The Exercise

Two qubits are in a state $\rho$, pure or mixed. A measurement in the Bell basis finds them in one of the four Bell states $|\beta_{xy}\rangle$ and so returns two bits, the phase flip label $x$ and the bit flip label $y$. Measurements on the qubits one at a time give each label’s statistics. Measure $Z$ on each qubit and add the two outcome bits modulo $2$: the sum is $k$ with the same probability as $y = k$ in the Bell measurement.^✦ Measure $X$ on each qubit instead, and the sum is distributed as $x$.^✦

A Quantum System

A quantum system $S$ is given by its state space, a finite dimensional complex Hilbert space^✦, and any such space $H$ is the state space of a system.^✦ The qubit $Q$ is the system with $H = \mathbb{C}^2$^✦; its computational basis states $|0\rangle$ and $|1\rangle$ are the standard basis vectors $e_0$ and $e_1$.^✦ The composite system $S \otimes T$ of two systems has for its state space the tensor product of theirs, and two qubits form $Q \otimes Q$.^✦

A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$.^✦ 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.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$, pure and mixed states alike.^✦

Measurements

A measurement $M$ on $S$ with a finite set $I$ of outcomes is a family of measurement operators $M_i$ on the state space of $S$, one for each outcome $i \in I$, that satisfies the completeness relation, where $1$ is the identity operator on the state space of $S$, $$\sum_{i \in I} M_i^\dagger M_i = 1 .$$(1)^✦ The positive operator valued measure (POVM) of $M$ has the effects $E_i = M_i^\dagger M_i$.^✦ When $M$ is performed on a state $\rho$, outcome $i$ occurs with the Born probability $$p_M(i \,|\, \rho) = \operatorname{Tr}(E_i \rho) = \operatorname{Tr}(M_i \rho M_i^\dagger) .$$(2)^✦

The projective measurement in an orthonormal basis $\{|b_i\rangle\}$ has the projectors $|b_i\rangle\langle b_i|$ as its operators.^✦ The square root measurement of a POVM with effects $E_i$ has the operators $\sqrt{E_i}$, so that $M_i^\dagger M_i = E_i$.^✦ A measurement $M$ on $S$ acts on the first factor of a composite $S \otimes T$ through the operators $M_i \otimes 1_T$^✦, and a measurement $N$ on $T$ acts on its second factor through $1_S \otimes N_j$.^✦ The cascade of two measurements $L$ and $M$ on one system, first $L$ and then $M$, has the pairs $(l, m)$ as outcomes and the composites $M_m L_l$ as operators.^✦

To coarse grain a measurement $M$ with outcomes in $I$ along a map $g$ from $I$ to a set $K$ is to merge the outcomes that $g$ sends to the same value. The result $M^g$ is the square root measurement, with outcomes in $K$, of the POVM whose effects are $$F_k = \sum_{i \,:\, g(i) = k} E_i .$$(3)^✦

Local Pauli Measurements

An observable $A$ on $S$ is a self adjoint operator on the state space of $S$.^✦ With it come the operators $$P_\pm = \tfrac{1}{2}(1 \pm A) ,$$(4) the projectors onto the $\pm 1$ eigenspaces of $A$ when $A^2 = 1$. For an observable $A$ on $S$ with $A^2 = 1$ the sign measurement of $A$ is the projective measurement with two outcomes and operators $M_0 = P_+$ and $M_1 = P_-$: outcome $0$ stands for the eigenvalue $+1$ and outcome $1$ for $-1$.^✦

The Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(5) For a real vector $\vec{n}$ in three dimensions the matrix $\vec{n} \cdot \vec{\sigma} = n_x X + n_y Y + n_z Z$^✦ is Hermitian, and as an operator on $\mathbb{C}^2$ it is an observable of the qubit.^✦ For a unit vector $\hat{n}$ the observable $\hat{n} \cdot \vec{\sigma}$ squares to $1$; $X$ and $Z$ are those for $\hat{n} = (1, 0, 0)$ and $\hat{n} = (0, 0, 1)$. The sign measurement of $X$ is the $X$ measurement of the qubit, a projective measurement whose outcomes $0$ and $1$ stand for $|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$ and $|-\rangle = (|0\rangle - |1\rangle)/\sqrt{2}$, and the sign measurement of $Z$ is the $Z$ measurement of the qubit, a projective measurement whose outcomes $0$ and $1$ stand for $|0\rangle$ and $|1\rangle$.

The local $X$ measurement $L_X$ on two qubits is the cascade of the $X$ measurement on the first qubit and the $X$ measurement on the second.^✦ Write $P^X_0$ and $P^X_1$ for the projectors $P_+$ and $P_-$ of $X$. An outcome of $L_X$ is a pair $(a, b)$ of bits, one for each qubit, with operator $$(1 \otimes P^X_b)(P^X_a \otimes 1) = P^X_a \otimes P^X_b .$$(6) The local $Z$ measurement $L_Z$ is built in the same way from the $Z$ measurement.^✦ The parity of an outcome $(a, b)$ of either is $a \oplus b$, the sum of the bits modulo $2$.

The Bell Measurement

For bits $x$ and $y$, with $\bar{y} = y \oplus 1$ the flip of $y$, the Bell vector is $$|\beta_{xy}\rangle = \frac{1}{\sqrt{2}} \bigl( |0\rangle|y\rangle + (-1)^x |1\rangle|\bar{y}\rangle \bigr) ,$$(7)^✦ and the Bell state $|\beta_{xy}\rangle$ is the pure state of $Q \otimes Q$ with this vector.^✦ The four are $|\beta_{00}\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$, $|\beta_{01}\rangle = (|01\rangle + |10\rangle)/\sqrt{2}$, $|\beta_{10}\rangle = (|00\rangle - |11\rangle)/\sqrt{2}$ and $|\beta_{11}\rangle = (|01\rangle - |10\rangle)/\sqrt{2}$.^✦ The four Bell states form the Bell basis, an orthonormal basis of the state space of $Q \otimes Q$.^✦ The Bell measurement $B$ is the projective measurement in this basis, with the pairs $(x, y)$ as outcomes and the projectors $|\beta_{xy}\rangle\langle\beta_{xy}|$ as operators.^✦

The Claim

Four coarse grainings enter the claim, each a measurement with outcomes $k \in \{0, 1\}$. Along the parity $(a, b) \mapsto a \oplus b$ the local measurements give $L_X^{\oplus}$ and $L_Z^{\oplus}$. Along $(x, y) \mapsto x$ and $(x, y) \mapsto y$, which keep the phase flip label and the bit flip label of a Bell outcome, the Bell measurement gives $B_{\mathrm{pf}}$ and $B_{\mathrm{bf}}$.

What Is Checked

The claim is stated over every state $\rho$ of two qubits and both values of $k$.

The library proves that each Bell vector (7) has unit norm^✦ and that the four are orthonormal.^✦ Their number equals the dimension $4$ of the state space of $Q \otimes Q$^✦, so they form a basis.^✦ The relations $X^2 = 1$^✦ and $Z^2 = 1$^✦ make the projectors (4) of $X$ and of $Z$ sign measurements. Every construction above is a measurement: a projective and a square root measurement satisfy the completeness relation (1) because their projectors or effects sum to the identity, and a measurement on one factor and a cascade inherit it from the measurements they are built from.