Covariant Channels and Symmetry-Restricted Recovery

Recovery under a symmetry constraint asks whether unknown quantum information can be reconstructed by an allowed covariant channel. This is stricter than preparing a known asymmetric state with an external phase standard. Twirling can make a decoder covariant when its objective is symmetric, and a finite reference can enable approximate relational decoding, but neither operation reverses information that the noise has irretrievably placed in an inaccessible environment.

Required background. Symmetry-constrained operations defines covariance; recovery and approximate Markovianity supplies unrestricted recovery bounds.

Helpful background. Superselection and accessible entanglement supplies the sector-by-sector operational decomposition.

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

Covariant noise, recovery, and objectives

Let $U_g$ act on the input and $V_g$ on the noisy output. A channel $\mathcal N$ is covariant when

$$\mathcal N(U_g\rho U_g^\dagger) =V_g\mathcal N(\rho)V_g^\dagger.$$

A reverse-direction recovery $\mathcal R$ is covariant when

$$\mathcal R(V_gXV_g^\dagger) =U_g\mathcal R(X)U_g^\dagger.$$

For compact $G$, any CPTP map can be group averaged:

$$\mathcal R_G(X) :=\int_Gdg\, U_g\mathcal R(V_g^\dagger XV_g)U_g^\dagger.$$

The result is CPTP and covariant. For a group-invariant code ensemble, a group-invariant performance functional that is affine in the channel has exactly the same value on $\mathcal R_G$ and $\mathcal R$. If instead one minimizes a group-invariant convex loss, Jensen’s inequality says that the twirled decoder is no worse, but equality need not hold. Neither statement preserves pointwise fidelity on one asymmetric state. Bartlett, Rudolph, and Spekkens 2007, § II.C, Eqs. (2.14)–(2.18) give the state and operation twirls and the covariance condition.

For a compact group, the relative entropy of asymmetry is

$$A_G(\rho) :=D(\rho\|\mathcal G(\rho)) =S(\mathcal G(\rho))-S(\rho), \qquad \mathcal G(\rho):=\int_Gdg\,U_g\rho U_g^\dagger,$$

whenever the entropies are finite. It cannot increase under a $G$-covariant channel. The group-orbit encoding and data-processing argument is Marvian and Spekkens 2014, § III, Eq. (3.1). Their Theorem 1 is instead a no-go result for continuous asymmetry monotones built only from characteristic functions or Noether moments; it is not the monotonicity theorem.

Covariance of Petz recovery

For a positive reference $\sigma$, the Petz map on the support of $\mathcal N(\sigma)$ is

$$\mathcal P_{\sigma,\mathcal N}(X) =\sigma^{1/2}\mathcal N^\dagger\!\left( \mathcal N(\sigma)^{-1/2}X \mathcal N(\sigma)^{-1/2} \right)\sigma^{1/2}.$$

Assume $\sigma$ is invariant and $\mathcal N$ is covariant. Then $\mathcal N(\sigma)$ is invariant, the support inverses commute with the relevant group action, and the adjoint intertwines the representations:

$$\mathcal N^\dagger(V_gXV_g^\dagger) =U_g\mathcal N^\dagger(X)U_g^\dagger.$$

Substitution gives

$$\mathcal P_{\sigma,\mathcal N}(V_gXV_g^\dagger) =U_g\mathcal P_{\sigma,\mathcal N}(X)U_g^\dagger.$$

The modular rotations are also invariant, so every rotated Petz map and their average are covariant. For fermionic Gaussian $\sigma$ and Gaussian $\mathcal N$, the Petz and rotated Petz maps remain Gaussian; Swingle and Wang 2019, Eqs. (3), (7)–(8), and §§ III.2–III.3 give their covariance-matrix construction and support qualifications.

