Lecture 5 — Constraints, regional algebras, and electric sectors

In the previous lecture we chose an algebra with a center and calculated its entropy. We now want the choice to follow from a physical restriction. A gauge constraint provides an elementary mechanism: an operation confined to one region may change a quantity that must be matched by the complementary region. Such an operation is unavailable unless it is accompanied by another operation across the division.

We will implement this idea with four qubits and one parity constraint. The model is deliberately small enough that the physical Hilbert space, regional algebras, central projectors, and entropy can all be written explicitly. The resulting center is a property of the specified regional observables. It is not an assertion that every algebra in a gauge theory has a nontrivial center.

How to use this lecture

Classroom core. Meeting 1: define the constraint and count physical states (30 minutes), construct a basis adapted to the division (35), calculate the regional observables and their center (40), and discuss the first two checkpoints (15). Meeting 2: work the entropy example (40), compare regional and global operations (30), explain what the gauge analogy requires (25), and solve a core problem together (25). These are provisional allocations.

Self-study. Check the basis map on all eight vectors. Derive the complete commutant using matrix units within each sector. Follow the coherent example through both the physical partial trace and the algebraic restriction.

Research extension. Add a second constraint and investigate how the regional algebra changes. Specify the new algebra before computing any entropy.

Prerequisites. Lectures 1 and 4: restrictions, commutants, centers, and block entropy. Pauli matrices are the only new computational ingredients.

Reading and imported claims. Casini, Huerta, and Rosabal, Remarks on entanglement entropy for gauge fields, arXiv:1312.1183, discusses the dependence on the chosen regional algebra. The four-qubit calculation below is self-contained. It illustrates one constraint and one algebra choice, not the full continuum gauge-field problem.

1. A constraint selects the allowed preparations

Start from four qubits with kinematical Hilbert space

$${\mathcal{H}}_{\mathrm{kin}}=({\mathbb{C}}^2{)}^{\otimes 4}.$$

The adjective kinematical means that we have listed candidate states before imposing the constraint. Let

$$G=Z_1{Z}_2{Z}_3{Z}_4.$$

We retain vectors invariant under the two-element group generated by $G$:

$${\mathcal{H}}_{\mathrm{phys}}=\{|\psi ⟩:G|\psi ⟩=|\psi ⟩\},P_{\mathrm{phys}}=\frac{I+G}{2}.$$

Because $G^2=I$, this is an orthogonal projector. On a computational basis vector,

$$G|n_1{n}_2{n}_3{n}_4⟩=(-1{)}^{n_1+n_2+n_3+n_4}|n_1{n}_2{n}_3{n}_4⟩.$$

Exactly the eight even-parity strings survive. The physical space has dimension eight.

A physical operation must map allowed states to allowed states. We will use operators commuting with $G$, which is a sufficient condition. For example, $Z_1$ commutes with $G$, while $X_1$ anticommutes and flips even parity to odd parity. The pair $X_1{X}_2$ commutes: two sign changes cancel.

This distinction concerns the operator’s action on the constraint surface. It does not say that a qubit flip is impossible in every apparatus. An apparatus implementing such a flip would have to supply the missing charge or enlarge the system we are describing.

Checkpoint 1. Is $P_{\mathrm{phys}}{X}_1{P}_{\mathrm{phys}}=0$ the statement that $X_1$ annihilates every physical vector?

Answer. No. It takes a physical vector into the odd-parity subspace. Projection back onto the physical space gives zero. Compression and actual action must again be distinguished.

2. Divide the links before choosing the regional algebra

Let $A$ contain qubits 1 and 2, and $B$ contain qubits 3 and 4. Define their parity operators

$$Q_A=Z_1{Z}_2,Q_B=Z_3{Z}_4.$$

The constraint is $Q_A{Q}_B=1$, so their eigenvalues must agree on every physical state. Denote the common eigenvalue by $(-1{)}^f$, with $f=0,1$.

Each regional parity eigenspace has dimension two. An explicit basis is

$$|f,j{⟩}_A=|j,j\oplus f{⟩}_{12},|f,k{⟩}_B=|k,k\oplus f{⟩}_{34},$$

where $j,k=0,1$ and $\oplus$ is addition modulo two. For $f=0$, the regional strings are $00,11$; for $f=1$, they are $01,10$.

