The Exercise

Two coupled nuclear spins of one half serve as the qubits of a nuclear magnetic resonance (NMR) experiment. They start in $|00\rangle$, both spins in $|0\rangle$. A pulse rotates each spin by $\pi/2$ about the $\hat x$ axis. The pair then evolves for a time $t$ under the coupling $J Z_1 Z_2$, where $Z_k$ is the Pauli $Z$ of spin $k$, $J$ is the coupling constant and $Jt = \pi/4$. A last pulse rotates the second spin by $-\pi/2$ about the $\hat y$ axis. The exercise is to show that the pair is then in the Bell state $(|00\rangle - |11\rangle)/\sqrt 2$.

Two Qubits

A quantum system $S$ is given by its state space, a finite dimensional complex Hilbert space.^✦ Any such space can be taken as the state space of a system, and the qubit is the two level system whose state space is $\mathbb{C}^2$. The composite system $S \otimes T$ of two systems $S$ and $T$ is the system whose state space is the tensor product of theirs; the two spins form the composite system of two qubits, with state space $\mathbb{C}^2 \otimes \mathbb{C}^2$.^✦

A pure state $\psi$ of $S$ is a unit vector $|\psi\rangle$ in the state space of $S$.^✦ The basis states $|0\rangle$ and $|1\rangle$ of the qubit are the standard basis vectors $e_0$ and $e_1$ of $\mathbb{C}^2$.^✦ The product $\psi \otimes \varphi$ of pure states $\psi$ of $S$ and $\varphi$ of $T$ is the pure state of $S \otimes T$ with vector $|\psi\rangle \otimes |\varphi\rangle$; the spins start in $|00\rangle = |0\rangle \otimes |0\rangle$.^✦

A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$.^✦ A pure state $\psi$ of $S$ determines the state $|\psi\rangle\langle\psi|$ of $S$, the projector onto its vector.^✦

For bits $x$ and $y$, the Bell vector of two qubits is $$|\beta_{xy}\rangle = \frac{1}{\sqrt 2}\bigl(|0\rangle|y\rangle + (-1)^x |1\rangle|\bar y\rangle\bigr),$$(1) with $\bar y$ the flip of $y$, and the Bell state $|\beta_{xy}\rangle$ is the pure state of two qubits with this vector. The target of the exercise is $|\beta_{10}\rangle$, the case $x = 1$, $y = 0$ of (1).

Observables and Rotations

An observable $A$ on $S$ is a self adjoint operator on the state space of $S$, $A^\dagger = A$.^✦ The tensor product $A \otimes B$ of observables $A$ of $S$ and $B$ of $T$ is the observable of $S \otimes T$ whose operator is the tensor product of their operators.^✦

In the basis $|0\rangle, |1\rangle$ the Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(2) For a real vector $\vec n$ in three dimensions, the matrix $\vec n \cdot \vec\sigma$ is $n_x X + n_y Y + n_z Z$, and the Pauli observable $\vec n \cdot \vec\sigma$ is the observable of the qubit whose operator has that matrix. The observable $Z$ is the Pauli observable for the unit vector $\hat n = (0, 0, 1)$.^✦ The coupling Hamiltonian $J Z_1 Z_2$ of the two spins, for a real coupling constant $J$, is the observable of two qubits $$H = J\, Z \otimes Z.$$(3)^✦

For a real vector $\vec n$ and an angle $\theta$, the rotation $R_{\vec n}(\theta)$ is the $2 \times 2$ complex matrix $$R_{\vec n}(\theta) = \exp\Bigl(-\frac{i\theta}{2}\, \vec n \cdot \vec\sigma\Bigr),$$(4)^✦ written $R_{\hat n}(\theta)$ when the axis is a unit vector $\hat n$. The rotations about the $\hat x$ and $\hat y$ axes are $R_x(\theta) = e^{-i\theta X/2}$, with $\hat n = (1, 0, 0)$, and $R_y(\theta) = e^{-i\theta Y/2}$, with $\hat n = (0, 1, 0)$.

Unitaries and the Circuit

A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$.^✦ It carries a pure state $\psi$ of $S$ to the pure state $U\psi$, with vector $U|\psi\rangle$.^✦ For unitaries $U$ and $V$ of $S$, $U$ after $V$ is the unitary $UV$, first $V$ and then $U$; the identity evolution of $S$ is the unitary $1$. The product unitary $U \otimes V$ of unitaries $U$ of $S$ and $V$ of $T$ is the unitary of $S \otimes T$ whose operator is the tensor product of their operators.^✦ A gate $U$ on the first factor alone is $U \otimes 1$, and a gate $V$ on the second alone is $1 \otimes V$.

The propagator of an observable $H$ of $S$, with Planck’s constant $\hbar$, from time $t_1$ to time $t_2$ is the unitary $$U(t_1, t_2) = \exp\bigl[-iH(t_2 - t_1)/\hbar\bigr],$$(5)^✦ which carries the state of a system with Hamiltonian $H$ at time $t_1$ to its state at $t_2$, $|\psi(t_2)\rangle = U(t_1, t_2)|\psi(t_1)\rangle$.

The pulse sequence of the exercise is the NMR Bell circuit, the unitary of two qubits $$U_{J,t} = \bigl(1 \otimes R_y(-\pi/2)\bigr)\, U(0, t)\, \bigl(R_x(\pi/2) \otimes 1\bigr)\bigl(1 \otimes R_x(\pi/2)\bigr),$$(6) where $U(0, t) = e^{-iHt}$ is the propagator (5) of the coupling Hamiltonian (3) with $\hbar = 1$. The two $\pi/2$ rotations about $\hat x$ commute, and the second spin is turned first.

The Claim

Only the product $Jt$ enters the propagator, $U(0, t) = e^{-iJt\, Z_1 Z_2}$, which for $Jt = \pi/4$ is $e^{-i\pi Z_1 Z_2/4}$.

What Is Checked

The claim is stated over the circuit (6), made of rotation gates on one qubit and the propagator of the coupling Hamiltonian, and over the input $|00\rangle$ and the Bell state $|\beta_{10}\rangle$, compared through the states they determine.

About these constructions the library proves that every Bell vector (1) is a unit vector, so that $|\beta_{xy}\rangle$ is a pure state.^✦ Each rotation gate is unitary because its matrix is. The Pauli observables are self adjoint because the matrix $\vec n \cdot \vec\sigma$ is Hermitian, and so is the coupling Hamiltonian, a real multiple of a tensor product of Hermitian operators.