^2-1}{3n} - \frac{K \alpha^2}{\pi^2 n} }\mathcal{F}_n\left(x\right) . \end{align} \tag{ 52 }$$

In the two limiting cases $L_A \gg \beta v$ and $L_A \ll \beta v$, where the four-point function factorises into a product of two two-point functions, we see that the result reduces to the expected one,

$$\begin{align} L_A \ll \beta v:\,\, &\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(0\right) \mathcal{T}^{\alpha}_{\sigma}\left(i\beta v\right) \mathcal{T}^{\alpha}_{\sigma} \left(L_A\right)\mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(i\beta v +L_A\right)\rangle \nonumber\\ \simeq &\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(0\right)\mathcal{T}^{\alpha}_{\sigma} \left(L_A\right)\rangle\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(i\beta v +L_A\right)\mathcal{T}^{\alpha}_{\sigma}\left(i\beta v\right)\rangle\propto \frac{1}{L_A^{4\Delta_{\mathcal{T}_{\sigma}^{\alpha}}}}, \nonumber\\ L_A \gg \beta v: \,\, &\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(0\right) \mathcal{T}^{\alpha}_{\sigma}\left(i\beta v\right) \mathcal{T}^{\alpha}_{\sigma} \left(L_A\right)\mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(i\beta v +L_A\right)\rangle\nonumber\\ \simeq &\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(0\right)\mathcal{T}^{\alpha}_{\sigma} \left(i\beta v\right)\rangle\langle \mathcal{T}^{-\alpha}_{\sigma^{-1}}\left(i\beta v +L_A\right)\mathcal{T}^{\alpha}_{\sigma}\left(L_A\right)\rangle\propto \frac{1}{\left[\frac{2\beta v}{\pi}\right]^{4\Delta_{\mathcal{T}_{\sigma}^{\alpha}}}}. \end{align} \tag{ 53 }$$

These results for the charged moments have the same dependence on the parameters $\beta v, L_A$ as the one of the OE found in [22], with the scaling dimension $\Delta_{\mathcal{T}_\sigma}$ replaced by $\Delta_{\mathcal{T}_\sigma^\alpha}$. Moreover, when K = 1, $\mathcal{F}_n(x) = 1$ for all x and the logarithm of equation (46) coincides with equation (43).

To go from the charged moment (52) to the SROE, we need to take the Fourier transform with respect to α. To do this, it is important to know the non-universal constant $c_{n,\alpha}$. We start from the free fermion chain (40), where $c_{n,\alpha}$ is known analytically, and later turn to the interacting case.

3.3.1. SROE in the free fermion chain.

The constant $c_{n,\alpha}$ is known in the free fermion chain (40) from the charged moments of the reduced density matrix computed in [69]. The ratio $c_{n,\alpha} / c_{n,0}$ behaves quadratically at small α as [69]

$$\begin{align} \frac{c_{n,\alpha}}{c_{n,0}} \, \underset{\alpha \rightarrow 0}{\simeq} \, e^{\alpha^2\left(2\gamma\left(n\right)-\frac{\log 2}{\pi^2 n}\right)} , \end{align} \tag{ 54 }$$

where

$$\begin{align} \gamma\left(n\right) = \frac{n}{4}\displaystyle \int_{-\infty}^{\infty} dw \left[\tanh^3\left(\pi n w \right)-\tanh\left(\pi n w\right)\right] \, i \,\log \dfrac{\Gamma\left(\frac{1}{2}+iw\right)}{\Gamma\left(\frac{1}{2}-iw\right)}. \end{align} \tag{ 55 }$$

(This expression is real, as can be checked easily.) For later convenience, we define $h_n = 4\gamma(n)-\frac{2\log 2}{\pi^2 n}$ so that $c_{n,\alpha}/c_{n,0} \simeq e^{\frac{\alpha^2}{2} h_n}$. The constant $c_{n,0}$ was computed by Korepin and Jin in [91]:

$$\begin{align} c_{n,0}& = e^{2\Upsilon_n+\frac{1-n^2}{3n}\log 2} \nonumber\\ \Upsilon_n& = n \int_{-\infty}^\infty dw \left[\tanh\left( \pi w\right)-\tanh \left(\pi n w\right)\right] \, i \, \log \frac{\Gamma\left(\frac12 +iw\right)}{\Gamma\left(\frac12 -iw\right)} \,. \end{align} \tag{ 56 }$$