With root fidelity $F(\rho,\tau)=\|\sqrt\rho\sqrt\tau\|_1$, the single averaged universal map obeys

$$D(\rho\|\sigma) -D(\mathcal N(\rho)\|\mathcal N(\sigma)) \ge -2\log F\!\left( \rho, (\mathcal R_{\sigma,\mathcal N}\circ\mathcal N)(\rho) \right),$$

where

$$\mathcal R_{\sigma,\mathcal N} =\int_{-\infty}^{\infty}dt\, \beta_0(t)\mathcal R_{\sigma,\mathcal N}^{t/2}, \qquad \beta_0(t)=\frac{\pi}{2[\cosh(\pi t)+1]}.$$

This is Junge et al. 2018, Remark 2.2, Eqs. (20)–(21). Their Theorem 2.1, Eqs. (17)–(19), is the stronger statement with an integral of the logarithm of the fidelity of each rotated map; it is not the single-map formula displayed here.

A two-mode Gaussian code

Let $f_1,f_2$ be fermionic modes with charge

$$Q=f_1^\dagger f_1+f_2^\dagger f_2.$$

Use the even-parity code

$$|0_C\rangle:=|00\rangle, \qquad |1_C\rangle:=|11\rangle.$$

Every normalized state $u|00\rangle+v|11\rangle$ is a two-mode BCS Gaussian state, but its two logical basis components carry charges $0$ and $2$. The desired output is a neutral logical qubit on which $U(1)$ acts trivially.

An unrestricted decoder can map $|0_C\rangle\mapsto|0_L\rangle$ and $|1_C\rangle\mapsto|1_L\rangle$ coherently. A strictly covariant decoder without a reference cannot map the charge-two operator $|0_C\rangle\langle1_C|$ to a neutral output operator: covariance forces that mode to vanish. The optimal decoder that preserves both logical basis states therefore induces complete dephasing,

$$\mathcal D_L(\rho) =|0\rangle\langle0|\rho|0\rangle\langle0| +|1\rangle\langle1|\rho|1\rangle\langle1|,$$

whose entanglement fidelity on the maximally mixed logical input is $1/2$.

Executed finite-reference alignment benchmark

Supply the phase reference with finite initial support

$$|R_M\rangle :=\frac1{\sqrt{M+1}} \sum_{n=0}^{M}|n\rangle, \qquad M\ge2.$$

$M$ is an initial-support parameter, not a hard Hilbert-space cutoff. The shifted branch below occupies number levels through $M+2$, so the reference carrier must include those levels or be unbounded.

A charge-conserving joint decoder transfers the code’s two units of charge into the reference. The two logical branches leave reference states $|R_M\rangle$ and

$$|R_M^{(2)}\rangle :=\frac1{\sqrt{M+1}} \sum_{n=2}^{M+2}|n\rangle.$$

Their exact overlap is

$$c_M :=\langle R_M|R_M^{(2)}\rangle =\frac{M-1}{M+1}.$$

The induced logical channel preserves populations and multiplies the off-diagonal elements by $c_M$. Its entanglement fidelity is therefore

$$F_e(M)=\frac{1+c_M}{2}=\frac{M}{M+1}.$$

For an equal logical mixture, the reference after one use is

$$\tau_M =\frac12\left( |R_M\rangle\langle R_M| +|R_M^{(2)}\rangle\langle R_M^{(2)}| \right).$$

The initial reference asymmetry is $A_{U(1)}(|R_M\rangle)=\log_2(M+1)$. Since $\tau_M$ has eigenvalues $(1\pm c_M)/2$ and its number-twirled distribution has four boundary weights $1/[2(M+1)]$ and $M-1$ interior weights $1/(M+1)$, the one-use asymmetry loss is

$$\Delta A_R(M) =H_2\!\left(\frac{M}{M+1}\right) -\frac{2}{M+1}.$$

Initial-support parameter $M$	Overlap $c_M$	$F_e(M)$	Initial asymmetry	One-use asymmetry loss
$2$	$1/3$	$0.666667$	$1.584963$ bits	$0.251629$ bits
$4$	$3/5$	$0.800000$	$2.321928$ bits	$0.321928$ bits
$9$	$4/5$	$0.900000$	$3.321928$ bits	$0.268996$ bits
$99$	$49/50$	$0.990000$	$6.643856$ bits	$0.060793$ bits

For this declared family and one-use protocol, the reference-assisted decoder approaches the unrestricted decoder as $M\to\infty$. That convergence is not a universal property of every reference state or repeated-use protocol.

