Symmetry enforces entanglement at high temperatures

TL;DR

This paper proves that certain symmetric quantum states can maintain entanglement at arbitrarily high temperatures, challenging the common notion that thermal states become classical and unentangled.

Contribution

It demonstrates that symmetry constraints can prevent the sudden death of entanglement in thermal states, extending previous understanding to fermionic systems and superselection rules.

Findings

Strongly symmetric thermal states remain entangled at high temperatures.

Fermionic Gibbs states exhibit persistent fermionic negativity at high temperatures.

Symmetry correlations can preserve quantum entanglement despite thermal decoherence.

Abstract

Many-body quantum systems with local interactions undergo ``sudden death of entanglement" at high temperatures, whereby thermal states become classical mixtures of product states. We investigate whether symmetry constraints can prevent this phenomenon. We prove that strongly symmetric thermal states (canonical ensemble) of generic Hamiltonians with on-site Abelian symmetries remain entangled with non-zero entanglement negativity at arbitrarily high temperatures, under mild conditions on the symmetry actions and the charge sector of the strong symmetry. Our results extend to weakly symmetric thermal states (Gibbs ensemble) under superselection rules, which restrict state decompositions to be symmetric. In particular, we show that fermionic Gibbs states evade sudden death of entanglement and have persistent fermionic negativity at high temperatures, proving along the way some existing conjectures about fermionic entanglement. These findings demonstrate that global symmetry correlations can preserve quantum entanglement despite thermal decoherence, providing new insights into the interplay between symmetry and quantum information in thermal equilibrium.

Tables

Table 1. Table 1: Typical behavior of entanglement measures ( E E , E S E_{S} ) as functions of the inverse temperature β = T − 1 \beta=T^{-1} for the Gibbs and canonical ensembles in the cases with and without superselection. E E ( E S E_{S} ) is a measure of (symmetric) entanglement, such as (fermionic) negativity. There is sudden death of entanglement for the Gibbs ensemble with no superselection rule if H H is local [ 6 ] , while there is not for the other cases if their respective entangling conditions ( EC and SEC ) are satisfied. For an example of the above, see Fig. 2 .

	No superselection	Superselection (e.g. fermions)
Gibbs ensemble $e^{-\beta H}$
Canonical ensemble $e^{-\beta H}{\Pi }_{\Lambda }$

Equations

$$\displaystyle\forall g\in G,\$$

$$\forall g\in G,[U_A(g),H]$$

$$\left(1-{\Pi }_{\Lambda }\right)(H{\Pi }_{\Lambda }{)}^{T_A}$$

$$(H^{{\mathcal{T}}_A}{)}^{†}\ne H^{{\mathcal{T}}_A}.$$

$${\Pi }_{\Lambda }{\rho }_f{\Pi }_{\Lambda }$$

$$F(\rho )=\mathrm{Tr}\rho H-{\beta }^{-1}S(\rho ).$$

$$F({\rho }_{\beta ,\Lambda })-F(\rho )\le 0.$$

$$\mathrm{Tr}\rho \ln {\rho }_{\beta ,\Lambda }=\mathrm{Tr}\rho \ln {\Pi }_{\Lambda }-\beta \mathrm{Tr}\rho H-\ln \mathrm{Tr}{\Pi }_{\Lambda }{e}^{-\beta H}.$$

$$F({\rho }_{\beta ,\Lambda })=\mathrm{Tr}{\rho }_{\beta ,\Lambda }H+{\beta }^{-1}\mathrm{Tr}{\rho }_{\beta ,\Lambda }\ln {\rho }_{\beta ,\Lambda }=-{\beta }^{-1}\ln \mathrm{Tr}{\Pi }_{\Lambda }{e}^{-\beta H}.$$

$$\mathrm{Tr}\rho \ln {\rho }_{\beta ,\Lambda }$$

$$=-\beta (F(\rho )-{\beta }^{-1}\mathrm{Tr}\rho \ln \rho )-\ln \mathrm{Tr}{\Pi }_{\Lambda }{e}^{-\beta H}$$

$$=\beta \left(F({\rho }_{\beta ,\Lambda })-F(\rho )\right)+\mathrm{Tr}\rho \ln \rho .$$

$$F({\rho }_{\beta ,\Lambda })-F(\rho )$$

$$=-{\beta }^{-1}D(\rho ||{\rho }_{\beta ,\Lambda }).$$

$${\rho }_{\Lambda }\propto \rho {\Pi }_{\Lambda }=\sum _{i,\lambda }{p}_i({\rho }_{A,i}{\Pi }_{\lambda }^A)\otimes ({\rho }_{B,i}{\Pi }_{\Lambda \overline{\lambda }}^B).$$

$${\Pi }_{\lambda }$$

$$\sum _{\lambda }{\Pi }_{\lambda }$$

$${\mathcal{H}}_A\simeq \underset{\lambda }{⨁}{V}_{\lambda }^{\otimes m_{\lambda }^A};\text{penalty}10000{\mathcal{H}}_B\simeq \underset{\mu }{⨁}{V}_{\lambda }^{\otimes m_{\mu }^A}$$

$$\mathcal{H}$$

$$M_{\Lambda }=\sum _{\lambda ,\mu }{\delta }_{\Lambda ,\lambda \mu }(m_{\lambda }^A+m_{\mu }^B)=\sum _{\lambda }(m_{\lambda }^A+m_{\Lambda \overline{\lambda }}^B).$$

$${\Pi }_{\Lambda }$$

$$=\frac{1}{|G{|}^2}\sum _{g,h\in G}\left(\sum _{\lambda }\overline{\lambda (g)}\lambda (h)\right)\overline{\Lambda (h)}{U}_A(g)\otimes U_B(h)$$

$$=\frac{1}{|G|}\sum _{g\in G}\overline{\Lambda (g)}U(g),$$

$$\mathrm{Tr}[{\Pi }_{\Lambda }]=\frac{d^N}{|G/K|}(1+O(e^{-N/\xi })),$$

$$\mathrm{Tr}[{\Pi }_{\Lambda }]=\frac{1}{|G|}\sum _{g\in G}\overline{\Lambda (g)}\mathrm{Tr}[u(g){]}^N.$$

$$\mathrm{Tr}[{\Pi }_{\Lambda }]$$

$$=\frac{d^N}{|G/K|}(1+O[(r/d{)}^N]).$$

$${\Vert {\mathrm{Tr}}_{A^C}[{\rho }_{\beta =0,\Lambda }]-1_A/d^{|A|}\Vert }_1=O(e^{-(N-|A|)/\xi }),$$

$${\mathrm{Tr}}_B[{\Pi }_{\Lambda }]$$

$$=1_A\frac{d^{|B|}}{|G/K|}(1+O(e^{-|B|/\xi })),$$

$${\mathrm{Tr}}_B[{\rho }_{\beta =0,\Lambda }]$$

$$=\frac{1_A}{d^{|A|}}(1+O(e^{-|B|/\xi })),$$

$${\Vert {\mathrm{Tr}}_B[{\rho }_{\beta =0,\Lambda }]-1_A/d^{|A|}\Vert }_1=O\left({\Vert 1_A/d^{|A|}\Vert }_1{e}^{-|B|/\xi }\right)=O(e^{-|B|/\xi }).$$

$${\Vert {\mathrm{Tr}}_{A^C}[{\rho }_{\beta ,\Lambda }]-{\mathrm{Tr}}_{A^C}[{\rho }_{\beta }]\Vert }_1=O(e^{-(c_{\beta }N-|A|)/{\xi }_{\beta }}),$$

Full text

Many-body systems equilibrate to their Gibbs distribution at finite temperatures by exchanging energy and information with a bath. The resulting state $\rho_{\beta}\propto e^{-\beta H}$ for system Hamiltonian $H$ at inverse temperature $\beta=T^{-1}$ is a mixed state that maximizes entropy under energy constraints, reflecting an observer’s ignorance of information leaked to the bath. For physical, local quantum Hamiltonians, the Gibbs state has structural properties at finite non-zero temperatures that reflect its low complexity, such as area law of mutual information [1] and of entanglement [2], and local Markov property [3, 4, 5]. Importantly, it was recently proven [6] that Gibbs states of local Hamiltonians undergo “sudden death of entanglement” (SDOE) at sufficiently high but finite temperatures in the thermodynamic limit, becoming exactly representable by a classical distribution of pure product states. This result confirmed evidence from previously studied few-body and solvable models [7, 8, 9, 10, 11, 12, 13, 14, 15] (See top left plot of Table 1).

Meanwhile, recent developments in mixed-state phases of matter have confirmed the essential role of symmetries [16, 17, 18, 19, 20, 21, 22, 23, 24] and uncovered novel connections to their global entanglement properties [25, 26, 27, 28, 29, 30, 31, 32, 33, 34]. In this Letter, we prove that symmetry restrictions on thermal states generically prevent SDOE. For that, we consider a symmetry group $G$, and a Hamiltonian $H$ that is symmetric under $G$: $\forall g\in G,[U(g),H]=0$, where $U:G\to U(\mathcal{H})$ is a symmetry representation of $G$ in the total Hilbert space $\mathcal{H}$. Gibbs states of local Hamiltonians $H$ that are symmetric also undergo SDOE, so we must impose a stronger symmetry constraint on the Gibbs state to see meaningful effects on entanglement. We consider two physically motivated ways of doing that:

One way is to impose strong symmetry in the thermal ensemble $\rho_{\beta}=\sum_{i}p_{i}\mathinner{\ket{\psi_{i}}\!\bra{\psi_{i}}}$ by limiting the mixture only to states $\ket{\psi}$ that have a fixed symmetry charge $\Lambda:G\to U(1)$, i.e. $U(g)\ket{\psi}=\Lambda(g)\ket{\psi}$. This constraint results in the canonical ensemble $\rho_{\beta,\Lambda}\propto e^{-\beta H}\Pi_{\Lambda}$, where $\Pi_{\Lambda}$ projects onto the charge sector $\Lambda$. By definition, $\rho_{\beta,\Lambda}$ is strongly symmetric under $G$, satisfying $U(g)\rho_{\beta,\Lambda}=\Lambda(g)\rho_{\beta,\Lambda}$, which is a stronger requirement than the usual weak symmetry condition $U(g)\rho U(g)^{\dagger}=\rho$ [35, 36, 37]. The canonical ensemble naturally arises when an initial strongly symmetric system is put into contact with a thermal bath in a way that preserves the symmetry of the system alone (See [38] for a proof).

Alternatively, we can impose a superselection rule by considering a restricted quantum theory consisting only of states that are (weakly) symmetric under $G$ [39, 40, 41]. This restriction can either be fundamental to physical reality, as in the case of the fermion parity superselection rule [42], or be put by hand, as in real quantum theory [43, 44, 45]. Crucially, the superselection rule alters the definition of entanglement itself [46, 47, 48, 49, 50]. Namely, we call a bipartite state $\rho_{AB}$ symmetrically separable if it is the mixture of tensor product states $\rho_{A}\otimes\rho_{B}$ whose parts, $\rho_{A}$ and $\rho_{B}$, are also valid states within the superselected theory, and thus weakly symmetric by themselves [51]. This is generally stronger than the usual separability condition [52], since there can be separable states that are not symmetrically separable.