This leads to the following result for the charged moments in the free fermion chain:

$$\begin{align} Z_n\left(\alpha\right) = c_{n,0} \left[\frac{2 \beta v}{\pi}\tanh\left(\frac{L_A\pi}{2\beta v}\right)\right]^{-\frac{n^2-1}{3n} - \frac{\alpha^2}{\pi^2 n} } e^{\frac{\alpha^2}{2} h_n} . \end{align} \tag{ 57 }$$

We check the result (57) against numerics in figure 2. The symbols represent the numerical data obtained through exact lattice computations, using the expression for the charged moments described in section 2.3. The data present finite size corrections, which become larger as n and α increase. In the right panel, in order to achieve the correct scaling limit, we have computed $\log Z_n(\alpha)$ at fixed $\alpha, n = 2$ and for fixed $L_A/\beta = 0.5$. The leading corrections to the scaling behave as $L_A^{-2(1-\alpha/\pi)/n}$, $L_A^{-2(1+\alpha/\pi)/n}$, $L_A^{-2/n}$, $L_A^{-2/n(2-\alpha/\pi)}$ (see also [92] for a similar analysis). We then perform a fit of the finite L_A data (fox fixed α, n = 2), keeping the first two power-law corrections and extrapolating at $L_A \to \infty$. The data obtained following this procedure are reported as black symbols in the right panel of figure 2.

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Plugging the expression (57) of the charged moments in the Fourier transform (24) and applying the saddle-point approximation for $\beta, L_A \gg 1$, we get

$$\begin{align} S^{\left(n\right)}_{q}\left(\rho_{\beta}\right) & = S^{\left(n\right)}\left(\rho_{\beta}\right)-\frac{1}{2} \log \left(\frac{4}{\pi} \log \delta_n \left( \frac{2\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right) \right)+ \frac{\log n}{2\left(1-n\right)}\nonumber\\ & \quad -\frac{\pi^4n\left(h_1-nh_n\right)^2}{4\left(1-n\right)^2\left(\log \left( \frac{2\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right)\right)^2} \nonumber\\ & \quad + q^2n \pi^4\frac{h_1-nh_n}{4\left(1-n\right)\left(\log \left( \frac{2\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right) \kappa_n\right)^2} \nonumber\\ & \quad + o\left(\log^{-2} \left( \beta \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right)\right), \end{align} \tag{ 58 }$$

where

$$\begin{align} \log \delta_n = -\dfrac{\pi^2 n \left(h_n-h_1\right)}{\left(1-n\right)},\qquad \log \kappa_n = -\pi^2\frac{\left(h_1+n h_n\right)}{2}, \end{align} \tag{ 59 }$$

and $S^{(n)}(\rho_\beta)$ is the total OE

$$\begin{align} S^{\left(n\right)}\left(\rho_{\beta}\right) = \frac{n+1}{3n} \log \left( \frac{4\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right) +\frac{2}{1-n}\Upsilon_n . \end{align} \tag{ 60 }$$

This formula is also valid for the symmetry resolved von Neumann OE taking properly the limits of the various pieces when n → 1. The expression (58) is checked against numerics in the free fermion chain in figure 3. The small deviations between data and analytics are a consequence of the finite size corrections already present in the charged moments (figure 2).

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3.3.2. Interacting case.

