Pauli Group and Stabilizers

Pauli stabilizers describe a quantum subspace by the measurements that always return $+1$ on it. The signs matter: $XX$ and $ZZ$ can both have eigenvalue $+1$, but their product is $-YY$, so adding $+YY$ as another $+1$ constraint makes the equations impossible. Keeping this minus sign is the first step toward understanding quantum codes.

This page develops small examples by direct multiplication. The stabilizer formalism extends these ideas to binary linear algebra, code projectors, logical operators and Clifford circuits.

Required background. Pauli matrices supplies the Hermitian matrices $X,Y,Z$ and tensor products. Subscripts label physical qubits; an omitted factor is the identity.

Helpful background. Why quantum error correction is possible explains what a code must preserve when errors occur.

Pauli strings, phases and checks

A Pauli string is a tensor product of $I,X,Y,Z$. The full $n$-qubit Pauli group also includes scalar phases:

$$\mathcal P_n=\{i^\ell P_1\otimes\cdots\otimes P_n\},$$

where $\ell\in\{0,1,2,3\}$ and each $P_j\in\{I,X,Y,Z\}$. These phases make the set closed under multiplication. For example, $XY=iZ$ is not one of the four phase-free single-qubit matrices.

Use the standard Hermitian convention $Y=\sigma_y$. A binary check is then a signed Hermitian Pauli $g=\pm P$, with $g^\dagger=g$ and $g^2=I$. The nonidentity checks have eigenvalues $\pm1$. The exceptional checks $+I$ and $-I$ have only one eigenvalue: the former imposes no constraint, while the latter has no $+1$ eigenvector. Elements $\pm iP$ are anti-Hermitian and cannot themselves be binary observables.

To stabilize $|\psi\rangle$ means to satisfy $g|\psi\rangle=|\psi\rangle$. Changing $g$ to $-g$ selects the opposite eigenspace of the underlying observable. This use of a sign is different from an unobservable overall phase multiplying a state vector.

Multiply locally, then combine the phases

The cyclic products are

$XY=iZ,\qquad YZ=iX,\qquad ZX=iY.$

Reversing either pair gives a minus sign. Multiply tensor factors in their original order and keep every scalar phase:

$$\begin{aligned} (X_1Y_2)(Z_1X_2) &=(XZ)_1(YX)_2\\ &=(-iY_1)(-iZ_2)=-Y_1Z_2. \end{aligned}$$

For commutation, a shorter calculation suffices. At a given qubit, distinct nonidentity Pauli letters anticommute; equal letters or a pair containing $I$ commute. If $\nu(P,Q)$ counts the anticommuting positions, then

$PQ=(-1)^{\nu(P,Q)}QP.$

Thus two strings commute when this count is even. For example, $X_1X_2X_3$ commutes with $Z_1Z_2$ because qubits 1 and 2 each contribute a minus sign. But $X_1X_2$ anticommutes with $Z_2Z_3$. Overall signs do not change commutation, although they do change products and selected eigenspaces.

Compatible checks and independent constraints

A stabilizer group $S$ is a commuting subgroup of $\mathcal P_n$ that excludes $-I$. Its common $+1$ eigenspace is

$\mathcal C=\{|\psi\rangle:g|\psi\rangle=|\psi\rangle\text{ for every }g\in S\}.$

The two conditions have different jobs. Anticommuting checks cannot share a nonzero eigenvector: if $g|\psi\rangle=h|\psi\rangle=|\psi\rangle$, then $gh|\psi\rangle=|\psi\rangle$ but also $gh|\psi\rangle=-hg|\psi\rangle=-|\psi\rangle$. Commuting checks can still be inconsistent if a product equals $-I$, because their $+1$ equations would imply $-|\psi\rangle=|\psi\rangle$.

By contrast, a product equal to $+I$ indicates a repeated constraint. For a consistent list, independence means that no nonempty product of the generators equals $I$. If $r$ independent generators act on $n$ qubits, then

$|S|=2^r,\qquad \dim\mathcal C=2^{n-r}.$

Each independent commuting Pauli condition halves the dimension. Merely appending a product of existing generators does not halve it again. The projector derivation proves this count and explains why $r\le n$.