These two symmetry conditions on $\rho_{\beta}$ have similar entanglement properties in the following sense: $\rho_{\beta}$ is symmetrically entangled (i.e. not symmetrically separable) if, and only if, the canonical ensemble $\rho_{\beta,\Lambda}$ is also entangled for some charge $\Lambda$ (See [38] for a simple proof). Then, we are able to prove the persistence of entanglement 111Not to be confused with the robustness measure of multipartite entanglement called “persistency of entanglement” [75, 76]. for both cases at high temperatures (See Fig. 1):

Gibbs ensemble $\rho_{\beta}$ is symmetrically entangled, and 2. 2.

canonical ensemble $\rho_{\beta,\Lambda}$ is entangled with non-zero entanglement negativity, for almost all symmetry actions and irreps $\Lambda$ in the thermodynamic limit.

The second part of the theorem above is stronger than the first in two ways: first, we provide necessary and sufficient conditions for $\rho_{\beta,\Lambda}$ to be entangled, with $\Lambda$ being arbitrarily chosen; and, second, $\rho_{\beta,\Lambda}$ is not only entangled, but has nonzero negativity, which is a computable measure of entanglement [54, 55, 56, 57]. The phrase “for almost all symmetry actions and irreps” excludes the condition of semiuniformity (Def. 1), which is absent in most examples of interest, such as finite $G$ at sufficiently large system size.

We only consider Abelian symmetry groups with on-site action because their Gibbs states are fully symmetrically separable at infinite temperature. In contrast, it has been found that non-Abelian symmetries [25, 26, 27] and anomalous non-on-site symmetries [28, 30, 31, 32, 33, 34] can guarantee entanglement even at infinite temperature, so it would not be as surprising if the entanglement of their symmetric thermal ensembles persists at high temperatures.

Recent works report similar findings on the persistence of entanglement [58] and negativity [59] in canonical ensembles at high temperatures. Our work provides significant generalization through distinct proof techniques and complete characterization of symmetries and irreps with persistent entanglement. Furthermore, we unify the “no SDOE” result for canonical ensembles with the persistence of symmetric entanglement in thermal states under superselection rule, which were previously noted for fermionic systems [49, 60, 61, 62, 63] and $U(1)$-symmetric systems [51]. More specifically for fermions, we are also able to show an equivalent version of Theorem 1 where the entanglement negativity measure used is the appropriate one for fermionic systems [64, 65], along the way proving existing conjectures in [65, 61].

Setup and notation— We consider a multipartite Hilbert space with $N<\infty$ identical sites, $\mathcal{H}=\bigotimes_{i=1}^{N}\mathcal{H}_{i}$, $\forall i,j.\mathcal{H}_{i}\simeq\mathcal{H}_{j}$. The symmetry group $G$ is Abelian and acts as a on-site unitary $U:G\to\mathcal{L}(\mathcal{H})$, $U=\bigotimes_{i}u_{i}$ in a non-trivial way on each site, i.e. $\exists g\in G,u(g)\not\propto\mathbb{1}$. Sometimes, we will specialize to homogeneous symmetry actions $U=u^{\otimes N}$.

From the symmetry action of $G$ on the Hilbert space $\mathcal{H}$, it is decomposed into $\mathcal{H}=\bigoplus_{\lambda}V_{\lambda}^{\otimes m_{\lambda}}$, where the direct sum is over all irreps $\lambda$ of $G$, with $m_{\lambda}$ being the multiplicities. Since $G$ is Abelian, each irrep space $V_{\lambda}$ is one-dimensional, and so the irreps are identified with their characters, $\lambda:G\to U(1)$. By Schur’s lemma, a weakly symmetric state $\rho$ can be decomposed into strongly symmetric block diagonal components, $\rho=\sum_{\lambda}q_{\lambda}\rho_{\lambda}$, where $\rho_{\lambda}=\rho\Pi_{\lambda}/q_{\lambda}$ with $\Pi_{\lambda}=\sum_{g\in G}\overline{\lambda(g)}U(g)$ being the projector onto the isotypic component $V_{\lambda}^{\otimes m_{\lambda}}$, and $q_{\lambda}=\operatorname{Tr}[\rho\Pi_{\lambda}]$. We shall call $\sum_{\lambda}q_{\lambda}\rho_{\lambda}$ the isotypic decomposition of $\rho$.

A state $\rho$ is separable ($\rho\in\textsf{SEP}$) with respect to a bipartition $A|B$ if it can be represented as a convex mixture of tensor product states, $\rho=\sum_{i}p_{i}\rho_{A,i}\otimes\rho_{B,i}$ [52]. If a separable state $\rho$ is weakly symmetric under a bipartite symmetry representation $U=U_{A}\otimes U_{B}$, and each $\rho_{A,i}$ and $\rho_{B,i}$ is weakly symmetric under $U_{A}$ and $U_{B}$ respectively, then we say $\rho$ is symmetrically separable ($\rho\in\textsf{SymSEP}$) 222For on-site symmetries this is in fact equivalent to the following definition of symmetric separability [49]: $\rho$ is symmetrically separable for a symmetry $U$ if there exists a decomposition $\rho=\sum_{i}p_{i}\rho_{i}$, such that $[\rho_{i},U]=0$ and $\rho_{i}$ is a tensor product state..

Persistence of entanglement— Now consider the Gibbs state $\rho_{\beta}\propto e^{-\beta H}$. If $H$ is local, $\rho_{\beta}$ becomes fully separable at high temperatures. We will now give conditions under which not only symmetry constraints prevent SDOE, but also leads to the persistence of entanglement (POE) at high temperature, by which we mean that there exists a $\beta^{*}>0$ such that $\rho_{\beta}$ is symmetrically entangled if $\beta<\beta^{*}$, with an analogous statement for $\rho_{\beta,\Lambda}$.

We proceed by assuming that the entanglement of $\rho_{\beta}$ does not persist. That is equivalent to $\rho_{\beta}$ being symmetrically separable for all $\beta$ in a decreasing sequence $(\beta_{n})_{n\in\mathbb{N}}$ converging to zero: $\rho_{\beta}=\sum_{i}p_{i}\rho_{\beta,A,i}\otimes\rho_{\beta,B,i}$ for a bipartition $A|B$. Since the $\rho_{\beta,A,i}$ states are individually symmetric, we have $\forall g\in G,[U_{A}(g),\rho_{\beta}]=\sum_{i}p_{i}[U_{A}(g),\rho_{\beta,A,i}]\otimes\rho_{\beta,B,i}=0$. Since each matrix element of $[U_{A}(g),\rho_{\beta}]$ is a holomorphic function of $\beta$ that is zero on the set $\{\beta_{n}\}$ with the accumulation point $\beta=0$, it is identically zero for all $\beta\in\mathbb{C}$. In particular, its derivative at $\beta=0$ must also be zero, which leads to a necessary condition for symmetric separability at high temperatures: $\forall g\in G,\ [U_{A}(g),H]=0$.

By the converse of the reasoning above, we arrive at a sufficient condition for the Gibbs state to be symmetrically entangled at arbitrarily high temperatures $\beta\in(0,\beta^{*})$ 333An equivalent symmetric entangling condition was found in [51] for $U(1)$ symmetries.,

[TABLE]

By the correspondence between the symmetric separability of $\rho_{\beta}$ and the separability of $\rho_{\beta,\Lambda}$, SEC also implies the existence of at least one irrep $\Lambda$ whose canonical state exhibits POE. Beyond this, following the same argument above for $\rho_{\beta,\Lambda}\propto e^{-\beta H}\Pi_{\Lambda}$, we can reach a different sufficient condition for POE for any given $\Lambda$,

[TABLE]

Combining these results, we arrive at the following theorem, which is summarized in Table 1

For a symmetric Hamiltonian $H$ that satisfies SEC, the Gibbs ensemble is symmetrically entangled, and $\exists$ an irrep $\Lambda$ such that the canonical ensemble is entangled at arbitrarily high temperatures. 2. 2.

For a symmetric Hamiltonian $H$ and an irrep $\Lambda$ that satisfies EC, the canonical ensemble is entangled at arbitrarily high temperatures.

Genericness and local indistinguishability — We now intend to prove that SEC and EC are generically satisfied. The crucial property of these conditions is that they are linear in $H$; for any $H$ that violates $\textsf{SEC}/\textsf{EC}$, adding a perturbation $V$ that satisfies $\textsf{SEC}/\textsf{EC}$ ensures that $H+\epsilon V$ also satisfies for all $\epsilon\neq 0$. Thus, proving the existence of entangling perturbations suffices to prove that fine-tuning $H$ is required to violate $\textsf{SEC}/\textsf{EC}$.

For example, consider a qubit chain with $G=\mathbb{Z}_{2}$ symmetry generated by $\bigotimes_{i}Z_{i}$. A fine-tuned Hamiltonian $H_{Z}$ consisting solely of $Z$ terms yields separable thermal states $\rho_{\beta}$ and $\rho_{\beta,\lambda}$. However, the operators $X_{i}$ and $Y_{i}$ carry charge $-1$, so $V_{ij}=X_{i}X_{j}$ or $Y_{i}Y_{j}$ is an entangling perturbation when $i\in A$ and $j\in B$, making $H_{Z}+\epsilon V_{ij}$ satisfy both EC and SEC, even for infinitesimal $\epsilon$.

For other symmetry groups, an entangling $V$ that satisfies SEC can be similarly constructed by pairing a charged operator in $A$ with an oppositely charged one in $B$, so that $[U(g),V]=0$ but $[U_{A}(g),V]\neq 0$. The EC  however, is harder to establish to in general, due to the projector $\Pi_{\Lambda}$. In fact, there are some charge sectors for which EC is not satisfiable: e.g. the maximal or the minimal charge irrep of a $U(1)$ action on a system of qubits as $\theta\mapsto e^{i\theta\sum_{i}Z_{i}}$ leads to an unentangled canonical ensemble, as the projector is rank one, negating ECregardless of $V$.

We can avoid such scenarios if we specialize to finite Abelian symmetries with homogeneous symmetry action, $U=u^{\otimes N}$, which allows us to use local indistinguishability of symmetric subspaces to justify the existence of $V$ satisfying both EC and SEC. Consider a large tripartition $A|B|C$, where $B$ is a buffer region that separates $A$ and $C$, taken to be large enough such that there are no local Hamiltonian terms in $H$ that is supported on $AC$. Since $\Pi_{\Lambda}$ is maximally mixed with just a global charge constraint, we expect \operatorname{Tr}_{C}\Pi_{\Lambda}\mathchoice{\mathrel{\raise 1.07639pt\hbox{\hbox to0.0pt{\hbox{\displaystyle\propto}\hss}\lower 4.03563pt\hbox{\displaystyle\sim}}}}{\mathrel{\raise 1.07639pt\hbox{\hbox to0.0pt{\hbox{\textstyle\propto}\hss}\lower 4.03563pt\hbox{\textstyle\sim}}}}{\mathrel{\raise 0.75346pt\hbox{\hbox to0.0pt{\hbox{\scriptstyle\propto}\hss}\lower 2.95741pt\hbox{\scriptstyle\sim}}}}{\mathrel{\raise 0.5382pt\hbox{\hbox to0.0pt{\hbox{\scriptscriptstyle\propto}\hss}\lower 2.18024pt\hbox{\scriptscriptstyle\sim}}}}\mathbb{1}_{AB}, which we can quantitatively establish for finite Abelian groups (See [38]). This implies that tracing out a sufficiently large region disjoint from the support of $V$ and $A$ reduces $[U_{A}(g),V]\Pi_{\Lambda}=0$ to $[U_{A}(g),V]\approx 0$, with equality in the thermodynamic limit. In fact, for finite Abelian groups, we can further show that SEC and EC are actually equivalent for large enough regions in the thermodynamic limit (See [38]).