For K ≠ 1, we do not know a microscopic model where the non-universal constant $c_{n,\alpha}$ is known. Nevertheless, we can generalise the previous results, assuming that $c_{n,\alpha}$ behaves smoothly as a function of α near α = 0, i.e.

$$\begin{align} c_{n,\alpha} = c_{n,0} - \frac{\alpha^2}{2} c_{n,0} h_n + o\left(\alpha^2\right) = c_{n,0} e^{\frac{\alpha^2}{2} h_n } + o\left(\alpha^2\right) \end{align} \tag{ 61 }$$

for some constants $c_{n,0}$ and h_n that depend on the microscopic model and are, in general, not equal to the ones in the free fermion chain. By repeating the same steps as before, the SROE reads

$$\begin{align} S^{\left(n\right)}_{q}\left(\rho_{\beta}\right) & = S^{\left(n\right)}\left(\rho_{\beta}\right)-\frac{1}{2} \log \left(\frac{4K}{\pi} \log \delta_n \left( \frac{2\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right) \right)+ \frac{\log n}{2\left(1-n\right)}\nonumber\\ & \quad -\frac{\pi^4n\left(h_1-nh_n\right)^2}{4K^2\left(1-n\right)^2\left(\log \left( \frac{2\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right)\right)^2} \nonumber\\ & \quad + q^2n \pi^4\frac{h_1-nh_n}{4K^2\left(1-n\right)\left(\log \left( \frac{2\beta v}{\pi } \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right) \kappa_n\right)^2} \nonumber\\ & \quad + o\left(\log^{-2} \left( \beta \tanh\left(\frac{\pi L_A}{2\beta v} \right)\right)\right). \end{align} \tag{ 62 }$$

3.3.3. Discussion.

To get insight into the meaning of equations (58) and (62), we study the following two asymptotic regimes, which are valid for any value of K:

$$\begin{align} L_A \ll \beta v:\,\, & \, S^{\left(n\right)}_{q}\left(\rho_{\beta}\right) = \frac{n+1}{3n} \log L_A -\frac{1}{2}\log\left(K \log L_A\right)+O\left(1\right), \nonumber\\ L_A \gg \beta v:\,\, & \, S^{\left(n\right)}_{q}\left(\rho_{\beta}\right) = \frac{n+1}{3n}\log\frac{2\beta v}{\pi}-\frac{1}{2}\log\left(K\log\frac{2\beta v}{\pi}\right)+O\left(1\right), \end{align} \tag{ 63 }$$

We find that the leading order term corresponds to the total OE, which diverges logarithmically in L_A at low temperatures, i.e. when the reduced density matrix ρ_A is very close to the one of the ground state, while it is bounded (in L_A ) at finite temperature β. This is the main striking difference with respect to the usual entanglement, which at finite temperature has an extensive behaviour. Indeed, the symmetry resolved Rényi entropy (19) of a thermal state in a bipartition $A = [0,L_A]$, $B = (-\infty,0)\cup (L_A,\infty)$ for a c = 1 CFT is [59, 93]

$$\begin{align} L_A \ll \beta v:\,\, & \, S^{\left(n\right)}_{q^{^{\prime}}}\left(\vert \psi\rangle\right) = \frac{n+1}{6n} \log L_A-\frac{1}{2}\log\left(K \log L_A\right)+O\left(1\right),\nonumber\\ L_A \gg \beta v:\,\, & \, S^{\left(n\right)}_{q^{^{\prime}}}\left(\vert \psi\rangle\right) = \frac{n+1}{6n}\frac{\pi L_A}{\beta v}-\frac{1}{2}\log\left(K\frac{L_A}{\beta v}\right)+O\left(1\right), \end{align} \tag{ 64 }$$

