Superselection Sectors Preview

A superselection rule says more than “there is a conserved quantum number.” It says that the allowed observables, preparations, or operations cannot reveal relative phases between certain sectors of Hilbert space. In that operational sense, a coherent superposition across those sectors behaves like an incoherent mixture for all accessible measurements.

The key is to specify which observables and operations are allowed. Particle-number examples are treated in Particle-Number Superselection Preview. Deeper charge, locality, and gauge-theory versions belong to field-theory and foundations treatments.

Sector Decomposition

For a finite or countable set of discrete sectors, start with a direct-sum decomposition

$$\mathcal H = \bigoplus_q \mathcal H_q,$$

where $q$ labels charge, particle number, fermion parity, or another sector label. Continuous sector labels need a direct-integral formulation beyond the sums used here. Let $\Pi_q$ be the projector onto $\mathcal H_q$.

A state vector can formally have components in several sectors:

$$\lvert\psi\rangle = \sum_q c_q\lvert\psi_q\rangle, \qquad \lvert\psi_q\rangle\in\mathcal H_q.$$

The direct sum alone does not create a superselection rule. It only says the Hilbert space has orthogonal sectors. A superselection rule adds a restriction on what observables and operations are physically available.

Allowed Observables

The clean operational condition is that every allowed observable is block diagonal in the sector decomposition:

$$A = \sum_q \Pi_q A\Pi_q.$$

Equivalently,

$$\Pi_q A\Pi_{q'} = 0 \qquad q\ne q'.$$

If all accessible observables have this form, then they cannot detect off-diagonal density-matrix blocks between sectors.

Define the sector-dephased state

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

For every allowed bounded observable $A$ (and for unbounded ones whenever the expectations exist),

$$\operatorname{Tr}(\rho A) = \operatorname{Tr}(\mathcal D(\rho)A).$$

Thus $\rho$ and $\mathcal D(\rho)$ are operationally indistinguishable under the allowed observables. This is the practical content of superselection: relative phases between distinct sectors cannot be measured by this observable algebra. Enlarging the available reference systems can reveal relational phases in some settings, without removing a fundamental constraint on the total system.

Charge Operator Viewpoint

Often the sectors are eigenspaces of a self-adjoint operator $Q$:

$$Q\Pi_q = q\Pi_q.$$

If every allowed observable commutes with $Q$,

$$[A,Q]=0,$$

then the observables are block diagonal in the $Q$ eigenspaces. In algebraic language, $Q$ behaves like a central label for the representation of the observable algebra.

This is stronger than saying a particular Hamiltonian commutes with $Q$. The condition

$$[H,Q]=0$$

says that the corresponding spectral probabilities of $Q$ are unchanged under closed-system evolution; it need not make $Q$ sharp in an arbitrary state. A superselection claim says the whole allowed observable algebra cannot compare phases between different $Q$ values.

Conservation Is Not Superselection

Conservation and superselection are often confused because both use commuting operators.

If

$$[H,Q]=0,$$

then time evolution generated by $H$ preserves each $Q$ sector. A state that begins in one sector remains in that sector. But this statement alone does not decide whether one can prepare, measure, or use coherent superpositions across sectors by coupling to a reference, reservoir, or larger system.

For a superselection rule, the relevant statement is closer to

$$[A,Q]=0 \qquad \text{for all allowed observables }A.$$

That is a statement about the accessible operations, not only about one Hamiltonian.

Selection Rules Are Not Superselection Rules

A selection rule usually says that a particular operator has vanishing matrix elements between certain states. For example, if an operator $O$ has a definite charge under a symmetry, then matrix elements may vanish unless the initial and final quantum numbers differ in the allowed way.

A superselection rule is more global. It says the entire allowed observable algebra lacks the off-diagonal blocks that would compare or create relative phases between sectors.

The hierarchy is:

$$\begin{array}{c|c} \text{Statement} & \text{Meaning}\\ \hline [H,Q]=0 & \text{time evolution preserves }Q\\ \langle q|O|q'\rangle=0 & \text{one operator has a selection rule}\\ [A,Q]=0\ \text{for all allowed }A & \text{sector coherences are inaccessible} \end{array}$$

Only the last row is a superselection statement.

Examples and Nonexamples

Electric charge is the traditional example. In ordinary settings, coherent superpositions of states with different total electric charge are not described as observable pure-state phase relations. In a more complete field-theoretic treatment, this is tied to gauge constraints, locality, and the algebra of observables. A finite matrix model alone does not establish this field-theoretic restriction.

Particle number is subtler. A Fock space has fixed-$N$ sectors,

$$\mathcal F = \bigoplus_{N=0}^{\infty} \mathcal H_N,$$

but the mere existence of this direct sum is not a superselection rule. Number conservation, coherent states, reservoirs, and phase references must be discussed carefully. The roles of these ingredients are explained in Particle-Number Superselection Preview.

Fermion parity often behaves as a robust sector label in fermionic systems. Physical observables are commonly even in fermionic creation and annihilation operators, so they commute with parity

$$P_F = (-1)^N.$$

This does not mean every model detail follows from parity alone. It means that even and odd fermion-parity sectors often require special care when discussing observables, boundary modes, and superconducting effective Hamiltonians.

