Let $\rho$ and $\sigma$ be two density operators such that probability of obtaining $a$ is $tr(\rho E_a)$ if the state before measurement was $\rho$ and $tr(\sigma E_a)$ if the state before measurement was $\sigma$, where $E_a$'s are 1-dim orthogonal projectors with $\sum_a E_a=I$. In that case we have \begin{equation} d(p,q)=\sum_a|p_a-q_a|\leq||ρ-σ||_1=d(ρ,σ) \end{equation} How should I proceed to find the optimal set of projectors $E_a$'s such that the inequality in the above equation becomes an equality?


1 Answer 1


You're essentially asking how to prove one of the characterisations of the trace distance: given any pair of states $\rho,\sigma$, you have $$\|\rho-\sigma\|_1 = \max_{P}\operatorname{tr}[P(\rho-\sigma)],$$ with the maximisation performed with respect to all Hermitian operators $P$ such that $0\le P\le I$. Equivalently, the optimisation is performed with respect to all projections $P$.

You can find the complete proof in the Wikipedia article. For what concerns finding the maximum, you simply need to use as $P$ the projection onto the subspace spanning the positive eigenvalues of $\rho-\sigma$. With this choice, call it $P^*$, you have $$\operatorname{tr}[P^*(\rho-\sigma)] = \sum_k \lambda_k^+$$ with $\lambda_k^+$ the positive eigenvalues of $\rho-\sigma$. Because $\operatorname{tr}(\rho-\sigma)=0$, the sum of all eigenvalues vanishes, hence $\sum_k\lambda_k^+=-\sum_k\lambda_k^-$ with $\lambda_k^-<0$ the negative eigenvalues, and therefore the trace distance, which equals in general (for Hermitian operators) the sum of the absolute values of the eigenvalues, reads $$\|\rho-\sigma\|_1 = \sum_k |\lambda_k|= \sum_k \lambda_k^+ - \sum_k \lambda_k^- = 2 \sum_k \lambda_k^+ = 2\operatorname{tr}[P^*(\rho-\sigma)].$$

The answer to your question follows immediately from this, because the total variation distance, which is (modulo a factor of 2) the LHS in your equation, equals half of $\operatorname{tr}[P^*(\rho-\sigma)]$, again because $\sum_a (p_a-q_a)=0$.

I also just realised that this is probably already answered in What is the relationship between trace distance and total variation distance?.


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