The Exercise

Grover’s algorithm on two qubits searches the four basis states $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$ for the one that an oracle marks, and a single iteration finds it. The iteration applies the oracle $O$, which changes the sign of the marked state, then Hadamard gates on both qubits, then the conditional phase shift $P$, which changes the sign of every basis state except $|00\rangle$, and Hadamard gates on both qubits once more.^✦

In a nuclear magnetic resonance experiment the qubits are two coupled nuclear spins. Every gate becomes a sequence of radio frequency pulses, each a rotation of one spin about the $x$ axis or the $y$ axis, and of periods in which the spins evolve freely under their coupling. The exercise is to show that, for each of the four marked states, the iteration shortens to two periods of free evolution, each followed on every spin by a pulse about the $y$ axis and then a pulse about the $x$ axis.

Pulses on Two Spins

The library defines each gate of the iteration as a pulse sequence and states the shortening as four identities between operators on the two spins, one for each marked state.

For a real vector $\vec n = (n_x, n_y, n_z)$ in three dimensions and a real angle $\theta$, the rotation $R_{\vec n}(\theta)$ about $\vec n$ is the operator on one qubit $$R_{\vec n}(\theta) = \exp\Bigl(-\frac{i\theta}{2}\, \vec n \cdot \vec\sigma\Bigr), \qquad \vec n \cdot \vec\sigma = n_x X + n_y Y + n_z Z,$$(1) where $X$, $Y$ and $Z$ are the Pauli matrices. The pulses are the rotations about $\hat x = (1, 0, 0)$ and $\hat y = (0, 1, 0)$, written $R_x(\theta) = e^{-i\theta X/2}$ and $R_y(\theta) = e^{-i\theta Y/2}$.

In a product of operators the rightmost factor acts first. For a real angle $s$, the sandwich $V(s)$ is the operator on one spin that places a rotation by $s$ about $\hat x$ between two rotations about $\hat y$, $$V(s) = R_y(\pi/2)\, R_x(s)\, R_y(-\pi/2).$$(2)^✦ On one spin the Hadamard pulse is the operator $$H = R_x(\pi)\, R_y(-\pi/2),$$(3) and on the two spins the Hadamard pair is $$H^{\otimes 2} = H \otimes H,$$(4) the first factor of a tensor product acting on spin 1 and the second on spin 2.

Fix an operator $\tau$ on the two spins, in the experiment the free evolution over one period. For real angles $s_1$ and $s_2$ the oracle pulse $O(s_1, s_2)$ and the conditional phase shift $P$ are the operators on the two spins $$O(s_1, s_2) = \bigl(V(s_1) \otimes V(s_2)\bigr)\, \tau,$$(5) $$P = \bigl(V(\pi/2) \otimes V(\pi/2)\bigr)\, \tau.$$(6) With the oracle pulse $O(s_1, s_2)$, the Grover iteration in pulse form $G(s_1, s_2)$ is the operator on the two spins $$G(s_1, s_2) = H^{\otimes 2}\, P\, H^{\otimes 2}\, O(s_1, s_2),$$(7)

For real angles $a_1$ and $a_2$ the shortened sequence $\tilde G(a_1, a_2)$ is the operator on the two spins $$\begin{aligned} \tilde G(a_1, a_2) &= \Bigl[\bigl(R_x(-\pi/2)\, R_y(-\pi/2)\bigr) \otimes \bigl(R_x(-\pi/2)\, R_y(-\pi/2)\bigr)\Bigr]\, \tau \\ &\quad \Bigl[\bigl(R_x(a_1)\, R_y(-\pi/2)\bigr) \otimes \bigl(R_x(a_2)\, R_y(-\pi/2)\bigr)\Bigr]\, \tau. \end{aligned}$$(8)^✦

The Claim

The oracle that marks $|x_1 x_2\rangle$ takes $s_i = \pi/2$ where $x_i = 0$ and $s_i = -\pi/2$ where $x_i = 1$.

What Is Checked

The claim is stated over the Grover iteration in pulse form (7), assembled from the pulses (2) to (6), and over the shortened sequence (8).