Angular momentum quantum numbers provide a useful nonexample. A rotationally invariant Hamiltonian may decompose into angular-momentum sectors, but ordinary observables can prepare and measure coherent superpositions of different $m$ values or even different total angular momenta when the experimental setup supplies the needed orientation and couplings. A representation decomposition is not automatically superselection.

Symmetry Interpretation

Measure the charge $Q$ in dimensionless units and suppose it generates a one-parameter unitary transformation

$$U(\theta) = e^{-i\theta Q}.$$

If all allowed observables are invariant,

$$U(\theta)AU(\theta)^\dagger = A,$$

then they commute with $Q$. A relative phase between two charge sectors changes under $U(\theta)$, but no invariant observable can detect that phase.

This is the symmetry-theoretic origin of the restriction: invariant observables cannot connect or compare distinct sectors. A state can still have a classical probability distribution over sector labels; superselection does not require every density matrix to lie in one sharp sector. The physical origin of that restriction may be a gauge constraint, a missing reference frame, locality, or an imposed operational rule.

Reference Frames and Enlarged Descriptions

Some apparent superselection restrictions soften when a suitable reference system is included. A phase that is meaningless for one isolated subsystem can become meaningful as a relational phase between subsystem and reference.

This is why cautious language matters. Instead of saying

$$\lvert q_1\rangle+\lvert q_2\rangle$$

is always meaningless, say what the allowed observables are. If all allowed observables preserve $q$, then the relative phase is inaccessible. If the physical description includes a reference carrying compensating charge or number, a relational coherence may be observable without violating the total constraint.

Exercises

1. Prove sector-dephased equivalence.

Let $A=\sum_q\Pi_qA\Pi_q$ and $\mathcal D(\rho)=\sum_q\Pi_q\rho\Pi_q$. Show that $\operatorname{Tr}(\rho A)=\operatorname{Tr}(\mathcal D(\rho)A)$.

Solution

Insert $I=\sum_q\Pi_q$ around $\rho$:

$$\operatorname{Tr}(\rho A) = \sum_{q,q'} \operatorname{Tr}(\Pi_q\rho\Pi_{q'}A).$$

Because $A$ is block diagonal,

$$\Pi_{q'}A = \Pi_{q'}A\Pi_{q'},$$

so the trace terms with $q\ne q'$ vanish after using orthogonality of the projectors. Therefore

$$\operatorname{Tr}(\rho A) = \sum_q \operatorname{Tr}(\Pi_q\rho\Pi_q A) = \operatorname{Tr}(\mathcal D(\rho)A).$$

2. Conservation versus superselection.

A Hamiltonian obeys $[H,Q]=0$. Does this alone prove that no coherent superposition of different $Q$ eigenvalues can ever be prepared?

Solution

No. The commutator says that the closed-system dynamics generated by this Hamiltonian preserves the $Q$ eigenspaces. It does not classify all possible preparations, external references, reservoirs, measurements, or enlarged systems. A superselection claim requires a statement about allowed observables or operations, such as $[A,Q]=0$ for all accessible observables.

3. Identify a nonexample.

Why is the decomposition of a central-potential Hilbert space into angular-momentum sectors not automatically a superselection rule?

Solution

The decomposition into angular-momentum sectors follows from rotational symmetry and representation theory. It helps block diagonalize rotationally invariant Hamiltonians. But ordinary physical operations can select directions, apply fields, or measure observables that do not commute with all angular-momentum labels. Unless the allowed observable algebra is restricted to be block diagonal in those sectors, the decomposition is not a superselection rule.

4. Fermion parity.

Let $P_F=(-1)^N$. If an observable $A$ is even in fermionic creation and annihilation operators, what commutation property should it have with $P_F$?

Solution

An even observable changes particle number by an even integer or preserves it. Therefore it preserves the parity of $N$ and commutes with

$$P_F = (-1)^N.$$

Thus

$$[A,P_F]=0.$$

This is why even and odd fermion-parity sectors are often treated as distinct operational sectors.

5. Take $Q=|1\rangle\langle1|$ and $H=\hbar\omega Q$. Compare $|\psi_\varphi\rangle=(|0\rangle+e^{i\varphi}|1\rangle)/\sqrt2$ with the equal incoherent mixture. Which measurement detects their difference?

Solution

Every diagonal observable has the same expectation in both states. But $\sigma_x=|0\rangle\langle1|+|1\rangle\langle0|$ has expectation $\cos\varphi$ in the pure state and zero in the mixture. Since $[Q,\sigma_x]\ne0$, this measurement is excluded by a $Q$ superselection rule, even though $[H,Q]=0$ alone does not exclude it. Measuring $\sigma_y$ gives $\sin\varphi$, so the pair distinguishes every coherent phase from the mixture.

References

• Y. Aharonov and L. Susskind, “Charge superselection rule,” Physical Review 155, 1428–1431 (1967). doi:10.1103/PhysRev.155.1428.
• S. D. Bartlett, T. Rudolph, and R. W. Spekkens, “Reference frames, superselection rules, and quantum information,” Reviews of Modern Physics 79, 555–609 (2007). doi:10.1103/RevModPhys.79.555; author manuscript.
• D. Giulini, “Superselection rules,” in Compendium of Quantum Physics, Springer, 2009. Author manuscript, arXiv:0710.1516.
• G. C. Wick, A. S. Wightman, and E. P. Wigner, “The intrinsic parity of elementary particles,” Physical Review 88, 101–105 (1952). doi:10.1103/PhysRev.88.101.