Quantum Information Applications

The Pauli algebra turns a two-level quantum system into a geometric object and makes many-qubit constraints calculable. A qubit need not be a physical spin: its two states may be atomic levels, photon polarizations, or an encoded subspace. The spin-$1/2$ mathematics applies once that subspace and its basis have been specified. This page follows the algebra from rotations to Bell correlations and stabilizer codes; the wider treatment is in Quantum Information and Computation.

Helpful background. Pauli matrices supplies the multiplication rule, the Bloch sphere translates between states and axes, and tensor products explains how operators act on separate qubits.

Qubit rotations and measurable directions

In a chosen basis $|0\rangle,|1\rangle$, every density operator has the form

$$\rho=\frac12(I+\mathbf r\cdot\boldsymbol\sigma), \qquad |\mathbf r|\leq1.$$

The three real components are expectation values: $r_i=\operatorname{Tr}(\rho\sigma_i)$. A unit vector describes a pure state; an interior vector describes a mixed state. The density-operator treatment derives the positivity bound. In an encoded qubit, these axes need not be spatial directions in the laboratory.

For an active rotation of the state,

$$U(\hat{\mathbf n},\theta) =e^{-i\theta\hat{\mathbf n}\cdot\boldsymbol\sigma/2} =I\cos\frac\theta2-i\hat{\mathbf n}\cdot\boldsymbol\sigma\sin\frac\theta2.$$

This follows from $(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I$. Conjugation $\rho\mapsto U\rho U^\dagger$ rotates $\mathbf r$ through $\theta$. For example,

$$U_y(\theta)|0\rangle =\cos\frac\theta2|0\rangle+\sin\frac\theta2|1\rangle, \qquad \mathbf r=(\sin\theta,0,\cos\theta).$$

Thus a positive quarter turn about $y$ prepares the $+1$ eigenstate of $X$. Checking only the $Z$ populations would not distinguish it from the opposite rotation: both give equal populations, but their $X$ expectations have opposite signs.

An ideal projective measurement of $\hat{\mathbf m}\cdot\boldsymbol\sigma$ uses projectors $(I\pm\hat{\mathbf m}\cdot\boldsymbol\sigma)/2$. Taking their traces with $\rho$ gives

$$p_\pm=\frac12(1\pm\mathbf r\cdot\hat{\mathbf m}).$$

This connects the geometric arrow to an experiment. It also shows why $\mathbf r=0$ produces unbiased outcomes along every axis, without implying a hidden definite direction.

A $2\pi$ spinor rotation gives $-I$. Applied to an entire isolated qubit, this is a global phase; applied conditionally in one branch of a larger superposition, it becomes a relative phase. Consequently, matrices equivalent up to phase as standalone gates need not give equivalent controlled gates. See spin rotations and single-qubit gates.

Joint Pauli measurements and Bell correlations

For two qubits use the ordered basis $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ and write $X_A=X\otimes I$, $Z_B=I\otimes Z$. A Pauli string is a tensor product of such factors. Distinct local operators commute, but two strings can overlap on several qubits. Every overlap with anticommuting nonidentity factors contributes one minus sign: an even number gives commuting strings.

For example, $X\otimes X$ and $Z\otimes Z$ commute although $X$ and $Z$ anticommute on each individual qubit:

$$(X\otimes X)(Z\otimes Z) =(-iY)\otimes(-iY)=-Y\otimes Y.$$

The Bell state

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

has $XX=+1$, $ZZ=+1$, and therefore $YY=-1$. Each local marginal is $I/2$, so all individual Pauli outcomes are unbiased. The information is in the joint correlations. In contrast, the separable mixture $(|00\rangle\langle00|+|11\rangle\langle11|)/2$ shares its $ZZ$ correlation but has $\langle XX\rangle=\langle YY\rangle=0$. A single correlation or a single marginal cannot identify the Bell state. The full derivation belongs to Bell states.

Commuting joint observables admit a common projective measurement. Measuring every individual local factor can reveal additional information and cause a different disturbance: measuring $Z_A$ and $Z_B$ separately resolves $|00\rangle$ from $|11\rangle$, whereas measuring only $Z_AZ_B$ need not. See local and global observables.

Stabilizers as constraints on a state or code

Choose $r$ independent, commuting Hermitian Pauli generators $S_a$ on $n$ qubits, each with eigenvalues $\pm1$. Require that their generated group not contain $-I$, which would make a common $+1$ eigenvector impossible. The projector onto their common $+1$ subspace is

$$P=\prod_{a=1}^{r}\frac{I+S_a}{2}, \qquad \operatorname{Tr}P=2^{n-r}.$$

To see the dimension, expand the product. Independence makes every nonempty product a nonidentity Pauli string, whose trace is zero; only $I$ contributes $2^n/2^r$. When $r=n$ this gives one stabilizer state. When $r<n$ it gives a code subspace with $n-r$ logical qubits, whose possible states need not themselves be stabilizer states.

For $n=2$, $XX$ and $ZZ$ are independent and select $|\Phi^+\rangle$ uniquely. Adding $-YY$ changes nothing because it is their product. Counting it as a third independent condition would give a meaningless half-dimensional subspace.

A Pauli error $E$ commuting with a generator preserves its sign; an anticommuting error reverses it, since $S_aE|\psi\rangle=\pm E|\psi\rangle$. These measured signs form the error syndrome. They need not determine the error uniquely, and an undetected logical operation may commute with all generators while acting nontrivially within the code. The stabilizer introduction develops these ideas; stabilizer identities collects useful formulas.

Clifford gates send Pauli operators to Pauli operators under conjugation. This permits efficient simulation of the standard stabilizer circuit family described by Gottesman and by Aaronson–Gottesman, including Pauli measurements. The statement concerns that specified family; arbitrary non-Clifford gates or arbitrary input states are additional resources, not automatically covered by the simulation result.

Exercises

1. For the Bell state $|\Phi^+\rangle$, apply $Z_A$. Which of the $XX$ and $ZZ$ stabilizer signs changes? Identify the resulting state.

Solution

$Z_A$ anticommutes with $XX$ and commutes with $ZZ$, so the signs become $(-1,+1)$. Directly,

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

Both states have maximally mixed local marginals. The changed relative phase is visible in a joint $XX$ measurement.

2. The three-qubit repetition code has generators $Z_1Z_2$ and $Z_2Z_3$. Find its $+1$ subspace. Does a $Z_1$ error produce a nontrivial syndrome?

Solution

The constraints require all computational-basis bits to agree, so the subspace is $\operatorname{span}\{|000\rangle,|111\rangle\}$ and has dimension $2^{3-2}=2$. $Z_1$ commutes with both generators and produces no syndrome. Nevertheless it changes $\alpha|000\rangle+\beta|111\rangle$ to $\alpha|000\rangle-\beta|111\rangle$, a logical phase operation. This code detects single bit flips but does not protect against arbitrary single-qubit errors.

References

• S. Aaronson and D. Gottesman, “Improved simulation of stabilizer circuits,” Physical Review A 70, 052328 (2004). doi:10.1103/PhysRevA.70.052328.
• D. Gottesman, “The Heisenberg representation of quantum computers” (1998). arXiv:quant-ph/9807006.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010. See the discussions of qubits, quantum circuits, and quantum error correction.