where q' denotes the eigenvalues of the charge operator Q in (40) restricted to the subsystem A. The expression above displays a logarithmic growth at zero-temperature and an extensive form in opposite limit; the leading order in both regimes coincides with the total Rényi entropies [14]. By comparing equations (63) and (64), we observe that, in the low-temperature limit $L_A\ll \beta v$, the only difference is that the leading-order term of the SROE is twice the one of the symmetry resolved Rényi entropy. This is consistent with the result that, for pure states, the total OE is exactly twice the entanglement entropy [22]. In fact, we can explain the logarithmic growth of the SROE at low-temperature by observing that $\rho_\beta = e^{-\beta H} \xrightarrow{\beta \to +\infty}\vert \psi\rangle\langle \psi\vert$, where $\vert \psi\rangle$ is the pure ground state of H, and since at leading order $S_{q}^{(n)}(\rho_\beta) \sim S^{(n)}(\rho_\beta)$, $S_{q^{^{\prime}}}^{(n)}(\vert \psi\rangle) \sim S^{(n)}(\vert \psi\rangle)$, for $L_A\ll \beta v$ we expect that $S_{q}^{(n)}(\rho_\beta)\sim 2 S_{q^{^{\prime}}}^{(n)}(\vert \psi\rangle)$. Then, the fact that $\vert \psi\rangle$ is a critical state justifies the logarithmic behaviour of the entanglement. Conversely, as highlighted earlier, the the result is completely different at high temperatures: the state entanglement's extensive behavior contrasts sharply with the operator area law observed in the thermal density matrix ρ_β . The practical consequence of this result is that ρ_β can be efficiently represented by an MPO [22].

Then $S^{(n)}_{q}(\rho_{\beta})$ presents a double logarithmic correction in L_A at low temperature, while it remains bounded at finite temperature, with a double logarithmic correction depending on the inverse temperature β. This behaviour is analogous to the symmetry resolved entanglement only in the $L_A\ll \beta v$ regime, as we can deduce from equation (64). Finally, the fact that the leading order term of the SROE coincides with the total OE resembles the entanglement equipartition for the symmetry resolution of a $U(1)-$invariant theory [43], and it can be traced back to the conformal invariance of the system also in this case. A similar result is also shown in equation (64) for the usual symmetry resolution of the entanglement, since, at leading order, the formula does not exhibit any dependence on the charge sector q'.

The goal of this section is to study the SROE of $c_x^\dagger(t),\sigma^z_{x}(t)$. In the former case, we take advantage of the form in which equation (6) can be written, while in the latter we exploit the connection between the charged moments of the super-density-matrix and the correlation matrix, similarly to what has been done in section 2.3.

4.1.1. SROE of creation operator $c_x^\dagger(t)$.

We start by studying the SROE of a single creation operator $c_x^\dagger(t)$ (the analysis for $c_x(t)$ can be performed in a similar way). We observe that $[Q,c_x^\dagger(t)] = c_x^\dagger(t)$, with $Q = \sum_j c_j^\dagger c_j$, so $q_O = 1$ in equation (6). We choose x = 0 and we can use the Fourier transform, for a ring of L sites,

$$\begin{align} c^{\dagger}_0\left(t\right) = \sum_jU^*_{j0}\left(t\right)c^{\dagger}_j,\quad U_{0j}\left(t\right) = \frac{1}{L}\sum_ke^{ikj}e^{it\cos\left(k\right)}. \end{align} \tag{ 65 }$$

In the thermodynamic limit, the matrix elements $U_{0m}(t)$ of the unitary evolution operator can be written in terms of Bessel functions, $U_{0m}(t)\underset{t \rightarrow \infty}{\longrightarrow} i^{-m} J_m(t)$. We can directly recast $c^{\dagger}_0(t)$ into the form of equation (6), with $A = (-\infty,0)$, which reads

$$\begin{align} c^{\dagger}_0\left(t\right) = \left(\sum_{m\unicode{x2A7D} 0} J_m\left(t\right)\right)\otimes\unicode{x1D7D9}_B +\unicode{x1D7D9}_A\otimes \left(\sum_{m> 0} J_m\left(t\right)\right). \end{align} \tag{ 66 }$$

This implies that there are only two possible charge sectors, $q = 1,0$ and we can also directly write down the Schmidt values $\lambda_m(0) = \lambda_m(1) = J_m(t)$. At large times, the Bessel functions satisfy

$$\begin{align} \sum_{m\unicode{x2A7D} 0} J_m^2\left(t\right) \underset{t \rightarrow \infty}{\longrightarrow} \frac{1}{2} . \end{align} \tag{ 67 }$$