A minus sign selects a Bell state

For two qubits,

$(X_1X_2)(Z_1Z_2)=-Y_1Y_2.$

The checks $XX$ and $ZZ$ are independent and have a one-dimensional common $+1$ sector, spanned by

$|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}.$

Adding $-YY$ repeats their product and changes nothing. Adding $+YY$ is inconsistent: the product of all three proposed generators is $-I$. In terms of eigenvalues, any joint eigenstate obeys $y=-xz$, where $x,z,y$ are its $XX,ZZ,YY$ results. Only two of these signs are independent.

The four Bell states occupy the four simultaneous eigenspaces of $XX$ and $ZZ$. Here $|\Phi^\pm\rangle=(|00\rangle\pm|11\rangle)/\sqrt2$ and $|\Psi^\pm\rangle=(|01\rangle\pm|10\rangle)/\sqrt2$. The $YY$ sign is fixed by $XX\,ZZ=-YY$; it cannot be chosen as a third independent sign.

For small lists, generate all products while retaining their signs. A product $-I$ signals inconsistency; repeated products reveal dependence. For larger lists the number of products grows exponentially, so the binary methods in the next page become useful (Gottesman, 1997).

Commutation signs predict the syndrome

Let $g_1,\ldots,g_r$ be independent stabilizers. For a Pauli error $E$, define

$g_jE=\chi_E(g_j)Eg_j,\qquad\chi_E(g_j)\in\{+1,-1\}.$

If $|\psi\rangle$ lies in the code, then

$g_jE|\psi\rangle=\chi_E(g_j)E|\psi\rangle.$

The signs therefore give the ideal check outcomes after that Pauli error: its syndrome. The equivalent bit is $s_j(E)=(1-\chi_E(g_j))/2$, with zero for commutation and one for anticommutation. This algebra predicts ideal outcomes; syndrome measurement explains the circuits and their possible faults.

The signs respect products:

$\chi_E(gh)=\chi_E(g)\chi_E(h).$

To see this, move $E$ past $h$ and then $g$, collecting the two commutation signs. Thus a redundant check $g_3=g_1g_2$ must satisfy $\chi_E(g_3)=\chi_E(g_1)\chi_E(g_2)$. Arbitrary independent sign choices for all three would be inconsistent. Replacing $g$ by $-g$ does not change its commutation sign with $E$; it changes the code sector from which the error starts.

Different errors can have the same syndrome. If $E$ and $F$ do, $E^\dagger F$ commutes with every check. It may act trivially on the code or may change the logical information. A syndrome identifies an eigensector, not necessarily a unique error.

Three-qubit example with two mixed checks

Take $g_1=X_1X_2X_3$ and $g_2=Z_1Z_2$. They commute, and

$g_1g_2=(-iY_1)(-iY_2)X_3=-Y_1Y_2X_3.$

Their group contains exactly $I$, $g_1$, $g_2$ and $-Y_1Y_2X_3$. It excludes $-I$, so two independent checks on three qubits leave a two-dimensional code space.

Error	$g_1$ sign	$g_2$ sign
$I$, $g_1$, $X_3$	$+1$	$+1$
$X_1$, $X_2$	$+1$	$-1$
$Z_1$, $Z_2$, $Z_3$	$-1$	$+1$
$Y_1$, $Y_2$	$-1$	$-1$

The table groups the listed errors by syndrome; it is not an enumeration of all three-qubit Paulis. For example, $Y_1$ differs from $X_1Z_1$ only by a phase, so its two signs are the products of the $X_1$ and $Z_1$ signs.

To see why an all-plus result need not mean no error, choose the basis

$$\begin{aligned} |a\rangle&=(|000\rangle+|111\rangle)/\sqrt2,\\ |b\rangle&=(|001\rangle+|110\rangle)/\sqrt2. \end{aligned}$$

Both satisfy the two checks. But $X_3$ swaps $|a\rangle$ and $|b\rangle$, whereas $I$ and $g_1$ leave both unchanged. This is a valid one-logical-qubit code, but it does not protect against every single-qubit error: an undetected weight-one operator changes the encoded state. The general language for these actions is developed in logical operators.

