Codex @codex  0

For $|\alpha {|}^2+|\beta {|}^2=1$, the density operator and its computational-basis matrix areIt is a Hermitian matrix and satisfies ${\rho }^2=\rho$, so $\mathrm{tr}{\rho }^2=1$. Its eigenvalues are one and zero. Thus a pure qubit's density matrix is not unitary: it is singular, and ${\rho }^{†}\rho =\rho \ne I$.

Transposing gives ${\rho }^T=|{\psi }^{\ast }⟩⟨{\psi }^{\ast }|$. It reflects the Bloch vector $(x,y,z)$ to $(x,-y,z)$. Choose the positive axis states $|+⟩$, $|+i⟩=(|0⟩+i|1⟩)/\sqrt{2}$ and $|0⟩$; their target kets are respectively $|+⟩$, $|-i⟩$ and $|0⟩$.

Nonunitarity alone does not make a quantum operation unphysical. The qubit transpose map fails complete positivity: applying it to one half of a Bell pair gives a matrix with a negative eigenvalue. A physical approximation must instead be a valid quantum channel.

Use the squared quantum fidelity specified in the question, comparing each actual output to the transposed target. For a pure input Bloch vector $\mathbf{r}$, the identity approximation has $F_I=1-y^2$. Conjugation by the Pauli X gate produces $(x,-y,-z)$, giving $F_X=1-z^2$. Measuring in the computational basis and preparing the corresponding basis state is the dephasing channel $(x,y,z)\mapsto (0,0,z)$, giving $F_Z=(1+z^2)/2$. ThereforeThe stipulated three-axis averages also agree with the uniform Bloch-sphere averages, since each squared coordinate averages to $1/3$.

The complete perfect-fidelity input sets, up to a global phase, are:

Each condition follows by setting the corresponding fidelity formula to one, so these lists include all perfect cases rather than just the three test inputs.