# Existence of a two-outcome measurement $M$ such that the induced distributions differs between different density matrices

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.

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.
• 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? Mar 29 at 17:54
• Sure, but you can compute the difference between the induced distributions, e.g. for outcome "A", $Pr(\Pi_A\rho) - Pr(\Pi_A \sigma) = \text{Tr}(\Pi_A(\rho - \sigma))$ Mar 29 at 19:12