Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 66 / 3

The positive part of a Hermitian operator and negative part of a Hermitian operator are the positive operatorsThus $X=X_{+}-X_{-}$ and $X_{+}{X}_{-}=0$. In this convention the negative part itself is nonnegative. This is the positive-negative decomposition of the Hermitian operator.

The operator absolute value is defined by the unique positive operator square rootThe scalar identity $|\lambda |=\max (\lambda ,0)+\max (-\lambda ,0)$ applied in the spectral decomposition proves

For any linear operator, the trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$, the sum of its singular values. For a Hermitian operator,The Holevo–Helstrom theorem concerns minimum-error discrimination of two density operators $\rho ,\sigma$. If their prior probabilities are $p$ and $1-p$, the maximum success probability over all quantum measurements isAn optimal binary POVM declares $\rho$ on the positive spectral subspace of $p\rho -(1-p)\sigma$ and declares $\sigma$ on its negative spectral subspace; zero-eigenvalue vectors can be assigned either way. For equal priors the formula becomes $P_{\mathrm{succ}}^{\ast }=\frac{1}{2}+\frac{1}{4}\Vert \rho -\sigma {\Vert }_1$.

The constraints $-I\le T\le I$ are in the order of Hermitian operators. They imply $-1\le ⟨{\alpha }_i|T|{\alpha }_i⟩\le 1$. Using the spectral decomposition of $X$,Choose $T=\mathrm{sgn}(X)$, with eigenvalues $+1$ on the positive spectral subspace, $-1$ on the negative spectral subspace, and zero on the kernel. It satisfies the constraints and attains equality. The trace-norm variational principle for Hermitian operators is therefore

To prove the Holevo–Helstrom theorem, write a binary POVM as $\{E,I-E\}$, where $0\le E\le I$, and associate the first outcome with $\rho$. Any larger outcome set followed by a binary decision can be grouped into this form. Put $\Delta =p\rho -(1-p)\sigma$. Its success probability isThe substitution $T=2E-I$ bijects the allowed effects with the interval $-I\le T\le I$. Since $\mathrm{Tr}\Delta =2p-1$,Choosing $E$ as the positive spectral projection of $\Delta$ attains this value, with an arbitrary decision on its kernel. This proves the optimal success probability and constructs an optimal measurement.

For two density operators, the trace distance and the unsquared quantum fidelity areThe unsquared convention matters: if $P=|\psi ⟩⟨\psi |$ is a pure state, then $\sqrt{\rho }P\sqrt{\rho }$ is a rank-one positive operator with its sole nonzero eigenvalue equal to $⟨\psi |\rho |\psi ⟩$. Hence

Write $x_i=⟨{\psi }_i|X|{\psi }_i⟩$, which is real because $X$ is a Hermitian operator. Setwith $\mathrm{sgn}(0)=0$. This Hermitian operator satisfies $-I\le T\le I$. The trace-norm variational principle for Hermitian operators gives the diagonal absolute-sum bound for the trace norm:

For $X=\rho -P$, with $P=|\psi ⟩⟨\psi |$, extend $|\psi ⟩$ to an orthonormal basis $|{\psi }_1⟩=|\psi ⟩,|{\psi }_2⟩,\dots$. Put $r=⟨\psi |\rho |\psi ⟩$. The first diagonal entry is $r-1\le 0$, and all the others are nonnegative. Their sum is $1-r$, because $\mathrm{Tr}\rho =1$. The bound therefore yields $\Vert \rho -P{\Vert }_1\ge 2(1-r)$. Using the definitions of trace distance and quantum fidelity,This pure-target lower bound on trace distance is attained whenever $\rho$ has no coherence between $|\psi ⟩$ and its orthogonal complement. The proof used the trace-norm variational principle for Hermitian operators, together with positivity and normalization of a density operator.