The Exercise

In nuclear magnetic resonance, two qubits are two coupled spins, and two kinds of operation act on them. A pulse rotates one qubit about the $x$ axis or the $y$ axis by any angle. Between pulses the pair evolves under the coupling Hamiltonian $J Z_1 Z_2$, with $J$ the coupling constant and $Z_1$, $Z_2$ the Pauli $Z$ operators of the two qubits.

The exercise is to build the SWAP gate, which exchanges the states of the two qubits, from these operations alone: a sequence of pulses and periods of coupled evolution, each period lasting a time $t$ with $Jt = \pi/4$, that acts on every state of the pair as SWAP does.

Two Qubits

A quantum system $S$ has as its state space a finite dimensional complex Hilbert space.^✦ The qubit is the system whose state space is $\mathbb{C}^2$, with computational basis $|0\rangle$, $|1\rangle$.^✦ The composite system $S \otimes T$ of two systems $S$ and $T$ has as its state space the tensor product of theirs; in the pair of qubits, qubit 1 is the first factor and qubit 2 the second.^✦

A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$.^✦ A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$.^✦ Under a unitary $U$ of $S$, a state $\rho$ of $S$ evolves into the state $$\rho \longmapsto U \rho\, U^\dagger .$$(1)^✦

For two unitaries $U$ and $V$ of $S$, $U$ after $V$ is the unitary $UV$: first $V$, then $U$.^✦ For a unitary $U$ of $S$ and a unitary $V$ of $T$, the product unitary $U \otimes V$ of $S \otimes T$ is the tensor product of the two operators.^✦ With the identity evolution $1$ on the other factor, a gate $U$ on the first factor is $U \otimes 1$ and a gate $V$ on the second is $1 \otimes V$.

The SWAP of two copies of a system $S$ is the unitary of $S \otimes S$ given by the isometry of its state space onto itself that exchanges the two factors: for vectors $x$ and $y$ of the state space of $S$, $$\mathrm{SWAP}\,(x \otimes y) = y \otimes x .$$(2)^✦

Pulses and the Coupling

The Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(3) For a real vector $\vec n = (n_x, n_y, n_z)$ in three dimensions, $\vec n \cdot \vec\sigma = n_x X + n_y Y + n_z Z$, and the rotation about $\vec n$ by an angle $\theta$ is the $2 \times 2$ matrix $$R_{\vec n}(\theta) = \exp\!\left( -\frac{i\theta}{2}\, \vec n \cdot \vec\sigma \right),$$(4) written $R_{\hat n}(\theta)$ when the axis is a unit vector $\hat n$.

The pulses are the rotation gates $R_x(\theta)$ and $R_y(\theta)$, the unitaries of the qubit whose matrices in the computational basis are $R_{\hat x}(\theta) = e^{-i\theta X/2}$ and $R_{\hat y}(\theta) = e^{-i\theta Y/2}$, with $\hat x = (1, 0, 0)$ and $\hat y = (0, 1, 0)$.

An observable $A$ of a system $S$ is a self adjoint operator on the state space of $S$.^✦ For observables $A$ of $S$ and $B$ of $T$, the tensor product $A \otimes B$ is an observable of $S \otimes T$.^✦ On the qubit every $\vec n \cdot \vec\sigma$ is an observable, the Pauli observable along $\vec n$, and along $\hat z = (0, 0, 1)$ it is the observable $\hat z \cdot \vec\sigma = Z$. The coupling Hamiltonian of the two spins, with a real coupling constant $J$, is the observable of the pair $$\mathcal{H}_J = J\, Z \otimes Z = J Z_1 Z_2 .$$(5)^✦

A Hamiltonian $\mathcal{H}$ of a system generates its propagator from time $t_1$ to time $t_2$, the unitary $$U(t_1, t_2) = \exp\!\left[ -\frac{i \mathcal{H} (t_2 - t_1)}{\hbar} \right]$$(6)^✦ that takes the state at $t_1$ to the state at $t_2$. Here $\hbar$ is the reduced Planck constant; in units with $\hbar = 1$, the coupling acting from time $0$ to time $t$ is $e^{-iJt Z_1 Z_2}$.

The Circuit

Pulses alone give a Hadamard gate and a rotation about $z$, $$H = R_x(\pi)\, R_y(\pi/2), \qquad R_z(\theta) = H\, R_x(\theta)\, H ,$$(7) with products read from right to left, so that $R_y(\pi/2)$ acts first. Up to a global phase, which the evolution (1) of a state does not see, these are the Hadamard gate of the book and the rotation $e^{-i\theta Z/2}$.

The coupling, run for a time $t$ after a rotation $R_z(-\pi/2)$ of each qubit, gives $$\mathrm{CZ} = e^{-iJt Z_1 Z_2} \big( R_z(-\pi/2) \otimes 1 \big) \big( 1 \otimes R_z(-\pi/2) \big).$$(8)^✦ For $Jt = \pi/4$ this is the controlled $Z$ gate, up to the phase $e^{i\pi/4}$. Hadamards on the target turn it into a controlled NOT gate, with the control on qubit 1 or on qubit 2, $$\mathrm{CNOT}_{12} = (1 \otimes H)\, \mathrm{CZ}\, (1 \otimes H), \qquad \mathrm{CNOT}_{21} = (H \otimes 1)\, \mathrm{CZ}\, (H \otimes 1),$$(9) the first index naming the control. The circuit $W$ is built from three controlled NOT gates, the middle one with control and target reversed, $$W = \mathrm{CNOT}_{12}\, \mathrm{CNOT}_{21}\, \mathrm{CNOT}_{12},$$(10)^✦ a unitary of the pair for every $J$ and $t$, which enter it through the three controlled $Z$ gates.

The Claim

What Is Checked

The claim is stated over the circuit $W$ of (10) and the SWAP (2) of two qubits, each acting on a state of the pair by (1).

For a unitary $U$ and a state $\rho$, the operator $U \rho\, U^\dagger$ is again a state.