Since $P|\psi\rangle=\lambda_P|\psi\rangle$, $P^2=I$, and $P$ <anticommutator>[anticommutes] with $Q$,
$$
\langle\psi|Q|\psi\rangle
=\langle\psi|PQP|\psi\rangle
=-\langle\psi|Q|\psi\rangle,
$$
so the <expectation value> of $Q$ is zero. The two <spectral projectors> of the Hermitian <Pauli operator> $Q$ are $(I\mathbin\pm Q)/2$, and hence the <Born rule> gives
$$
\Pr(\lambda_Q=\pm1)
=\left\langle\psi\middle|\frac{I\pm Q}{2}\middle|\psi\right\rangle
=\frac12.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 324 / 1 / b / i / Solution

{title2=$\{A,B\}$}
{wiki}

The anticommutator of two operators is $\{A,B\}=AB+BA$. They anticommute when this operator vanishes, equivalently when $AB=-BA$.


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