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 $\mu=2$.


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