Not only is $\rho_{\beta,\Lambda}$ entangled when EC is satisfied, but it is also locally equivalent to $\rho_{\beta}$. We prove this for $k$-local Hamiltonians symmetric under finite $G$ at high temperatures, in [38].

Classification of irreps— Using local indistinguishability to construct entangling perturbations has two limitations: it assumes large system sizes, and its rigorous formulation requires finite Abelian groups. As already seen, infinite groups like $U(1)$ have irreps that are not only locally distinguishable to identity, but also lead to unentangled canonical ensembles. We now present a construction of entangling perturbations that goes beyond the local indistinguishability argument, using the property of semiuniformity of irreps.

A small-sized example of semiuniform irrep is the trivial irrep $\Lambda=1$ of a $\mathbb{Z}_{3}$ action on $N=3$ qubits generated by $\tilde{Z}_{1}\tilde{Z}_{2}\tilde{Z}_{3}$, where $\tilde{Z}=\operatorname{diag}(1,e^{2\pi i/3})$. In this case, $\Pi_{\Lambda=1}$ projects onto the subspace generated by the uniform states $\ket{000}$ and $\ket{111}$. An example of an uniform irrep is the maximal or the minimal charge irrep of the aforementioned $U(1)$ symmetry on qubits.

By excluding semiuniform irreps, we are able to construct two-body entangling Hamiltonians:

is not semiuniform, then for any pair of sites $(i,j)$, there exists a two-body symmetric Hamiltonian $v_{i,j}$ supported on $\{i,j\}$ that entangles any bipartition that separates site $i$ from site $j$. Moreover, given a connected interaction graph, the uniform nearest-neighbor perturbation $\sum_{\langle i,j\rangle}v_{i,j}$ entangles all bipartitions. 2. 2.

is semiuniform, then there is no entangling Hamiltonian with support over $N-1$ sites or fewer, and for all $(N-1)$-local Hamiltonian, $\rho_{\beta,\Lambda}$ is fully separable when $\rho_{\beta}$ is fully separable.

We now sketch the construction of the entangling $v_{i,j}$, leaving a full proof to the Supplemental Materials (S.M.)[38]. If the irrep is not semiuniform, then for any pair of sites $i,j$ there exist at least two local irreps $\lambda^{\prime}\neq\lambda$. Then, let $\sigma_{i}^{+}\vcentcolon=\!\mathinner{\ket{\lambda^{\prime}}\!\bra{\lambda}}_{i}$ and $\sigma_{j}^{+}\vcentcolon=\!\mathinner{\ket{\lambda^{\prime}}\!\bra{\lambda}}_{j}$. Denoting their Hermitian conjugates by $\sigma_{i,j}^{-}\vcentcolon=(\sigma_{i,j}^{+})^{\dagger}$, we choose $v_{i,j}\vcentcolon=\sigma_{i}^{+}\sigma_{j}^{-}+\sigma_{i}^{-}\sigma_{j}^{+}$. It is easy to check that $v_{i,j}$ is symmetric and satisfies the entangling condition Eq. Symmetry enforces entanglement at high temperatures if $i\in A$, since it pairs a charged operator in $A$ with an opppositely charged one in $B$.

In fact, we can also show that semiuniform irreps are rare: if the symmetry is homogeneous and independent of the number of sites, the fraction of semiuniform global irreps goes to zero in the thermodynamic limit as $O(2^{d}/N)$, where $d$ is the on-site Hilbert space dimension. Furthermore, if $G$ is finite, there are no semiuniform global irreps for $N>|G|$ [38].

Persistence of negativity— We can also ask whether $\rho_{\beta,\Lambda}$ remains entangled while exhibiting zero negativity. This question was answered negatively in Ref. [51] for $U(1)$ symmetries and in Ref. [59] for the classical Ising Hamiltonian $H=\sum_{i}Z_{i}Z_{i+1}$ with $\mathbb{Z}_{2}$ symmetry generated by $\prod_{i}X_{i}$. In S.M. [38], we report the same conclusion for the cluster chain Hamiltonian $H=\sum_{i}Z_{i-1}X_{i}Z_{i+1}$ with the same symmetry (See Fig. 2).

We generalize these examples by showing that the entanglement condition of Eq. (Symmetry enforces entanglement at high temperatures) also implies nonzero negativity, and thus the absence of bound entanglement with positive partial transpose (PPT) 444The logarithmic negativity upper bounds distillable entanglement; however non-zero negativity does not necessarily imply no bound or undistillable entanglement, as there may exist bound entanglement which is nevertheless non-PPT.. To establish this, we begin by defining the basis with respect to which the partial transpose is taken.

For the case of on-site Abelian symmetries $U=\bigotimes_{i}u_{i}$, we choose to do the partial transpose in a region $A$ with respect to a basis that diagonalizes $U_{A}$, such that $\forall g\in G,U_{A}(g)^{T_{A}}=U_{A}(g)$, and thus we would have $\Pi_{\Lambda}^{T_{A}}=\Pi_{\Lambda}$. Now consider the operator $\tilde{\Pi}_{\Lambda}\,\rho_{\beta,\Lambda}^{T_{A}}\,\tilde{\Pi}_{\Lambda}$, with $\tilde{\Pi}_{\Lambda}\vcentcolon=\mathbb{1}-\Pi_{\Lambda}$. It is traceless, as $\operatorname{Tr}(\tilde{\Pi}_{\Lambda}\,\rho_{\beta,\Lambda}^{T_{A}}\,\tilde{\Pi}_{\Lambda})\propto\operatorname{Tr}(e^{-\beta H}\Pi_{\Lambda}\tilde{\Pi}_{\Lambda}^{T_{A}})=0$, and if $\rho_{\beta,\Lambda}$ has PPT, then $\tilde{\Pi}_{\Lambda}\,\rho_{\beta,\Lambda}^{T_{A}}\,\tilde{\Pi}_{\Lambda}$ is also positive semidefinite. Hence, it must be identically zero. Just like before, assuming $\rho_{\beta,\Lambda}$ has PPT in a sequence of $\beta$ converging to zero implies the derivative of $\tilde{\Pi}_{\Lambda}\,\rho_{\beta,\Lambda}^{T_{A}}\,\tilde{\Pi}_{\Lambda}$ at $\beta=0$ must vanish, which gives a necessary condition for zero negativity at high temperatures: $(\mathbb{1}-\Pi_{\Lambda})(H\Pi_{\Lambda})^{T_{A}}(\mathbb{1}-\Pi_{\Lambda})=0$.

Negating this necessary condition gives a sufficient one for POE with non-zero negativity for a bipartition $A|B$:

[TABLE]

In S.M. [38], we show the equivalence of the entangling and negativity conditions, $\text{EC}\iff\text{NC}$. Combining the results so far, we have the general implication for POE in canonical ensembles, and all the results quoted in Theorem 1.

Fermions— We now discuss fermionic systems, where the physically relevant states are always (weakly) symmetric under fermion parity. Consider a local Hilbert space $\mathcal{H}$ and an operator algebra spanned by mutually anticommuting Majorana operators $\{c_{j}\}$ with $\{c_{j},c_{k}\}=2\delta_{jk}$, with the fermion parity operator $P\equiv\prod_{i}c_{i}$. Physical density operators preserve $P$ and their set is referred to as $\mathcal{S}(\mathcal{H})$. Any (symmetrically) separable state under a bipartition $A|B$ must also preserve the local parity operators $P_{A},P_{B}$, where $P_{A(B)}\equiv\prod_{i\in A(B)}c_{i}$ [49]. Our previous results thus imply that the fermion parity superselection rule generically protects entanglement from sudden death at high temperatures, agreeing with previous expectation [14]. Here, we establish a stronger result: even a computable entanglement measure, the fermionic negativity [64, 65], generically persists at arbitrarily high temperatures for both the Gibbs and the canonical ensembles.

Any density matrix $\rho\in\mathcal{S}(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$ can be expanded in Majorana monomials $a_{p_{1}}\cdots a_{p_{k_{1}}}b_{q_{1}}\cdots b_{q_{k_{2}}}$, where $a_{p_{j}}$ and $b_{q_{j}}$ act on $A$ and $B$ respectively, and $k_{1}+k_{2}$ is even by fermion parity. The fermionic partial transpose $\mathcal{T}_{A}$ multiplies such monomial by the phase $i^{k_{1}}$. The fermionic negativity, defined as $E_{N}^{f}=\log||\rho^{\mathcal{T}_{A}}||_{1}$ is a more sensitive measure of fermionic entanglement [65] than its “bosonic” counterpart [69, 70], which is defined without the phase factor $i^{k_{1}}$.

We proceed by contradiction to show that $E_{N}^{f}$ persists generically for the Gibbs ensemble $\rho_{\beta}\propto e^{-\beta H}$, conjectured in [61]. Assume $\rho_{\beta}$ has zero fermionic negativity, i.e. $||\rho_{\beta}^{\mathcal{T}_{A}}||_{1}=1$, for a sequence of $\beta$ converging to [math]. This, combined with the fact that $\operatorname{Tr}{\rho_{\beta}^{\mathcal{T}_{A}}}=\operatorname{Tr}{\rho_{\beta}}=1$, implies that $\rho_{\beta}^{\mathcal{T}_{A}}$ is Hermitian 555This is proved in Lemma 2.2. of [77]: $||A||_{1}=\operatorname{Tr}{A}$ if, and only if, $A\geq 0$.. By taking the derivative of $\rho_{\beta}^{\mathcal{T}_{A}}-\left(\rho_{\beta}^{\mathcal{T}_{A}}\right)^{\dagger}=0$ at $\beta=0$, we have a necessary condition for zero $E_{N}^{f}$ at high temperature: $H^{\mathcal{T}_{A}}-(H^{\mathcal{T}_{A}})^{\dagger}=0,$ whose negation gives a sufficient condition for the persistence of $E_{N}^{f}$,

[TABLE]

This is satisfied by the two-Majorana term $V=ic_{i}c_{j}$, with $i\in A$ and $j\in B$, since $(ic_{i}c_{j})^{\mathcal{T}_{A}}=-c_{i}c_{j}$ is not Hermitian. Note that this entangling condition never occurs for the “bosonic” negativity, since $H^{T_{A}}$ is always Hermitian. The same entangling perturbation also leads to the persistence of $E^{f}_{N}$ for the canonical ensemble with fermion parity. A similar idea also resolves a conjecture in [65], that fermionic states mixing local fermion parity have non-vanishing fermionic negativity, as their partial transposed state is necessarily not Hermitian. We explain both in S.M. [38].

Discussions— Our results imply the persistence of entanglement at high temperatures in two related scenarios: when the thermal state is enforced to be strongly symmetric (canonical ensemble), and when the symmetry is enforced at the level of decomposition of the mixed state into pure states (Gibbs ensemble with superselection). We also discuss the generic nature of this entanglement, showing the existence of local Hamiltonian perturbations which lead to entanglement at arbitrarily high temperatures, as well as a complete classification of charge sectors with POE for homogeneous on-site global Abelian symmetries. Ultimately, genericness follows because the set of symmetrically separable states is contained in a lower dimensional subspace given by linear constraints.