Combining equation (16) and (67), we get the following result at large time:

$$\begin{align} p\left(0\right) = p\left(1\right) = 1/2,\qquad S_0^{\left(n\right)}\left(c^{\dagger}_0\left(t\right)\right) = S_1^{\left(n\right)}\left(c^{\dagger}_0\left(t\right)\right) = 0 \end{align} \tag{ 68 }$$

and

$$\begin{align} \displaystyle \sum_q p\left(q\right) S_{q}\left(c^{\dagger}_0\left(t\right)\right)- \displaystyle \sum_q p\left(q\right) \log p\left(q\right) = 2\frac{\log 2}{2} = \log 2. \end{align} \tag{ 69 }$$

This implies that the SROE vanishes in all the charge sectors and the total OE is given only by the number/fluctuation entanglement (the second term of equation (15)) i.e. by the probability of finding $q = 1,0$ as an outcome of $\mathcal{Q}_A$.

4.1.2. SROE of local density operator $c_x^\dagger (t) c_x(t)$.

We now consider a less trivial example, which is the symmetry resolution of a pair of creation and annihilation operators. In the spin language (up to additive constants) this corresponds to

$$\begin{align} c^{\dagger}_x c_x \leftrightarrow \sigma^z_x. \end{align} \tag{ 70 }$$

Let us focus on $\sigma^z_0(t)$. From the state-operator correspondence, we want to study the entanglement due to the state $c^{\dagger}_0(t) \tilde{c}^{\dagger}_0(t)\vert 0\rangle$ [22]. Before reporting the details of the computations, we report the main result here. We observe that there are 3 non-trivial charge sectors, $q = 0,\pm 1$, where

$$\begin{align} p\left(0\right) = 2 p\left(\pm 1\right) = \frac{1}{2}\quad S^{\left(n\right)}_{q = 0}\left(\sigma^z_0\left(t\right)\right) = \log 2, \quad S^{\left(n\right)}\left(q\neq 0\right) = 0, \end{align} \tag{ 71 }$$

and

$$\begin{align} \displaystyle \sum_q p\left(q\right) S_{q}\left(\sigma^z_0\left(t\right)\right)- \displaystyle \sum_q p\left(q\right) \log p\left(q\right) = \frac{\log 2}{2}+\frac{\log 2}{2}+2\frac{\log 4}{4} = 2\log 2, \end{align} \tag{ 72 }$$

i.e. equation (15) is satisfied. Therefore, differently from what we found before for one single creation operator $c^{\dagger}_0(t)$, there exists one charge sector q = 0 where the SROE is different from 0 and both terms in equation (15) contribute to the total OE.

Given the simplicity of the result, it might be possible to find it following the logic of the previous subsection about $c^{\dagger}(t)$. However, as an illustration of the techniques used in section 2.3, we follow an alternative path. First of all, we need to compute

$$\begin{align} \langle 0\vert\tilde{c}_0\left(t\right)c_0\left(t\right)c^{\dagger}_nc_mc^{\dagger}_0 \left(t\right)\tilde{c}^{\dagger}_0\left(t\right)\vert 0\rangle. \end{align} \tag{ 73 }$$

Therefore, plugging equation (65) into equation (73), we get

$$\begin{align} \langle 0\vert\tilde{c}_0\left(t\right)c_0\left(t\right)c^{\dagger}_nc_mc^{\dagger}_0 \left(t\right)\tilde{c}^{\dagger}_0\left(t\right)\vert 0\rangle = U_{0n}\left(t\right)U^*_{m0}\left(t\right). \end{align} \tag{ 74 }$$

By repeating the same computations for $c^{\dagger}_n\tilde{c}^{\dagger}_m$, $\tilde{c}_nc_m$, $\tilde{c}_n\tilde{c}^{\dagger}_m$, we can write down the $2L_A \times 2L_A$ correlation matrix for a subsystem $(-L_A+1,0)$ as

$$\begin{align} C_A\left(t\right) = \begin{pmatrix} &\mathcal{C}\left(t\right) & 0_{L_A}\\ & 0_{L_A} & \unicode{x1D7D9}_{L_A}-\mathcal{C}\left(t\right) \\ \end{pmatrix}, \end{align} \tag{ 75 }$$

