The Exercise

Three nuclear spins with equal Zeeman energies are in thermal equilibrium at high temperature, as in a nuclear magnetic resonance experiment on a liquid. Their density matrix is close to a multiple of the identity, and what departs from it is diagonal: each spin in $|0\rangle$ raises the population of a level and each spin in $|1\rangle$ lowers it by the same amount. A pure state would have one level standing above seven equal ones; here the populations take four values.

A circuit of three gates, each exchanging two basis states, rearranges the populations. The exercise is to show that afterwards the first spin labels two subspaces, and in each of them the second and third spins carry, up to sign, the part of the density matrix of the pure state $|00\rangle$ that departs from a multiple of the identity. This is logical labeling.

The Deviation Matrix

The basis states of the three spins are ordered $|000\rangle, |001\rangle, \dots, |111\rangle$, as the binary numbers $0$ to $7$. The population pattern $D$ of the three spins is the diagonal matrix $$D = \operatorname{diag}(6,\, 2,\, 2,\, -2,\, 2,\, -2,\, -2,\, -6),$$(1)^✦ whose entry at the basis state $|b_1 b_2 b_3\rangle$, with each $b_i$ equal to $0$ or $1$, is $2(n_0 - n_1)$, where $n_0$ and $n_1$ are the numbers of spins in $|0\rangle$ and in $|1\rangle$. For complex numbers $\delta$ and $\alpha$, the deviation matrix $\rho$ of the three spins is the operator on their eight dimensional state space $$\rho = \delta I + \alpha D,$$(2)^✦ with $I$ the identity.

The relabeled pattern $D'$ is the diagonal matrix $$D' = \operatorname{diag}(6,\, -2,\, -2,\, -2,\, -6,\, 2,\, 2,\, 2)$$(3) in the same basis, and the relabeled deviation matrix $\rho'$, for the same $\delta$ and $\alpha$, is $$\rho' = \delta I + \alpha D'.$$(4)

The Circuit

Each gate of the circuit exchanges two basis states and fixes the other six. The doubly controlled NOT $T_{01}$ flips the first spin when the second and third are in $|01\rangle$: $$T_{01} = |101\rangle\langle 001| + |001\rangle\langle 101| + \sum_{x \neq 001,\, 101} |x\rangle\langle x|,$$(5) The doubly controlled NOT $T_{10}$ on the three spins flips the first spin when the second and third are in $|10\rangle$: $$T_{10} = |110\rangle\langle 010| + |010\rangle\langle 110| + \sum_{x \neq 010,\, 110} |x\rangle\langle x|.$$(6) The controlled exchange $F$ acts when the first spin is in $|1\rangle$, and then exchanges $|00\rangle$ and $|11\rangle$ of the other two: $$F = |111\rangle\langle 100| + |100\rangle\langle 111| + \sum_{x \neq 100,\, 111} |x\rangle\langle x|.$$(7)^✦ The logical labeling circuit $P$ applies $T_{01}$, then $T_{10}$, then $F$: $$P = F\, T_{10}\, T_{01}.$$(8)^✦

The Claim

What Is Checked

The claim is stated over the deviation matrices $\rho$ and $\rho'$ of (2) and (4), and over the circuit $P$ of (8). The library proves that $P$ is unitary, $$P^\dagger P = P P^\dagger = I,$$(10)^✦ so that (9) is the evolution of the three spins under the circuit.