Reference Frames, Asymmetry, and Charged Resources

A missing reference frame makes coherence between symmetry sectors inaccessible to an agent’s declared operations. Group twirling gives the effective state seen without that frame, while asymmetry monotones quantify some of the tasks a physical reference can enable. An ideal phase standard is a limiting resource, not free background structure: its charge support, accuracy, localization, and behavior under reuse belong in the protocol.

Required background. Symmetry-constrained operations defines covariant operations and admissible ancillas.

Helpful background. Superselection and accessible entanglement shows how a reference changes sector accessibility.

Chapter map. See the chapter-wide path from a task specification to an information resource, claim-validity table, and three gates before an operational claim.

Twirling and irreducible sectors

Let $G$ be compact, let $U(g)$ be a finite-dimensional unitary representation, and normalize Haar measure by $\int_Gdg=1$. The twirling channel is

$$\mathcal G(\rho) =\int_Gdg\,U(g)\rho U(g)^\dagger.$$

Decompose the representation as

$$\mathcal H =\bigoplus_\lambda \left(V_\lambda\otimes M_\lambda\right), \qquad U(g)=\bigoplus_\lambda \left(U_\lambda(g)\otimes I_{M_\lambda}\right).$$

Here $V_\lambda$ is an irreducible carrier of dimension $d_\lambda$, and $M_\lambda$ is the multiplicity space. Schur orthogonality gives

$$\mathcal G(\rho) =\bigoplus_\lambda \frac{I_{V_\lambda}}{d_\lambda} \otimes \operatorname{Tr}_{V_\lambda} \!\left(\Pi_\lambda\rho\Pi_\lambda\right).$$

The twirl removes coherence between inequivalent irreducible sectors, depolarizes each representation carrier, and preserves information in multiplicity spaces. For $U(1)$, irreducible representations are one dimensional and the formula reduces to charge dephasing,

$$\mathcal G(\rho)=\sum_q\Pi_q\rho\Pi_q.$$

Twirling may represent ignorance of a frame, an actual randomization channel, or averaging over an uncontrolled group element. Those interpretations have the same density operator but different physical histories Bartlett, Rudolph, and Spekkens 2007, §§ II.B–II.C.

Relative entropy of asymmetry—and its limit

For a finite-dimensional state, define

$$A_G(\rho) =S(\mathcal G\rho)-S(\rho).$$

Because $\log(\mathcal G\rho)$ is invariant under $G$,

$$\operatorname{Tr}\!\left[\rho\log(\mathcal G\rho)\right] =\operatorname{Tr}\!\left[\mathcal G(\rho) \log(\mathcal G\rho)\right],$$

and therefore

$$A_G(\rho) =D\!\left(\rho\middle\|\mathcal G\rho\right).$$

A $G$-covariant channel commutes with the twirl. Data processing for relative entropy then proves $A_G(\mathcal E\rho)\le A_G(\rho)$. These identities and the reference-token interpretation are established in Gour, Marvian, and Spekkens 2009, §§ II–III.

This scalar does not determine every conversion task. In particular,

$$A_G^\infty(\rho) =\lim_{n\to\infty}\frac1n A_G(\rho^{\otimes n})=0$$

for finite groups and compact Lie groups in the setting analyzed in Gour, Marvian, and Spekkens 2009, § IV.B–IV.E, especially Eq. (14). Relative entropy of asymmetry is consequently not a universal asymptotic conversion rate. Charge variance, quantum Fisher information, or the full set of asymmetry modes can govern other tasks.

Relational encoding

An absolute phase is unavailable without a reference, but a relative phase can live inside a fixed total-charge sector. For two unit-charge modes,

$$|0_L\rangle=|0\rangle_S|1\rangle_R, \qquad |1_L\rangle=|1\rangle_S|0\rangle_R.$$

Both kets have total charge one. Under the diagonal $U(1)$ action they acquire the same phase $e^{i\theta}$, so every logical density operator is invariant. Symmetric operations can therefore manipulate coherent superpositions $\alpha|0_L\rangle+\beta|1_L\rangle$ without violating the rule.

Locality remains separate. If $S$ and $R$ are in distant laboratories, distributing and synchronizing the reference is itself a physical resource. A formal global relational state does not give either laboratory instantaneous access to the other mode.

