1. Frame duality

The most general kind of measurement in quantum theory is modeled by a positive operator valued measure or POVM. If the number of outcomes of the measurement is a finite number \(n\) and the Hilbert space used to model the system of interest is \(d\) dimensional, then a POVM consists of a set of \(n\) \(d \times d\) positive semidefinite Hermitian matrices called effects which sum to the identity: \[\tag{1.1}\begin{aligned}\forall i: E_i \ge 0 && \sum_i E_i = I.\end{aligned}\]

If the \(\{E_i\}\) span the \(d^2\)-dimensional operator space, we’ll call it a reference measurement. In particular, if \(n=d^2\), the measurement is called a minimal informationally complete (MIC) POVM [1]. Suppose that conditional on the outcome \(E_i\) occuring, the state \(\sigma_i = E_i/\text{tr}(E_i)\) is prepared. In fact, we will assume that the measurement is unbiased so that \(\forall i: \text{tr}(E_i) = d/n\), which is \(1/d\) for a MIC. Then the conditional probability matrix \[\tag{1.2}\begin{aligned}P_{ij} = P(E_i|E_j) = \text{tr}(E_i \sigma_j)\end{aligned}\]

characterizes the reference measurement in its own terms: the reference effects characterizing the reference states, and the reference states characterizing the reference effects. Operationally, you can think of \(P(E_i|E_j)\) as the probability of getting reference outcome \(E_i\) given that you got reference outcome \(E_j\) last time.

Since the reference states and effects are spanning, we can write for any quantum state \(\rho\), \[\tag{1.3}\begin{aligned}\rho = \sum_i w_i \sigma_i,\end{aligned}\]

for some vector of expansion coefficients \(w\). Since \[\tag{1.4}\begin{aligned}p_i \equiv P(E_i|\rho) = \text{tr}(E_i \rho) = \sum_j w_j \text{tr}(E_i \sigma_j),\end{aligned}\]

we have \(p = Pw\). If the reference measurement is a MIC, then the reference effects (and states) are not just spanning, but linearly independent. Since the columns of \(P\) just represent \(\sigma_j\) in a different basis, they must be linearly independent themselves, and so \(P\) must be invertible. We conclude that \(w = P^{-1}p\), and we call \(\Phi \equiv P^{-1}\) the Born matrix. It takes the probabilities \(p\) characterizing a state to the expansion coefficients \(w\). We can therefore write \[\tag{1.5}\begin{aligned}\rho = \sum_{ij} \Phi_{ij}P(E_j|\rho) \sigma_i,\end{aligned}\]

and so for any other measurement outcome represented by an effect \(A\), \[\tag{1.6}\begin{aligned}P(A|\rho) &= \text{tr}(A \rho) = \sum_{ij} \Phi_{ij}P(E_j|\rho) \text{tr}(A \sigma_i)\end{aligned}\] \[\tag{1.7}\begin{aligned}&=\sum_{ij} P(A|E_i)\Phi_{ij}P(E_i|\rho),\end{aligned}\]

since we can think of \(\text{tr}(A\sigma_i)\) as the probability of getting outcome \(A\) given that the reference measurement gave outcome \(E_i\) first. This is the QBist reformulation of the Born rule as a coherence condition analogous to the law of total probability [2, 3]. In fact, a rule of this form can be derived on very general grounds (without assuming quantum theory) [4, 5].

In the language of frame theory [6], \(p_i = \text{tr}(E_i\rho)\) is a representation of \(\rho\) in frame coordinates. There exists a dual frame \(D_j = \sum_{j} \Phi_{ij}\sigma_i\) so that \(\rho = \sum_j p_j D_j\). Conversely, \(w_i = \frac{1}{d}\text{tr}(D_j \rho)\) is a representation of \(\rho\) in dual frame coordinates. (In fact, \(w\) gives us a quasiprobability representation [7].) In turns out we can take another point of view on this notion of frame-theoric duality.