These results can be generalized in various ways. Firstly, the results of the sufficient conditions for POE also extend to Abelian higher form and subsystem symmetries. In fact, strongly symmetric versions of toric codes (with Abelian one form symmetries) already show a persistence of negativity, as can be inferred from [13]. A more careful study of higher form and other symmetries, including antiunitary symmetries, is left for a future study.

Some open questions about the quantification of entanglement at high temperatures remain: our results only imply that a faithful measure of entanglement (and negativity) will remain non-zero at arbitrarily high temperatures for any fixed system size $N$, but it could vanish as $N\to\infty$. Calculations on stabilizer models, however, suggest that the entanglement negativity does not vanish at non-zero $\beta$ even in the thermodynamic limit $N\to\infty$, which will be useful to establish more generally. This is also relevant to the question of local detectability of the entanglement in $\rho_{\beta,\Lambda}$ for temperatures past SDOE of $\rho_{\beta}$.

It is also interesting to explore whether quantum resources beyond entanglement, such as magic and Gaussianity [63], undergo sudden death for the canonical ensemble. Lastly, it would be useful to investigate the complexity of preparation of the canonical ensemble, in view of the recent interest in similar questions about the Gibbs ensemble [72, 73, 74, 6].

I Emergence of canonical ensemble

Consider a Hamiltonian $H$ with a symmetry described by a group $G$, where $H$ commutes with the unitary representations $U(g)$ for all $g\in G$. Consider a one dimensional irrep $\Lambda$ of the symmetry, labeled by the eigenvalues of the group generators, $U(g)\ket{\Lambda,\alpha}=\Lambda(g)\ket{\Lambda,\alpha}$, with $\alpha$ labeling the multiple representations of the irrep in the local tensor product Hilbert space. Note that the requirement of a one dimensional irrep is naturally satisfied for an Abelian group.

The projector onto the irrep space of $\Lambda$ is given by $\Pi_{\Lambda}=\sum_{\alpha}\ket{\Lambda,\alpha}\bra{\Lambda,\alpha}$. A strongly symmetric mixed state $\rho$ in the symmetric subspace is one that satisfies the constraint $\rho=\Pi_{\Lambda}\rho\Pi_{\Lambda}$. A strongly symmetric quantum channel $\mathcal{E}=\sum_{a}K_{a}(\cdot)K_{a}^{\dagger}$ is one whose Kraus operators $\{K_{a}\}$ commute with the symmetry: $\forall g\in G,[K_{a},U(g)]=0$ [35, 36, 37]. Importantly, a strongly symmetric quantum channel preserved the strong symmetry:

Lemma 1**.**

A strongly symmetric mixed state remains in the symmetric subspace when evolved with a strongly symmetric quantum channel.

Proof.

Suppose the initial state $\rho_{i}$ is in the symmetric subspace, i.e., $\rho_{i}=\Pi_{\Lambda}\rho_{i}\Pi_{\Lambda}$. Then, the evolved state $\rho_{f}=\mathcal{E}(\rho_{0})$ under a strongly symmetric channel $\mathcal{E}=\sum_{a}K_{a}(\cdot)K_{a}^{\dagger}$ also satisfies the condition:

[TABLE]

∎

Now, we want to show that the canonical ensemble is a state that minimizes the free energy associated with a local Hamiltonian when the states are restricted to a symmetric subspace. This would imply that such a state would naturally be the steady state of a symmetric mixing quantum process in the presence of a bath.

The free energy associated with a state at a temperature $\beta^{-1}$ is given by,

[TABLE]

We can prove the following result,

Lemma 2**.**

The canonical ensemble minimizes the free energy for all mixed states in the symmetric subspace. Specifically, for any state $\rho$ such that $\rho=\Pi_{\Lambda}\rho\Pi_{\Lambda}$, we have,

[TABLE]

Proof.

We study the quantity,

[TABLE]

For strongly symmetric states $\rho=\Pi_{\Lambda}\rho\Pi_{\Lambda}$, the first term above, $\operatorname{Tr}\rho\ln\Pi_{\Lambda}$, is zero, since $\ln\Pi_{\Lambda}$ is zero on the support of $\rho$.

For the canonical ensemble, we have,

[TABLE]

For any strongly symmetric state $\rho$, we have,

[TABLE]

Thus,

[TABLE]

By the positivity of relative entropy, we have the desired result. ∎

II Two qubit example with $\mathbb{Z}_{2}$ symmetry

In this section, we pedagogically examine the case of two qubits with the $\mathbb{Z}_{2}$ symmetry generated by $X_{1}X_{2}$. The persistence of entanglement results of Theorem 2 are valid for such symmetry action, but some others are not due to the small system size. More specifically, the projection term $\Pi_{\lambda}$ cannot be removed from the EC $[U_{A}(g),H]\neq 0$.

Indeed, consider the 2-qubit XYZ Hamiltonian $H=X_{1}X_{2}+J(\frac{1+\gamma}{2}Y_{1}Y_{2}+\frac{1-\gamma}{2}Z_{1}Z_{2})=X_{1}X_{2}+JY_{1}Y_{2}(\gamma\Pi_{+1}+\Pi_{-1})$, with added restriction to the $\mathbb{Z}_{2}$-symmetric sector $X_{1}X_{2}=+1$. The negativity of its canonical ensemble state $\rho_{\beta,+1}\propto e^{-\beta H}\Pi_{+1}$ can be computed to be $\mathcal{N}(\rho_{\beta,+1})=\frac{1}{2}|\tanh(\beta J\gamma)|$. Thus, for all temperatures, $\rho_{\beta,+1}$ is separable if $\gamma=0$. However, if $J\neq 0$, $H$ does not commute with the partial symmetry action and the SEC is satisfied: $[X_{1},H]=iJ(Z_{1}Y_{2}-Y_{1}Z_{2})\neq 0$, implying persistence of symmetric entanglement. If one instead includes the projection $\Pi_{+1}$, then we have $[X_{1},H\Pi_{+1}]=2iJ\gamma Z_{1}Y_{2}\Pi_{+1}=0$, as expected from the separability of $\rho_{\beta,+1}$. One might notice that this counterexample relies on the projections $\Pi_{\pm 1}$ directly appearing in the expression of the Hamiltonian, which should not happen for a system of many parties and a local Hamiltonian.

We can visually represent this example in the case of $\gamma=0$ with the diagram in Fig. 3. It is an accurate representation of a slice in the space of quantum states akin to Fig. 1. We can see that the Gibbs state (black dashed line, $J=1$) becomes symmetrically entangled as soon as $\beta>0$, but has sudden death of (non-symmetric) entanglement at low enough $\beta$. At the same time, the canonical ensemble (thick line from $\frac{1}{2}\Pi_{-}$ to $\ket{\Psi_{-}}$) becomes entangled as soon as $\beta>0$.

III Symmetric entanglement

We briefly recall the definition of separability and symmetric separability. A state $\rho$ is separable for a bipartition $A|B$ if it can be represented as a convex mixture of tensor product states, $\rho=\sum_{i}p_{i}\rho_{A,i}\otimes\rho_{B,i}$. A separable state $\rho$ is also symmetrically separable if all $\rho_{A,i}\otimes\rho_{B,i}$ are also (weakly) symmetric.

Following these definitions, and the definitions of weak and strong symmetries, we prove the following lemmas,

Lemma 3**.**

A strongly symmetric state $\rho$ is symmetrically separable if, and only if, it is separable.

Proof.

$(\Rightarrow)$ is true for any state. For the $(\Leftarrow)$ direction, given a separable and strongly symmetric state $\rho=\sum_{i}p_{i}\rho_{A,i}\otimes\rho_{B,i}$, each tensor product state $\rho_{A,i}\otimes\rho_{B,i}$ is also strongly symmetric due to the inheritance property of the strong symmetry [29]. In particular, they are also weakly symmetric. ∎

Lemma 4**.**

If the symmetry action $U:G\to U(\mathcal{H})$ is on-site, $U=U_{A}\otimes U_{B}$, then a weakly symmetric state $\rho$ is symmetrically separable if, and only if, each of its strongly symmetric parts $\rho_{\lambda}\propto\rho\Pi_{\Lambda}$ is separable.

Proof.

$(\Rightarrow)$ Since the irrep projector $\Pi_{\Lambda}$ decomposes as $\Pi_{\Lambda}=\sum_{\lambda}\Pi_{\lambda}^{A}\Pi^{B}_{\Lambda\overline{\lambda}}$, then if $\rho=\sum_{i}p_{i}\rho_{A,i}\otimes\rho_{B,i}$, we have the following separable decomposition of $\rho_{\lambda}$:

[TABLE]

$(\Leftarrow)$ follows from $\rho$ being a convex mixture of its strongly symmetric parts $\rho_{\lambda}$, each of which is separable, and thus symmetrically separable by Lemma 3. ∎

These two lemmas prove our assertion in the main text: $\rho$ is symmetrically entangled (i.e. not symmetrically separable) if, and only if, its strongly symmetric component $\rho_{\Lambda}$ is also entangled for some charge $\Lambda$.

IV Local indistinguishability of the canonical and Gibbs ensembles

In this section, we explore whether the canonical ensemble $\rho_{\beta,\Lambda}$ is locally indistinguishable from the Gibbs ensemble $\rho_{\beta}$, up to an exponentially small error. We establish this for $\beta=0$ for finite Abelian groups — which we use to develop intuition about the equivalence of EC and SEC in the main text — and for small enough $\beta<\beta_{*}$ if $H$ is a $k$-local Hamiltonian.

We start by collecting some known facts about the symmetries and their representations. From the symmetry action of $G$ on the Hilbert space $\mathcal{H}$, it gets decomposed into $\mathcal{H}=\bigoplus_{\lambda}V_{\lambda}^{\otimes m_{\lambda}}$, where the direct sum is over all irreps $\lambda$ of $G$, with $m_{\lambda}$ being the multiplicities. The projectors onto $V_{\lambda}^{\otimes m_{\lambda}}\subseteq\mathcal{H}$ can be expressed as

[TABLE]

where $\lambda(g)$ is the character. They form an orthogonal resolution of identity. Namely,

[TABLE]

For a bipartition $\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, each Hilbert space $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ also decomposes into a direct sum of irrep spaces of $U_{A}$ and $U_{B}$:

[TABLE]

where both direct sums are over all 1d irreps $\lambda$ and $\mu$ of $G$, with multiplicities $m_{\lambda}^{A}$ and $m_{\mu}^{B}$, respectively. Note that the multiplicities can be zero. Similar constructions as 13 can be done for projectors onto $V_{\lambda}^{\otimes m^{A}_{\lambda}}\subset\mathcal{H}_{A}$ and $V_{\lambda}^{\otimes m^{B}_{\lambda}}\subset\mathcal{H}_{B}$.

