# How to find the POVM that optimally distinguishes between two given states?

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?

• – glS
Apr 18, 2022 at 20:30

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.
• That's not correct. $E_0$ is the projector onto the positive part of $\tfrac23 \rho_0 - \tfrac13 \rho_1$. Mar 2, 2021 at 10:43
• @Rammus, given that I have $[X]_+$, what is the construction for finding the projector $E_0$? Mar 2, 2021 at 17:06
• 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 $\sum_{i \,:\, \lambda_i \neq 0} |v_i\rangle\langle v_i|$. Mar 2, 2021 at 18:09
• The eigenvectors $|v_i\rangle$ need all be normalized? Mar 2, 2021 at 18:11