Stabilizer States Preview

A stabilizer description specifies a quantum state by naming observables for which the state has definite eigenvalue $+1$. For many qubit states, especially Bell, GHZ, cluster, and graph states, this is more compact than writing all amplitudes in the computational basis.

For a pure state $\lvert\psi\rangle$, an operator $S$ stabilizes the state if

$$S\lvert\psi\rangle = \lvert\psi\rangle.$$

Here a Pauli stabilizer state uses signed Hermitian products of Pauli matrices and identities. A general state need not admit a complete description of this form.

Helpful background. Pauli matrices provides their products and signs; graph states supplies a family of examples. The full stabilizer formalism develops binary representations, circuits, and error-correcting codes.

Stabilizers as Commuting Observables

The basic building blocks are Pauli operators acting on named qubits:

$$X_j, \qquad Y_j, \qquad Z_j.$$

A Pauli product is a tensor product of single-qubit Pauli matrices and identities, possibly with an overall sign. For example,

$$Z_1X_2Z_3$$

means

$$Z\otimes X\otimes Z$$

on qubits $1,2,3$, with identities on any other qubits left implicit.

To specify a state by simultaneous eigenvalue equations, the chosen observables must commute. If

$$[S_a,S_b]=0,$$

then the two observables can have simultaneous eigenstates. A stabilizer state is a pure state specified as the unique common $+1$ eigenstate of a sufficiently large commuting family of Pauli products.

For $n$ qubits, $n$ independent commuting signed Pauli generators specify exactly one pure stabilizer state, provided the generated group contains no $-I$. Independence means no generator is a product of the others, and the absence of $-I$ excludes contradictory eigenvalue requirements:

$$S_j\lvert\psi\rangle = \lvert\psi\rangle, \qquad j=1,\ldots,n.$$

If there are fewer independent generators, the common $+1$ eigenspace is larger. That larger space is the entry point to stabilizer codes.

More generally, for $r$ independent compatible generators $S_1,\ldots,S_r$, the projector onto their common $+1$ space is

$$P=\prod_{j=1}^r\frac{I+S_j}{2} =\frac1{2^r}\sum_{g\in\mathcal S}g, \qquad \operatorname{Tr}P=2^{n-r}.$$

Each factor is a projector and the factors commute. In the group sum, only $I$ has nonzero trace, giving the dimension directly. For $r=n$, $P=|\psi\rangle\langle\psi|$. For $r<n$, $P$ projects onto a larger subspace; $P/2^{n-r}$ is the maximally mixed state in that subspace, not every possible encoded state.

Simple One-Qubit Examples

The state $\lvert0\rangle$ is stabilized by $Z$:

$$Z\lvert0\rangle = \lvert0\rangle.$$

The state $\lvert1\rangle$ is stabilized by $-Z$:

$$(-Z)\lvert1\rangle = \lvert1\rangle.$$

The state $\lvert+\rangle$ is stabilized by $X$:

$$X\lvert+\rangle = \lvert+\rangle, \qquad \lvert+\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle+\lvert1\rangle \bigr).$$

The state $\lvert-\rangle$ is stabilized by $-X$.

For two qubits, the product state $\lvert00\rangle$ is the unique simultaneous $+1$ eigenstate of

$$Z_1, \qquad Z_2.$$

Thus stabilizer language is not only for entangled states. It is a way of recording exact Pauli measurement outcomes.

Bell States as Stabilizer States

The Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr)$$

is stabilized by two commuting Pauli products:

$$X_1X_2\lvert\Phi^+\rangle = \lvert\Phi^+\rangle, \qquad Z_1Z_2\lvert\Phi^+\rangle = \lvert\Phi^+\rangle.$$

The first equation says local $X$ measurements always agree. The second says local $Z$ measurements always agree. These are alternative measurement choices, not simultaneous pre-existing values for each qubit. These two statements are compatible because $X_1X_2$ and $Z_1Z_2$ commute: there is one $X$-$Z$ anticommutation on each qubit, so the two minus signs cancel.

The four Bell states can be labeled by their eigenvalues under the commuting pair $X_1X_2$ and $Z_1Z_2$:

$$\begin{array}{ccc} \text{state} & X_1X_2 & Z_1Z_2 \\ \lvert\Phi^+\rangle & +1 & +1\\ \lvert\Phi^-\rangle & -1 & +1\\ \lvert\Psi^+\rangle & +1 & -1\\ \lvert\Psi^-\rangle & -1 & -1 \end{array}$$

This is the stabilizer version of saying that Bell states are distinguished by joint correlations rather than by one-qubit reduced states.

The generator product has a sign that matters: $(XX)(ZZ)=-YY$. The projector formula therefore gives

$$|\Phi^+\rangle\langle\Phi^+|=\frac14(I+XX+ZZ-YY).$$

Although the two parity observables commute, measuring each local $X$ and forgetting the individual outcomes is not the same state update as measuring only $XX$. Local and global observables shows why measuring extra information can destroy coherence.

GHZ States as Stabilizer States

The $n$-qubit GHZ plus state is

$$\lvert\mathrm{GHZ}_n^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle^{\otimes n} + \lvert1\rangle^{\otimes n} \bigr).$$

A convenient stabilizer generating set is

$$X_1X_2\cdots X_n, \qquad Z_1Z_k \quad (k=2,\ldots,n).$$

The $Z_1Z_k$ stabilizers encode equality of all computational-basis bits:

$$Z_1Z_k \lvert\mathrm{GHZ}_n^+\rangle = \lvert\mathrm{GHZ}_n^+\rangle.$$

The global $X$ stabilizer encodes the coherence between the all-zero and all-one branches:

$$X_1X_2\cdots X_n \lvert\mathrm{GHZ}_n^+\rangle = \lvert\mathrm{GHZ}_n^+\rangle.$$

Together these $n$ independent generators specify the pure GHZ plus state. Changing the sign of the global $X$ generator selects the minus-phase GHZ state.

