I have these two classical-quantum states:

$$\rho = \sum_{a} \lvert a\rangle \langle a\lvert \otimes q^a \\ \mu = \sum_{a} \lvert a\rangle \langle a\lvert \otimes r^a $$

Where $a$ are the classical basis vectors, $q^a, r^a$ are arbitrary matrices dependent on $a$.

Now, we can take the trace distance of these two classical-quantum states, which would be:

$$T(\rho, \mu) = \frac{1}{2} ||\rho - \mu||_1 \\ = \frac{1}{2} || \sum_a \lvert a\rangle \langle a\lvert \otimes(q^a - r^a)||_1$$

Now, my question is, can we rewrite the above expression in the following way?

$$T(\rho, \mu) = \frac{1}{2} \sum_a ||q^a - r^a||_1$$ I.e. just pulling the summation out of the norm.


1 Answer 1


Yes, since the trace norm is the sum of the absolute value of the singular values, and the singular values can be found for each of the $a$ blocks independently.

  • $\begingroup$ is it like saying absolute(eig(A+B)) = abs(eig(A)) + abs(eig(B)) ? $\endgroup$ Commented May 18, 2019 at 1:01
  • 1
    $\begingroup$ @HasanIqbal No, this is wrong, just take B=-A. It is like saying $\mathrm{abs}(\mathrm{eig}(A\oplus B)) = \mathrm{abs}(\mathrm{eig}(A))+\mathrm{abs}(\mathrm{eig}(B))$. $\endgroup$ Commented May 18, 2019 at 12:23
  • $\begingroup$ this is perfect... :) thanks Norbert $\endgroup$ Commented May 18, 2019 at 16:36

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