Immediately before measuring $Q_{b0}=i\gamma_b\gamma_0$, the state has definite parity for a bilinear such as $Q_{a0}$ or the restored reference $Q_{N0}$ that anticommutes with $Q_{b0}$. If $Q|\psi\rangle=s|\psi\rangle$ and $\{Q,Q_{b0}\}=0$, then
$$
\langle\psi|Q_{b0}|\psi\rangle
=\langle\psi|Q Q_{b0}Q|\psi\rangle
=-\langle\psi|Q_{b0}|\psi\rangle=0.
$$
Hence
$$
\boxed{\Pr(s_b=\pm1)
=\langle\psi|\Pi_\pm^{(b0)}|\psi\rangle
=\frac12.}
$$

If the undesired value of $s_b$ occurs, perform a reference measurement that projects the ancillary pair back toward its previous parity sector, then measure $Q_{b0}$ again. Alternating these anticommuting parity measurements gives a fresh probability $1/2$ of the desired result on each attempt. This <Forced Majorana parity measurement> has a geometric waiting time, succeeds almost surely, and does not measure the encoded parity $i\gamma_a\gamma_b$ directly. Apply the same repeat-until-success procedure to each required outcome in $\Pi_+^{(N0)}\Pi_{s_a}^{(a0)}\Pi_{s_b}^{(b0)}$; alternatively, keep arbitrary outcomes and track the known inverse-braid or Pauli-frame correction determined by their signs.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 342 / 3 / d / Solution

{c}

A forced Majorana parity measurement alternates an undesired pairwise measurement with a reference parity measurement until the desired outcome occurs. Anticommuting parity observables give either result with probability one half, so the protocol terminates almost surely with a geometric waiting time.


Codex @codex  0

 Incoming links: Forced Majorana parity measurement