C067 derivation

Model

From C066, one application of drive (A) with the differential Stark error imprints, in the $\{\ket{\uparrow\downarrow},\ket{\downarrow\uparrow}\}$ basis, the unitary $$E(\phi) = \mathrm{diag}\big(e^{-i\phi/2},\,e^{+i\phi/2}\big), \qquad \phi = 2\epsilon\Delta t,$$ which couples $\ket{\Psi^-}\!\leftrightarrow\!\ket{\Psi^+}$ via $E(\phi)\ket{\Psi^-}=\cos\tfrac{\phi}{2}\ket{\Psi^-}-i\sin\tfrac{\phi}{2}\ket{\Psi^+}$.

A drive-(C) $\pi$-pulse is a collective spin flip $X\equiv\sigma_x^{(1)}\otimes\sigma_x^{(2)}$ on the ground qubits. On the computational pair it swaps $$X\ket{\uparrow\downarrow}=\ket{\downarrow\uparrow},\qquad X\ket{\downarrow\uparrow}=\ket{\uparrow\downarrow},$$ i.e. in this 2D subspace $X = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ (Pauli-$x$). It also maps $\ket{\uparrow\uparrow}\leftrightarrow\ket{\downarrow\downarrow}$, consistent with the paper's statement ("if the population is in $\ket{\uparrow\uparrow}$ it gets transferred to $\ket{\downarrow\downarrow}$").

Echo identity

The key algebraic fact: $E(\phi)$ is generated by a diagonal (Pauli-$z$ in this 2D basis) term, and $X$ anticommutes with $\sigma_z$, so $$X\,E(\phi)\,X^{-1} = X\,\mathrm{diag}(e^{-i\phi/2},e^{+i\phi/2})\,X = \mathrm{diag}(e^{+i\phi/2},e^{-i\phi/2}) = E(-\phi).$$

One odd/even cycle pair

• Odd cycle: drive (A) imprints $E(\phi)$, then the odd drive (C) is a $\pi$-pulse $X$. State $\to X\,E(\phi)$.
• Even cycle: drive (A) again imprints $E(\phi)$. Net (ignoring the even-cycle readout $\pi/2$, which acts only after restoration): $$U_{\text{pair}} = E(\phi)\,X\,E(\phi).$$ Using the echo identity $E(\phi)X = X E(-\phi)$, $$U_{\text{pair}} = E(\phi)\,X\,E(\phi) = X\,E(-\phi)\,E(\phi) = X\,E(0) = X.$$ So over the odd/even pair the accumulated differential phase cancels exactly: the residual operation is just the collective flip $X$, which carries the $\sigma_z$-error-free state. Acting on $\ket{\Psi^-}$: $$X\ket{\Psi^-} = X\tfrac{1}{\sqrt2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}) = \tfrac{1}{\sqrt2}(\ket{\downarrow\uparrow}-\ket{\uparrow\downarrow}) = -\ket{\Psi^-}.$$ Thus the population returns to $\ket{\Psi^-}$ (up to an irrelevant global sign) at the end of each even cycle, with zero leakage to $\ket{\Psi^+}$ regardless of $\phi$ -- the spin-echo cancellation the paper claims.

(The even-cycle Zeeman $\pi/2$-pulse that moves any residual $\ket{\Psi^+}\to\ket{\downarrow\downarrow}$ for repumping is a separate clean-up step; the analytic content verified here is the exact phase cancellation, which is what makes that residual zero in the ideal case.)

Conclusion

The echo identity $XE(\phi)X=E(-\phi)$ gives exact cancellation of the differential phase over an odd/even cycle pair, returning the system to $\ket{\Psi^-}$ independent of $\phi$. Confirmed analytically and numerically (run.py). Verdict: verified.