Superselection Rules and Accessible Entanglement

A superselection rule can make part of a formal bipartite entropy inaccessible to a specified protocol. For a fixed total charge, the reduced state decomposes into local charge sectors: one entropy term records uncertainty in the sector label, while another records entanglement within sectors. Only after the local operation class and reference budget are fixed can either term be given an operational meaning.

Required background. Symmetry-constrained operations defines the covariant operation class and reference budget.

Helpful background. Measure selection distinguishes total correlations from task-specific entanglement.

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.

Charge sectors and the entropy identity

Work first with a regulated bipartite Hilbert space $\mathcal H_A\otimes\mathcal H_B$ and an additive Abelian charge

$$Q_{\rm tot}=Q_A+Q_B.$$

If the global state $\rho_{AB}$ commutes with $Q_{\rm tot}$, then its reduced state commutes with $Q_A$. To see this, conjugate $\rho_{AB}$ by $e^{i\theta(Q_A+Q_B)}$ and use invariance of the partial trace under a unitary on $B$. Thus

$$[\rho_A,Q_A]=0, \qquad \rho_A=\bigoplus_q p_q\rho_{A,q},$$

where

$$p_q=\operatorname{Tr}(\Pi_q^A\rho_A), \qquad \rho_{A,q}=\frac{\Pi_q^A\rho_A\Pi_q^A}{p_q}$$

when $p_q>0$. The normalized sector states have $\operatorname{Tr}\rho_{A,q}=1$. In the algebra of all operators on $A$, $\Pi_q^A$ need not be central. It is central in the invariant accessible algebra, whose operators commute with $Q_A$.

The spectrum of the $q$ block is $p_q$ times the spectrum of $\rho_{A,q}$. Therefore

$$S(\rho_A) =H(\{p_q\})+\sum_qp_qS(\rho_{A,q}),$$

with natural logarithms and

$$H(\{p_q\})=-\sum_qp_q\ln p_q.$$

This direct-sum identity is kinematic. It holds before any decision about distillation, metrology, or reference-assisted operations. The same sector normalization and entropy split appear in the QFT charged-moment construction Murciano, Di Giulio, and Calabrese 2020, § 2.1, Eqs. (4)–(10).

The precise scope of accessible entanglement

Now impose all of the following hypotheses:

• $\rho_{AB}=|\psi\rangle\langle\psi|$ is pure and has fixed total particle number $Q_0$;
• Alice and Bob use local operations and classical communication that preserve their local particle numbers;
• no shared phase reference or charge reservoir is supplied; and
• one evaluates the sector-resolved, one-copy accessible-entanglement functional.

The normalized projected state in local sector $q$ is

$$|\psi_q\rangle =\frac{(\Pi_q^A\otimes\Pi_{Q_0-q}^B)|\psi\rangle}{\sqrt{p_q}}.$$

Alice may measure $q$ without violating the rule. Conditioned on the outcome, the usual pure-state entanglement is $S(\rho_{A,q})$, so

$$E_{\rm acc}(|\psi\rangle) =\sum_qp_qS(\rho_{A,q}).$$

This is the operational quantity introduced by Wiseman and Vaccaro 2003, Eqs. (1)–(4) and the examples following Eq. (4). It is not a universal formula for mixed-state distillable entanglement, for non-Abelian groups, or for protocols supplied with a reference.

Under these hypotheses,

$$S(\rho_A)-E_{\rm acc}=H(\{p_q\}).$$

The right-hand side is the number or fluctuation entropy. It is not, by itself, a complete frameness monotone or a universally convertible supply of Bell pairs. For particle-number superselection, superselection-induced variance is an additional asymptotic pure-state resource; collective processing can trade that relational number information against otherwise inaccessible tasks Schuch, Verstraete, and Cirac 2004, §§ II–III.

For a non-Abelian compact group, the local representation decomposes as

$$\mathcal H_A =\bigoplus_\lambda \left(V_\lambda\otimes M_\lambda\right).$$

Here $V_\lambda$ carries an irreducible representation of dimension $d_\lambda$, while $M_\lambda$ is its multiplicity space. Invariant local operators act nontrivially on multiplicity spaces but as the identity on an irreducible carrier. Irrep-label uncertainty, carrier dimensions, and multiplicity entanglement are therefore distinct contributions. The Abelian formula above should not be extended by merely inserting a factor $d_\lambda$.