Two irreps, $\lambda$ and $\mu$, fuse together in the tensor product to an irrep $\lambda\otimes\mu\simeq\lambda\cdot\mu$ of $U=U_{A}\otimes U_{B}$, such that

[TABLE]

where

[TABLE]

Similarly, the projector onto the irrep $\Lambda$ of $\mathcal{H}$ decomposes as,

[TABLE]

where, in the third line, we used Schur’s orthogonality relations for Abelian groups: $\sum_{\lambda}\overline{\lambda(g)}\lambda(h)=|G|\delta_{g,h}$. Similar identities can be derived for multipartitions.

With these tools we can prove the following lemma:

Lemma 5**.**

Given a finite Abelian group $G$, and an homogeneous symmetry action $U=u^{\otimes N}$, with $u$ acting on a $d$-dimensional on-site Hilbert space, then, as $N\to\infty$,

[TABLE]

where $K=\{g\in G\mid u(g)\propto\mathbb{1}\}$, and $\xi=-1/\log(\max_{g\notin K}\left|\operatorname{Tr}[u(g)]\right|/d)$.

Proof.

Using that $\Pi_{\Lambda}=\sum_{g\in G}\overline{\Lambda(g)}U(g)$, we have

[TABLE]

From the above, we can see that the behavior of $\operatorname{Tr}[\Pi_{\Lambda}]$ as $N\to\infty$ is dictated by the largest values of $\left|\operatorname{Tr}[u(g)]\right|$. Since $u(g)$ is unitary, then $\left|\operatorname{Tr}[u(g)]\right|\leq d$ for all $g\in G$, and $\left|\operatorname{Tr}[u(g)]\right|=d$ if, and only if, $u(g)\propto\mathbb{1}$. The set of elements satisfying this relation form the normal subgroup $K=\{g\in G\mid u(g)\propto\mathbb{1}\}$, and the value of $\Lambda$ there is fixed by the equation $\Lambda(h)\Pi_{\Lambda}=U(h)\Pi_{\Lambda}=(\operatorname{Tr}[u(h)]/d)^{N}\Pi_{\Lambda}$, where $h\in K$. In the last step, we have used that $u(h)=\operatorname{Tr}[u(h)]\mathbb{I}_{d}/d$, from the definition of the group $K$.

Finally, let $r=\max_{g\notin K}\left|\operatorname{Tr}[u(g)]\right|<d$ be the second largest absolute value of $\operatorname{Tr}[u(g)]$ in the following:

[TABLE]

∎

If $\operatorname{Tr}[u(g)]=0$ for all $g\in G$ satisfying $u(g)\not\propto\mathbb{1}$, then $\xi=+\infty$, and the exponentially small error above is actually zero. This is the case of $\mathbb{Z}_{d}$ acting on a qudit with $u(1)=Z$ or any other qudit Pauli.

Theorem 4**.**

The maximally mixed symmetric state $\rho_{\beta=0,\Lambda}$ of a finite abelian group $G$ is locally indistinguishable from the identity state, up to an exponentially small error in the trace norm:

[TABLE]

with $\xi$ given in Lemma 5.

Proof.

Let $B=A^{C}$. First note that

[TABLE]

where in the second line we used Lemma 5 and $\sum_{\lambda}\Pi^{A}_{\lambda}=\mathbb{1}_{A}$.

Then, we have

[TABLE]

Hence,

[TABLE]

∎

Finally, we end with a theorem that a similar result for local indistinguishability of canonical and Gibbs ensembles holds true even at finite temperatures,

Theorem 5**.**

For $G$ a finite Abelian group and $H=\sum_{i=1}^{N}h_{i}$ a local Hamiltonian with $\left\lVert h_{i}\right\rVert\leq J$ and $|\operatorname{supp}h_{i}|\leq k$, then

[TABLE]

where $\xi_{\beta},c_{\beta}>0$ if $\beta<\beta_{*}$, with $\xi_{\beta}$, $c_{\beta}$ and $\beta_{*}$ independent of system size $N$.

Proof.

We first rewrite the difference of states above in terms of a difference in partition functions and a difference of Gibbs exponentials. For that, let $B=A^{C}$, and $Z_{\beta}\vcentcolon=\operatorname{Tr}[e^{-\beta H}]$, $Z_{\beta,\Lambda}=\operatorname{Tr}[e^{-\beta H}\Pi_{\Lambda}]$:

[TABLE]

The first term above is nothing but the second term specialized to $A=\emptyset$. Hence, we will focus on bounding the latter. Without loss of generality, we can assume $H\leq 0$ by adding to it a constant term proportional to the identity operator that will not change $\rho_{\beta}$ nor $\rho_{\beta,\Lambda}$. In that case, $Z_{\beta}=\operatorname{Tr}[e^{-\beta H}]\geq d^{N}$. By applying a high-temperature expansion of the exponentials, we arrive at

[TABLE]

We now focus on each term of the sum above,

[TABLE]

At this stage, it is useful to introduce a notation for the region with non-trivial action of $\prod_{\ell=1}^{n}h_{i_{\ell}}$, $S_{\{i_{\ell}\}}=\cup_{\ell=1}^{n}\operatorname{supp}h_{i_{\ell}}$. We decompose the trace over $B$ into trace over $B\cap S_{\{i_{\ell}\}}$ and $B\cap S_{\{i_{\ell}\}}^{C}$. We also decompose $\Pi_{\Lambda}=\sum_{\lambda}\Pi_{\lambda}^{R}\Pi_{\Lambda\overline{\lambda}}^{R^{C}}$ for $R=A\cup S_{\{i_{\ell}\}}$, such that $R^{C}\equiv B\cap S_{\{i_{\ell}\}}^{C}$. We also use the fact that $\sum_{\lambda}\Pi_{\lambda}^{R}=\mathbb{1}_{R}$ to simplify the second term in Eq. (35):

[TABLE]

where in the third line we have used Lemma 5 in the form $||G/K|\operatorname{Tr}[\Pi^{R}_{\lambda}]-d^{|R|}|<Cr^{|R|}$, with $C$ a constant and, crucially, $r<d$; and in the fifth line we used that $\prod_{\ell=1}^{n}h_{i_{\ell}}$ is block diagonal with respect to the subspaces projected by $\Pi_{\lambda}^{A\cup S_{\{i_{\ell}\}}}$ because the product commutes with the symmetry acting on $A\cup S_{\{i_{\ell}\}}$. Putting it back in the original sum (Eq. (34)),

[TABLE]

where the values of $\xi_{\beta}$ and $c_{\beta}$ depend if $r=0$ or not: if $r=0$, then the first term above vanishes, and so $\xi_{\beta}^{-1}=1-\ln(e\beta Jk)$ and $c_{\beta}=-\xi_{\beta}(\beta J+\ln(e\beta Jk))$; and if $r>0$, then, at small enough $\beta$, the first term dominates the second, and we have $\xi_{\beta}^{-1}=\ln(d/r)$ and $c_{\beta}=1-\xi_{\beta}\beta Jd^{k}/r^{k}$. Importantly, in both cases we have $\xi_{\beta},c_{\beta}>0$ if $\beta<\beta_{*}$ for $\beta_{*}$ small enough.

Since this bound is tighter the smaller region $A$ is, then the first term of Eq. (33) is smaller than the second one, and we arrive at the conclusion. ∎

V Equivalence of SEC and EC for finite Abelian groups

Here, we find a sufficient condition on the symmetry multiplicities that allows us to establish equivalence between the entangling condition $(\textsf{EC})$ and $(\textsf{SEC})$ for finite Abelian groups. First, a technical lemma.

Lemma 6**.**

Consider a Hamiltonian defined on a Hilbert space with the tripartition $A|B|C$, $\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}\otimes\mathcal{H}_{C}$, $H=H_{AB}\otimes\mathbb{1}_{C}+\mathbb{1}_{A}\otimes H_{BC}$, such that $B$ acts as a buffer region between $A$ and $C$ with no global interaction term nor any local interaction term between $A$ and $C$. Furthermore, each term $H_{AB}$ and $H_{BC}$ is assumed to be individually symmetric: $\forall g\in G,[U(g),H_{AB}]=[U(g),H_{BC}]=0$. If all irreps $\lambda$ that appear in $AB$ (i.e. $m^{AB}_{\lambda}\neq 0$) contribute to the total irrep $\Lambda$ (i.e. $m^{C}_{\Lambda\overline{\lambda}}\neq 0$), then we have an equivalence between the EC and SEC,

[TABLE]

Proof.

The ($\Rightarrow$) direction is trivial. For the ($\Leftarrow$) direction, we prove their negations are equivalent. We first right-multiply the RHS $[U_{A}(g),H]\Pi_{\Lambda}=0$ by $\Pi^{AB}_{\lambda}$ where $\lambda$ is any irrep that appears in $\mathcal{H}_{AB}$, i.e. $m_{\lambda}^{AB}\neq 0$:

[TABLE]

Then, we trace out the $C$ subsystem:

[TABLE]

By assumption, $m_{\lambda}^{AB}\neq 0$ implies $m^{C}_{\Lambda\overline{\lambda}}\neq 0$, which allows us to get rid of $m^{C}_{\Lambda\overline{\lambda}}$ from the equation above. Summing over $\lambda$, we finally arrive at

[TABLE]

which, after restoring the $C$ system by tensoring with $\mathbb{1}_{C}$, we have $[U_{A}(g),H]=0$. ∎

In a multipartite system $\mathcal{H}=\bigotimes_{i\in M}\mathcal{H}_{i}$ with a Hamiltonian $H=\sum_{X\subseteq M}h_{X}$ that is a sum of local terms $h_{X}$, $\operatorname{supp}(h_{X})=X$, we usually require them to be individually symmetric, $\forall X\subseteq M,\forall g\in G,[U(g),h_{X}]=[U_{X}(g),h_{X}]=0$. Here, locality means that for all regions $A$ with size $|A|\leq N/2$, there exists a buffer region $B$ with size $|B|<N/2$ (normally, $|B|\ll N$) for which $H$ has no direct interaction between $A$ and $C=(AB)^{C}$, i.e. $X\cap A\cap C\neq 0\Rightarrow h_{X}=0$. Then the Lemma above can be applied for such $A$ and $B$ regions if its conditions are met. In such case, we know that $\forall A\subseteq M,|A|<N/2\Rightarrow\exists g\in G,[U_{A}(g),H]\neq 0$ is equivalent to $\rho_{\beta,\Lambda}$ having nonzero negativity for all bipartitions at high enough temperatures.

For finite Abelian symmetry groups $G$ acting faithfully, then every irrep of $G$ should appear for large enough regions $C$. Moreover, “large enough” here should depend only on the group itself, and not on the representation or system size. For such cases, then the above lemma is useful to prove POE purposes for any finite-range Hamiltonian. Let us make this intuition more precise, starting with the following lemma:

Lemma 7** (Stable power of generating set).**

Let $H$ be a finite group (not necessarily Abelian) and $S\subseteq H$ a generating set of $H=\langle S\rangle$ containing the identity element $e\in S$. With $S^{n}$ being the set of all products of $n$ elements of $S$, we have

[TABLE]

Proof.

First, note that $m\leq n\implies S^{m}\subseteq S^{n}$ since $S^{n}=S^{m}\cdot S^{n-m}\supseteq S^{m}\cdot\{e\}=S^{m}$. Moreover, since $H$ is finite and $S$ generates it, then, for large enough $n$, $S^{n}=H$. Hence, it only remains to prove that $\forall n\geq|H|,S^{n-1}=S^{n}$.