Exact finite-phase-reference benchmark

Let $|1\rangle_S=a_f^\dagger|0\rangle_S$ be a charged wavepacket and, with equal prior probabilities, define the two possible target phases

$$|\pm\rangle_S =\frac{|0\rangle_S\pm|1\rangle_S}{\sqrt2}.$$

Without a reference, $U(1)$ twirling gives the same state for either sign:

$$\mathcal G_S(|\pm\rangle\langle\pm|) =\frac12\left(|0\rangle\langle0|+|1\rangle\langle1|\right).$$

No invariant measurement can then distinguish the signs better than guessing, so $P_{\rm succ}=1/2$.

Now supply the finite reference

$$|R_N\rangle =\frac1{\sqrt{N+1}} \sum_{n=0}^N|n\rangle_R.$$

The joint states are still assessed by invariant measurements, equivalently after twirling their total charge. For each interior total charge $k=1,\ldots,N$, define

$$|\chi_k^\pm\rangle =\frac{|0\rangle_S|k\rangle_R \pm|1\rangle_S|k-1\rangle_R}{\sqrt2}.$$

The twirled joint states are

$$\begin{aligned} \rho_\pm^{(N)} ={}&\frac{|0,0\rangle\langle0,0| +|1,N\rangle\langle1,N|}{2(N+1)}\\ &+\frac1{N+1}\sum_{k=1}^N |\chi_k^\pm\rangle\langle\chi_k^\pm|. \end{aligned}$$

For every interior $k$, $|\chi_k^+\rangle$ and $|\chi_k^-\rangle$ are orthogonal. The two boundary sectors contain only one basis state each and therefore carry no sign information. It follows that the trace distance is

$$\frac12\left\lVert \rho_+^{(N)}-\rho_-^{(N)} \right\rVert_1 =\frac{N}{N+1}.$$

The invariant Helstrom measurement succeeds with

$$P_{\rm succ}(N) =\frac12\left(1+\frac{N}{N+1}\right) =1-\frac1{2(N+1)}.$$

This is a complete one-shot task: the two target states, prior probabilities, reference, allowed measurements, and error are fixed. The reference resources are also explicit:

$$\begin{aligned} A_{U(1)}(|R_N\rangle)&=\ln(N+1),\\ \langle Q_R\rangle&=\frac N2,\\ \operatorname{Var}(Q_R)&=\frac{N(N+2)}{12}. \end{aligned}$$

For example, $N=9$ gives

$$P_{\rm succ}=0.95, \qquad A_{U(1)}=\ln10, \qquad \langle Q_R\rangle=4.5, \qquad \operatorname{Var}(Q_R)=8.25.$$

To achieve error at most $\epsilon$,

$$N+1\ge\frac1{2\epsilon},$$

and hence

$$A_{U(1)}(|R_N\rangle) \ge\ln\frac1{2\epsilon}, \qquad \operatorname{Var}(Q_R) \ge\frac1{48\epsilon^2}-\frac1{12}.$$

Replacing $|R_N\rangle$ by an implicit perfect phase standard therefore hides an unbounded-support limit. Conversely, removing the reference returns the exact success probability to $1/2$. This executes both the charged-wavepacket application and its adversarial perfect-frame replacement.

Reuse is a protocol-dependent claim

The one-shot Helstrom benchmark does not specify a postmeasurement reference state. It therefore proves an accuracy–resource tradeoff, not a repeated-use law. To study reuse, one must choose a charge-conserving interaction and an instrument, then record after every use:

1. the reference marginal;
2. correlations and mutual information between the reference and all targets;
3. performance on a fresh target; and
4. total asymmetry, charge support, and energy budget.

Specific finite phase and direction references do degrade under repeated measurement, with longevity determined by the chosen task Bartlett et al. 2006, §§ 3–4. More generally, a finite-dimensional reference cannot broadcast asymmetry for a connected Lie group while keeping its marginal unchanged, even when correlations are allowed; weaker indefinite repeatability requires an infinite-dimensional system Lostaglio and Müller 2019, Theorems 1–3. These results rule out a free finite catalyst but do not imply that every reference quality measure decreases monotonically in every protocol.