Exercises

Exercise 1: Multiply a Signed Pauli String

Calculate

$$\begin{aligned} A&=(-X_1Y_2)(Z_1X_2),\\ B&=(X_1I_2)(Y_1Z_2). \end{aligned}$$

State whether each result is Hermitian. Identify the phase error made by replacing every local product with only its resulting Pauli letter.

Solution

For $A$, keep the leading minus sign and use $XZ=-iY$ and $YX=-iZ$:

$$\begin{aligned} A&=-(-iY_1)(-iZ_2)\\ &=-(-i)^2Y_1Z_2=Y_1Z_2. \end{aligned}$$

The two factors of $-i$ multiply to $-1$, which cancels the supplied minus sign. Thus $A$ is Hermitian. For $B$,

$$B=(XY)_1Z_2=iZ_1Z_2.$$

This element is anti-Hermitian and is not a binary check. Ignoring local phases loses the sign cancellation in $A$ and the decisive factor $i$ in $B$.

Exercise 2: Test Overlap Parity

Let $P=X_1Y_2Z_4$ and $Q=Z_1Y_2X_3X_4$ on four qubits. Compute $\nu(P,Q)$ and decide whether $P$ and $Q$ commute. Then replace $Q$ by $Q'=Z_1X_2X_3X_4$ and repeat.

Solution

For $P$ and $Q$, qubit 1 contributes one anticommutation because $X$ and $Z$ differ, qubit 2 contributes none because both letters are $Y$, qubit 3 contributes none because $P$ has $I$, and qubit 4 contributes one because $Z$ and $X$ differ. Hence $\nu(P,Q)=2$ and

$$PQ=(-1)^2QP=QP.$$

For $Q'$, qubits 1, 2, and 4 each contain distinct nonidentity letters. Thus $\nu(P,Q')=3$ and

$$PQ'=(-1)^3Q'P=-Q'P.$$

The first pair commutes and the second anticommutes. No overall phase calculation is needed for this binary decision.

Exercise 3: Separate inconsistent and redundant checks

Compare the two signed two-qubit lists

$$\begin{aligned} L_-&=(X_1X_2,\;Z_1Z_2,\;-Y_1Y_2),\\ L_+&=(X_1X_2,\;Z_1Z_2,\;+Y_1Y_2). \end{aligned}$$

Both lists are pairwise commuting. Decide which third entry is redundant and which list is inconsistent. Give the independent rank and stabilized dimension wherever those quantities exist.

Solution

The product of the first two checks is

$$\begin{aligned} (X_1X_2)(Z_1Z_2) &=(-iY_1)(-iY_2)\\ &=-Y_1Y_2. \end{aligned}$$

In $L_-$, the third check is exactly the product of the first two. It is redundant, so the raw list length is three but the independent rank is $r=2$. The group excludes $-I$, and for $n=2$ its all-plus sector has dimension $2^{2-2}=1$.

In $L_+$, multiplying all three listed checks gives $(-Y_1Y_2)(+Y_1Y_2)=-I$. The list is therefore inconsistent and has no nonzero common $+1$ vector. The phase sign distinguishes harmless redundancy from a fatal obstruction.

Exercise 4: Replace Generators Without Changing the Sector

For the worked group generated by $g_1=X_1X_2X_3$ and $g_2=Z_1Z_2$, replace $g_2$ with $g'_2=g_1g_2=-Y_1Y_2X_3$. Show directly that $(g_1,g'_2)$ generates the same group, has the same independent rank, and stabilizes the same sector.

Solution

The new generator $g'_2=g_1g_2$ belongs to the original group, so every product of $g_1$ and $g'_2$ belongs to $\langle g_1,g_2\rangle$. Conversely,

$$g_1g'_2=g_1(g_1g_2)=g_1^2g_2=g_2,$$

where commutation and $g_1^2=I$ were used. Thus each old generator is generated by the new pair, and the groups coincide. Neither new generator is $I$, and their product is $g_2\ne\pm I$, so the pair remains independent with $r=2$ and $\dim\mathcal C=2$.