Thus

$$\boxed{{\mathcal{H}}_{\mathrm{phys}}=\underset{f=0}{\overset{1}{⨁}}\left({\mathbb{C}}_{a_f}^2\otimes {\mathbb{C}}_{b_f}^2\right).}$$

The basis map realizing this decomposition is

$$W|f,j,k⟩=|j,j\oplus f,k,k\oplus f⟩.$$

Its images are the eight distinct even-parity strings, so $W^{†}W=I_8$ and $W{W}^{†}=P_{\mathrm{phys}}$.

The physical space is an abstract eight-dimensional Hilbert space and can be factorized in many arbitrary ways. What fails is the particular interpretation as independently variable regional physical spaces. Alice cannot choose $f=0$ while Bob chooses $f=1$. The constraint correlates their sector labels.

Physical picture. Think of two rooms connected by a bookkeeping rule: the charge recorded at one side of the dividing surface must agree with the charge recorded at the other. Each room still contains quantum degrees of freedom. The shared label organizes the allowed combinations; it does not replace all quantum information by a classical variable.

3. Derive the regional algebra

We define Alice’s observables to be operators supported on qubits 1 and 2 that commute with the constraint. For $O_A\otimes I_B$,

$$[O_A\otimes I_B,G]=[O_A,Q_A]\otimes Q_B.$$

Since $Q_B$ is invertible, gauge invariance is equivalent to

$$[O_A,Q_A]=0.$$

Such an operator cannot mix the two parity eigenspaces. Within either eigenspace it is an arbitrary two-by-two matrix. Restricted to the physical space, the regional algebra is therefore

$${\mathcal{M}}_A=\underset{f=0}{\overset{1}{⨁}}\left(M_2(\mathbb{C}{)}_{a_f}\otimes I_{b_f}\right).$$

This conclusion comes from an explicit support condition and a specified gauge constraint. It is not inferred from a diagram alone.

Useful generators are

$$Q_A,Z_1,X_1{X}_2.$$

In the basis $|f,j{⟩}_A$,

$$Q_A|f,j{⟩}_A=(-1{)}^f|f,j{⟩}_A,$$ $$Z_1|f,j{⟩}_A=(-1{)}^j|f,j{⟩}_A,X_1{X}_2|f,j{⟩}_A=|f,j\oplus 1{⟩}_A.$$

The last two operators act as logical $Z$ and $X$ within each sector. They anticommute, so the accessible information is genuinely quantum inside a fixed sector. Multiplying by

$$P_f=\frac{I+(-1{)}^f{Q}_A}{2}$$

allows different matrix coefficients in the two sectors. These generators therefore span the full displayed algebra.

Alice can measure the sector and a qubit within that sector. She cannot create a coherent transition between the sectors using this algebra.

3.1 The commutant and the center

An operator commuting with ${\mathcal{M}}_A$ must commute with both $P_f$, hence be block diagonal. Within a block it must commute with all matrices on $a_f$, so the tensor-factor calculation from Lecture 1 gives

$${\mathcal{M}}_A^{\prime }=\underset{f}{⨁}\left(I_{a_f}\otimes M_2(\mathbb{C}{)}_{b_f}\right).$$

This is precisely the regional $B$ algebra for our choice of observables. Their common part is

$$Z({\mathcal{M}}_A)=\mathrm{span}\{P_0,P_1\}.$$

The equality of the regional $Q_A$ and $Q_B$ actions on the physical space explains why both sides can read the same central label.

Checkpoint 2. If both sides can measure $f$, have we reconstructed an unknown qubit twice?

Answer. No. They share a commuting two-outcome variable. The noncommuting pair $Z_1,X_1{X}_2$ belongs to Alice’s within-sector qubit, while Bob has a different commuting-with-Alice pair.

4. Restriction of a general physical state

Let $\rho$ be a density matrix on ${\mathcal{H}}_{\mathrm{phys}}$. Write it in sector blocks,

$$\rho =\left(\begin{array}{cc}{\rho }_{00}& {\rho }_{01}\\ {\rho }_{10}& {\rho }_{11}\end{array}\right).$$

Every diagonal block acts on $a_f\otimes b_f$. Set

$$p_f=\mathrm{Tr}{\rho }_{ff},{\rho }_{a_f}=\frac{{\mathrm{Tr}}_{b_f}{\rho }_{ff}}{p_f}$$

when $p_f>0$. For $O={⨁}_f(O_f\otimes I)$,

$$\mathrm{Tr}(\rho O)=\sum _f{p}_f\mathrm{Tr}({\rho }_{a_f}{O}_f).$$

The off-diagonal blocks do not enter. They may still be present in the physical state; the chosen algebra cannot test them.

