The Exercise

Two qubits are to be measured in the Bell basis, the orthonormal basis of the four Bell states. A quantum circuit reads its qubits in the computational basis, so a measurement in another orthonormal basis is made by first applying a unitary that carries that basis onto the computational one. For the Bell basis the unitary is a controlled NOT (CNOT), with the first qubit as control, followed by a Hadamard on the first qubit. The exercise is to show that this circuit, followed by a reading of both qubits in the computational basis, performs the measurement in the Bell basis: the elements of its positive operator valued measure (POVM) are the four projectors onto the Bell states.

Two Qubits

A quantum system $S$ is a physical object whose state space is a finite dimensional complex Hilbert space.^✦ A pure state $\psi$ of $S$ is a unit vector in the state space of $S$, $\lVert \psi \rVert = 1$, determined physically up to a global phase.^✦ The composite system $S \otimes T$ of two quantum systems $S$ and $T$ is the quantum system whose state space is the tensor product of their state spaces.^✦ For pure states $\psi$ of $S$ and $\varphi$ of $T$, the product state $\psi \otimes \varphi$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors.^✦

The qubit is the quantum system whose state space is $\mathbb{C}^2$.^✦ For a bit $i$, the basis state $|i\rangle$ of the qubit is the pure state whose vector is the standard basis vector $e_i$ of $\mathbb{C}^2$; $|0\rangle$ and $|1\rangle$ are the computational basis states.^✦ On two qubits the product of $|x\rangle$ and $|y\rangle$ is written $|xy\rangle$.

Gates

A unitary $U$ of a quantum system $S$ is an evolution of $S$ as a closed system: an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$.^✦ For unitaries $U$ and $V$ of the same system, $U$ after $V$ is the unitary $UV$, in which $V$ acts first.^✦ For a unitary $U$ of $S$ and a unitary $V$ of a system $T$, the product unitary $U \otimes V$ is the unitary of $S \otimes T$ whose operator is the tensor product of the two operators.^✦ The gate $U$ on the first wire of $S \otimes T$ is the product $U \otimes 1$ of $U$ with the identity evolution of $T$, the unitary whose operator is the identity.^✦

The Hadamard $H$ and the Pauli $X$, the quantum NOT, are the unitaries of the qubit whose matrices in the computational basis are $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$(1)

For a unitary $V$ of a system $S$, the controlled gate $C(V)$ is the unitary of the composite of a control qubit and the target $S$ given by $$C(V) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes V.$$(2)^✦ The CNOT is the controlled gate $C(X)$ on two qubits: it flips the second qubit exactly when the first is $|1\rangle$.^✦ The circuit of the exercise is the unitary of two qubits $$U = (H \otimes 1)\,\mathrm{CNOT}.$$(3)^✦

The Bell Basis

For bits $x$ and $y$, the Bell vector $|\beta_{xy}\rangle$ is the vector of the state space of two qubits $$|\beta_{xy}\rangle = \frac{1}{\sqrt{2}} \big( |0y\rangle + (-1)^x |1\bar{y}\rangle \big),$$(4)^✦ where $\bar{y} = y + 1 \bmod 2$ is the flip of $y$. For bits $x$ and $y$, the Bell state $|\beta_{xy}\rangle$ is the pure state of two qubits whose vector is the Bell vector $|\beta_{xy}\rangle$ of (4).^✦ Written out, the four Bell states are $$\begin{aligned} |\beta_{00}\rangle &= \tfrac{1}{\sqrt{2}} \big( |00\rangle + |11\rangle \big), & |\beta_{01}\rangle &= \tfrac{1}{\sqrt{2}} \big( |01\rangle + |10\rangle \big), \\ |\beta_{10}\rangle &= \tfrac{1}{\sqrt{2}} \big( |00\rangle - |11\rangle \big), & |\beta_{11}\rangle &= \tfrac{1}{\sqrt{2}} \big( |01\rangle - |10\rangle \big). \end{aligned}$$(5) The Bell basis is the orthonormal basis of the state space of two qubits formed by the four Bell states, indexed by the pairs $(x, y)$.^✦

Measurements

A measurement on a quantum system $S$ with a finite set of outcomes is a family of measurement operators $M_i$ on the state space of $S$, one for each outcome $i$, that satisfies the completeness relation $$\sum_i M_i^\dagger M_i = 1.$$(6)^✦ The POVM of a measurement assigns to the outcome $i$ the element $E_i = M_i^\dagger M_i$.^✦ The projective measurement in an orthonormal basis $\{|b_i\rangle\}$ of the state space of $S$ is the measurement whose operator for the outcome $i$ is the projector $|b_i\rangle\langle b_i|$.^✦ The Bell measurement is the projective measurement in the Bell basis, with the pairs of bits as outcomes: $$M_{xy} = |\beta_{xy}\rangle\langle \beta_{xy}|.$$(7)^✦

The Claim

What Is Checked

The claim is stated over the Bell measurement (7) with its POVM, and over the circuit $U$ of (3).

For these constructions the library proves that each Bell vector (4) has unit norm and that the four Bell states are orthonormal, $\langle \beta_{xy} | \beta_{x'y'} \rangle = \delta_{xx'}\,\delta_{yy'}$. The projectors onto the vectors of an orthonormal basis satisfy the completeness relation (6), each projector $P$ having $P^\dagger P = P$ and the projectors summing to the identity. On the side of the circuit, the controlled gate (2) is unitary for every unitary $V$, and so are the matrices (1). Tensor products and composites of unitaries are again unitary.