where $\mathcal{C}_{nm}(t) = i^{m-n}J_m(t)J_n(t)$, while $0_{L_A}$ and $\unicode{x1D7D9}_{L_A}$ denote the $L_A\times L_A$ null and identity matrix, respectively. In the scaling limit $t, L_A \to \infty$, we can compute $\mathrm{Tr}(2C_A(t)-1)^j$, observing that

$$\begin{align} \mathrm{Tr}\left(2C_A\left(t\right)-1\right)^{2j} = 2L_A-2, \quad \mathrm{Tr}\left(2C_A\left(t\right)-1\right)^0 = 2L_A,\quad \mathrm{Tr}\left(2C_A\left(t\right)-1\right)^{2j-1} = 0. \end{align} \tag{ 76 }$$

This can be understood as follows: first of all, because of the construction of $C_A(t)$ with the blocks $\mathcal{C}(t)$ and $1-\mathcal{C}(t)$, $\mathrm{Tr}(2C_A(t)-1)^j$ vanishes for any odd values of j. For the even powers, we can start by studying the case j = 2. We restrict to the upper $L_A \times L_A$ block and we compute the following trace (the trace of the square of the lower block is the same)

$$\begin{align} \mathrm{Tr}\left(2\mathcal{C}\left(t\right)-1\right)^2 = \sum_{kj}\left(2J_k\left(t\right)J_j\left(t\right)i^{k-j}-\delta_{kj}\right)\left(2J_k\left(t\right)J_j\left(t\right)i^{j-k}-\delta_{kj}\right). \end{align} \tag{ 77 }$$

Therefore, equation (77) and (67) give

$$\begin{align} 4\sum_{kj}J^2_k\left(t\right)J^2_j\left(t\right)-4\sum_jJ^2_j\left(t\right)+\sum_j 1 = L_A-1. \end{align} \tag{ 78 }$$

The generalisation to higher powers of j follows the same logic, since all the powers involving terms like $\sum_j J_j^{2k}(t)$ vanish for k > 1 in the scaling limit.

We can now come back to our main focus of the section, i.e. the evaluation of the charged moments built from $\sigma^z_0(t)$. The charged moments in equation (39) can be rewritten using [94, 95]

$$\begin{align} h_{n,\alpha}\left(x\right) = &\log \left[ \left( \frac{1+x}{2} \right)^n e^{i\alpha}+\left(\frac{1-x}{2} \right)^n \right] = \sum_{m = 0}^{\infty}s_{n,\alpha}\left(m\right)x^m, \nonumber\\ \log Z_n\left(\alpha\right) = & -i\alpha L_A+\sum_{m = 0}^{\infty}s_{n,\alpha}\left(m\right)\mathrm{Tr}\left(2C_A\left(t\right)-1\right)^m. \end{align} \tag{ 79 }$$

The coefficients $s_{n,\alpha}(m)$ correspond to the function $h_{n,\alpha}(x)$ evaluated in certain simple points:

$$\begin{align} s_n\left(0\right) & = h_{n,\alpha}\left(0\right) = \left(1-n\right)\log 2+\log \cos \frac{\alpha}{2}+i\frac{\alpha}{2}, \nonumber\\ \sum_{m = 1}^{\infty}s_{n,\alpha}\left(2m\right) & = \left(\frac{h_{n,\alpha}\left(1\right)+h_{n,\alpha}\left(-1\right)}{2}-s_{n,\alpha}\left(0\right) \right) = \left(n-1\right)\log 2-\log \cos \frac{\alpha}{2}. \end{align} \tag{ 80 }$$

It follows that

$$\begin{align} \log Z_n\left(\alpha\right) = -i\alpha L_A+s_n\left(0\right)\left(2L_A\right)+2\left(L_A-1\right)\sum_{m = 1}^{\infty}s_{n,\alpha}\left(2m\right) = 2\left(1-n\right)\log 2+2\log \cos \frac{\alpha}{2}. \end{align} \tag{ 81 }$$

In figure 4 we compare the prediction (81) for the logarithm of the charged moments $\log Z_n(\alpha)$ with the exact lattice computations done using the correlation matrix in equation (75). The dependence of $Z_n(\alpha)$ in t and α is perfectly reproduced by the exact result.

