Does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable?

Question

I believe the answer to this question is well-known; but, unfortunately, I don't know.

In quantum computing, we know that mixed states are represented by density matrices. And the trace norm of the difference of two density matrices characterizes the distinguishability of the two corresponding mixed states. Here, the definition of trace norm is the sum of all eigenvalues of the density matrix, with an extra multiplicative factor 1/2 (in accordance with statistical difference of two distributions). It is well-known that, when the the difference of two density matrices is one, then the corresponding two mixed states are totally distinguishable, while when the difference is zero, the two mixed states are totally indistinguishable.

My question is, does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable? If this is the case, then taking the optimal measurement to distinguish these two mixed states will behave like to distinguish two distributions over the same domain with disjoint support.

Answer 1

Here is one way to prove the fact you are interested in.

Suppose $\rho_0$ and $\rho_1$ are density matrices. Like every other Hermitian matrix, it is possible to express the difference $\rho_0-\rho_1$ as $$\rho_0-\rho_1 = P_0-P_1$$ for $P_0$ and $P_1$ being positive semidefinite and having orthogonal images. (Sometimes this is called a Jordan-Hahn decomposition; it is unique and easily obtained from a spectral decomposition of $\rho_0-\rho_1$.) Note that the fact that $P_0$ and $P_1$ have orthogonal images implies that they are simultaneously diagonalizable, which I interpret is the property you are interested in.

The trace norm of the difference $\rho_0-\rho_1$ (as you define it, with the multiplicative factor 1/2), is given by $$\|\rho_0-\rho_1\|_{\text{tr}} = \frac{1}{2}\operatorname{Tr}(P_0) + \frac{1}{2}\operatorname{Tr}(P_1).$$ Under the assumption that this quantity is 1, we will conclude that $P_0=\rho_0$ and $P_1=\rho_1$, which proves what you want to prove.

To draw this conclusion, note first that $\operatorname{Tr}(P_0)-\operatorname{Tr}(P_1)=0$ and $\operatorname{Tr}(P_0)+\operatorname{Tr}(P_1)=2$, so $\operatorname{Tr}(P_0)=\operatorname{Tr}(P_1)=1$. Next, take $\Pi_0$ and $\Pi_1$ to be the orthogonal projections onto the images of $P_0$ and $P_1$, respectively. We have $$\Pi_0 (\rho_0 - \rho_1) = \Pi_0 (P_0 - P_1) = P_0$$ so $$\operatorname{Tr}(\Pi_0 \rho_0) - \operatorname{Tr}(\Pi_0 \rho_1) = 1.$$ Both $\operatorname{Tr}(\Pi_0 \rho_0)$ and $\operatorname{Tr}(\Pi_0 \rho_1)$ must be contained in the interval [0,1], from which we conclude that $\operatorname{Tr}(\Pi_0\rho_0)=1$ and $\operatorname{Tr}(\Pi_0\rho_1) = 0$. From these equations it is not difficult to conclude $\Pi_0\rho_0=\rho_0$ and $\Pi_0\rho_1=0$, and therefore $P_0=\rho_0$ by the equation above. A similar argument shows $P_1=\rho_1$.

Answer 2

Yes. If the trace distance of two density matrices is equal to 1, then they have orthogonal supports, and therefore they are simultaneously diagonalizable.