Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 58 / 4

For a positive operator $A$, the spectral theorem writes $A={\sum }_j{a}_j|j⟩⟨j|$ with $a_j\ge 0$. Hence its unique positive square root of an operator is $\sqrt{A}={\sum }_j\sqrt{a_j}|j⟩⟨j|$. Conjugating this spectral decomposition shows $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$ for a unitary operator $U$.

Let $\Delta ={\rho }_1-{\rho }_2$ and ${\rho }_i^{\prime }=U{\rho }_i{U}^{†}$. Then ${\Delta }^{\prime }=U\Delta U^{†}$ and $({\Delta }^{\prime }{)}^{†}{\Delta }^{\prime }=U{\Delta }^{†}\Delta U^{†}$. The square-root covariance and the cyclic property of the trace giveFor the quantum fidelity, covariance first gives $({\rho }_1^{\prime }{)}^{1/2}=U{\rho }_1^{1/2}{U}^{†}$. ThereforeApply the same square-root identity and the cyclic property of the trace to obtainThus unitary invariance of trace distance and fidelity givesThe same $U$ acts on both quantum states: these quantities depend on the relative distinguishability of the quantum states, not on a choice of basis.

Write ${\rho }_1=|\psi ⟩⟨\psi |$ and ${\rho }_2=|\phi ⟩⟨\phi |$. A pure-state density matrix is a rank-one orthogonal projector, so ${\rho }_1^{1/2}={\rho }_1$ andTaking its square root and trace yields the unsquared quantum fidelityUse unitary invariance of trace distance and fidelity to choose a basis of the span of the two quantum states in which $|\psi ⟩=|0⟩$ and $|\phi ⟩=c|0⟩+s|1⟩$, with $c=F\ge 0$ and $s=\sqrt{1-c^2}\ge 0$. Choosing state-vector phases to make the coefficients nonnegative does not change either density matrix. On that span,Its eigenvalues are $s$ and $-s$, with zero eigenvalues on the orthogonal complement. The positive square root of ${\Delta }^{†}\Delta$ therefore has two eigenvalues $s$, so the trace distance is $(s+s)/2=s$. Thus the pure-state trace distance and fidelity identity isIdentical pure quantum states give $F=1,D=0$; orthogonal pure quantum states give $F=0,D=1$. If the span is one-dimensional, the identical-state result follows directly, so no second basis vector is required. This formula uses the unsquared fidelity convention in the paper; if fidelity denotes squared overlap instead, the square inside the formula is omitted.

The Pauli X gate interchanges the two basis quantum states. Because the bit-flip outcome is a classical unobserved error rather than a coherent superposition, the quantum bit-flip channel givesFor $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its explicit density matrix isFor a pure reference quantum state, the square of the paper's quantum fidelity is $F^2=⟨\psi |{\rho }_2|\psi ⟩$. ThereforeWith $|\psi ⟩=\cos (\theta /2)|0⟩+e^{i\varphi }\sin (\theta /2)|1⟩$, the expectation is $⟨X⟩=\sin \theta \cos \varphi$, the $x$ component of the Bloch vector. Hence“All possible quantum states” is interpreted as the rotationally invariant uniform pure-qubit average, with measure $d\mu =\sin \theta d\theta d\varphi /(4\pi )$. Thusso the average pure-qubit fidelity of a bit-flip channel isEquivalently, rotational symmetry gives $⟨r_x^2⟩=⟨r_y^2⟩=⟨r_z^2⟩=1/3$, since a pure-state Bloch vector has unit length. The $X$ eigenstates are unchanged as density matrices and have fidelity one, while a computational basis quantum state has $F^2=1-p$; these provide checks on the angular expression. A uniform distribution in $\theta$ rather than in solid angle would produce the wrong ensemble average. The requested quantity is the mean squared fidelity, not the square of the mean unsquared fidelity.