# 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.

## 1 Answer

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.

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