Let $\rho_{ABC}$ and $\sigma_{C}$ be arbitrary quantum states and $\lambda\in \mathbb{R}$ be minimal such that

$$\rho_{ABC}\leq \lambda \rho_{AB}\otimes\sigma_C$$

We assume there are no issues with support in the above statement to avoid infinities. Now, one traces out the $B$ register. Let $\mu\in \mathbb{R}$ be minimal so that

$$\rho_{AC}\leq \mu\rho_A\otimes \sigma_C$$

Clearly, $\lambda\geq\mu$ since partial tracing is a completely positive quantum operation but in this case, since the traced out register had the same state on both the lhs and rhs, is $\lambda = \mu$?


1 Answer 1


No, not necessarily. For example, take $\rho$ to be a GHZ state and let $\sigma$ be the completely mixed state of one qubit. We then have $\lambda=4$ and $\mu=2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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