The Exercise

Two spins one half, each a qubit, are coupled by the Heisenberg exchange interaction. The exercise is to show that letting the coupling act for a time $\pi$ exchanges the states of the two qubits, up to a global phase, and that letting it act for half that time gives a square root of the exchange. It also asks for a controlled NOT gate (CNOT) built from two such square roots, with single qubit gates around and between them.

Two Coupled Spins

Each spin has the computational basis $|0\rangle, |1\rangle$. The Pauli matrices $\sigma_1, \sigma_2, \sigma_3$ are the complex $2 \times 2$ matrices $$\sigma_1 = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad \sigma_2 = Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad \sigma_3 = Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$(1) in this basis, the components of the Pauli vector $\vec{\sigma}$.

The pair has the state space $\mathbb{C}^2 \otimes \mathbb{C}^2$, with product basis $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, the first label for spin 1 and the second for spin 2. For spin $k = 1, 2$, the Pauli vector $\vec{\sigma}_k$ has the matrices $\sigma_a$ of (1) acting on spin $k$ as its components. The isotropic coupling $\vec{\sigma}_1 \cdot \vec{\sigma}_2$ of the two spins is the operator $$\vec{\sigma}_1 \cdot \vec{\sigma}_2 = \sum_{a=1}^{3} \sigma_a \otimes \sigma_a$$(2)^✦ on this space, the first factor of each term acting on spin 1 and the second on spin 2.

The Heisenberg coupling $\vec{S}_1 \cdot \vec{S}_2$ of the two spins is the scalar product of their spin vectors $\vec{S}_k = \tfrac{1}{2} \vec{\sigma}_k$, in units with $\hbar = 1$: $$\vec{S}_1 \cdot \vec{S}_2 = \tfrac{1}{4} \, \vec{\sigma}_1 \cdot \vec{\sigma}_2 .$$(3)^✦

The exchange evolution $U(\varphi)$ of the pair for a time $\varphi$ is the operator that the Heisenberg coupling generates as a Hamiltonian: $$U(\varphi) = \exp\bigl( -i \varphi \, \vec{S}_1 \cdot \vec{S}_2 \bigr) .$$(4)^✦

The Gates

The swap gate $\mathrm{SWAP}$ is the operator on the pair that exchanges the states of the two qubits; in the product basis, in the order above, it is the permutation matrix $$\mathrm{SWAP} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} .$$(5)^✦ The square root of swap gate $\sqrt{\mathrm{SWAP}}$ is the exchange evolution for a time $\pi/2$: $$\sqrt{\mathrm{SWAP}} = U(\pi/2) .$$(6)^✦

The Hadamard gate $H$ and the phase gate $S$ of a qubit are the matrices $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$$(7) in the computational basis, and $S^\dagger$ is the adjoint of $S$.

The controlled NOT gate $\mathrm{CNOT}$ of the pair, with spin 1 as control and spin 2 as target, flips the target when the control is in $|1\rangle$: $$\mathrm{CNOT} \, |a\rangle |b\rangle = |a\rangle |a \oplus b\rangle , \qquad a, b \in \{0, 1\},$$(8)^✦ with $\oplus$ addition modulo 2.

The circuit $V$ of the exercise is the operator on the pair $$V = (I \otimes H) \, (S \otimes S^\dagger) \, \sqrt{\mathrm{SWAP}} \, (Z \otimes I) \, \sqrt{\mathrm{SWAP}} \, (I \otimes H)$$(9)^✦ with $I$ the identity on a single qubit; the gate on the right acts first.

The Claim

What Is Checked

The claim is stated over the exchange evolution $U(\varphi)$ of (4), generated by the Heisenberg coupling (3) built from the Pauli matrices (1), and over the swap and controlled NOT gates and the gates of the circuit (9), each a fixed matrix in the product basis.