Exact charged-Gaussian activation benchmark

Let $a_j^\dagger$ and $b_j^\dagger$ create orthogonal fermionic wavepacket modes in Alice’s and Bob’s laboratories. The one-particle Slater determinant

$$|\psi_1\rangle =\frac{a_1^\dagger+b_1^\dagger}{\sqrt2}|0\rangle$$

is a number-conserving Gaussian state with one particle. Alice’s correlation matrix is the one-by-one matrix $C_A=(1/2)$, so

$$\rho_A =\frac12|0\rangle\langle0| +\frac12|1\rangle\langle1|, \qquad S(\rho_A)=\ln2.$$

The sectors $q_A=0$ and $q_A=1$ are each pure. Consequently

$$H(p)=\ln2, \qquad E_{\rm acc}^{(1)}=0.$$

The formal mode entropy is one ebit if logarithms are converted to base two, but Alice and Bob cannot distill it from one copy using local number-preserving operations.

Now supply a second delocalized particle in an independent pair of modes:

$$|\psi_2\rangle =\frac{(a_1^\dagger+b_1^\dagger) (a_2^\dagger+b_2^\dagger)}{2}|0\rangle.$$

This is again a Slater determinant. The local number distribution is

$$(p_0,p_1,p_2) =\left(\frac14,\frac12,\frac14\right).$$

The $q_A=0$ and $q_A=2$ conditional states are products. Up to the fixed fermionic ordering sign, the $q_A=1$ state is

$$|\psi_{q_A=1}\rangle =\frac{|10\rangle_A|01\rangle_B +|01\rangle_A|10\rangle_B}{\sqrt2},$$

which has entropy $\ln2$. Hence

$$E_{\rm acc}^{(2)}=\frac12\ln2, \qquad H(p)=\frac32\ln2, \qquad S(\rho_A)=2\ln2.$$

In bits these are, respectively, $0.5$, $1.5$, and $2$. The second delocalized particle supplies a finite relational charge resource.

Hidden-reference control. Removing that second particle returns the accessible entanglement to zero; treating it as an unrecorded shared reference overstates the protocol’s yield. For $C$ identical copies the binomial sectors give

$$E_{\rm acc}^{(C)} =C\ln2 -H\!\left[\operatorname{Bin}\!\left(C,\frac12\right)\right].$$

The deficit grows only as $\tfrac12\ln C+O(1)$, so the accessible entanglement per copy approaches $\ln2$. This is collective activation, not additivity of the one-copy quantity.

How to resolve a Gaussian state into sectors

For a number-conserving Gaussian reduced state, diagonalize the correlation matrix. If its eigenvalues are $\nu_1,\ldots,\nu_m$, then in the corresponding natural modes

$$\rho_A =\bigotimes_{j=1}^m \left[(1-\nu_j)|0_j\rangle\langle0_j| +\nu_j|1_j\rangle\langle1_j|\right].$$

The probability-generating function is

$$\sum_qp_qz^q =\prod_{j=1}^m(1-\nu_j+z\nu_j),$$

so $p_q$ is the coefficient of $z^q$. A configuration $\boldsymbol n=(n_1,\ldots,n_m)$ has weight

$$w_{\boldsymbol n} =\prod_j\nu_j^{n_j}(1-\nu_j)^{1-n_j}.$$

Conditioning on $\sum_jn_j=q$ gives

$$S(\rho_{A,q}) =-\sum_{|\boldsymbol n|=q} \frac{w_{\boldsymbol n}}{p_q} \ln\frac{w_{\boldsymbol n}}{p_q}.$$

Fixed-number projection generally produces a non-Gaussian state. The correlation eigenvalues determine its configuration weights, but simply projecting the correlation matrix does not determine the sector density operator. Correlation-function methods and their limits are described in Monkman and Sirker 2023, §§ 2–3.

For a truncated numerical sector window, report

$$\epsilon_{\rm tail}=1-\sum_{q\in\mathcal W}p_q$$

and enlarge $\mathcal W$ until normalization, the direct-sum identity, and the requested entropy precision are simultaneously stable.

QFT and gauge-theory boundary

In continuum QFT, the decomposition presupposes a regulated tensor factor, a split inclusion, or a specified regional algebra. In gauge theory, boundary-flux labels arise from Gauss law and the center of that algebra; representation-edge factors depend on the chosen extension. Their formal direct sums resemble particle-number sectors, but their allowed local operations and distillable content must be derived from the gauge-region prescription rather than imported from this benchmark.