Frameness is not entanglement

A single-system reference can be asymmetric without being bipartite entangled, and an entangled state can be globally symmetric. Frameness enables transformations forbidden by covariance; entanglement enables nonlocal tasks under local operations. When a shared reference unlocks entanglement across superselection sectors, the gain combines two resources rather than identifying them. Quantitative tradeoffs among work, accessible entanglement, and reference quality are analyzed in Vaccaro et al. 2008, §§ III–IV.

Common pitfalls

Treating twirling as only physical noise. It can instead encode lack of a reference. State the operational interpretation.

Calling $A_G$ the conversion rate. It is a monotone with a useful relative-entropy meaning, but its regularized value vanishes in the compact-group setting above.

Claiming degradation without an instrument. One-shot accuracy does not determine the reference’s postmeasurement state. Specify the reuse protocol and track correlations.

Exercises

1. Derive the non-Abelian twirl

Use Schur orthogonality to derive the block formula for $\mathcal G(\rho)$.

Solution

Write each block as $\Pi_\lambda\rho\Pi_\mu$. Averaging an off-diagonal block intertwines inequivalent irreducible representations, so Schur’s lemma makes it zero when $\lambda\ne\mu$. Inside one $\lambda$ block, expand the operator in a basis of multiplicity-space matrix elements. The group average acts only on $V_\lambda$ and maps any carrier operator $X$ to

$$\int_Gdg\,U_\lambda(g)XU_\lambda(g)^\dagger =\frac{\operatorname{Tr}X}{d_\lambda}I_{V_\lambda}.$$

Applying this to every multiplicity matrix element gives

$$\mathcal G(\rho) =\bigoplus_\lambda \frac{I_{V_\lambda}}{d_\lambda} \otimes\operatorname{Tr}_{V_\lambda} \!\left(\Pi_\lambda\rho\Pi_\lambda\right).$$

2. Relative entropy identity and monotonicity

Prove $A_G(\rho)=D(\rho\|\mathcal G\rho)$ and show that it cannot increase under a $G$-covariant channel.

Solution

The operator $\log(\mathcal G\rho)$ is invariant. Averaging $\rho$ inside its trace therefore changes nothing:

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

Substitution in $D(\rho\|\sigma)=\operatorname{Tr}\rho(\log\rho-\log\sigma)$ gives

$$D(\rho\|\mathcal G\rho) =-S(\rho)+S(\mathcal G\rho)=A_G(\rho).$$

If $\mathcal E$ is covariant, then $\mathcal E\mathcal G=\mathcal G\mathcal E$. Data processing yields

$$\begin{aligned} A_G(\mathcal E\rho) &=D(\mathcal E\rho\|\mathcal E\mathcal G\rho)\\ &\le D(\rho\|\mathcal G\rho) =A_G(\rho). \end{aligned}$$

3. Accuracy cost of the finite phase reference

Derive $P_{\rm succ}(N)$ and determine the smallest $N$ that guarantees error at most $1\%$.

Solution

The $N$ interior sectors have total probability $N/(N+1)$ and contain orthogonal sign states, so they can be distinguished perfectly. The two boundary sectors have total probability $1/(N+1)$ and identical density operators, so their best success probability is $1/2$. Thus

$$P_{\rm succ} =\frac{N}{N+1} +\frac1{N+1}\frac12 =1-\frac1{2(N+1)}.$$

For error at most $0.01$,

$$\frac1{2(N+1)}\le0.01 \quad\Longrightarrow\quad N+1\ge50.$$

The smallest integer is $N=49$. It requires

$$A_{U(1)}=\ln50, \qquad \operatorname{Var}(Q_R) =\frac{49\cdot51}{12}=208.25.$$

References

• Bartlett, Stephen D., Terry Rudolph, Robert W. Spekkens, and Peter S. Turner. “Degradation of a Quantum Reference Frame.” New Journal of Physics 8 (2006): 58. DOI.
• 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.
• 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.
• Lostaglio, Matteo, and Markus P. Müller. “Coherence and Asymmetry Cannot Be Broadcast.” Physical Review Letters 123 (2019): 020403. DOI.
• Vaccaro, Joan A., F. Anselmi, Howard M. Wiseman, and Kurt Jacobs. “Tradeoff between Extractable Mechanical Work, Accessible Entanglement, and Ability to Act as a Reference System, under Arbitrary Superselection Rules.” Physical Review A 77 (2008): 032114. DOI.