Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 66 / 5

For $t<1$, normalize the remaining probabilities by $q_j=P_X(j)/(1-t)$, $j=2,\dots ,k$. Directly splitting the Shannon entropy sum givesHere $h$ is the binary entropy. The Shannon entropy of a distribution on $k-1$ points is at most ${\log }_2(k-1)$. For example, nonnegativity of its Kullback-Leibler divergence from the uniform distribution gives ${\log }_2(k-1)-H(q)\ge 0$. Thus the entropy bound with one prescribed probability isFor $t<1$, equality holds precisely when the remaining probabilities are all $(1-t)/(k-1)$. For $t=1$, the distribution is deterministic and both sides are zero. The displayed formula is for $k\ge 2$; a one-point alphabet simply has zero Shannon entropy.

Put $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩$. The rank-one dephasing is ${\rho }^{\prime }={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$. If $p_i=0$, positivity givesso $\rho |{\psi }_i⟩=0$. The kernel of ${\rho }^{\prime }$ is exactly the span of these zero-probability basis vectors and is therefore contained in the kernel of $\rho$. Taking orthogonal complements proves support inclusion under rank-one dephasing:Here the support of a positive operator is the orthogonal complement of its kernel.

On that support, ${\log }_2{\rho }^{\prime }$ is diagonal in the measurement basis, givingConsequently the relative-entropy identity for rank-one dephasing isKlein's inequality gives $D_{\mathrm{rel}}(\rho \Vert {\rho }^{\prime })\ge 0$, since both density operators have trace one. For singular $\rho$, first restrict to $\mathrm{supp}{\rho }^{\prime }$ where ${\rho }^{\prime }$ is positive definite, replace $\rho$ by $(\rho +\eta I)/(1+\eta \dim \mathrm{supp}{\rho }^{\prime })$, and let $\eta \downarrow 0$. The support inclusion ensures that the limit is finite. ThusThis proves entropy increase under nonselective projective measurement using Klein's inequality. Equality holds exactly when $\rho ={\rho }^{\prime }$, so the original density operator was already diagonal in the chosen basis.

Choose an orthonormal basis of the joint $d^2$-dimensional Hilbert space whose first vector is the given purification of a density operator $|\rho {⟩}_{QR}$. Apply rank-one dephasing to ${\sigma }_{QR}$ in this basis, and denote the resulting diagonal probabilities by $p_1,\dots ,p_{d^2}$. Their first entry isThe preceding entropy increase under nonselective projective measurement and entropy bound with one prescribed probability yieldTherefore the quantum Fano inequality isThe quantity $f^2$ is the entanglement fidelity of $\mathcal{N}$ on ${\rho }_Q$; it is already a squared overlap, so it is $f^2$, rather than $f$, that enters the binary entropy. The argument is an instance of the entropy bound from overlap with a pure state in dimension $d^2$. At $f^2=1$ the output is the original pure state and its Von Neumann entropy is zero. For $d=1$ the system is trivial and the same zero-entropy conclusion holds without evaluating ${\log }_2(d^2-1)$.