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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.