2. Legendre duality

Physicists who know their classical mechanics (as well as thermodynamics) will be familiar with the idea of a Legendre dual. Let \(f:S \rightarrow \mathbb{R}\) be a convex function on a convex set \(S \subseteq \mathbb{R}^m\). The Legendre dual [8] of \(f\) is \[\tag{2.1}\begin{aligned}f^*(x^*) = \sup_{x \in S}\Bigg\{ \langle x^*, x\rangle - f(x)\Bigg\}.\end{aligned}\]

Its domain is \(S^*\) which consists of all \(x^*\) for which \(f(x^*)\) doesn’t blow up to infinity. For example, in classical physics, the Hamiltonian function is the Legendre dual of the Lagrangian function [9]. In thermodynamics, the Helmholtz free energy as a function of temperature is the Legendre dual of the internal energy as a function of entropy [10] (up to a sign). It comes up all over the place. It was surprising to me, however, that this notion of duality can coincide with the frame-theoretic notion.

Consider a potential function defined on quantum states \[\tag{2.2}\begin{aligned}\Psi(\rho) = \frac{1}{d} \frac{1}{2}\text{tr}(\rho^2),\end{aligned}\]

where both prefactors have been cleverly chosen. We note that \(\text{tr}(\rho^2)\) is called the purity of the state: it is 1 iff \(\rho\) is a pure state, and reaches its minimum of \(1/d\) for the maximally mixed state [11]. At the same time, we can also view \(\Psi\) as a function of dual frame coordinates, \[\tag{2.3}\begin{aligned}\Psi(w) = \frac{1}{d} \frac{1}{2} \text{tr}\left(\sum_i w_i \sigma_i \sum_j w_j \sigma_j\right)= \frac{1}{2}\sum_{ij} w_i \text{tr}(E_i \sigma_j) w_j = \frac{1}{2}w^T P w,\end{aligned}\]

where we’ve used the fact that \(E_i = \sigma_i/d\). Now let’s take its first derivatives of this potential, that is, the gradient. One kind of slick way of doing this is to recall that the differential is defined as \(d\Psi(w) = (\nabla_w \Psi(w))^T dw\), where \(dw=(dw_1, \dots, dw_n)\). At the same time, using the chain rule for differentials, \[\tag{2.4}\begin{aligned}d\Psi(w) = \frac{1}{2}\Big\{(dw)^T P w + w^T P dw\Big\} = \frac{1}{2}w^T(P^T + P)dw = w^T P dw,\end{aligned}\]

where we’ve used that fact that \(P=P^T\). Matching this expression to the definition, we conclude \[\tag{2.5}\begin{aligned}\nabla_w \Psi(w) = P w.\end{aligned}\]

Meanwhile, let us consider the Legendre dual of \(\Psi(w)\) whose argument we’ll denote (...for no particular reason...) \(p\): \[\tag{2.6}\begin{aligned}\Psi^*(p) = \sup_w \Bigg\{ p^Tw - \Psi(w)\Bigg\}.\end{aligned}\]

To evaluate the supremum, let’s take the gradient with respect to \(w\) and set it to zero. \(\nabla_w p^T w = p\) and we just worked out that \(\nabla_w \Psi(w) = Pw\) so that \[\tag{2.7}\begin{aligned}\nabla_w\Big\{ p^Tw - \Psi(w)\Big\} = p - Pw = 0,\end{aligned}\]

so that \(Pw = p\) or \(w = P^{-1}p\). Indeed, we see that the relation between \(w\) and \(p\) is exactly what we derived earlier in the frame-theoretic setting: \(p\) gives the probability representation of \(\rho\). (We note that if the POVM had not been a MIC, that is, if it had been overcomplete, then the same construction can be used with \(\Phi = P^+\), the Moore-Penrose pseudoinverse, since this acts as the usual inverse on the range of \(P\): we need not spell out the details here [12, 8].) In any case, plugging this back into the expression for \(\Psi^*(p)\), we have \[\tag{2.8}\begin{aligned}\Psi^*(p) & = p^T P^{-1} p - \frac{1}{2}(p P^{-1})^T P (P^{-1}p) = \frac{1}{2} p^T P^{-1} p,\end{aligned}\]

