Suppose there are two fixed density matrices $\rho_1$ and $\rho_2$ are prepared for equal probability. Can we say something about the minimum number of measurements required to distinguish the two states?

One approach I was thinking of is using the fact that the maximum success probability of discriminating correctly is $p_{success} = 1/2 + \Vert \rho_1 - \rho_2 \Vert_1/2$ (from trace distance). Since the trace distance between two random density matrices is generically exponentially small, the success probability will also be exponentially small. From here, I'm lost how to connect to the number of measurements. Is there a rigorous way to say something about the (minimum) number of measurements? This is a single-copy measurement setting by the way.


1 Answer 1



In your setting, you assume you are given $n$ copies of some state $\rho$, which is either $\rho_1$ or $\rho_2$ with equal probability. The setting you have chosen is also symmetric in another way - you treat the error of misidentifying $\rho_1$ as $\rho_2$ to be the same as the error of misidentifying $\rho_2$ as $\rho_1$. You are not allowed joint measurements on $\rho^{\otimes n}$. You could consider sequential measurements where the result of the previous measurements is used for subsequent measurements but in my answer, I assume this is not allowed either. So you make $n$ independent measurements.

Many of these assumptions can be changed and some possibilities are covered in Chapter 7 of Tomamichel's book.

Experimental procedure

You perform the Helstrom measurement $n$ independent times and you obtain $s$ measurements where you get the correct result and $n-s$ measurements where you get the incorrect result.

Probability of correct guess

Your probability of correctly identifying the state is $P_w$. The probability of losing this is $P_l = 1 - P_w$. You have that

$$P_{w} = \frac{p^s(1-p)^{n-s}}{p^s(1-p)^{n-s} + (1-p)^sp^{n-s}}.$$

How many measurements?

We know that for sufficiently large $n$, $s = pn$. Substituting that and doing some algebra, you obtain

\begin{align} \frac{1 - P_{w}}{P_{w}} &= \left(\frac{(1-p)^pp^{(1-p)}}{p^p(1-p)^{(1-p)}}\right)^n\\ \log\frac{P_{l}}{P_{w}} &= n\log\left(\frac{(1-p)^pp^{(1-p)}}{p^p(1-p)^{(1-p)}}\right) = -n\left(D(p\|q) + D(q\|p)\right)\\ P_{w} &= \frac{1}{1+2^{-n\left(D(p\|q) + D(q\|p)\right)}} \end{align}

I have set $q = 1-p$ and used the quantity $D(p\|q)$, the Kullback-Leibler divergence.

  • $\begingroup$ I am little bit confused about the ability of performing the Helstrom measurement. Since $\rho_1$ and $\rho_2$ are unknown, isn't the Helstrom measurement is also unknown? Because as I remember correctly, the Helstrom measurement is a projection onto the positive eigenspace of $\rho_1 - \rho_2$, which can only be identified when we know what these two states are. $\endgroup$
    – Jon Megan
    Nov 20, 2022 at 20:36
  • $\begingroup$ The state you are given is unknown but you must be given a promise that the state is either $\rho_1$ or $\rho_2$. The $\rho_i$ are not unknown states, because then the setting doesn't make sense. $\endgroup$
    – rnva
    Nov 20, 2022 at 20:40
  • $\begingroup$ I think it does still make sense. Consider a setup where you start with $\lvert 0 \rangle^{\otimes n}$ and act a fixed random unitary gate, and then you can measure in some POVM multiple times. On the other hand, consider a different case where everytime you put a new random unitary gate to the initial state. For instance of $1$-qubit system, the first case is putting a fixed random $1$-qubit unitary, whereas in the second case you uniformly sample from all possible $1$-qubit unitary gate everytime. The problem now is to distinguish between two cases by performing some POVM measurements. $\endgroup$
    – Jon Megan
    Nov 20, 2022 at 20:48
  • $\begingroup$ I think you also have to specify some kind of rule to decide which of the two states is "correct" from the string of outcomes. Being the protocol probabilistic, you'll generally get a number of incorrect results, so you have to set some rule such as "I conclude the state is $\rho_i$ if the measurements told me it is $\rho_i$ the majority of the times". And with this you can work out the probability of making a mistake with such protocol when using $n$ measurements. Though for large $n$ the probability of getting $s\neq pn$ successes is exponentially small, so the problem simplifies $\endgroup$
    – glS
    Nov 21, 2022 at 1:01
  • $\begingroup$ @gIS yes, I didn't include this because here, it is simply whichever outcome you see more times in the $n$ tests. $\rho_1$ and $\rho_2$ are chosen with equal probability and your test succeeds with $p\geq 0.5$ so you do the obvious thing. Assuming you do that, the error you make in your conclusion is given by $P_w$ $\endgroup$
    – rnva
    Nov 21, 2022 at 3:55

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