# Are projective measurements the only optimal measurements to discriminate between two states?

Consider two density matrices $$\rho$$ and $$\sigma$$. The task is to distinguish between these two states, given one of them --- you do not know beforehand which one.

There is an optimal measurement to distinguish between these two states --- the Helstrom measurement. Note that it is an orthogonal projector.

Are orthonormal/orthogonal projectors the only optimal measurement possible?

In other words, can we say that if we had chosen any non-orthogonal POVM as our measurement, we would not have achieved the optimal distinguishing probability?

• the bit about entanglement is a separate and rather different question, which merits a post of its own imo. The very rough answer to which is that yes, measuring on multiple copies you can do better. The intuitive reason being that $\rho_i^{\otimes n}$ become more orthogonal/distinguishable for increasing $n$
– glS
Commented Sep 26, 2021 at 15:48
• I asked the two questions separately. Commented Sep 27, 2021 at 8:02

Suppose you are given either $$\rho_1$$ or $$\rho_2$$, and you also know that the probabilities you got one or the other are $$p_1$$ and $$p_2$$, respectively. If you have no prior knowledge of the frequencies with which I'm going to give you one state or the other, you just use $$p_1=p_2=1/2$$.
You are asking what's the measurement that optimally distinguishes between $$\rho_1$$ and $$\rho_2$$, given the priors $$p_1,p_2$$. Consider a simple scheme where the measurement outcome is directly used to decide which one was the input state. We are thus looking at two-outcome measurements, i.e. POVMs with elements $$\boldsymbol\mu\equiv \{\mu_1,\mu_2\}$$, and use the strategy of guessing $$\rho_i$$ if the $$i$$-th outcome is found. Note that there is no loss of generality in fixing this strategy: you could consider a more general scenario where upon finding the outcome $$i$$, some function $$f:\{0,1\}\to\{0,1\}$$ is used to determine the input state, but any such strategy just amounts to a relabelling of the POVM elements.
The (average) success probability of this strategy, when using the measurement $$\boldsymbol\mu$$, reads $$p_{\rm success}(\boldsymbol\mu) \equiv \sum_i \langle\mu_i, p_i\rho_i\rangle = \langle\mu_1, p_1\rho_1\rangle + \langle\mu_2, p_2\rho_2\rangle.$$ Using the normalisation condition on the POVM, $$\mu_1+\mu_2=I$$, we get $$p_{\rm success}(\boldsymbol\mu) = p_2 + \langle\mu_1, p_1\rho_1- p_2\rho_2\rangle.$$ We get the overall optimal success probability maximising over all such POVMs: $$\mu_{\rm optimal} \equiv \operatorname*{argmax}_{\boldsymbol\mu} p_{\rm success}(\boldsymbol\mu) = \operatorname*{argmax}_{\boldsymbol\mu} \langle\mu_1, p_1\rho_1- p_2\rho_2\rangle.$$ To get the optimal success probability, observe that $$\max_{\boldsymbol\mu} p_{\rm succ}(\boldsymbol\mu) = p_2 + \max_{0\le \mu_1\le I} \langle \mu_1,p_1 \rho_1-p_2\rho_2\rangle.$$ To compute this maximum, observe that for any Hermitian $$P$$, $$\max_{0\le \mu_1\le I} \langle \mu_1,P\rangle = \operatorname{tr}(P_+) = \frac12(\|P\|_1 + \operatorname{tr}(P)),$$ where $$P_\pm$$ are the unique positive definite operators such that $$P_+ P_-=0$$ in the decomposition $$P=P_+ - P_-$$, and $$\|P\|_1=\operatorname{tr}(P_+)+\operatorname{tr}(P_-)$$. The maximum is achieved with $$\mu_1$$ the projection onto the support of $$P$$: $$\mu_1=\Pi_{\operatorname{supp}(P)}$$. We thus have $$\max_{0\le \mu_1\le I} \langle \mu_1,p_1 \rho_1-p_2\rho_2\rangle = \frac12((p_1-p_2) + \|p_1 \rho_1-p_2\rho_2\|_1),$$ and $$\max_{\boldsymbol\mu} p_{\rm succ}(\boldsymbol\mu) = \frac12 + \frac12\|p_1 \rho_1-p_2\rho_2\|_1,$$ with the optimal measurement being the POVM with $$\mu_1$$ projection onto the support of $$p_1\rho_1-p_2\rho_2$$, and $$\mu_2=I-\mu_1$$ projection onto its orthogonal complement. More explicitly, this means that the optimal measurement can always be chosen to be projective, and corresponding to measuring in the eigenbasis of $$p_1\rho_1-p_2\rho_2$$.
It is still possible to have non-projective optimal measurements, for example by adding projections onto the ker of $$p_1\rho_1-p_2\rho_2$$ to the POVM elements. Any such change won't affect the state discrimination procedure in any way, but it can give non-projective optimal POVMs (if arguably in a somewhat trivial manner).
As an example, let $$p_1=p_2=1/2$$ and $$\rho_1=\frac23\begin{pmatrix}1/2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0&0\end{pmatrix}, \qquad \rho_2=\frac23\begin{pmatrix}1/2 & 0 & 0 \\ 0 & 1/2 & 1/2 \\ 0 & 1/2 & 1/2\end{pmatrix}.$$ Then, the nonzero eigenvalues of $$\Delta\rho$$ are $$\lambda_\pm=\pm\sqrt2/3$$, and a possible non-projective optimally distinguishing two-outcome POVM is $$\mu_1 = |\lambda_+\rangle\!\langle\lambda_+| + \frac12 |0\rangle\!\langle0|, \\ \mu_2 = |\lambda_-\rangle\!\langle\lambda_-| + \frac12 |0\rangle\!\langle0|.$$