# Helstrom Measurement when two quantum states are close

I've been reading a paper about Entangled-quantum GAN (see this PDF) and wondering why descriptions below Eq.(3) in the paper are in fact true.

To summarize the description, suppose we have two quantum states $$\sigma$$ and $$\rho$$ and a Helstrom measurement (optimal measurement with the minimum error to distinguish two states) $$T$$. (Note that $$T$$ is a projector to a positive eigenspace of $$\sigma - \rho$$). Then, the description in paper says: "when $$\sigma$$ and $$\rho$$ are close, $$T$$ is close to orthogonal to $$\sigma$$ and opposite to $$\rho$$". Why is this indeed true?

Heuristically, when $$\sigma$$ and $$\rho$$ are "close," we can write $$\sigma=(1-\epsilon)\rho+\epsilon\varrho$$ for some small positive number $$\epsilon$$ and some other normalized state $$\varrho$$. Then $$\sigma-\rho=\epsilon(\varrho-\rho),$$ so $$T$$ is the projector onto the positive eigenspace of $$\varrho-\rho$$.
If we take the extreme case that $$\rho$$ and $$\varrho$$ have completely different eigenspaces, then the positive eigenspace is just the eigenspace of $$\varrho$$. Then we can compute the two projected operators $$T\sigma T=(1-\epsilon)T\rho T+\epsilon T\varrho T\approx0+\epsilon \varrho$$ and $$T\rho T\approx 0,$$ so the projector is seen to be approximately orthogonal to both $$\sigma$$ and $$\rho$$. I don't know in what sense this can be written as the "opposite" of $$\rho$$.
If we take the more realistic case where the eigenspaces of $$\rho$$ and $$\varrho$$ can overlap slightly, through something like $$\rho=\delta\varrho+(1-\delta)\varrho^\prime$$ for small $$\delta>0$$, then the positive eigenspace of $$\varrho-\rho$$ is the positive eigenspace of $$\varrho-\varrho^\prime$$. With orthogonal $$\varrho$$ and $$\varrho^\prime$$, $$T$$ again projects onto the eigenspace of $$\varrho$$, such that we now have $$T\sigma T=(1-\epsilon)\delta\varrho+\epsilon \varrho$$ and $$T\rho T=\delta \varrho$$. Again, the projector seems approximately orthogonal to both $$\sigma$$ and $$\rho$$ without being "opposite" to $$\rho$$ in any sense. In fact, it is impossible to have $$T\rho T\approx - \rho$$ because $$T$$ is a positive operator, so the word "opposite" is probably not defined in a common sense of the word.