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Let $\rho \neq \sigma$ be density matrices.

I want to show that there exists a two-outcome measurement $M$ such that the induced distributions $M(\rho)$ and $M(\sigma)$ differ.

From what I learned, for a density matrix $\rho$, the probability of measuring the state $|k\rangle$ is $\langle k| \rho |k\rangle\,.$

That means that $$\text{Pr}[\text{Measure} |k\rangle] = \langle k| \rho |k\rangle = \text{Tr}(|k\rangle\langle k|\rho)\,.$$

From here, I think that the difference between $\rho$ and $\sigma$ should play a part, but I'm not sure how to proceed. Is my idea correct?

Help would be appreciated.

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Yes, you can use $\rho - \sigma$ to construct a measurement where the measurement statistics differ for $\rho$ compared to $\sigma$. The issue is that $\rho - \sigma$ is not necessarily positive, while a measurement needs to consist of positive operators (no negative eigenvalues) that sum to the identity.

Here is a place to start: Consider the Hermitian operator $\Lambda = (\rho - \sigma)$ and then take its spectral decomposition to be $\Lambda = A - B$ where $A$ and $B$ are both positive operators. You can write these operators in terms of the eigenvalues/eigenvectors of $\Lambda$: \begin{align} A &= \sum_{i: \lambda_i > 0} \lambda_i |\lambda_i\rangle \langle \lambda_i| \\ B &= \sum_{i: \lambda_i < 0} (-\lambda_i) |\lambda_i\rangle \langle \lambda_i| \end{align}

Now think of the projector $\Pi_A$ onto the support of $A$, which can be written $\Pi_A = \sum_{i: \lambda_i>0} |\lambda_i \rangle \langle \lambda_i|$. Then, $\Pi_A$ is orthogonal to the projector $\Pi_B$ onto the support of $B$ (defined analogously), and $\Pi_A + \Pi_B = I$. You should be able to get a measurement out of this and compute the corresponding measurement statistics.

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    $\begingroup$ In fact, this is the best measurement for distinguishing the two states. (And the resulting distance, or distinguishability, is precisely the trace norm distance.) $\endgroup$ Mar 28 at 5:54
  • $\begingroup$ I'm not so rue how to continue from here. If I'm not mistaken then the measure for the density matrix $\rho$ would be $Pr[\Pi_A] = Tr(\Pi_A\rho)$ and $Pr[\Pi_B] = Tr(\Pi_B\rho)$, and similarly for $\sigma$ and then the trace computation should differ? $\endgroup$
    – Gabi G
    Mar 29 at 17:54
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    $\begingroup$ Sure, but you can compute the difference between the induced distributions, e.g. for outcome "A", $Pr(\Pi_A\rho) - Pr(\Pi_A \sigma) = \text{Tr}(\Pi_A(\rho - \sigma))$ $\endgroup$
    – forky40
    Mar 29 at 19:12
  • $\begingroup$ Thanks! That clears what I was missing. $\endgroup$
    – Gabi G
    Mar 31 at 21:45

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