The density matrix representing this restricted state with one copy of each irreducible matrix block is

$${\rho }_{{\mathcal{M}}_A}=\underset{f}{⨁}{p}_f{\rho }_{a_f}.$$

Its eigenvalues are $p_f{r}_{fj}$, where $r_{fj}$ are the eigenvalues of ${\rho }_{a_f}$. Expanding their logarithms yields

$$\boxed{S_{{\mathcal{M}}_A}(\rho )=H(p_0,p_1)+\sum _f{p}_{f}S({\rho }_{a_f}).}$$

The first term records uncertainty about the common sector. The second records quantum uncertainty in the accessible within-sector matrices.

4.1 Compare with the kinematical partial trace

Embed the physical state into the four-qubit space as $W\rho W^{†}$ and trace qubits 3 and 4. Different $f$ sectors occupy orthogonal subspaces of $B$, so the partial trace eliminates ${\rho }_{01}$ and ${\rho }_{10}$. In the regional parity basis,

$${\rho }_A^{\mathrm{kin}}=\underset{f}{⨁}{p}_f{\rho }_{a_f}.$$

Therefore

$$S({\rho }_A^{\mathrm{kin}})=S_{{\mathcal{M}}_A}(\rho )$$

in this particular model and algebra choice.

There is no extra auxiliary entropy here. A constraint can produce a center without producing a nonzero analogue of the central entropy operator from Lecture 4. The existence of a center and the value of an entropy contribution are separate questions.

If instead we extend the restricted state to the full physical space by making the invisible $b_f$ qubit maximally mixed, the resulting matrix is

$$\hat{\rho }=\underset{f}{⨁}{p}_f{\rho }_{a_f}\otimes I_{b_f}/2.$$

Its entropy is $S_{{\mathcal{M}}_A}+\log 2$. This extra term comes from that chosen extension. It is not the entropy of the actual kinematical $A$ reduction.

5. A coherent state with a checkable spectrum

Consider

$$|{\Psi }_p⟩=\sqrt{p}|f=0⟩|{\Phi }_{+}{⟩}_{a_0{b}_0}+\sqrt{1-p}|f=1⟩|0,0{⟩}_{a_1{b}_1},0<p<1.$$

The sector notation refers to the direct sum; the two terms are orthogonal. In the original qubits,

$$W|{\Psi }_p⟩=\sqrt{\frac{p}{2}}(|0000⟩+|1111⟩)+\sqrt{1-p}|0101⟩.$$

The three $B$ strings $00,11,01$ are orthogonal. Alice’s reduced eigenvalues are therefore

$$\frac{p}{2},\frac{p}{2},1-p,0.$$

Direct substitution gives

$$S_A=h(p)+p\log 2,h(p)=-p\log p-(1-p)\log (1-p).$$

The full physical state is pure. Within sector 0 there is one Bell pair; within sector 1 the two regional qubits are in a product state.

At $p=1/4$, the spectrum is $1/8,1/8,3/4,0$. One can check its entropy without diagonalizing any larger matrix. Measuring the sector on both sides produces a Bell pair with probability $1/4$ and a product state otherwise. This is a concrete conditional protocol. We are not claiming a universal distillation theorem for arbitrary gauge constraints or operations involving many copies.

Checkpoint 3. Does replacing this pure state by its incoherent sector mixture change any regional measurement?

Answer. No. Both regional algebras are block diagonal. It does change suitable joint gauge-invariant measurements.

6. The center is relative to the regional restriction

The joint operator

$$T=X_2{X}_4$$

commutes with $G$ and acts as

$$TW|f,j,k⟩=W|f\oplus 1,j,k⟩.$$

It changes the sector coherently. In particular, it fails to commute with $Q_A$ on the physical space. Thus $Q_A$ is central in ${\mathcal{M}}_A$ but not central in the full physical operator algebra.

For the state above,

$$⟨T{⟩}_{{\Psi }_p}=\sqrt{2p(1-p)}.$$

Only the overlap between the $j=k=0$ components contributes: $T$ connects $0000$ with $0101$, while $1111$ is sent to $1010$, absent from the state. The incoherent sector mixture gives zero.

This calculation separates a regional center from a fundamental global superselection rule. If we further prohibited $T$ and all sector-changing global operators, we would be specifying a different observable theory.

