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 $T_n$?


1 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

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


Your Answer

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

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