Entanglement Asymmetry and Symmetry Restoration

Entanglement asymmetry measures charge coherence in a reduced state: it compares that state with the result of forgetting its symmetry phase. Under symmetry-preserving dynamics, a finite subsystem can lose this coherence even though the full state retains exactly the same global asymmetry. A restoration claim must therefore name the subsystem, twirl, regulator, norm, and order of limits.

Required background. Charge-resolved entanglement supplies charge projectors and Fourier resolution for states that already commute with the subsystem charge.

Helpful background. Information measures along RG flows supplies the distinction between a finite crossover and an asymptotic theorem.

Chapter map. The overview gives the task-to-resource map, the comparison table, and the three gates for an operational claim.

Twirling away charge coherence

For a compact group with subsystem representation $U_A(g)$ and normalized Haar measure, define

$$\mathcal G_A(\rho_A) :=\int_G dg\,U_A(g)\rho_AU_A(g)^\dagger.$$

This is a trace-preserving conditional expectation onto the invariant operator algebra. The von Neumann asymmetry is

$$\Delta S_A :=S(\mathcal G_A\rho_A)-S(\rho_A) =D(\rho_A\Vert\mathcal G_A\rho_A) \ge0.$$

The equality follows because $\log(\mathcal G_A\rho_A)$ is invariant and the twirl is self-adjoint in the trace inner product:

$$\operatorname{Tr}\!\left[\rho_A\log(\mathcal G_A\rho_A)\right] =\operatorname{Tr}\!\left[(\mathcal G_A\rho_A)\log(\mathcal G_A\rho_A)\right].$$

For $U(1)$ with integer charge,

$$\mathcal G_A(\rho_A) =\int_{-\pi}^{\pi}\frac{d\alpha}{2\pi} e^{i\alpha Q_A}\rho_Ae^{-i\alpha Q_A} =\sum_q\Pi_q\rho_A\Pi_q.$$

Hence $\Delta S_A=0$ exactly when $[\rho_A,Q_A]=0$ in a finite-dimensional regulator. Quantum Pinsker gives the quantitative implication

$$\left\|\rho_A-\mathcal G_A\rho_A\right\|_1 \le\sqrt{2\Delta S_A}$$

when logarithms are natural. A small asymmetry therefore controls the entire reduced state in trace norm, not merely one symmetry-odd expectation value. The $U(1)$ definition and relative-entropy identity are Ares, Murciano, and Calabrese 2023, Eq. (1); the general compact-group resource-theory construction is given in Gour, Marvian, and Spekkens 2009, §§ II–III.

Despite its name, entanglement asymmetry is not an entanglement measure between $A$ and its complement. A pure product state can have nonzero $\Delta S_A$ whenever it coherently superposes subsystem charges.

Rényi asymmetry needs multi-angle moments

For $n>0$, $n\ne1$, the standard Rényi asymmetry is

$$\Delta S_A^{(n)} :=S_n(\mathcal G_A\rho_A)-S_n(\rho_A) =\frac{1}{1-n} \log\frac{\operatorname{Tr}[(\mathcal G_A\rho_A)^n]} {\operatorname{Tr}(\rho_A^n)}.$$

Because a compact-group twirl is a mixture of unitaries, it makes the spectrum more mixed; the standard Rényi entropy is Schur concave for $n>0$, so this difference is nonnegative. Only the $n\to1$ quantity has the relative-entropy identity above.

When $[\rho_A,Q_A]\ne0$, the one-angle moment $\operatorname{Tr}(\rho_A^n e^{i\alpha Q_A})$ does not reconstruct $\operatorname{Tr}[(\mathcal G_A\rho_A)^n]$. Expanding the $U(1)$ twirl instead gives

$$\begin{aligned} \operatorname{Tr}[(\mathcal G_A\rho_A)^n] =\int\prod_{j=1}^n\frac{d\alpha_j}{2\pi}\, \operatorname{Tr}\!\big(&\rho_Ae^{i(\alpha_2-\alpha_1)Q_A} \rho_Ae^{i(\alpha_3-\alpha_2)Q_A}\cdots\\ &\rho_Ae^{i(\alpha_1-\alpha_n)Q_A}\big). \end{aligned}$$

These are multi-angle charged moments; their phase differences sum to zero. This construction is Ares, Murciano, and Calabrese 2023, Eqs. (2)–(5). Ordinary charge-resolved moments are recovered only after the state is already block diagonal.

Symmetric dynamics and orders of limits

Let $[H,Q]=0$ but $[\rho(0),Q]\ne0$. Covariance implies

$$\mathcal G\!\left(e^{-itH}\rho(0)e^{itH}\right) =e^{-itH}\mathcal G(\rho(0))e^{itH},$$

so unitary invariance of entropy gives $\Delta S_{\rm full}(t)=\Delta S_{\rm full}(0)$. Regional asymmetry can nevertheless move into correlations between $A$, its complement, and increasingly nonlocal observables.

