8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $$ \frac12+\frac14\text{Tr}((P_1-P_2)(\rho_1-\rho_2))=\frac12+\frac14\text{Tr}|\rho_1-\rho_2|. $$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Josu Etxezarreta Martinez Sep 6 '18 at 8:48
  • $\begingroup$ @JosuEtxezarretaMartinez Yes. $\endgroup$ – DaftWullie Sep 6 '18 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 $Tr|\rho_1-\rho_2|$? $\endgroup$ – Josu Etxezarreta Martinez Sep 6 '18 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 '18 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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