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I came across a problem in which the trace distance is maximum if the Bloch vectors of the two density matrices are perpendicular to each other. What is the physical interpretation of this?

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    $\begingroup$ If the trace distance (which is a metric on the space of density matrices) tells us the degree of distinguishability between two states, it means that if the states are orthogonal, this distance is maximized. So, if this distance is a minimum, the states are completely indistinguishable and if it's a maximum, the states are perfectly distinguishable, and this happens when the states are orthogonal. $\endgroup$
    – joseph h
    Oct 31, 2022 at 8:00
  • $\begingroup$ also related: quantumcomputing.stackexchange.com/q/8709/55. For qubits, the trace distance is equivalent to the Euclidean distance in the Bloch sphere, though this stops being the case in higher dimensions $\endgroup$
    – glS
    Nov 2, 2022 at 9:04

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The trace distance, which is a metric on the space of density matrices i.e., $$D(\rho_1,\rho_2)=\frac{1}{2}\mathrm{Tr}||\rho_1-\rho_2||$$ for two density operators $(\rho_1,\rho_2)$, tells$^1$ us the degree of distinguishability between two states, and if the states are orthogonal ("perpendicular"), this distance is maximized.

So, if this distance is a minimum, the states are completely indistinguishable and if it's a maximum, the states are perfectly distinguishable (in principle), and as stated, this happens when the states are orthogonal.

$^1$ To get a more intuitive feel for this, you can compare the Trace distance to its classical analogue, called the Kolmogorov distance applicable to classical probability distributions, and gives a measure of the similarity between two classical probability distributions.

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