# Trace distance of two classical-quantum states

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.

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.
• @HasanIqbal No, this is wrong, just take B=-A. It is like saying $\mathrm{abs}(\mathrm{eig}(A\oplus B)) = \mathrm{abs}(\mathrm{eig}(A))+\mathrm{abs}(\mathrm{eig}(B))$. – Norbert Schuch May 18 '19 at 12:23