Why is state discrimination possible to infidelity $\delta$ using $n=\Theta(1/\delta)$ states?

Question

In (Haah et al. 2015), in the first section, the authors study the asymptotic behaviours of fidelity and trace distance between $\rho^{\otimes n}$ and $\sigma^{\otimes n}$ for some given pair of states $\rho,\sigma$. More precisely, given $F(\rho,\sigma)\equiv \|\sqrt\rho\sqrt\sigma\|_1$ and $T(\rho,\sigma)\equiv \frac12\|\rho-\sigma\|_1$, defining $T_n\equiv T(\rho^{\otimes n},\sigma^{\otimes n})$, observing that $F(\rho^{\otimes n},\sigma^{\otimes n})=F(\rho,\sigma)^n$, and using the well-known relation $$1-F \le T \le \sqrt{1-F^2},$$ we can write the asymptotic behaviour of $T_n$ as bounded by $F$ (assuming $F$ is sufficiently small) as follows: $$\frac12 F^{2n} \le 1-T_n \le F^n.$$ From this they infer that:

This means that $\ln(1/F)$ or infidelity gives nearly sharp bounds on the rate at which $T_n$ converges to $1$; the actual rate is between $\ln(1/F)$ and $2\ln(1/F)$. In particular, for fixed [state dimension] $d$, the state discrimination is possible to infidelity $\delta$ using $n=\Theta(1/\delta)$ copies.

I'm trying to understand more precisely what is meant by this, and how the conclusion is reached. I think the first part refers to what we get taking the log of the above relation: $$2n \log F -\log2\le \log(1-T_n) \le n \log F \\ \sim 2\log F \le \frac{\log(1-T_n)}{n} \le \log F.$$ This looks close enough to the "rate of convergence of $T_n$ to $1$ is $O(\ln(1/F))$", but I'm unsure about how this "rate" is defined, precisely.

The next step might follow immediately from understanding of this, but the additional unclear point about the above sentence is: what does "state discrimination is possible to infidelity $\delta$ with $n$ copies" mean exactly? Are they referring to a bound of the form: the probability that the infidelity is larger than $\delta$ is smaller than some constant error threshold using $n\sim 1/\delta$ input states? How is this connected with the previous statement about the trace distance $T_n$?

Answer 1

The statement

state discrimination is possible to infidelity 𝛿 with 𝑛 copies

is probably related to minimum error state discrimination. The minimum probability of error (while discriminating between states $\rho$ and $\sigma$) can be shown to be

$P_{e}^{min}=\frac{1}{2}\big(1-\frac{1}{2}||\rho-\sigma||_{tr}\big)=\frac{1}{2}\big(1-T(\rho, \sigma)\big)$,

assuming a uniform prior on $\{\rho, \sigma\}$.

For n-copies of the unknown state, the corresponding expression can be obtained by replacing $\rho$ by $\rho^{\otimes n}$ and $\sigma$ by $\sigma^{\otimes n}$. So, we'd get

$P_{n,e}^{min}=\frac{1}{2}\big(1-T_n\big)$, as per your notation. Now, infidelity $\delta = 1-F$, but $1-F^n \leq T_n$. So,

$1-(1-\delta)^n \leq T_n = 1-2P_{n,e}^{min}$.

Using the approximation $(1-\delta)^n \approx 1-n\delta$, the inequality above becomes

$n\delta \leq 1-2P_{n,e}^{min} \equiv P_{n,e}^{min} \leq \frac12(1-n\delta)$.