For $a\ne b$, the operator $Q_{ab}=i\gamma_a\gamma_b$ is Hermitian and
$$
Q_{ab}^2=-\gamma_a\gamma_b\gamma_a\gamma_b=1.
$$
Therefore
$$
(\Pi_\pm^{(ab)})^2
=\frac14(1\pm2Q_{ab}+Q_{ab}^2)
=\Pi_\pm^{(ab)},
\qquad
\Pi_+^{(ab)}\Pi_-^{(ab)}=0.
$$
For $f=(\gamma_a+i\gamma_b)/2$, one has $Q_{ab}=2f^\dagger f-1$, so these projectors distinguish the two occupations, equivalently the two values of pair <fermion parity>. This proves that
$$
\boxed{\Pi_\pm^{(ab)}=\frac12(1\pm i\gamma_a\gamma_b)}
$$
describe a fermion-parity measurement.

Let $P=\Pi_+^{(N0)}$ and suppose $P|\psi\rangle=|\psi\rangle$. The bilinears $i\gamma_a\gamma_0$ and $i\gamma_b\gamma_0$ each anticommute with $i\gamma_N\gamma_0$, whereas their product commutes with it. Expanding the projectors and sandwiching by $P$ therefore gives
$$
\begin{aligned}
P\Pi_{s_a}^{(a0)}\Pi_{s_b}^{(b0)}P
&=\frac14P(1+i s_a\gamma_a\gamma_0)
(1+i s_b\gamma_b\gamma_0)P\\
&=\frac14(1+s_as_b\gamma_a\gamma_b)P\\
&=\frac1{2\sqrt2}
R_{ab}^{s_as_b}P.
\end{aligned}
$$
After normalizing the post-measurement state, the sequence consequently implements
$$
\boxed{R_{ab}^{s_as_b}}
$$
on the encoded ground space. This is <Measurement-only Majorana braiding>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-342/3/c/solution