Bob has a black-box, with the label "V-Wade", which he has been promised prepares a qubit which he would like to know the state of. He asks Alice, who happens also to be an experimental physicist, to determine the state of his qubit. Alice reports $\sigma$ but Bob would like to know her honest opinion $\rho$ for the state of the qubit. To ensure her honesty, Bob performs a measurement $\{E_i\}$ and will pay Alice $R(q_i)$ if he obtains outcome $E_i$, where $q_i={\rm Tr}(\sigma E_i)$. Denote the honest probabilities of Alice by $p_i={\rm Tr}(\rho E_i)$. Then her honesty is ensured if
$$
\sum_i p_i R(p_i)\leq\sum_i p_i R(q_i).
$$
The Aczel-Pfanzagl theorem holds and thus $R(p)=C\log p + B$. Thus, Alice's expected loss is (up to a constant $C$)
$$
\sum_i p_i(\log{p_i}-\log{q_i})=\sum_i {\rm Tr}(\rho E_i)\log\left[\frac{{\rm Tr}(\rho E_i)}{{\rm Tr}(\sigma E_i)}\right].
$$
Blume-Kohout and Hayden showed that if Bob performs a measurement in the diagonal basis of Alice's reported state, then her expected loss is the quantum relative entropy
$$
D(\rho\|\sigma)={\rm Tr}(\rho\log\rho)-{\rm Tr}(\rho\log\sigma).
$$
Clearly in this example Alice is constrained to be honest since the minimum of $D(\rho\|\sigma)$ is uniquely obtained at $\sigma$. This is not true for any measurement Bob can make (take the trivial measurement for example). So, we naturally have the question: which measurements can Bob make to ensure Alice's honesty? That is, which measurement schemes are characterized by
$$
\sum_i {\rm Tr}(\rho E_i)\log\left[\frac{{\rm Tr}(\rho E_i)}{{\rm Tr}(\sigma E_i)}\right]=0\Leftrightarrow \sigma=\rho?
$$

Note that $\{E_i\}$ can depend on $\sigma$ (what Alice reports) but not $\rho$ (her true beliefs).

Partial answer: Projective measurement in the eigenbasis of $\sigma$ $\Rightarrow$ yes, Blume-Kohout/Hayden showed that this is the unique scheme constraining Alice to be honest for a projective measurement.

Informationally complete $\Rightarrow$ yes, clearly this constrains Alice to be honest since the measurement will uniquely specify the state (moreover, the measurement can be chosen independent of $\sigma$).

Trivial measurement $\Rightarrow$ no, Alice can say whatever she wants without impunity.

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