Common pitfalls

Calling $H(p)$ distillable entanglement. Under the declared one-copy SSR task, the Bell-pair contribution is the average within-sector entropy. Number uncertainty can become useful only after an additional collective or reference-assisted protocol is specified.

Projecting only the correlation matrix. The fixed-number conditional state is generally non-Gaussian. Reconstruct its configuration weights or use charged moments.

Hiding the activating copy. The second delocalized particle is part of the resource input. Compare the protocol with and without it.

Exercises

1. Entropy of a direct sum

Prove that $\rho=\bigoplus_qp_q\rho_q$ implies $S(\rho)=H(p)+\sum_qp_qS(\rho_q)$.

Solution

Let $\lambda_{q,k}$ be the eigenvalues of $\rho_q$. The eigenvalues of $\rho$ are $p_q\lambda_{q,k}$, so

$$\begin{aligned} S(\rho) &=-\sum_{q,k}p_q\lambda_{q,k} \ln(p_q\lambda_{q,k})\\ &=-\sum_qp_q\ln p_q -\sum_qp_q\sum_k\lambda_{q,k}\ln\lambda_{q,k}\\ &=H(p)+\sum_qp_qS(\rho_q). \end{aligned}$$

We used $\sum_k\lambda_{q,k}=1$ in the first term.

2. Two-copy activation

Expand $|\psi_2\rangle$, project onto Alice’s three number sectors, and recover $E_{\rm acc}^{(2)}=\tfrac12\ln2$.

Solution

There are four equal-probability occupation configurations. One has $q_A=0$, two have $q_A=1$, and one has $q_A=2$, giving probabilities $(1/4,1/2,1/4)$. The endpoint sectors contain one product configuration each and have zero entropy. The middle sector is an equal superposition of two orthogonal Alice configurations correlated with two orthogonal Bob configurations, so its Schmidt coefficients are $(1/\sqrt2,1/\sqrt2)$ and its entropy is $\ln2$. Averaging gives

$$E_{\rm acc}^{(2)} =\frac14(0)+\frac12\ln2+\frac14(0) =\frac12\ln2.$$

3. Two unequal Gaussian modes

Take correlation eigenvalues $\nu_1=a$ and $\nu_2=b$. Find the sector probabilities and the entropy in the one-particle sector.

Solution

Expanding $(1-a+za)(1-b+zb)$ gives

$$\begin{aligned} p_0&=(1-a)(1-b),\\ p_1&=a(1-b)+(1-a)b,\\ p_2&=ab. \end{aligned}$$

The two normalized weights in the $q=1$ sector are

$$r=\frac{a(1-b)}{p_1}, \qquad 1-r=\frac{(1-a)b}{p_1}.$$

Therefore

$$S(\rho_{A,1}) =-r\ln r-(1-r)\ln(1-r).$$

The $q=0$ and $q=2$ states are pure. Substitution verifies

$$H(p)+p_1S(\rho_{A,1}) =h_2(a)+h_2(b),$$

which is the entropy of the original product of two natural-mode density matrices.

For the reproducible choice $(a,b)=(1/4,3/4)$,

$$(p_0,p_1,p_2) =\left(\frac3{16},\frac{10}{16},\frac3{16}\right),$$

and the unnormalized projected purities are

$$\operatorname{Tr}\!\left[(\Pi_q\rho_A)^2\right] =\left(\frac9{256},\frac{82}{256},\frac9{256}\right).$$

Consequently the conditional second Rényi entropy of the $q=1$ block is $-\ln(0.82)$.

References

• Monkman, Kyle, and Jesko Sirker. “Symmetry-Resolved Entanglement: General Considerations, Calculation from Correlation Functions, and Bounds for Symmetry-Protected Topological Phases.” Journal of Physics A: Mathematical and Theoretical 56 (2023): 495001. DOI.
• Murciano, Sara, Giuseppe Di Giulio, and Pasquale Calabrese. “Entanglement and Symmetry Resolution in Two Dimensional Free Quantum Field Theories.” Journal of High Energy Physics 2020, no. 8 (2020): 073. DOI.
• Schuch, Norbert, Frank Verstraete, and J. Ignacio Cirac. “Quantum Entanglement Theory in the Presence of Superselection Rules.” Physical Review A 70 (2004): 042310. DOI.
• Wiseman, Howard M., and John A. Vaccaro. “Entanglement of Indistinguishable Particles Shared between Two Parties.” Physical Review Letters 91 (2003): 097902. DOI.