since the inverse of a symmetric matrix is also symmetric. In fact, writing \(\Phi\) for \(P^{-1}\), we have \[\tag{2.9}\begin{aligned}\Psi^*(p) &= \frac{1}{2} p^T \Phi p = \frac{1}{2}\sum_{ij} \text{tr}(E_i \rho) \Phi_{ij} \text{tr}(E_j\rho)\end{aligned}\] \[\tag{2.10}\begin{aligned}&= \frac{1}{d}\frac{1}{2}\text{tr}\left( \left[\sum_{ij} \Phi_{ij}\text{tr}(E_j \rho)\sigma_i\right] \rho \right)= \frac{1}{d}\frac{1}{2}\text{tr}(\rho^2),\end{aligned}\]

so that actually the potential is self-dual in this sense. In fact, we can already see this from \(p^T \Phi p = (w P)^T \Phi (Pw) = w^T P w\). Finally, by the same argument as above, \(\nabla_p \Psi^*(p) = \Phi p = w\).

So let us take stock. We defined a “purity potential” \[\tag{2.11}\begin{aligned}\Psi(\rho) = \frac{1}{d} \frac{1}{2}\text{tr}(\rho^2).\end{aligned}\]

Since \(\rho = \sum_i w_i \sigma_i\) for some reference states \(\sigma_i = E_i/d\), we can express the potential in terms of dual frame coordinates \(w\) or frame coordinates \(p\), \[\tag{2.12}\begin{aligned}\Psi(w) = \frac{1}{2}w^T P w && \Psi^*(p) = \frac{1}{2} p^T \Phi p,\end{aligned}\]

where \(\Phi = P^{-1}\) and \(w = \Phi p\). In fact, \(\Psi(w)\) and \(\Psi^*(p)\) are Legendre dual to each other, and \[\tag{2.13}\begin{aligned}\nabla_w \Psi(w) = p && \nabla_p \Psi^*(p) = w.\end{aligned}\]

Thus the first derivatives of the potentials switch us between the quasiprobability and probability representations. Moreover, taking another derivative gives us the Hessian matrices, \[\tag{2.14}\begin{aligned}\nabla^2 \Psi(w) = \nabla_w (Pw) = P && \nabla^2 \Psi^*(p) = \nabla_p (\Phi p) = \Phi.\end{aligned}\]

In other words, \[\tag{2.15}\begin{aligned}\frac{\partial \Psi(w)}{\partial w_i \partial w_j} = P_{ij} && \frac{\partial \Psi^*(p)}{\partial p_i \partial p_j} = \Phi_{ij}.\end{aligned}\]

For a MIC, these will both be positive definite and symmetric (and in fact this implies the potential functions are strictly convex), and so we can interpret them as metrics [13, 14].

For one thing, they encode the curvature of the potential. Expanding \(\Psi(w+dw)\) around \(w\), we have \[\tag{2.16}\begin{aligned}\Psi(w+ dw) &= \Psi(w) + (\nabla \Psi(w))^T dw + \frac{1}{2} dw^T \nabla^2 \Psi(w) dw\end{aligned}\] \[\tag{2.17}\begin{aligned}&= \frac{1}{2}w^T P w + w^TP dw + \frac{1}{2}dw^T P dw:\end{aligned}\]

the Taylor series terminates since the Hessian is a constant. The second derivative term represents the curvature of the function around \(w\). Similarly, \[\tag{2.18}\begin{aligned}\Psi^*(p + dp) &= \frac{1}{2} p^T \Phi p + p^T \Phi dp + \frac{1}{2} dp^T \Phi dp.\end{aligned}\]

We can then define a metric e.g., \[\tag{2.19}\begin{aligned}ds^2 = dp^T \Phi dp,\end{aligned}\]

so that the distance between two points will be larger when the curvature of the potential is greater. Thus if you are moving at a constant rate, you’ll go over curved parts more slowly, and flat parts faster. Although in this case, the metric is constant everywhere.

