In the context of quantum state discrimination, the task of finding the POVM $\mu$ that optimally discriminates between the elements of an ensemble $a\mapsto (p_a,\rho_a)$, amounts to maximising the quantity $\sum_a p_a \langle\mu_a,\rho_a\rangle$ with respect to all POVMs $\mu$. Given the nontriviality of said optimisation, an alternative approach is to employ the pretty good measurement, which involves instead using the POVM $$\mu_a \equiv \left( \sum_b p_b \rho_b\right)^{-\frac12} \!\!\!\!p_a\rho_a \left( \sum_b p_b \rho_b\right)^{-\frac12}, $$ with straightforward generalisation using pseudo-inverses when $\sum_b p_b \rho_b$ is singular.

This measurement results in a success probability in discriminating inputs such that $$p_{\rm succ}^{\rm PGM}\equiv\sum_a p_a \langle\mu_a ,\rho_a\rangle \ge \mathrm{opt}(\{(p_a,\rho_a)\}),$$ where the RHS is the optimal discrimination probability associated to the ensemble. That is, the one obtained solving the maximisation problem mentioned at the beginning.

Is there a way to understand the intuition behind this choice? Where does this particular structure come from, and why should it result in a "pretty good" discrimination probability?

  • 1
    $\begingroup$ One of the early papers on PGM makes some arguments that the gradient of the mutual information between measurement results and the input state is very small for the PGM (compared to zero for an optimal measurement) as a result of a kind of symmetry in the measurement with respect to input states. They suggest this means that the PGM is nearly optimal in the special case where the input state is approaching an orthogonal ensemble - maybe worth a look: tandfonline.com/doi/abs/10.1080/09500349414552221. $\endgroup$
    – forky40
    May 31, 2022 at 15:40

2 Answers 2


I'll reproduce here a standard argument used to prove the fundamental bound for pretty good measurements (PGMs), the the most part taken from Watrous' book, with some minor changes in notation, presentation, and an attempt at deriving the structure of PGMs without knowing it a priori (that however still requires to know part of the structure of the solution, so this approach only goes a little bit in that direction). I don't find this to be a particularly satisfactory answer to the question about "the idea behind" pretty good measurements, but being this bound arguably the main reason PGMs are interesting, I figured it's still good to have this derivation here, for reference.

Proof idea - The main trick of the derivation, the way I see it, is applying a "double CS inequality" after rewriting the inner product that corresponds to the outcome probabilities. It's a bit like saying: we want to estimate the inner product $\langle \mathbf v,\mathbf w\rangle\equiv \sum_i v_i w_i$. We can rewrite it as $\langle\mathbf v,\mathbf w\rangle = \langle P \mathbf v,P^{-1}\mathbf w\rangle$ for some invertible Hermitian $P$. Using the CS inequality then gives a different bound for each choice of $P$: $$\langle \mathbf v,\mathbf w\rangle \le \|P\mathbf v\| \|P^{-1}\mathbf w\|.$$

Problem setting - Say we want to discriminate between an ensemble of states $a\mapsto (p_a,\rho_a)$, meaning the possible states are $\{\rho_a\}_a$, and the $a$-th state is known to occur with probability $p_a$. Define the ensemble operators as $\eta_a\equiv p_a \rho_a$, and $\eta\equiv \sum_a \eta_a$, so that $\operatorname{tr}(\eta)=\sum_a \operatorname{tr}(\eta_a)=1$.

Let $a\mapsto \mu_a$ be an arbitrary POVM with number of outcomes equal to the number of elements in the ensemble (we can assume this wlog). A standard way to use the POVM to discriminate between the states, is to guess the input state to being $\rho_a$ if the $a$-th outcome is found. This happens with probability $p_a=\langle\mu_a,\eta_a\rangle\equiv \operatorname{tr}(\mu_a\eta_a)=p_a \operatorname{tr}(\mu_a\rho_a)$. It follows that the overall probability of correctly guessing the input state using this POVM and strategy is $$p_{\rm guess}(\mu) = \sum_a \langle \mu_a,\eta_a\rangle = \sum_a p_a \langle \mu_a,\rho_a\rangle.$$ Ideally, we'd like to find the optimal discrimination strategy, that is, maximise $p_{\rm guess}(\mu)$ over the set of possible POVMs. This amounts to a semidefinite program. However, the point of pretty good measurements (PGMs) is to find a recipe to find a measurement strategy that works well enough, but doesn't require performing such optimisation to find it.