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By doing the Fourier transform in equation (24), we can compute

$$\begin{align} \mathcal{Z}^{\left(n\right)}_{q}\left(\sigma^z_0\left(t\right)\right) = 2^{2\left(1-n\right)}\int_{-\pi}^{\pi}\frac{d\alpha}{2\pi}e^{-iq\alpha}Z_n\left(\alpha\right) = 2^{2\left(1-n\right)}\frac{\sin\left(\pi q\right)}{2\pi\left(q-q^3\right)}. \end{align} \tag{ 82 }$$

We recognise that $\mathcal{Z}^{(n)}_{q = 0}(\sigma^z_0(t)) = \frac{2^{2(1-n)}}{2}$, $\mathcal{Z}^{(n)}_{q = \pm 1}(\sigma^z_0(t)) = \frac{2^{2(1-n)}}{4}$, otherwise $\mathcal{Z}^{(n)}_{q}(\sigma^z_0(t)) = 0$. This has been checked in the left panel of figure 5. Finally, we deduce that the SROE behaves as reported in equation (71): The fact that the OE is different from 0 only in one charge sector is a strong violation of the equipartition that we have found in section 3. This lack of equipartition is a direct consequence of the fact that the operator entanglement stays always finite.

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We now turn to the calculation of the SROE of a Jordan–Wigner string in the tight-binding chain with periodic boundary conditions, i.e.

$$\begin{align} \mathrm{JW}_x\left(t\right)\equiv \prod_{y\unicode{x2A7D} x}\left(1-2c^{\dagger}_y\left(t\right)c_y\left(t\right)\right). \end{align} \tag{ 83 }$$

We place the endpoint of the string at the origin without loss of generality. Therefore, we are interested in the correlation matrix built from

$$\begin{align} \mathrm{JW}_0\left(t\right)\to \vert \mathrm{JW}_0\left(t\right)\rangle = \prod_{y \unicode{x2A7D} 0}\left(1-c^{\dagger}_y\left(t\right)\tilde{c}^{\dagger}_y\left(t\right)\right)\vert 0\rangle, \end{align} \tag{ 84 }$$

by replacing c(t) with $\tilde{c}^{\dagger}(t)$ which anticommutes with all the c's. It is convenient to extract the time dependence as

$$\begin{align} \vert \mathrm{JW}_0\left(t\right)\rangle = e^{it\left(H\otimes \unicode{x1D7D9}-\unicode{x1D7D9}\otimes \tilde{H}\right)}\vert \mathrm{JW}_0\left(0\right)\rangle, \end{align} \tag{ 85 }$$

and to perform a Bogoliubov transformation (exactly as done for the total OE in [22]),

$$\begin{align} \begin{pmatrix} b^{\dagger}_x\\ d_x \end{pmatrix} = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 & -1 \\ 1 &1 \\ \end{pmatrix} \begin{pmatrix} c^{\dagger}_x \\ \tilde{c}^{\dagger}_x \end{pmatrix}. \end{align} \tag{ 86 }$$

Thus, the initial state becomes

$$\begin{align} \vert \mathrm{JW}_0\left(t\right)\rangle = \frac{1}{5}\prod_{y\unicode{x2A7D} 0}d_y^{\dagger}b_y^{\dagger}\vert 0_{d,b}\rangle, \end{align} \tag{ 87 }$$

where now $\vert 0_{d,b}\rangle$ is the vacuum for the d and b modes. This transformation does not do anything to the Hamiltonian $H\otimes \unicode{x1D7D9}-\unicode{x1D7D9}\otimes \tilde{H}$ and to the charge operator $Q\otimes \unicode{x1D7D9}-\unicode{x1D7D9}\otimes Q^T$ in equation (37), which already were diagonal. However, due to this simplification on the initial state, we can compute the correlation matrix as

$$\begin{align} C_A\left(t\right) = \begin{pmatrix} &\mathcal{C}\left(t\right) & 0_{L_A}\\ & 0_{L_A} & \unicode{x1D7D9}_{L_A}-\mathcal{C}\left(t\right), \\ \end{pmatrix} \end{align} \tag{ 88 }$$

