Help understanding the proof of a theorem on entanglement monotones

Question

I currently read the paper of Guifre Vidal about Entanglement Monotones (link). On page 5 he states the following theorem:

Theorem 1 The maximal probability of success $P(\rho \rightarrow \rho*)$ in the converion of $\rho \rightarrow \rho*$ that the parties can achieve by means of a strategy involving only LQCC [local quantum operations assisted with classical communication] is given by $$P(\rho \rightarrow \rho*)=\min \frac{\mu(\rho)}{\mu(\rho*)}$$ where the minimization is performed over the hole set of all Entanglement Monotones $\mu$.

The proof of this theorem is very short and it seems obvious from Vidal's point of view, but I don't understand it. Can anyone help me to understand this?

Answer 1

Since the probability of success for any strategy is upper bounded by $\mu(\rho)/\mu(\rho^*)$ as seen in eq (14), this will in particular also hold for the strategy that achieve the maximal probability of success $P[\rho\rightarrow\rho^*]$. Note that eq. (14) holds for any entanglement monotone $\mu$ so the entanglement monotone that minimise $\mu(\rho)/\mu(\rho^*)$ is therefore also an upper bound. This means we have the following inequality, $$P[\rho\rightarrow\rho^*]\leq\min_{\mu}\frac{\mu(\rho)}{\mu(\rho^*)}.\tag{I}$$ We now need to show that this is actually an equality. One way to see this is to first note that $$\min_{\mu}\frac{\mu(\rho)}{\mu(\rho^*)}\leq\frac{\mu'(\rho)}{\mu'(\rho^*)}\tag{II}$$ for any fixed entanglement monotone $\mu'$. Then we can use the fact that $P[\rho\rightarrow\rho^*]$ defines an entanglement monotone given by $$\mu_{\rho^*}(\rho)=P[\rho\rightarrow\rho^*].\tag{III}$$ Putting this into equation (II) we have that $$\min_{\mu}\frac{\mu(\rho)}{\mu(\rho^*)}\leq\frac{\mu_{\rho^*}(\rho)}{\mu_{\rho^*}(\rho^*)}=\frac{P[\rho\rightarrow\rho^*]}{P[\rho^*\rightarrow\rho^*]}=P[\rho\rightarrow\rho^*]\tag{IV}$$ where we used that $P[\rho^*\rightarrow\rho^*]=1$. Finally putting this into equation (I) we have that $$\min_{\mu}\frac{\mu(\rho)}{\mu(\rho^*)}\leq P[\rho\rightarrow\rho^*]\leq \min_{\mu}\frac{\mu(\rho)}{\mu(\rho^*)}\tag{V}$$ which implies that $$P[\rho\rightarrow\rho^*]=\min_{\mu}\frac{\mu(\rho)}{\mu(\rho^*)}\tag{VI}$$