In the <computational basis>,
$$
|\psi\rangle\langle\psi|
=\begin{pmatrix}
|\alpha|^2&\alpha\beta^*\\
\alpha^*\beta&|\beta|^2
\end{pmatrix}.
$$
At $p=1/2$, the two opposite off-diagonal contributions cancel, giving
$$
\boxed{\widetilde\rho
=\Lambda_{1/2}(\rho)
=\begin{pmatrix}|\alpha|^2&0\\0&|\beta|^2\end{pmatrix}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 3 / a / ii / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/3/a/ii/2/solution