There are three distinct late-time questions:

1. take the thermodynamic limit and then let $t\to\infty$ at fixed finite $A$;
2. take $\ell,t\to\infty$ at fixed $t/\ell$; or
3. keep the total regulator finite and study a time average, a finite observation window, or a limsup.

A finite closed system generally has recurrences, so its strict pointwise late-time limit need not exist. Likewise, integrability can retain information in conserved mode occupations without implying that every finite subsystem remains asymmetric. The convergence criterion should be stated in terms of $\Delta S_A$, trace distance, or a complete family of charged observables—not one selected odd operator.

Exact two-site XX quench

An exactly solvable regulated example already separates local disappearance from global conservation. Let

$$H=J(\sigma_1^+\sigma_2^-+\sigma_1^-\sigma_2^+), \qquad Q=n_1+n_2, \qquad n_j=\frac{1-\sigma_j^z}{2},$$

and start from the symmetry-breaking spin product

$$|\psi(0)\rangle=|+\rangle_1|0\rangle_2, \qquad |+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}.$$

This XX Hamiltonian maps to a number-conserving free-fermion hopping problem. The spin state is used directly, so no fermion-parity superselection assumption is hidden. Evolution gives

$$|\psi(t)\rangle =\frac1{\sqrt2}\left[ |00\rangle+\cos(Jt)|10\rangle-i\sin(Jt)|01\rangle \right].$$

For the one-site subsystem, write $c:=\cos(Jt)$. Direct partial trace gives

$$\rho_1(t) = \begin{pmatrix} 1-\tfrac12c^2&\tfrac12c\\ \tfrac12c&\tfrac12c^2 \end{pmatrix}, \qquad \mathcal G_1\rho_1= \begin{pmatrix} 1-\tfrac12c^2&0\\ 0&\tfrac12c^2 \end{pmatrix}.$$

The smaller eigenvalue of $\rho_1$ is

$$\lambda_-(c) :=\frac{1-\sqrt{1-c^2+c^4}}{2}.$$

Writing $h_2(x):=-x\log x-(1-x)\log(1-x)$, the exact asymmetry is

$$\Delta S_1(t) =h_2\!\left(\frac{c^2}{2}\right)-h_2(\lambda_-(c)).$$

$Jt$	$\Delta S_1(t)$ in nats	$\Delta S_1(t)$ in bits
$0$	$0.6931471806$	$1$
$\pi/4$	$0.3165597780$	$0.4566992218$
$\pi/2$	$0$	$0$
$3\pi/4$	$0.3165597780$	$0.4566992218$
$\pi$	$0.6931471806$	$1$

At $Jt=\pi/2$, the state factorizes as

$$|\psi(\pi/2J)\rangle =|0\rangle_1\frac{|0\rangle_2-i|1\rangle_2}{\sqrt2}.$$

Charge coherence has moved completely from site 1 to site 2. At $Jt=\pi$ it has returned to site 1, so the zero is not irreversible restoration.

For the two-site subsystem, which is the full regulated system, the state remains pure and the exact charge weights are

$$(p_0,p_1)=\left(\frac12,\frac12\right).$$

Each projected state is pure, so

$$\Delta S_{12}(t)=H(p_0,p_1)=\log2=0.6931471806$$

for all times. Comparing subsystem sizes one and two therefore demonstrates the precise licensed statement: local charge coherence can vanish while global asymmetry is conserved.

The tilted-state XX quenches of Ares, Murciano, and Calabrese 2023, Eqs. (10)–(14) extend this mechanism to large systems, with an explicit ballistic scaling function and a late-time power law. Their calculation uses Gaussian components and parity-resolved combinations; a generic tilted spin state should not be replaced silently by an ordinary number-conserving correlation matrix. A genuinely fermionic BCS initial state instead requires the full Nambu covariance.

Adversarial controls: resolution and time window

First change the discrete approximation to the global $U(1)$ twirl. With

$$\mathcal G_M(\rho) :=\frac1M\sum_{m=0}^{M-1} e^{2\pi i mQ/M}\rho e^{-2\pi i mQ/M},$$

charges are separated only modulo $M$. In the two-site benchmark:

Phase samples	Coherences removed	Reported global asymmetry
$M=1$	none	$0$
$M\ge2$	the supported $q=0$ and $q=1$ sectors	$\log2=0.6931471806$

Thus an under-resolved twirl can falsely report zero asymmetry or underestimate it while remaining a normalized quantum channel.

Next define a first-crossing time $t_\star(\delta)$ by $\Delta S_1(t_\star)=\delta$ on $0\le Jt\le\pi/2$. Direct solution gives

$$Jt_\star(0.1)=1.2225181291, \qquad Jt_\star(0.01)=1.4893814071.$$

Both appear to approach the zero at $\pi/2$ if the observation window ends there. Extending the same calculation to $Jt=\pi$ reveals the full revival. The manifest adversary therefore fails both proposed shortcuts: a restoration time depends on the threshold and window, while the inferred magnitude depends on sector resolution. A thermodynamic restoration claim must show convergence under increasing total size, subsystem sizes, phase resolution, and observation window.