Suppose, by contradiction, that there is an element $h\in S^{n}\setminus S^{n-1}$. Since $h\in S^{n}$, there are $s_{1},\ldots,s_{n}\in S$ such that $h=s_{1}s_{2}\cdots s_{n}$. Now, recursively define the partial product sequence $(h_{k})_{k=0}^{n}$ by $h_{0}=e$ and $h_{k}=h_{k-1}s_{k}$. Crucially, such sequence satisfies $h_{k}\in S^{k}$ and $h_{k}^{-1}h\in S^{n-k}$.

We will prove that the assumption $h\notin S^{n-1}$ implies all elements $h_{k}$ are distinct from each other. Otherwise, if $h_{k}=h_{l}$ for $0\leq k<l\leq n$, then

[TABLE]

contradicting the assumption $h\notin S^{n-1}$. However, $H$ has only $|H|$ elements, so the $n+1\geq|H|+1$ elements of $(h_{k})_{k=0}^{n}$ cannot be all distinct. Thus, the original hypothesis of $\exists h\in H,h\in S^{n}\setminus S^{n-1}$ is incorrect, and we have $S^{n-1}=S^{n}$. ∎

The condition $e\in S$ is important. For a counterexample, consider $H=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and $S=\{(1,0),(1,1)\}$. Even though $S$ generates $H$, its powers alternate between $S=S^{3}=\cdots=\{(1,0),(1,1)\}$ and $S^{2}=S^{4}=\cdots=\{(0,0),(0,1)\}$, never stabilizing to the full group $H$.

We can apply the lemma above to a subgroup of $\widehat{G}$, the group of 1d irreps of $G$ (also called the Pontryagin dual of $G$), generated by the set of irreps appearing in the on-site representation $u$, here assumed to be site-independent. Let us frame it in the context of spin chains for simplicity, but a similar result will be valid for any finite-range Hamiltonian in any bounded-degree graph (which includes higher dimensional lattices).

Theorem 6**.**

Let $G$ be a finite Abelian group. Consider a spin chain with periodic boundary conditions and homogeneous symmetry action $U=u^{\otimes N}:G\to U(\mathcal{H})$, with $u$ containing the trivial irrep in its decomposition. Further assume that the terms $h_{i}$ of the Hamiltonian $H=\sum_{i}h_{i}$ are supported around site $i\ni\operatorname{supp}(h_{i})$, have finite range $d$, $\operatorname{diam}(\operatorname{supp}(h_{i}))\leq d$, and are individually symmetric under $U$. For any global irrep $\Lambda$ and any connected region $A$ (an interval) separating $H=H_{A}\otimes\mathbb{1}_{A^{C}}+\mathbb{1}_{A}\otimes H_{A^{C}}+H_{AA^{C}}$ with $|A|\leq N-(2d+|G|-3)=N-O(1)$, we have

[TABLE]

Proof.

Let $S=\{\lambda\in\widehat{G}\mid\lambda\in u\}$, where “$\lambda\in u$” means that $\lambda$ appears in the irrep decomposition of the on-site representation $u$ with nonzero multiplicity. By assumption $e\in S$. Then, it is easy to see that $S^{n}=\{\lambda\in\widehat{G}\mid\lambda\in u^{\otimes n}\}$. Lemma 7 guarantees that all irreps that will ever appear in any tensor product of $u$ form the subgroup $\langle S\rangle\leq\widehat{G}$, and that all regions $C$ with size $|C|\geq|\widehat{G}|-1=|G|-1$ will contain them, i.e. $S^{|C|}=\langle S\rangle$. Here, we used that if $G$ is a finite Abelian group, then $G\simeq\widehat{G}$. That follows from $\widehat{\mathbb{Z}_{n}}\simeq\mathbb{Z}_{n}$ and the fact that $G$ is a direct product of cyclic groups. For such regions $C$, we can apply Lemma 6 if $B$ is large enough to separate the interval $A$ from $C$. That is because both the irrep $\lambda$ of $AB$ with multiplicity $m_{\lambda}^{AB}\neq 0$ is an element of $S^{|AB|}\subseteq\langle S\rangle$, and the total irrep $\Lambda$ is an element of $S^{N}=\langle S\rangle$, which means that $\Lambda\overline{\lambda}\in\langle S\rangle=S^{|C|}$ appears in $C$. For that, we choose $B=\{i\in A^{C}\mid\text{dist}(A,i)\leq d-1\}$, and $C=(AB)^{C}$. Given that $A$ is connected, these three regions separate Hamiltonian into $H=\mathbb{1}_{A}\otimes H_{BC}+H_{AB}\otimes\mathbb{1}_{C}$, thus satisfy all of the assumptions of Lemma 6. Finally, the condition $|C|\geq|G|-1$ translates to $|A|=N-|B|-|C|\leq N-2(d-1)-(|G|-1)$. ∎

As an example, let us consider a nearest-neighbor Hamiltonian $H=\sum_{i}h_{i,i+1}$ with homogeneous on-site $\mathbb{Z}_{2}$ symmetry given by $\prod_{i}u_{i}$, $u_{i}^{2}=\mathbb{1}_{i}$ (so $d=2$ and $|G|=2$). If the terms $h_{i,i+1}$ are $\mathbb{Z}_{2}$ symmetric, but not diagonal in the $u_{i}$ basis, then they will not commute with the partial action of the symmetry, i.e. $[u_{i},h_{i,i+1}]\neq 0$. Under these assumptions, the corollary above implies that any canonical ensemble state will not have SDOE for all bipartitions into two intervals, if the total system size $N$ is larger than $5$. That is because $\lfloor N/2\rfloor\leq N-(2d+|G|-3)=N-3$ for $N\geq 5$.

Note that the finiteness of $G$ is important for the above. For an example of an infinite Abelian symmetry with tighter constraints on the regions $A$ for which Lemma 6 can be applied, consider an homogeneous $G=U(1)$ symmetry action with on-site charges [math] and $+1$, which is the case for $U(1)$ acting on qubits as $\theta\mapsto\operatorname{diag}(1,e^{i\theta})$. Then, the condition of Lemma 6 is equivalent to the following condition on the total charge $n$ ($\Lambda(\theta)=\exp(in\theta)$): $|AB|\leq n\leq N-|AB|$. In particular, the Lemma can be applied to any finite-range Hamiltonian constrained to a total charge $n\approx N-O(1)$, if only entanglement of finite regions $|A|=O(1)$ is relevant. For larger regions, such as $|A|=N/2-O(1)$, the effect of the global charge constraint can be noticeable, which does not happen for finite Abelian groups (if $N$ is large enough).

VI Classification of irreps

Here we provide a detailed proof of Theorem 3 presented in the main text. The definition of semiuniform irreps and the theorem is quoted below for the ease of readers.

Definition 2**.**

A global irrep $\Lambda$ of a $N$-partite system is uniform if its projector $\Pi_{\lambda}$ is a uniform tensor product of an on-site irrep $\lambda$: $\Pi_{\Lambda}=\bigotimes_{i=1}^{N}\Pi_{\lambda}^{(i)}$. $\Lambda$ is semiuniform if its projector $\Pi_{\Lambda}$ is a direct sum of uniform tensor products: $\Pi_{\Lambda}=\sum_{\alpha}\bigotimes_{i=1}^{N}\Pi_{\lambda_{\alpha}}^{(i)}$.

Theorem 7**.**

Consider a multipartite system with homogeneous symmetry action $U$. If $\Lambda$

1. (A)

is not semiuniform, then for any pair of sites $(i,j)$, there exists a two-body symmetric Hamiltonian $v_{i,j}$ supported on $\{i,j\}$ that entangles any bipartition that separates site $i$ from site $j$. Moreover, given a connected interaction graph, the uniform nearest-neighbor perturbation $\sum_{\langle i,j\rangle}v_{i,j}$ entangling all bipartitions $A|B$, with $A,B\neq\emptyset$. 2. (B)

is semiuniform, then there is no entangling Hamiltonian with support over $N-1$ sites or fewer, and for all $(N-1)$-local Hamiltonian, $\rho_{\beta,\Lambda}$ is fully separable when $\rho_{\beta}$ is fully separable.

Proof.

(A) Since $\Lambda$ is not semiuniform, then there is a tensor product projector term $\bigotimes_{i=1}^{N}\Pi^{(i)}_{\lambda_{i}}$ in the decomposition of $\Pi_{\Lambda}$ with at least two distinct irreps over sites $k$ and $l$: $\lambda\neq\lambda^{\prime}$. Hence,

[TABLE]

Since $\Pi_{\Lambda}$ has weak permutation symmetry from the homogeneity of the symmetry action, then we can permute sites $(k,l)$ with $(i,j)$ and $(j,i)$ on the equation above, thus proving that $\Pi_{\lambda}^{(i)}\Pi_{\lambda^{\prime}}^{(j)}\Pi_{\Lambda},\Pi_{\lambda^{\prime}}^{(i)}\Pi_{\lambda}^{(j)}\Pi_{\Lambda}\neq 0$.

Thus there exist states $\ket{\lambda}_{i}$ and $\ket{\lambda^{\prime}}_{i}$ of $\mathcal{H}_{i}$ that transform as irreps $\lambda$ and $\lambda^{\prime}$ under the symmetry, respectively, and similarly with two states $\ket{\lambda}_{j}$ and $\ket{\lambda^{\prime}}_{j}$ of $\mathcal{H}_{j}$. Then, let $\sigma_{i}^{+}\vcentcolon={}_{i}\!\mathinner{\ket{\lambda^{\prime}}\!\bra{\lambda}}_{i}\in\mathcal{L}(\mathcal{H}_{i})$ and $\sigma_{j}^{+}\vcentcolon={}_{j}\!\mathinner{\ket{\lambda^{\prime}}\!\bra{\lambda}}_{j}\in\mathcal{L}(\mathcal{H}_{j})$. Denoting their Hermitian conjugates by $\sigma_{i}^{-}\vcentcolon=(\sigma_{i}^{+})^{\dagger}$ and $\sigma_{j}^{-}\vcentcolon=(\sigma_{j}^{+})^{\dagger}$, we choose $v_{i,j}\vcentcolon=\sigma_{i}^{+}\sigma_{j}^{-}+\sigma_{i}^{-}\sigma_{j}^{+}$, which is an entangling Hamiltonian because:

• •

$v_{i,j}$ is symmetric under $U=U_{i}\otimes U_{j}\otimes_{k\neq i,j}U_{k}$:

[TABLE]

• •

$v_{i,j}$ satisfies the entangling condition EC of Theorem 2. To see this, note that the assumptions imply the existence of two states, $\ket{\Psi}=\ket{\lambda}_{i}\otimes\ket{\lambda^{\prime}}_{j}\otimes\ket{\psi}_{\setminus{i,j}}$ and $\ket{\Psi^{\prime}}=\ket{\lambda^{\prime}}_{i}\otimes\ket{\lambda}_{j}\otimes\ket{\psi}_{\setminus{i,j}}$ that transform as the global irrep $\Lambda$. Moreover, since $\lambda\neq\lambda^{\prime}$, there exists $g\in G$ such that $\lambda(g)\neq\lambda^{\prime}(g)$. Now we establish a nonzero matrix element of $[U_{i}(g),H]\Pi_{\Lambda}$:

[TABLE]

We now turn to the uniform nearest-neighbors perturbation $V=\sum_{\{i,j\}\in E}v_{i,j}$, where $E$ is the set of edges of a connected graph $\mathcal{G}=(V,E)$. By contradiction, let us assume it is not entangling for a certain region $A\neq\emptyset,V$.