Genuine full-dephasing no-go control

Now let the noise act before decoding as the physical charge dephasing

$$\mathcal D_C(\rho) =\Pi_0\rho\Pi_0+\Pi_2\rho\Pi_2.$$

Apply it to half of the maximally entangled code state

$$|\Phi\rangle_{RC} =\frac1{\sqrt2} \left( |0\rangle_R|0_C\rangle +|1\rangle_R|1_C\rangle \right).$$

The result is

$$(\operatorname{id}_R\otimes\mathcal D_C) (|\Phi\rangle\langle\Phi|) =\frac12 \sum_{j=0}^{1} |j\rangle\langle j|_R \otimes|j_C\rangle\langle j_C|,$$

which is separable across $R:C$. Any decoder acting on $C$ and an independent phase reference is local with respect to the inaccessible $R$ and cannot create $R:L$ entanglement. The largest overlap of a separable two-qubit state with $|\Phi\rangle$ is $1/2$, so

$$F_e^{\rm full\ dephasing}\le\frac12$$

for unrestricted, covariant, and independently reference-assisted decoders alike. A phase standard can align or prepare coherence; it cannot reconstruct unknown coherence whose which-charge record has been discarded.

Executed covariance-breaking adversary

Let $\mathcal R_{\rm cov}$ induce complete logical dephasing and let $\mathcal R_{\rm ideal}$ be the unrestricted coherent decoder. Define

$$\mathcal R_\epsilon =(1-\epsilon)\mathcal R_{\rm cov} +\epsilon\mathcal R_{\rm ideal}, \qquad 0\le\epsilon\le1.$$

Its entanglement fidelity is

$$F_e(\mathcal R_\epsilon) =\frac12+\frac\epsilon2.$$

For the neutral output, measure covariance breaking by

$$\delta_{\rm cov}(\mathcal R) :=\sup_{\theta,\rho} \frac12\left\| \mathcal R(U_\theta\rho U_\theta^\dagger) -\mathcal R(\rho) \right\|_1.$$

The inputs $|+_C\rangle$ and $U_{\pi/2}|+_C\rangle=|-_C\rangle$ attain the supremum, giving

$$\delta_{\rm cov}(\mathcal R_\epsilon) =\epsilon =2F_e(\mathcal R_\epsilon)-1.$$

Thus even a small concealed noncovariant component produces a linearly improved score. A charged ancilla that supplies this component is a reference resource; omitting its charge distribution and degradation from the channel specification invalidates the restricted-recovery comparison.

Capacity and QFT boundaries

The strict decoder’s induced logical channel is entanglement breaking, so its quantum capacity for this fixed neutral-output code is zero even though it transmits the classical charge populations. This is a statement about the specified code and resource theory, not a no-go theorem for every covariant encoding. An asymptotic capacity claim must also state the allowed reference consumption per use, energy constraint, localization, and error criterion.

Group covariance alone does not imply spacetime locality. A globally twirled recovery can require access to the entire output. In gauge theory, the channel must act on the gauge-invariant algebra and respect its center; an anomaly can obstruct a local symmetric realization even when a formal group average exists.

Common pitfalls

Citing the wrong recovery statement. The single averaged universal map is Junge et al. Remark 2.2, Eqs. (20)–(21); Theorem 2.1 is the stronger integral inequality.

Repreparing instead of recovering. An independent phase standard cannot restore entanglement with an inaccessible reference after full dephasing.

Hiding a charged ancilla. A joint covariant protocol can look noncovariant after its reference is omitted. Include the reference state, charge range, and degradation.