What current evidence establishes

As of 26 August 2026, large-system spin-chain results establish model-dependent local relaxation and Mpemba-like order reversals, not a theorem that every symmetric QFT dynamics restores every subsystem. The primary XX-chain result and its scaling qualifications are Ares, Murciano, and Calabrese 2023, Eqs. (10)–(14) and Fig. 3.

Florio and Murciano 2026 is a valid published gauge-theory application, but it is not a restoration quench. It computes the chiral-charge asymmetry of the full massless Schwinger-model ground or Gibbs state: Eqs. (7)–(10) define the quantity, Eq. (18) gives the zero-temperature result, and Eqs. (23)–(28) give the finite-temperature expression, limits, variance, and bound in the author manuscript. Its conclusion identifies dynamical symmetry restoration in gauge theories as future work. It therefore supports a static anomaly-sensitive asymmetry diagnostic, not evidence for late-time subsystem restoration.

Anomaly, conserved-charge, finite-density, and infrared effects must be examined model by model. Any broader current-evidence survey should carry a dated source set, contrary evidence, and explicit alternative explanations.

Common pitfalls

Using one-angle moments for a noncommuting state. Rényi asymmetry requires the multi-angle sequence generated by powers of the twirled state.

Inferring global restoration from a subsystem. Global asymmetry is conserved by the symmetric unitary even when a finite reduced state becomes charge diagonal.

Changing limits silently. Fixed-subsystem late time, ballistic scaling, and a finite-system observation window are different statements.

Using one odd observable as a complete test. One vanishing expectation value does not imply $[\rho_A,Q_A]=0$.

Exercises

1. Prove the relative-entropy identity

Show that $D(\rho\Vert\mathcal G\rho)=S(\mathcal G\rho)-S(\rho)$ for a compact-group twirl.

Solution

The twirled state is invariant, so $\log(\mathcal G\rho)$ is invariant as well. Haar invariance and cyclicity give

$$\operatorname{Tr}[\rho\log(\mathcal G\rho)] =\operatorname{Tr}[(\mathcal G\rho)\log(\mathcal G\rho)].$$

Consequently,

$$\begin{aligned} D(\rho\Vert\mathcal G\rho) &=\operatorname{Tr}(\rho\log\rho) -\operatorname{Tr}[\rho\log(\mathcal G\rho)]\\ &=-S(\rho)+S(\mathcal G\rho). \end{aligned}$$

Nonnegativity and equality only at $\rho=\mathcal G\rho$ follow from the corresponding properties of relative entropy.

2. Recover the two-angle second moment

For a $U(1)$ charge, show directly that the multi-angle formula at $n=2$ equals $\sum_q\operatorname{Tr}[(\Pi_q\rho\Pi_q)^2]$.

Solution

At $n=2$, only the difference $\beta=\alpha_2-\alpha_1$ remains after one redundant group integral:

$$\operatorname{Tr}[(\mathcal G\rho)^2] =\int_{-\pi}^{\pi}\frac{d\beta}{2\pi} \operatorname{Tr}(\rho e^{i\beta Q}\rho e^{-i\beta Q}).$$

Insert $I=\sum_q\Pi_q$ on both sides of each $\rho$. The phase integral gives $\delta_{q q'}$, leaving

$$\sum_q\operatorname{Tr}(\Pi_q\rho\Pi_q\rho) =\sum_q\operatorname{Tr}[(\Pi_q\rho\Pi_q)^2] =\operatorname{Tr}[(\mathcal G\rho)^2].$$

A one-angle moment $\operatorname{Tr}(\rho^2e^{i\beta Q})$ has a different operator ordering when $[\rho,Q]\ne0$.

3. Test a false restoration claim

Suppose only the samples $0\le Jt\le\pi/2$ of the two-site quench are retained and $M=1$ is used for the global twirl. What conclusions would be reported, and which exact controls refute them?

Solution

With $M=1$, $\mathcal G_1$ is the identity channel, so the reported global asymmetry is incorrectly zero at every time. On the truncated one-site window, $\Delta S_1$ decreases monotonically to zero, suggesting permanent restoration near $Jt=\pi/2$.

The exact $M=2$ twirl separates the two supported global charges and gives the conserved value $\log2$. Extending the time window to $Jt=\pi$ gives $\Delta S_1(\pi/J)=\log2$, an exact revival. The data license only a finite-size transfer of coherence, not global or irreversible restoration.

References

• Ares, Filiberto, Sara Murciano, and Pasquale Calabrese. “Entanglement Asymmetry as a Probe of Symmetry Breaking.” Nature Communications 14 (2023): 2036. DOI. Open PDF.
• Florio, Adrien, and Sara Murciano. “Entanglement Asymmetry in Gauge Theories: Chiral Anomaly in the Finite Temperature Massless Schwinger Model.” Physical Review D 113 (2026): L091901. DOI. Open PDF.
• Gour, Gilad, Iman Marvian, and Robert W. Spekkens. “Measuring the Quality of a Quantum Reference Frame: The Relative Entropy of Frameness.” Physical Review A 80 (2009): 012307. DOI.