Physical picture. A shared classical label can arise because the two observers are restricted to operations within their own regions. A joint operation may restore sensitivity to the relative phase. The phrase “classical sector” must always say which algebra regards it as classical.

7. What the gauge analogy supplies

The four factors can be regarded as links incident on one vertex, with $G$ the only imposed ${\mathbb{Z}}_2$ gauge transformation. We chose $Z$ as the electric basis convention. The outer endpoints are not additional constrained vertices in this model. Adding their Gauss laws would change the physical space and must be done explicitly.

A regional flux label appears because each side carries part of the one constraint. This captures a reason that physical gauge-invariant states need not factorize according to a spatial cut. It does not determine the unique correct algebra for every region in every theory.

In lattice gauge theory, different treatments of boundary observables or added boundary degrees of freedom can change the algebra and the entropy bookkeeping. The source cited above develops that issue in greater generality. Our calculation gives one fully specified case on which to test the definitions.

For holography, the useful lesson is that “the bulk subsystem” may require an algebra with sectors, rather than an assumed tensor factor. Its commutant, entropy, and recoverable observables must be computed for that choice. No continuum area divergence or gravitational edge-mode formula has been derived from four qubits.

8. What to take away

The constraint selects allowed states; the support and invariance conditions select regional observables. Together they produce a direct sum of quantum tensor factors. Both regions can access the common central label while retaining different noncommutative within-sector algebras. Entropy follows from the restricted state’s spectrum, with its trace convention specified. Global physical operations can reveal coherences invisible to both regional algebras.

The next lecture replaces a single encoding by a network of encodings. We will ask when connectivity controls recovery and when a cut actually computes entropy.

9. Problem set

Classroom core

1. Count and construct. List the eight even-parity computational strings and match them with $(f,j,k)$. Verify $W^{†}W=I$.

2. The regional algebra. Show that $Z_2$ acts as $(-1{)}^{f}Z$ within sector $f$. Use $P_f,Z_1,X_1{X}_2$ to construct a matrix unit $|f,0⟩⟨f,1|$.

3. Entropy. Evaluate the three nonzero reduced eigenvalues and the entropy of $|{\Psi }_p⟩$ at $p=1/2$. Identify the contribution of the sector uncertainty.

Self-study consolidation

4. Phase-sensitive measurement. Insert a phase $e^{i\theta }$ in the $f=1$ component of $|{\Psi }_p⟩$. Compute $⟨X_2{X}_4⟩$ and every regional state’s dependence on $\theta$.

5. Change the algebra. Retain only the commuting observables generated by $Z_1$ and $Q_A$. For a fixed-sector logical $|+⟩$ state on $a_f$, compare its entropy on the full regional algebra and on this smaller diagonal algebra.

Research extension

6. A second Gauss constraint. Add $Z_1{Z}_2=1$ to the original constraint. Determine the physical space and the new regional centers. Explain why counting constraints and choosing a regional algebra must precede any entropy comparison.

10. Answer checkpoints

1. The strings are $0000,0011,0101,0110,1001,1010,1100,1111$. The map $W$ assigns each once. For example, $(f,j,k)=(1,1,0)$ gives $1001$.

2. The second bit is $j\oplus f$, hence its $Z$ eigenvalue is $(-1{)}^{j+f}$. The required matrix unit is $P_f(I+Z_1)X_1{X}_2/2$, restricted to Alice’s parity basis: the flip acts first and the projector retains output $j=0$.

3. The eigenvalues are $1/4,1/4,1/2$, giving $S_A=3\log 2/2$. The sector contribution is $\log 2$, and the averaged within-sector contribution is $\log 2/2$.

4. The expectation is $\sqrt{2p(1-p)}\cos \theta$. Both regional reductions are independent of $\theta$ because their partial traces remove cross-sector terms.

5. On $M_2$ inside the fixed sector, $|+⟩$ is pure and has entropy zero. Its restriction to the diagonal algebra assigns equal probabilities to $j=0,1$ and has entropy $\log 2$. Removing observables changes which distinctions enter the entropy.

6. Only $f=0$ remains, with two unconstrained regional qubits $j,k$. The physical space is ${\mathbb{C}}_a^2\otimes {\mathbb{C}}_b^2$; each full regional algebra is a factor and its center is scalar. A second constraint has removed the shared sector in this specified example, not all local quantum degrees of freedom.

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