Derivation - To find such strategy, we can try using the following trick: let $A$ be some invertible Hermitian positive definite matrix; then $$\langle \mu_a, \eta_a\rangle = \langle A\mu_a A,A^{-1}\eta_a A^{-1}\rangle.$$ This is immediate to verify writing the inner product with the trace. We also don't actually need $A$ to be invertible. It is sufficient to assume $\operatorname{im}(\eta_a)\subseteq \operatorname{im}(A)$, and use the pseudoinverse instead of the actual inverse. This ensures $A^{+}\eta_a A^{+}$ is well-defined even though $A$ is not invertible, where $A^+$ indicates the standard Moore-Pensores pseudoinverse. From this:

  1. Using CS, we get for every $a$, $$\langle A\mu_a A,A^{+}\eta_a A^{+}\rangle \le \| A\mu_a A\|_2 \|A^+ \eta_a A^+\|_2 = \langle A^2 \mu_a A^2,\mu_a\rangle \langle A^{+ 2}\eta_a A^{+2},\eta_a\rangle$$

  2. From the above we have $$\sum_a \langle \mu_a,\eta_a\rangle \le \sum_a \| A\mu_a A\|_2 \|A^+ \eta_a A^+\|_2.$$ We can think of this as again the inner product between the two vectors $(\| A\mu_a A\|_2 )_a$ and $( \|A^+ \eta_a A^+\|_2 )_a$, and thus using CS again, $$\sum_a \langle \mu_a,\eta_a\rangle \le \left(\sum_a \| A\mu_a A\|_2^2\right)^{1/2} \left(\sum_b \| A^+\eta_b A^+\|_2^2\right)^{1/2}.$$

  3. Now looking into the two terms we got, note that for the first one $$\sum_a \| A\mu_a A\|_2^2 = \sum_a \langle A\mu_a A,A\mu_a A\rangle = \sum_a \operatorname{tr}( A^2 \mu_a A^2 \mu_a) \\ \le \sum_a \operatorname{tr}(A^4 \mu_a) = \operatorname{tr}(A^4),$$ where we used $\operatorname{tr}(PQ)\le \operatorname{tr}(P)$ if $P,Q\ge0$ and $Q\le I$, and the standard normalisation condition $\sum_a \mu_a = I$.

    On the other hand, the other term reads $$ \sum_b \| A^+\eta_b A^+\|_2^2 = \sum_b \langle A^{+2} \eta_b A^{+2},\eta_b\rangle. $$

  4. Remember that what we are looking for is an upper bound for $p_{\rm guess}(\mu)$ in terms of the guessing probability $p_{\rm guess}(\mu^{\rm PGM})$ of some POVM $\mu^{\rm PGM}$ that only depends on the ensemble $a\mapsto \eta_a$ (and has a "sufficiently simple" expression).

    Looking at the bound we got for $p_{\rm guess}(\mu)$, we've got it, if we can find $A$ such that $\operatorname{tr}(A^4)=1$, and such that $(A^{+2}\eta_b A^{+2})_b$ is a POVM. Given that the only information we're allowed to use is the ensemble $a\mapsto \eta_a$, and that $\operatorname{tr}(\eta)=1$, it's pretty natural to try defining $A=\eta^{1/4}\equiv\left(\sum_a \eta_a\right)^{1/4}$, and thus $$\mu^{\rm PGM}_b \equiv \sqrt{\eta^+} \eta_b \sqrt{\eta^+},$$ and it's simple enough to check that this is indeed a valid POVM.

We thus showed that using the POVM $\mu^{\rm PGM}_b \equiv \sqrt{\eta^+} \eta_b \sqrt{\eta^+}$ the discrimination probability satisfies the bound $$p_{\rm guess}(\mu^{\rm PGM})^2 \ge p_{\rm guess}(\mu), \,\,\forall \mu,$$ and thus also $p_{\rm guess}(\mu^{\rm PGM})^2 \ge \max_\mu p_{\rm guess}(\mu)$.


I think a way to motivate pretty good measurements is to observe that given an ensemble of states $\mathscr{E}=\{p_i,\rho_i\}$, one can always define a POVM using the states of this ensemble.

Define $\rho = \sum_i p_i \rho_i$,

$\rho_i^{r} = p_i \rho_i$, and

then note that the operators

$P_i = \rho^{-1/2} \rho_i^{r} \rho^{-1/2}$ are positive and sum to identity.

Naturally, if these states are orthogonal then it's clear that this POVM would be optimal to distinguish states of the given ensemble. And it's also clear it won't be optimal in general. The best we can get is a kind of near-optimality for ensemble discrimination in terms of the probability of correctly guessing the outcome (Barnum-Knill, Theorem 3.10 in Watrous' book).


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