I have been reading the paper Belief propagation decoding of quantum channels by passing quantum messages by Joseph Renes for decoding Classical-Quantum channels and I have crossed with the concept of Helstrom Measurements.

I have some knowledge about quantum information theory and quantum error correction, but I had never read about such measurement until I worked on that paper. In such article, the author states that the measurement is optimal for this decoding procedure, so I would like to know what are such kind of measurements and how can they be done.


1 Answer 1


The Helstrom measurement is the measurement that has the minimum error probability when trying to distinguish between two states.

For example, let's imagine you have two pure states $|\psi\rangle$ and $|\phi\rangle$, and you wish to know which it is that you have. If $\langle\psi|\phi\rangle=0$, then you can specify a measurement with three projectors $$ P_{\psi}=|\psi\rangle\langle\psi|\qquad P_{\phi}=|\phi\rangle\langle\phi|\qquad \bar P=\mathbb{I}-P_{\psi}-P_{\phi}. $$ (For a two-dimensional Hilbert space, $\bar P=0$.)

The question is what measurement should you perform in the case that $\langle\psi|\phi\rangle\neq0$? Specifically, let's assume that $\langle\psi|\phi\rangle=\cos(2\theta)$, and I'll concentrate just on projective measurements (IIRC, this is optimal). In that case, there is always a unitary $U$ such that $$ U|\psi\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle\qquad U|\phi\rangle=\cos\theta|0\rangle-\sin\theta|1\rangle. $$ Now, those states are optimally distinguished by $|+\rangle\langle +|$ and $|-\rangle\langle -|$ (you get $|+\rangle$, and you assume you had $U|\psi\rangle$). Hence, the optimal measurement is $$ P_{\psi}=U^\dagger|+\rangle\langle+|U\qquad P_{\phi}=U^\dagger|-\rangle\langle-|U\qquad \bar P=\mathbb{I}-P_{\psi}-P_{\phi}. $$ The success probability is $$ \left(\frac{\cos\theta+\sin\theta}{\sqrt{2}}\right)^2=\frac{1+\sin(2\theta)}{2}. $$

More generally, how do you distinguish between two density matrices $\rho_1$ and $\rho_2$? Start by calculating $$ \delta\rho=\rho_1-\rho_2, $$ and finding the eigenvalues $\{\lambda_i\}$ and corresponding eigenvectors $|\lambda_i\rangle$ of $\delta\rho$. You construct 3 measurement operators $$ P_1=\sum_{i:\lambda_i>0}|\lambda_i\rangle\langle\lambda_i|\qquad P_2=\sum_{i:\lambda_i<0}|\lambda_i\rangle\langle\lambda_i|\qquad P_0=\mathbb{I}-P_1-P_2. $$ If you get answer $P_1$, you assume you had $\rho_1$. If you get $P_2$, you had $\rho_2$, while if you get $P_0$ you simply guess which you had. You can verify that this reproduces the pure state strategy described above. What's the success probability of this strategy? $$ \frac12\text{Tr}((P_1+P_0/2)\rho_1)+\frac12\text{Tr}((P_2+P_0/2)\rho_2) $$ We can expand this as $$ \frac14\text{Tr}((P_1+P_2+P_0)(\rho_1+\rho_2))+\frac14\text{Tr}((P_1-P_2)(\rho_1-\rho_2)) $$ Since $P_1+P_2+P_0=\mathbb{I}$ and $\text{Tr}(\rho_1)=\text{Tr}(\rho_2)=1$, this is just $$ \frac12+\frac14\text{Tr}((P_1-P_2)(\rho_1-\rho_2))=\frac12+\frac14\text{Tr}|\rho_1-\rho_2|. $$

  • $\begingroup$ Thanks a lot for the complete answer, it does describe what I wanted to know. Just one slight point to clarify, are the $\lambda_i$ the eigenvalues of the $\delta\rho$ matrix? I assume that yes, but I just want to be sure. $\endgroup$ Sep 6, 2018 at 8:48
  • $\begingroup$ @JosuEtxezarretaMartinez Yes. $\endgroup$
    – DaftWullie
    Sep 6, 2018 at 8:49
  • $\begingroup$ Other little comment, in the last step how do you go from $Tr((P_1-P_2)(\rho_1-\rho_2))$ to $Tr|\rho_1-\rho_2|$? $\endgroup$ Sep 6, 2018 at 8:54
  • 1
    $\begingroup$ @JosuEtxezarretaMartinez $P_1$ projects on the positive eigenvalues of $\delta\rho$. $-P_2$ projects on the negative eigenvalues of $\delta\rho$, multiplying them by -1, thereby converting negative eigenvalues into positive ones. So $(P_1-P_2)\delta\rho=|\delta\rho|$. $\endgroup$
    – DaftWullie
    Sep 6, 2018 at 8:59

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