Exercises

1. Twirling a decoder

Prove that $\mathcal R_G$ is CPTP and covariant. For a group-invariant ensemble, show that a group-invariant affine score is unchanged, whereas a group-invariant convex loss is no greater. Explain why a general equality claim fails.

Solution

Each integrand is a composition of unitary channels and the CPTP map $\mathcal R$, hence is CPTP; a convex Haar integral remains CPTP. For $h\in G$,

$$\begin{aligned} \mathcal R_G(V_hXV_h^\dagger) &=\int dg\, U_g\mathcal R(V_{h^{-1}g}^\dagger X V_{h^{-1}g})U_g^\dagger\\ &=U_h\mathcal R_G(X)U_h^\dagger, \end{aligned}$$

after a Haar-invariant change of variable. Write $\mathcal R_g$ for the transformed decoder inside the integral. Ensemble and output invariance give the same value to every $\mathcal R_g$. Hence an affine score $F$ obeys

$$F(\mathcal R_G) =\int dg\,F(\mathcal R_g) =F(\mathcal R).$$

For a convex loss $L$, Jensen’s inequality instead gives

$$L(\mathcal R_G) \leq\int dg\,L(\mathcal R_g) =L(\mathcal R),$$

with possible strict inequality. Without the stated invariances, even these average comparisons can fail, and neither argument fixes a pointwise loss on one asymmetric input.

2. Finite-reference fidelity and degradation

Derive $c_M$, $F_e(M)$, and the spectrum of $\tau_M$.

Solution

The number supports of $|R_M\rangle$ and $|R_M^{(2)}\rangle$ overlap on $n=2,\ldots,M$, which contains $M-1$ basis states. Every matching term contributes $1/(M+1)$, so

$$c_M=\frac{M-1}{M+1}.$$

A qubit dephasing channel with coherence factor $c_M$ has

$$F_e=\langle\Phi|(\operatorname{id}\otimes\mathcal E)(|\Phi\rangle\langle\Phi|)|\Phi\rangle =\frac{1+c_M}{2} =\frac{M}{M+1}.$$

An equal mixture of two pure states with real overlap $c_M$ has eigenvalues $(1\pm c_M)/2$. Twirling $\tau_M$ in number gives the boundary/interior distribution stated above, from which subtracting $S(\tau_M)=H_2((1+c_M)/2)$ yields the displayed asymmetry loss.

3. Why an independent reference cannot undo dephasing

Prove the bound $F_e\le1/2$ after $\mathcal D_C$, even if the decoder receives an arbitrary independent reference state.

Solution

After dephasing, the reference–code state is separable. Tensoring an independent reference state preserves separability across the cut between the inaccessible system $R$ and everything controlled by the decoder. A local CPTP map on the decoder’s side also preserves separability.

For any product pure state $|a\rangle|b\rangle$, Cauchy–Schwarz gives

$$|\langle\Phi|a,b\rangle|^2 =\frac12|a_0b_0+a_1b_1|^2 \le\frac12.$$

Convexity extends the bound to every separable mixed state. Therefore no decoder with an independent phase reference can exceed entanglement fidelity $1/2$ after the full which-charge record has been lost.

References

• Bartlett, Stephen D., Terry Rudolph, and Robert W. Spekkens. “Reference Frames, Superselection Rules, and Quantum Information.” Reviews of Modern Physics 79 (2007): 555–609. DOI. Open PDF.
• Junge, Marius, Renato Renner, David Sutter, Mark M. Wilde, and Andreas Winter. “Universal Recovery Maps and Approximate Sufficiency of Quantum Relative Entropy.” Annales Henri Poincaré 19 (2018): 2955–2978. DOI. Open PDF.
• Marvian, Iman, and Robert W. Spekkens. “Extending Noether’s Theorem by Quantifying the Asymmetry of Quantum States.” Nature Communications 5 (2014): 3821. DOI. Open PDF.
• Swingle, Brian, and Yixu Wang. “Recovery Map for Fermionic Gaussian Channels.” Journal of Mathematical Physics 60 (2019): 072202. DOI. Open PDF.