In what follows, let $\partial A\vcentcolon=\{\{i,j\}\in E\mid i\in A,j\notin A\}$, and $g\in G$ be a group element for which $\lambda^{\prime}(g)\neq\lambda(g)$. Since $V$ is not entangling, then we must have,

[TABLE]

Since $A\neq\emptyset,V$ and $\mathcal{G}$ is connected, the sum above contains at least one term corresponding to an edge $\{i^{*},j^{*}\}\in\partial A$. Now, let $\ket{\Psi}=\ket{\lambda}_{i^{*}}\otimes\ket{\lambda^{\prime}}_{j^{*}}\bigotimes_{l\neq i^{*},j^{*}}\ket{\psi_{l}}_{l}$ and $\ket{\Psi^{\prime}}=\ket{\lambda^{\prime}}_{i^{*}}\otimes\ket{\lambda}_{j^{*}}\bigotimes_{l\neq i^{*},j^{*}}\ket{\psi_{l}}_{l}$ be tensor product states contained within $\Lambda$, i.e. $\Pi_{\Lambda}\ket{\Psi}=\ket{\Psi}$. Then,

[TABLE]

a contradiction. Here, only the term $\sigma^{+}_{i^{*}}\sigma^{-}_{j^{*}}$ survives because the others are either $\sigma^{+}_{j^{*}}\sigma^{-}_{i^{*}}$ or they contain at least one vertex, say $l\in V$, that is not $i^{*}$ or $j^{*}$, and thus will get a diagonal contribution corresponding to the state $\ket{\psi_{l}}_{l}$. In both cases, the matrix element is zero by construction.

Thus, $V=\sum_{\{i,j\}\in E}v_{i,j}$ is an entangling Hamiltonian for all regions $A\neq\emptyset,V$, as desired.

(B) Say $\Lambda$ is semiuniform with projector $\Pi_{\Lambda}=\sum_{\alpha}\Pi_{\alpha}=\sum_{\alpha}\bigotimes_{i=1}^{N}\Pi_{\lambda_{\alpha}}^{(i)}$, where $\Pi_{\alpha}=\bigotimes_{i=1}^{N}\Pi^{(i)}_{\lambda_{\alpha}}$ are orthogonal tensor product projectors. Let $H$ be an arbitrary Hamiltonian that is $G$-symmetric and supported over the $N-1$ first sites, without loss of generality. Then,

[TABLE]

From Eq. (71), we have that $\alpha\neq\beta\Rightarrow\Pi_{\alpha}(H\Pi_{\Lambda})\Pi_{\beta}=0$, which also implies $\Pi_{\alpha}e^{-\beta^{\prime}H\Pi_{\Lambda}}\Pi_{\beta}=0$. As such, if $\rho_{\beta^{\prime}}\propto e^{-\beta^{\prime}H}$ is separable with decomposition $\rho_{\beta^{\prime}}=\sum_{m}p_{m}\mathinner{\ket{\Psi_{m}}\!\bra{\Psi_{m}}}$, with $\ket{\Psi_{m}}=\bigotimes_{i=1}^{N}\ket{\psi^{(i)}_{m}}$ product state, then

[TABLE]

which gives $\rho_{\beta,\Lambda}$ a fully separable decomposition into the product states $\Pi_{\alpha}\ket{\Psi_{m}}=\bigotimes_{i=1}^{N}(\Pi^{(i)}_{\lambda_{\alpha}}\ket{\psi^{(i)}_{m}})$.

∎

Now, it remains to prove that semiuniform irreps are rare among the set of all global irreps:

Theorem 8**.**

Consider a $N$-partite system with an Abelian symmetry group $G$ that 1) does not depend on $N$, 2) acts homogeneously (i.e. $U=u^{\otimes N}$), and 3) acts nontrivially on each site (i.e. $\exists g\in G,u(g)\not\propto\mathbb{1}$, or equivalently there are at least two distinct on-site irreps). Then, the fraction of semiuniform global irreps goes to zero in the thermodynamic limit as $O(2^{d}/N)$, where $d$ is the on-site Hilbert space dimension. Furthermore, if $G$ is finite, there are no semiuniform global irreps for $N>|G|$.

Proof.

We separate the analysis in two cases: either $\lambda\overline{\mu}\in\widehat{G}$ has finite order for all on-site irreps $\lambda,\mu\in u$, or there is a pair $(\lambda_{\infty},\mu_{\infty})$ with $\lambda_{\infty}\overline{\mu_{\infty}}$ having infinite order.

In the first case, define $n<\infty$ be the least common multiple of the orders of all such $\lambda\overline{\mu}$. Note that this is always the case for finite symmetry groups $G$, with $n\leq|\widehat{G}|=|G|$. We will prove that there are no semiuniform global irreps for $N>n$, thus trivially satisfying the fraction bound.

Indeed, if $\Lambda$ is semiuniform with $\Pi_{\Lambda}\neq 0$, there is at least one $\lambda^{*}$ one of the irreps that appear as a nonzero uniform term $\bigotimes_{i=1}^{N}\Pi^{(i)}_{\lambda^{*}}$ in the sum. Furthermore, since $u$ acts nontrivally, let $\mu\neq\lambda^{*}$ be another on-site irrep, distinct from $\lambda^{*}$, with $\Pi_{\mu}^{(i)}\neq 0$. Then, we can construct a non-uniform irrep vector $\vec{\lambda}^{NU}$ given by $\lambda^{NU}_{i}=\mu$ if $1\leq i\leq n$ and $\lambda^{NU}_{i}=\lambda^{*}$ if $n<i\leq N$. This irrep appears in the decomposition of $\Pi_{\Lambda}$ because $\mu^{n}(\lambda^{*})^{N-n}=(\lambda^{*})^{N}=\Lambda$, where we have used that $(\lambda^{*})^{n}=\mu^{n}=1$, the trivial irrep. This, however, contradicts the assumption of semiuniformity of $\Lambda$. 2. 2.

In the latter case — of $\lambda_{\infty}\overline{\mu_{\infty}}$ having infinite order — we can lower bound the number of global irreps by choosing some of the on-site irreps to be $\lambda_{\infty}$, and some to be $\mu_{\infty}$. As a result, we form a set $\{\lambda_{\infty}^{m}\mu_{\infty}^{N-m}\}_{m=0}^{N}$ of $N+1$ global irreps that are pairwise distinct precisely because $\lambda_{\infty}\overline{\mu_{\infty}}$ has infinite order.

To upper bound the number of semiuniform global irreps, first note that there are at most $d$ distinct on-site irreps contained in the decomposition of $u$. Since a semiuniform irrep is a sum over a subset of on-site irreps, there are at most $2^{d}$ of them. Hence, the fraction of semiuniform global irreps is bounded by $2^{d}/(N+1)<2^{d}/N$.

∎

For some infinite groups, including $U(1)$, the fraction of irreps can be further constrained to be at most $d/(N+1)$ due to the following result:

Proposition 1**.**

If $\widehat{G}$ is torsion-less (e.g. $G=U(1)$), then all semiuniform global irreps $\Lambda$ are uniform.

Proof.

If there are two uniform tensor products, $\bigotimes_{i}\Pi^{(i)}_{\lambda}$ and $\bigotimes_{i}\Pi^{(i)}_{\mu}$, contributing to $\Pi_{\Lambda}$, then this means $(\lambda\overline{\mu})^{N}=\Lambda\overline{\Lambda}=1$. Since $\widehat{G}$ is torsion-less, then $\lambda=\mu$. By repeating this argument, we conclude that $\Lambda$ is uniform. ∎

VII Equivalence of EC and NC

Here we prove that the conditions EC (eq. Symmetry enforces entanglement at high temperatures) and NC (eq. Symmetry enforces entanglement at high temperatures) are equivalent. This shows that whenever a symmetric Hamiltonian and irrep satisfies EC to lead to entangled canonical ensembles, the latter must also be negative partial transpose, i.e., their entanglement negativity must be non-zero. We prove the equivalence through the following theorem:

Theorem 9**.**

The following are equivalent:

1. (i)

$(\mathbb{1}-\Pi_{\Lambda})(H\Pi_{\Lambda})^{T_{A}}(\mathbb{1}-\Pi_{\Lambda})=0$. 2. (ii)

$H\Pi_{\Lambda}=\sum_{\lambda}\Pi_{\lambda}^{A}(H\Pi_{\Lambda})\Pi^{A}_{\lambda}=\sum_{\lambda}\Pi_{\lambda}^{A}\Pi^{B}_{\Lambda\overline{\lambda}}H\Pi^{A}_{\lambda}\Pi^{B}_{\Lambda\overline{\lambda}}$. 3. (iii)

For all $g\in G$, $[U_{A}(g),H\Pi_{\Lambda}]=[U_{B}(g),H\Pi_{\Lambda}]=0$.

Proof.

• •

$(i)\Rightarrow(ii)$ First, we have

[TABLE]

where we have used $(\Pi^{A}_{\lambda})^{T}=\Pi^{A}_{\lambda}$, which is true as we choose to do the partial transpose for $A$ and $B$ with respect to a basis that diagonalizes $U_{A}$ and $U_{B}$, such that $\forall g\in G,U_{A}(g)^{T_{A}}=U_{A}(g)$ and $U_{B}(g)^{T_{B}}=U_{B}(g)$. Then,

[TABLE]

By multiplying the equation above with $\Pi_{\Lambda}$ and using that $[H,\Pi_{\Lambda}]=0$, we have

[TABLE]

• •

$(ii)\Rightarrow(iii)$ From assertion (ii), we know that $H\Pi_{\Lambda}$ is block diagonal with respect to the irreps of $U_{A}$. Thus, by the converse of Schur’s lemma, $H\Pi_{\Lambda}$ commutes with all symmetry operators $U_{A}(g)$, for all $g\in G$. More explicitly, from $U_{A}(g)\Pi_{\lambda}^{A}=\Pi_{\lambda}^{A}U_{A}(g)=\lambda(g)\Pi^{A}_{\lambda}$, we have

[TABLE]

• •

$(iii)\Rightarrow(i)$ Condition $(iii)$ also implies that for all irreps $\lambda$ and $\mu$, $[\Pi^{A}_{\lambda},H\Pi_{\Lambda}]=[\Pi^{B}_{\mu},H\Pi_{\Lambda}]=0$. Hence,

[TABLE]

By taking the partial transpose of the equation above, we get

[TABLE]

which, upon right multiplication with $(\mathbb{1}-\Pi_{\Lambda})$, results in zero.

∎

Note that the negation of the statements $(i)$ and $(iii)$ are the NC and EC, respectively, thereby proving their equivalence.

VIII Entanglement negativity of the thermal cluster chain state

We analytically compute the entanglement negativity of the Gibbs and canonical ensembles of the one-dimensional cluster chain Hamiltonian. This model was chosen for both its analytical simplicity, and for the Gibbs ensemble showing non-trivial SDOE, that is, it is entangled at low temperatures, and becomes separable at high temperatures. This last property makes the cluster chain a more realistic example of the generic case, and is not shared with the classical Ising model studied in [59].

