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3

The fidelity case was already worked in the other answer. Here is an idea for the trace distance one. The trace distance between $\rho$ and some $|\psi\rangle\!\langle\psi|$ is $$\|\rho - |\psi\rangle\!\langle\psi|\|_1 = \operatorname{Tr}\lvert \,\rho - |\psi\rangle\!\langle\psi|\,\rvert, $$ which is equal to the sum of the singular values of $\rho-|\psi\...


5

Recall that for any Hermitian operator $A$ and any unit vector $|\psi\rangle$ the real number $\langle \psi|A|\psi\rangle$, known as the Rayleigh quotient, is bounded by the largest eigenvalue $\lambda_{max}$ of $A$ $$ \langle \psi|A|\psi\rangle \le \lambda_{max}. $$ Moreover, the maximum is achieved when $|\psi\rangle$ is the unit norm eigenvector of $A$ ...


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