where $\mathcal{C}_{nm}(t) = \sum_{y = -L_A}^0i^{m-n}J_{m-y}(t)J_{n-y}(t)$ [96]. Rewriting the initial state as in equation (87) also allows us to have an analytical insight on the time evolution of the SROE, following the observations done in [22]. Indeed, equation (87) is a product state $\vert \mathrm{DWIS}\rangle_b\otimes \vert \mathrm{DWIS}\rangle_d = \prod_{y\unicode{x2A7D} 0}d_y^{\dagger}\vert 0_{d}\rangle\otimes \prod_{y\unicode{x2A7D} 0}b_y^{\dagger}\vert 0_{b}\rangle$. The two independent initial states for the b's and the d's operators are domain-wall initial states (DWIS). For a bipartition $A \cup B = (-L_A,0) \cup (0,L_A)$ and in the scaling limit $L_A \to \infty$, the charged moments of the JW string are exactly twice the charged moments of the domain wall computed in [97], i.e.

$$\begin{align} \log Z_n\left(\alpha\right) = -\left[ \frac{n^2-1}{3n}+\frac{\alpha^2}{\pi^2 n} \right]\log 4t+2\alpha^2\gamma\left(n\right). \end{align} \tag{ 89 }$$

Let us remark that the knowledge of the non-universal constants from [69] allows us to benchmark our analytical prediction against the exact lattice computations without fitting any parameter, as we have done in the left panel of figure 6. By doing a Fourier transform and applying the saddle point approximation in the limit $t \to \infty$, we get

$$\begin{align} S_{q}^{\left(n\right)}\left(\mathrm{JW}_0\left(t\right)\right) & = S^{\left(n\right)}\left(\mathrm{JW}_0\left(t\right)\right)-\frac{1}{2} \log \left(\frac{4}{\pi} \log \left(2t \delta_n\right)\right) \nonumber\\ & \quad + \frac{\log n}{2\left(1-n\right)}-\frac{\pi^4n\left(h_1-nh_n\right)^2}{4\left(1-n\right)^2\left(\log \left(2t\right)\right)^2} \nonumber\\ & \quad + q^2n \pi^4\frac{h_1-nh_n}{4\left(1-n\right)\left(\log \left(2t\kappa_n\right)\right)^2} + o\left(\log^{-2} t\right), \end{align} \tag{ 90 }$$

where the constants h_n , δ_n , κ_n have been defined in equation (59) and after equation (54), and

$$\begin{align} S^{\left(n\right)}\left(\mathrm{JW}_0\left(t\right)\right) = \frac{n+1}{3n}\log \left(4t\right)+\frac{2}{1-n}\Upsilon_n. \end{align} \tag{ 91 }$$

This is just the double of the entanglement entropy following a quench from a domain wall state [97]. Such a result is not surprising since it was already observed in [22] that the total OE of the JW string is exactly twice the entanglement entropy of a domain wall [98] and we find that the same is true in each symmetry sector. Hence, the equipartition of OE in the symmetry sectors is asymptotically restored with an asymptotic correction of the form $q^2/\log^2(t)$, with a non-trivial prefactor depending on non-universal quantities.

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The results of this and of the former subsection suggest that the breaking of the equipartition is a feature directly related to the finiteness of operator entanglement. The latter takes place for operators with a finite support (like σ_z , i.e. $c^{\dagger}_xc_x$) while extended operators, like the Jordan–Wigner strings, have diverging (in t) operator entanglement and restore equipartition for large times.

We conclude this section by commenting that one may also study the SROE or OE of operators like $Q = \sum_j c^{\dagger}_j c_j$ or the Hamiltonian itself, like the one in equation (40), even at time t = 0, i.e. before evolving them in the Heisenberg picture. For instance, we may study the SROE of Q by extending the results of section 4.1.2. We expect that, given that the charge is a sum of local operators, its SROE should satisfy an area law behaviour and a possible equipartition due to the presence of more charge sectors, and not only 3 as in the example reported here.