A quantum state preparation machine emits a state $\rho_0$ with probability $2/3$ and emits the state $\rho_1$ with probability $1/3$. We aim to make the best guess which one is it using a set of two POVM operators $\{E_0,E_1\}$. The probability of success is simply:

\begin{equation} p_{\text{succ}}=\frac{2}{3}\text{tr}(E_0\rho_0)+\frac{1}{3}\text{tr}(E_1\rho_1). \end{equation} How to find the POVMs such that $p_{\text{succ}}$ is maximized? Note that the states need not be necessarily qubits, they can be general qudits.

My guess is that we can take $E_0=\rho_0$ and $E_1=I-\rho_0$, in which case, if the density matrices have a corresponding pure orthonormal states, we have $p_{\text{succ}}=1$, which should be the case. But how to check if that is optimal?


1 Answer 1


The optimal probability of guessing correctly is $$ \frac12 + \frac12 \Big\|\frac23 \rho_0 - \frac13 \rho_1 \Big\|_1 $$ where $\| X \|_1 = \mathrm{Tr}[\sqrt{X^* X}]$ is the Schatten 1-norm. This success probability is achieved by the POVM with operator $$ E_0 = \Pi_{[\tfrac23 \rho_0 - \tfrac13 \rho_1]_+} \qquad E_1=I-E_0 $$ where $[X]_+$ denotes the positive part of the Hermitian matrix $X$ in the Hanh-Jordan decomposition and $\Pi_Y$ denotes the projector onto the image of $Y$. Recall the Hanh-Jordan decomposition says we can decompose any Hermitian matrix $X$ as $X= X_+ - X_-$ where $X_+,X_-$ are both positive-semidefinite matrices. This optimal distinguishing is known as the Holevo-Helstrom Theorem, see Watrous' TQI for a proof for example.

  • $\begingroup$ @Rammus, given that I have $[X]_+$, what is the construction for finding the projector $E_0$? $\endgroup$ Mar 2, 2021 at 17:06
  • $\begingroup$ By the spectral theorem we know $[X]_+ = \sum_i \lambda_i |v_i\rangle \langle v_i|$ where $|v_i\rangle$ is the orthonormal set of eigenvectors. So the image is $\mathrm{span} \{ |v_i\rangle : \lambda_i \neq 0\}$ and the projector onto this span is just $\sum_{i \,:\, \lambda_i \neq 0} |v_i\rangle\langle v_i|$. $\endgroup$
    – Rammus
    Mar 2, 2021 at 18:09
  • $\begingroup$ The eigenvectors $|v_i\rangle$ need all be normalized? $\endgroup$ Mar 2, 2021 at 18:11
  • $\begingroup$ @SiddhantSingh Yes, they are the columns of the unitary matrix that diagonalizes $[X]_+$. $\endgroup$
    – Rammus
    Mar 2, 2021 at 18:42
  • $\begingroup$ @Rammus very thanks. I had a mistake of normalization. Now I get the maximum probability of success for distinguishing, for this example to be ~ 0.8727. $\endgroup$ Mar 2, 2021 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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