The Exercise

Three qubits hold the bit flip code, and a flip of one of them is diagnosed by the syndrome measurement, the projective measurement with the four projectors $$\begin{aligned} P_0 &= |000\rangle\langle 000| + |111\rangle\langle 111|, & P_1 &= |100\rangle\langle 100| + |011\rangle\langle 011|, \\ P_2 &= |010\rangle\langle 010| + |101\rangle\langle 101|, & P_3 &= |001\rangle\langle 001| + |110\rangle\langle 110|. \end{aligned}$$(1) The outcome $j = 1, 2, 3$ signals a flip of qubit $j$ on either code word $|000\rangle$ or $|111\rangle$, and the outcome $0$ no flip.

The same diagnosis can be made with two parity measurements in turn, first of the observable $Z_1 Z_2$, the product of the Pauli operator $Z$, with $Z|0\rangle = |0\rangle$ and $Z|1\rangle = -|1\rangle$, on the first two qubits, then of $Z_2 Z_3$, each with the outcomes $\pm 1$. The exercise is to show that the two procedures agree up to a relabeling of the outcomes: each pair of parities occurs with the probability of one syndrome, and leaves the three qubits in the same state.

Three Qubits

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$^✦, and a state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$.^✦

The qubit is the system whose state space is $\mathbb{C}^2$, and its basis state $|i\rangle$, for a bit $i$, is the pure state given by the standard basis vector $e_i$.^✦ The composite system $S \otimes T$ of two systems $S$ and $T$ is the system whose state space is the tensor product of their state spaces.^✦ The three qubit register is the composite of three qubits, with state space $\mathbb{C}^2 \otimes (\mathbb{C}^2 \otimes \mathbb{C}^2)$.^✦

For pure states $|\psi\rangle$ of $S$ and $|\varphi\rangle$ of $T$, the product $|\psi\rangle \otimes |\varphi\rangle$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors.^✦ The basis state $|q_1 q_2 q_3\rangle$ of the register, for three bits $q_1, q_2, q_3$, is the product $|q_1\rangle \otimes (|q_2\rangle \otimes |q_3\rangle)$; these eight states are the computational basis in which (1) is written.

Measurements

A measurement on $S$ with a finite set of outcomes $\iota$ is a family of operators $M_i$ on the state space of $S$, one for each outcome $i$, satisfying the completeness relation $$\sum_{i} M_i^\dagger M_i = I.$$(2)^✦ The positive operator valued measure (POVM) of a measurement $M$ on $S$ is the family of effects $E_i = M_i^\dagger M_i$, one for each outcome $i$. The probability of the outcome $i$ when the measurement $M$ is performed on a state $\rho$ of $S$ is given by the Born rule, $$p(i) = \operatorname{Tr}\left(M_i \rho M_i^\dagger\right) = \operatorname{Tr}\left(E_i \rho\right),$$(3) and when $p(i) \neq 0$ the state after the outcome is $$\rho_i = \frac{M_i \rho M_i^\dagger}{p(i)}.$$(4)^✦

The cascaded measurement of two measurements $L$ and $M$ on a system $S$, $L$ performed first and $M$ second, is the measurement on $S$ whose outcomes are the pairs $(l, m)$ and whose operators are $$N_{(l,m)} = M_m L_l.$$(5)^✦

Parity Observables

An observable $A$ on $S$ is a self adjoint operator on the state space of $S$.^✦ The identity observable $I$ on $S$ is the observable whose operator is the identity.^✦ The tensor product $A \otimes B$ of an observable $A$ of $S$ and an observable $B$ of $T$ is the observable of $S \otimes T$ whose operator is the tensor product of theirs; it measures $A$ on the first factor and $B$ on the second.^✦

The Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(6) Write $\vec{\sigma} = (X, Y, Z)$ for the Pauli vector. For a real vector $\vec{n} = (n_x, n_y, n_z)$, the Pauli observable $\vec{n} \cdot \vec{\sigma}$ of the qubit is the observable whose operator on $\mathbb{C}^2$ is the matrix $n_x X + n_y Y + n_z Z$.^✦ The observable $Z$ is the Pauli observable for $\hat{n} = (0, 0, 1)$.^✦ On the three qubit register the two parity observables, built from the Pauli observable $Z$, are $$Z_1 Z_2 = Z \otimes (Z \otimes I), \qquad Z_2 Z_3 = I \otimes (Z \otimes Z),$$(7) and each squares to the identity, $(Z_1 Z_2)^2 = (Z_2 Z_3)^2 = I$.

For an observable $A$ with $A^2 = I$, the operators $$P_\pm = \tfrac{1}{2}\left(I \pm A\right)$$(8) are the projectors onto the eigenspaces of $A$ for $+1$ and $-1$. The sign measurement $\{P_+, P_-\}$ of such an $A$ is the measurement with two outcomes whose operators are $M_0 = P_+$ and $M_1 = P_-$, the outcome $0$ standing for the eigenvalue $+1$ and $1$ for $-1$; it is the projective measurement of $A$.^✦ The two parity measurements are the sign measurements $L$ of $Z_1 Z_2$ and $M$ of $Z_2 Z_3$. Measuring one after the other is their cascade (5), with outcomes $(a, b)$, $a$ for $Z_1 Z_2$ and $b$ for $Z_2 Z_3$.

The Syndrome Measurement

The syndrome measurement is the measurement on the three qubit register with the outcomes $0, 1, 2, 3$ whose operators are the projectors (1), one for each pair of basis strings listed there.^✦ The relabeling $r$ is the one to one map from pairs of parity outcomes to syndromes given by $$r(0,0) = 0, \qquad r(1,0) = 1, \qquad r(1,1) = 2, \qquad r(0,1) = 3.$$(9)^✦

The Claim

For every state $\rho$ of the register and every pair $(a, b)$, the cascade and the syndrome measurement give matched outcomes the same probability, $$p_{\mathrm{par}}(a, b) = p_{\mathrm{syn}}\bigl(r(a, b)\bigr),$$(10)^✦ where $p_{\mathrm{par}}$ and $p_{\mathrm{syn}}$ are the Born probabilities (3) of the cascade and of the syndrome measurement. A pair of nonzero probability is therefore matched to a syndrome of nonzero probability, for which the state (4) is defined.

What Is Checked

The claim is stated over the cascade of the sign measurements of $Z_1 Z_2$ and $Z_2 Z_3$ and over the syndrome measurement.

About these constructions the library proves that the cascade of two measurements satisfies the completeness relation (2) because its factors do, and that the sign measurement of an observable with $A^2 = I$ and the syndrome projectors satisfy it as well. It proves that the effects of a measurement are positive, that the state (4) after an outcome of nonzero probability is a state of $S$, a positive operator of unit trace, and that a product of pure states is a unit vector. The Pauli observable $\vec{n} \cdot \vec{\sigma}$ and the tensor product of two observables are self adjoint.