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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?

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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.

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    $\begingroup$ That's not correct. $E_0$ is the projector onto the positive part of $\tfrac23 \rho_0 - \tfrac13 \rho_1$. $\endgroup$ – Mateus Araújo Mar 2 at 10:43
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    $\begingroup$ @MateusAraújo Whoops, thanks for pointing that out! Fixed now. $\endgroup$ – Rammus Mar 2 at 11:04
  • $\begingroup$ @Rammus, given that I have $[X]_+$, what is the construction for finding the projector $E_0$? $\endgroup$ – Siddhant Singh Mar 2 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 at 18:09
  • $\begingroup$ The eigenvectors $|v_i\rangle$ need all be normalized? $\endgroup$ – Siddhant Singh Mar 2 at 18:11

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