Graph States as Stabilizer States

Graph States are stabilizer states with generators attached to graph vertices. For a graph $G=(V,E)$, the generator at vertex $v$ is

$$K_v = X_v \prod_{u\in N(v)} Z_u.$$

The graph state satisfies

$$K_v\lvert G\rangle = \lvert G\rangle \qquad \text{for every }v\in V.$$

For the three-qubit line $1-2-3$, the generators are

$$K_1=X_1Z_2, \qquad K_2=Z_1X_2Z_3, \qquad K_3=Z_2X_3.$$

This illustrates why stabilizer descriptions are economical. A state with eight computational-basis amplitudes is specified by three local Pauli constraints.

Reduced states from stabilizers

For a pure stabilizer state, $\rho=2^{-n}\sum_{g\in\mathcal S}g$. Tracing out a subset removes every term with a nonidentity Pauli factor on a discarded qubit, because $\operatorname{Tr}X=\operatorname{Tr}Y=\operatorname{Tr}Z=0$. Therefore

$$\rho_A=2^{-|A|}\sum_{g\in\mathcal S:\,g_{\bar A}=I}g_A.$$

For the three-qubit path, the only group terms supported on qubits $1,2$ are $I$ and $X_1Z_2$. Thus

$$\rho_{12}=\tfrac14(I+X_1Z_2) =\tfrac12|+,0\rangle\langle+,0|+\tfrac12|-,1\rangle\langle-,1|.$$

It is separable even though the graph has an edge $1-2$. This calculation illustrates how a compact stabilizer sum describes both global entanglement and local mixedness.

With fewer than $n$ compatible generators, the larger common eigenspace can encode information. Pauli errors that anticommute with a generator flip its eigenvalue, providing a syndrome. The general error-correction construction is developed in stabilizer formalism.

Stabilizers and Witnesses

Because stabilizers are measurable Pauli products, they often provide practical entanglement diagnostics. For a target stabilizer state, one can measure selected stabilizers or products of stabilizers and compare the results with bounds satisfied by separable states.

This is one route from the structural definition of entanglement to Entanglement Witnesses. The witness page treats the general convex-geometry idea; stabilizer witnesses are a structured family of examples.

References

• D. Gottesman, Stabilizer Codes and Quantum Error Correction, Ph.D. thesis, California Institute of Technology (1997).
• M. Hein, J. Eisert, and H. J. Briegel, “Multiparty entanglement in graph states”, Physical Review A 69, 062311 (2004).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Ch. 10.

Exercises

1. The three observables $Z_1Z_2$, $Z_2Z_3$, and $-Z_1Z_3$ commute. Can a state have eigenvalue $+1$ for all three?

Solution

Their product is $-I$. A simultaneous $+1$ eigenvector would therefore satisfy $-I|\psi\rangle=|\psi\rangle$, forcing $|\psi\rangle=0$. Commutation alone does not make the specified signs consistent. The third constraint contradicts the product of the first two.

2. Show that $X_1X_2$ and $Z_1Z_2$ commute.

Solution

On one qubit, $X$ and $Z$ anticommute:

$$XZ=-ZX.$$

For two qubits,

$$(X_1X_2)(Z_1Z_2) = (X_1Z_1)(X_2Z_2).$$

Moving each $Z$ past an $X$ gives two minus signs:

$$(X_1Z_1)(X_2Z_2) = (-Z_1X_1)(-Z_2X_2) = (Z_1Z_2)(X_1X_2).$$

Thus the two Pauli products commute.

3. Verify that $\lvert\Phi^-\rangle$ has eigenvalues $-1$ under $X_1X_2$ and $+1$ under $Z_1Z_2$.

Solution

For

$$\lvert\Phi^-\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle-\lvert11\rangle \bigr),$$

the operator $X_1X_2$ swaps $\lvert00\rangle$ and $\lvert11\rangle$:

$$X_1X_2\lvert\Phi^-\rangle = \frac{1}{\sqrt2} \bigl( \lvert11\rangle-\lvert00\rangle \bigr) = -\lvert\Phi^-\rangle.$$

The operator $Z_1Z_2$ gives eigenvalue $+1$ to both $\lvert00\rangle$ and $\lvert11\rangle$, so

$$Z_1Z_2\lvert\Phi^-\rangle = \lvert\Phi^-\rangle.$$

4. Use the stabilizer projector of $|\Phi^+\rangle$ to compute its one-qubit reduction without expanding the ket.

Solution

In $\rho=(I\otimes I+X\otimes X-Y\otimes Y+Z\otimes Z)/4$, the last three terms trace to zero on the second qubit. The first gives $2I/4=I/2$. Tracing retains only group elements that act trivially on the discarded subsystem.

5. Suppose $r$ independent commuting stabilizer generators act on $n$ qubits and generate no $-I$. Why is the common $+1$ eigenspace dimension $2^{n-r}$?

Solution

The full $n$-qubit Hilbert space has dimension $2^n$. A nontrivial Pauli-product stabilizer has eigenvalues $\pm1$ with equal-dimensional eigenspaces, so imposing one independent $+1$ constraint halves the dimension. Imposing $r$ independent commuting constraints halves the dimension $r$ times:

$$2^n \longmapsto \frac{2^n}{2^r} = 2^{n-r}.$$

Commutation is needed so the constraints can be imposed simultaneously.