If a vector has eigenvalue $+1$ under $g_1$ and $g_2$, it has eigenvalue $+1$ under their product $g'_2$. Conversely, eigenvalue $+1$ under $g_1$ and $g'_2$ implies eigenvalue $+1$ under $g_1g'_2=g_2$. The stabilized sectors are identical.

Exercise 5: Errors with the same syndrome

For the worked generators $g_1=X_1X_2X_3$ and $g_2=Z_1Z_2$, compute the signatures of $X_1$, $X_2$, $X_1X_2$, and $X_3$. Identify a pair with the same syndrome and two distinct all-plus operators. Use the explicit basis $|a\rangle,|b\rangle$ to decide whether the four errors are jointly correctable.

Solution

Both $X_1$ and $X_2$ commute with $g_1$ and anticommute with $g_2$, so each has signature $(+1,-1)$. They have the same syndrome. Character multiplication gives

$$\chi_{X_1X_2}(g_j) =\chi_{X_1}(g_j)\chi_{X_2}(g_j),$$

so $X_1X_2$ has $(+1,+1)$. Directly, its two overlaps with $g_2$ also give even parity. The operator $X_3$ commutes with $g_1$ because its only nonidentity letter equals the third $X$, and it has disjoint support from $g_2$, so it too has $(+1,+1)$.

Thus $X_1,X_2$ are a pair with the same syndrome, while $X_1X_2$ and $X_3$ are distinct undetected Paulis. The explicit basis above gives $X_1X_2|a\rangle=|b\rangle$ and $X_1X_2|b\rangle=|a\rangle$. Thus $X_1^\dagger X_2$ is not scalar on the code: the pair $X_1,X_2$ already fails the Knill–Laflamme test, and the four-error set is not jointly correctable.

Exercise 6: Find a state from three checks

Analyze the three-qubit list

$$h_1=X_1X_2, \qquad h_2=Z_1Z_2, \qquad h_3=Z_3.$$

Check Hermiticity, pairwise commutation, independence, and exclusion of $-I$; state the group size and stabilized dimension; and compute the signature of $E=X_1Y_3$. Identify the normalized state in the stabilized sector.

Solution

Each $h_j$ is a Hermitian signed Pauli string and squares to $I$. The first pair has two distinct nonidentity overlaps and therefore commutes; $h_3$ has disjoint support from both $h_1$ and $h_2$. The product of the first pair is

$$h_1h_2=(-iY_1)(-iY_2)=-Y_1Y_2.$$

The four products generated by $h_1,h_2$ are $I,h_1,h_2,-Y_1Y_2$, all acting trivially on qubit 3. Multiplying any of them by $h_3=Z_3$ gives a distinct string with nonidentity third support. Hence no nonempty product is $I$ or $-I$: the list is independent, has rank $r=3$, and generates $2^3=8$ elements without the minus identity. With $n=3$,

$$\dim\mathcal C=2^{3-3}=1.$$

For $E=X_1Y_3$, the $X_1$ factor commutes with $h_1$ and anticommutes with $h_2$, while $Y_3$ anticommutes with $h_3$. Therefore

$$(\chi_E(h_1),\chi_E(h_2),\chi_E(h_3)) =(+1,-1,-1).$$

The stabilized state is $|\Phi^+\rangle_{12}|0\rangle_3$. The first two checks fix the Bell pair and $Z_3$ fixes the third qubit. A one-dimensional code space specifies a known state; it has no freedom to encode an unknown logical qubit.

References

• D. Gottesman, Stabilizer Codes and Quantum Error Correction, Ph.D. thesis, California Institute of Technology (1997), chapter 3, arXiv:quant-ph/9705052 — signed Pauli groups, stabilized subspaces and error syndromes.
• D. Gottesman, “An introduction to quantum error correction and fault-tolerant quantum computation,” Proceedings of Symposia in Applied Mathematics 68, 13–58 (2010), arXiv:0904.2557 — introductory stabilizer calculations and their relation to error correction.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), chapter 10, doi:10.1017/CBO9780511976667 — textbook treatment and examples.