We confirm that the Gibbs state of the cluster chain exhibits sudden death of entanglement negativity, while the negativity of the canonical ensemble persists for arbitrarily high temperatures. Moreover, we find that there is no bound entanglement for the Gibbs ensemble.

VIII.1 Gibbs ensemble

Consider a connected region $A$ with size $|A|\geq 2$ and its complement $A^{C}=B$ with size $|B|\geq 2$, so that the total system size $N$ is even. Region $A$ has two endpoints: $a_{L}$ and $a_{R}$, and they are adjacent to the endpoints $b_{L}$ and $b_{R}$ of B, respectively. Define $A_{\text{int}}\vcentcolon=A\setminus\{a_{L},a_{R}\}$ to be the interior of region $A$, with endpoints $a_{L}$ and $a_{R}$ removed, and similarly with $B_{\text{int}}$.

To simplify the entanglement calculation, we will conjugate the cluster chain Hamiltonian $H_{\text{cc}}=-\sum_{i}Z_{i-1}X_{i}Z_{i+1}$ by $CZ_{\text{int}}$, defined as

[TABLE]

Since $CZ_{\text{int}}$ is a local unitary to the $A|B$ bipartition, it preserves the entanglement value of the thermal state. This results in the trivial paramagnetic Hamiltonian everywhere except on the two boundary links:

[TABLE]

where

[TABLE]

The Gibbs state $\rho_{\beta}^{\prime}=e^{-\beta H^{\prime}_{cc}}/\mathcal{Z_{\beta}}$ is, then,

[TABLE]

where $\lambda\vcentcolon=\tanh(\beta)$ and $N$ is the system size. Its partial transpose is, then

[TABLE]

and the logarithmic negativity $E_{N}=\log\left\lVert\cdot\right\rVert_{1}$ is

[TABLE]

As shown in the red curve of Fig. 2, the entanglement negativity is zero for small $\lambda\leq\lambda_{c}$. Indeed, it can be easily deduced from the expression above that $\lambda_{c}=\sqrt{2}-1$. Moreover, we can further argue that the state is exactly separable for $\lambda\leq\lambda_{c}$ by showing an explicit separable decomposition for it. Consider the density matrix $\rho_{AB}$ defined by

[TABLE]

where $H_{A}$ contains the terms of $H$ entirely supported on region $A$, and similarly for $H_{AB}$. If $\rho_{AB}$ can be decomposed into an ensemble of states that are separable between $A$ and $B$, then a separable decomposition for the full density matrix $\rho_{\beta}$ can also be constructed.

In the case of the cluster chain Hamiltonian, and expressing $\rho_{AB}$ only at the four closest sites near the rightmost boundary between $A$ and $B$ for simplicity, we have

[TABLE]

It is easy to check that $\rho_{AB}=p_{1}\rho_{1}+p_{2}\rho_{2}+p_{3}\rho_{3}$, with

[TABLE]

and probabilities $p_{1}=p_{2}=\lambda/(\lambda^{2}+2\lambda)$ and $p_{3}=\lambda^{2}/(\lambda^{2}+2\lambda)$.

Each individual $\rho_{i}$ is in fact a fully separable matrix since it has the form $\mathbb{1}+\alpha\hat{P}$, where $\hat{P}$ is a Pauli operator. However, $\rho_{i}$ is positive-semidefinite if and only if:

[TABLE]

Therefore the state is separable between left and right if and only if $\lambda\leq\lambda_{c}=\sqrt{2}-1$.

VIII.2 Canonical ensemble

Here, we will repeat the calculation for the canonical ensemble of the 1d cluster chain. Being a nontrivial SPT, the cluster chain exhibit edge modes that become entangled when global strong symmetry is imposed. Since we are interested in the bulk entanglement, we will disregard this contribution by considering periodic boundary conditions, forming a 1d ring. Hence, a nontrivial connected region $A$ will have two boundary points. Furthermore, contrary to the 1d Ising model, the stabilizers of the cluster chain and the symmetry operators are not independent. As we will see, this correlates the degrees of freedom inside the bulk with the ones at the boundary, and also between the boundary points.

Even though the cluster chain has $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ symmetry generated by $\prod_{i}X_{2i}$ and $\prod_{i}X_{2i-1}$, we will only impose the strong symmetry of the diagonal component $P\vcentcolon=\prod_{i}X_{i}=\Lambda\in\{\pm 1\}$ with projector $\Pi_{\Lambda}=(\mathbb{1}+\Lambda P)/2$, for simplicity. After conjugation by $CZ_{\text{int}}$, the symmetry becomes $P^{\prime}=P_{\text{int}}P_{LR}$, where $P_{\text{int}}=\prod_{i\in A_{\text{int}}}X_{i}\prod_{i\in B_{\text{int}}}X_{i}$ and $P_{LR}=Y_{a_{L}}Y_{b_{L}}Y_{a_{R}}Y_{b_{R}}$. The strongly symmetric thermal state $\rho_{\beta,\Lambda}^{\prime}=e^{-\beta H^{\prime}}\Pi_{\Lambda}^{\prime}/\mathcal{Z}_{\beta,\Lambda}$ is

[TABLE]

Now, $\rho_{\beta,\Lambda}^{\prime}$ does not factorize into a tensor product over the interior of each region and their boundaries. To deal of this, we decompose $\Pi_{\Lambda}^{\prime}=\Pi^{\text{int}}_{+1}\Pi^{LR}_{\Lambda}+\Pi^{\text{int}}_{-1}\Pi^{LR}_{-\Lambda}$. Then,

[TABLE]

The ratio $r_{N,\mu}$ converges to $1$ in the thermodynamic limit $N\to\infty$ if $\lambda<1$ ($\beta<\infty$). Hence, we will set it to 1 in what follows.

The partial transpose is

[TABLE]

so the logarithmic negativity is

[TABLE]

This result is plotted on the blue curve of Fig. 2, which shows the persistency of entanglement for the canonical ensemble for arbitrarily small $\beta$. Also note that the final expression is independent of the global charge $\Lambda$. For finite size, however, the entanglement will depend on $\Lambda$.

The analytical results for the entanglement negativity were validated with small size ($4\leq N\leq 6$) numerical calculations.

IX Persistence of fermionic negativity in the canonical ensemble

Consider the canonical ensemble for fermions with strong fermion parity, $\rho_{\beta,\Lambda}\propto e^{-\beta H}\Pi_{\Lambda}$, where $\Pi_{\Lambda}=\frac{1}{2}(\mathbb{1}+\Lambda P)$ projects onto a fixed fermion parity sector with $\Lambda=\pm 1$. Assume $\rho_{\beta,\Lambda}$ has zero fermionic negativity, i.e. $||\rho_{\beta,\Lambda}^{\mathcal{T}_{A}}||_{1}=1$, for a sequence of $\beta$ converging to [math]. This, combined with the fact that $\operatorname{Tr}{\rho_{\beta,\Lambda}^{\mathcal{T}_{A}}}=\operatorname{Tr}{\rho_{\beta,\Lambda}}=1$, imply that $\rho_{\beta,\Lambda}$ is Hermitian and positive semi-definite [77].

Under the fermionic partial transpose, $P^{\mathcal{T}_{A}}=\eta_{A}P$, with $\eta_{A}=(-1)^{|A|\bmod 2}$, implying $\Pi_{\Lambda}\Pi_{-\eta_{A}\Lambda}^{\mathcal{T}_{A}}=\Pi_{\Lambda}\Pi_{-\Lambda}=0$. Thus, the matrix $\Pi_{-\eta_{A}\Lambda}\rho_{\beta,\Lambda}^{\mathcal{T}_{A}}\Pi_{-\eta_{A}\Lambda}$, which is positive semidefinite, is also traceless $\operatorname{Tr}\!\left(\rho_{\beta,\Lambda}^{\mathcal{T}_{A}}\Pi_{-\eta_{A}\Lambda}\right)\propto\operatorname{Tr}\!\left(e^{-\beta H}\Pi_{\Lambda}\Pi_{-\eta_{A}\Lambda}^{\mathcal{T}_{A}}\right)=0$, and thus identically zero. Since the matrix is holomorphic function of $\beta$, its derivative at $\beta=0$ must also vanish. Negating the chain of arguments, we find that $E_{N}^{f}$ persists if the derivative at $\beta=0$ is nonvanishing:

[TABLE]

This is the fermionic analogue of the NC, which is also generically satisfied. The two-Majorana term $V=ic_{i}c_{j}$, with $c_{i}\in A$ and $c_{j}\in B$, obeys $(V\Pi_{\Lambda})^{\mathcal{T}_{A}}=iV\Pi_{-\eta_{A}\Lambda}$, yielding $\Pi_{-\eta_{A}\Lambda}(V\Pi_{\Lambda})^{\mathcal{T}_{A}}\Pi_{-\eta_{A}\Lambda}=iV\Pi_{-\eta_{A}\Lambda}\neq 0,$ thereby providing an explicit entangling perturbation.

X Proof of Conjecture made in Shapourian–Ryu [65]

Recall that the fermionic (logarithmic) negativity is

[TABLE]

Ref. [65] introduces two types of fermionic states (see their Remark 2):

• •

Type I: Fermion-number parity of subsystems is even, $[P_{A},\,\rho]=0$.

• •

Type II: Fermion-number parity of subsystems is mixed, $[P_{A},\,\rho]\neq 0$.

Type I states may be (symmetrically) separable or entangled, and, in both cases, the fermionic and bosonic negativities coincide. In contrast, Type II density matrices are always entangled. Ref. [65] further conjectured that no type II state has vanishing fermionic negativity (their Conjecture 1). Here, we confirm this expectation. We first introduce the following lemma, proved in [77].

Lemma 8**.**

For $A$ a square matrix, if $\|A\|_{1}\equiv\mathrm{Tr}\bigl|A\bigr|=\mathrm{Tr}\,A$, then $A\geq 0$. In particular, $A$ is Hermitian.

Theorem 10**.**

For any bipartite fermionic state $\rho$ of type II, if $\rho$ is symmetrically entangled, then $E^{f}_{N}(\rho)>0$.

Proof.

We prove the converse by assuming $E^{f}_{N}(\rho)=0$. By definition, this implies $\mathrm{Tr}\bigl|\rho^{\mathcal{T}_{A}}\bigr|=1$. Since the fermionic partial transpose preserves the trace, $\mathrm{Tr}\,\rho^{\mathcal{T}_{A}}=\mathrm{Tr}\,\rho=1$. By Lemma 8, $\rho^{\mathcal{T}_{A}}$ is therefore Hermitian (in fact, positive semi-definite).

We now show that Hermiticity of $\rho^{\mathcal{T}_{A}}$ contradicts the assumption that $\rho$ is type II. Expand $\rho$ in Majorana monomials $a_{p_{1}}\cdots a_{p_{k_{1}}}\,b_{q_{1}}\cdots b_{q_{k_{2}}}$, where $a_{(\cdot)}$ act on $A$ and $b_{(\cdot)}$ on $B$. Being type II means that in this expansion there exists at least one term with an odd number of Majorana operators on $A$. However, such a term acquires a factor of $\pm i$ under the fermionic partial transpose on $A$ (see Eq. (23) of Ref. [65]), which generates purely imaginary contributions in $\rho^{\mathcal{T}_{A}}$ that preclude its Hermiticity.

∎