# Accurate quantum state estimation via “Keeping the experimentalist honest”

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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Trivially for any set of measurements $\{E_i\}$ where $\rho$ and $\sigma$ have equal expectation value for each $E_i$, $$\sum_i\mbox{Tr}(\rho E_i) \log \left[ \frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}\right] = \sum_i\mbox{Tr}(\rho E_i) \times 0 = 0.$$

Note that the log-sum inequality theorem says that $$\sum_i a_i \log\left(\frac{a_i}{b_i}\right) \leq \big(\sum_i a_i\big)\times \log \left(\big(\sum_i a_i\big)/\big(\sum_i b_i\big)\right),$$ with equality only if $\frac{a_i}{b_i}$ is constant for all $i$. This implies that $$\sum_i\mbox{Tr}(\rho E_i) \log \left[ \frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}\right] = 0$$ if and only if $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}$ is constant. Since the sum of the traces over $i$ is 1 for both density matrices this implies that you must have $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)} = 1$.

If, however, the projections onto each subspace selected by $E_i$ aren't equal, then there exists some $i$ such that $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)} \neq 1$.

Hence for Alice to have zero expected loss, then $\mbox{Tr}(\rho E_i) = \mbox{Tr}(\sigma E_i) ~~~\forall i$. If expectation value for each $E_i$ uniquely identifies a density matrix, then necessarily $\rho = \sigma$ if Alice has zero expected loss. A sufficient condition for this is when $E_i$ project onto $(\dim \mathcal{H})^2 -1$ linearly independent density matrices, where $\mathcal{H}$ is the Hilbert space of the system.

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Joe, your sufficient condition is that $\{E_i\}$ is informationally complete -- but we already knew that was sufficient and not necessary by the Blume-Kohout/Hayden result. So is there something more interesting that can be said about the condition Tr($\rho E_i$) = Tr($\sigma E_i$), for all $i$? –  Chris Ferrie Nov 15 '11 at 2:15
@ChrisFerrie: No, that is not the same as the sufficient condition I gave since I only required linear independence and not orthogonality and only for a sufficiently large subset of $\{E_i\}$, so it includes a bunch of weak measurements. However that was only an example. The necessary and sufficient condition is that all expectation values are the same. This includes a huge range of measurements, such as measuring a randomly chosen Pauli operator, which yields only a single bit of information. –  Joe Fitzsimons Nov 15 '11 at 4:00
Also, if you take any complete or over-complete basis for tomography you can make the measurements and make them arbitrarily weak, you still satisfy the criterion (though Alice's expected loss tends towards zero as the measurement tends towards the identity). –  Joe Fitzsimons Nov 15 '11 at 4:01
This argument is for the case where the set of measurements is fixed and independent of $\sigma$. If the scheme need only work for certain classes of $\sigma$ then this imposes correlations between entries in the density matrix which reduces the number of linearly independent measurement operators required to uniquely identify it. An example of this is where $\sigma = |+\rangle\langle +|$, where purity implies that the state has expectation value 0 for $Z$ and $Y$ measurements, and if $\mbox{Tr}(\rho X) = \mbox{Tr}(\sigma X)$ then $\mbox{Tr}(\rho Y) = \mbox{Tr}(\sigma Y)$, etc. –  Joe Fitzsimons Nov 15 '11 at 4:13
And hence you need only measure in the $X$ basis, even though this does not have sufficiently many linearly independent outcomes to uniquely identify an arbitrary density matrix. –  Joe Fitzsimons Nov 15 '11 at 4:14