3. Bonus: a generating function for POVMs

Let us consider another potential, \[\tag{3.1}\begin{aligned}\Psi(X) = - \text{tr} \log (I - X) = - \log \det(I - X).\end{aligned}\]

You may know that for \(|x|<1\), \[\tag{3.2}\begin{aligned}- \log(1-x) = \sum_{m=1}^\infty \frac{x^m}{m}.\end{aligned}\]

The same holds if we replace \(1 \rightarrow I\) and \(x \rightarrow X\) as long as the largest absolute value of \(X\)’s eigenvalues is \(< 1\) [15]. (This will be true, for example, for a density matrix.) Thus \[\tag{3.3}\begin{aligned}\Psi(X) = \sum_{m=1}^\infty \frac{\text{tr}(X^m)}{m}.\end{aligned}\]

Our earlier potential (up to the prefactor) is just the second term in the Taylor expansion of this grand potential. Let’s expand \(X\) in terms of POVM elements \(X=\sum_i x_i E_i\) and substitute in (notice we’re using a slightly different convention from before), \[\tag{3.4}\begin{aligned}\Psi(x) &= \sum_{m=1}^\infty \frac{1}{m} \text{tr}\left(\left[\sum_i x_i E_i\right]^m\right)\end{aligned}\] \[\tag{3.5}\begin{aligned}&= \sum_i x_i \text{tr}(E_i) + \frac{1}{2} \sum_{ij} x_i x_j \text{tr}(E_i E_j) + \frac{1}{3}\sum_{ijk}x_i x_j x_k \text{tr}(E_i E_j E_k) + \dots.\end{aligned}\]

Taking the first derivatives, \[\tag{3.6}\begin{aligned}\frac{\partial \Psi(x)}{\partial x_a} &=\text{tr}(E_a) + \frac{1}{2}\sum_j x_j \Big\{\text{tr}(E_a E_j) + \text{tr}(E_a E_j)\Big\}\end{aligned}\] \[\tag{3.7}\begin{aligned}& + \frac{1}{3}\sum_{jk} x_j x_k \Big\{\text{tr}(E_a E_j E_k) + \text{tr}(E_j E_a E_k) + \text{tr}(E_j E_k E_a)\Big\} + \dots,\nonumber\end{aligned}\]

and so \[\tag{3.8}\begin{aligned}\frac{\partial \Psi(x)}{\partial x_a}\Big|_{x=0} = \text{tr}(E_a).\end{aligned}\]

Taking another derivative, we find \[\tag{3.9}\begin{aligned}\frac{\partial^2 \Psi(x)}{\partial x_a \partial x_b} &= \frac{1}{2} \Big\{\text{tr}(E_a E_b) + \text{tr}(E_b E_a)\Big\}\end{aligned}\] \[\tag{3.10}\begin{aligned}&+ \frac{1}{3}\sum_k x_k \Big\{\text{tr}(E_a E_b E_k) + \text{tr}(E_a E_k E_b) + \text{tr}(E_b E_a E_k) + \text{tr}(E_k E_a E_b) + \text{tr}(E_b E_k E_a) + \text{tr}(E_k E_b E_a) \Big\} \dots,\nonumber\end{aligned}\]

so that, by the cyclicity of the trace, \[\tag{3.11}\begin{aligned}\frac{\partial^2 \Psi(x)}{\partial x_a \partial x_b}\Big|_{x=0} = \text{tr}(E_a E_b).\end{aligned}\]

Taking another deriative, we find \[\tag{3.12}\begin{aligned}\frac{\partial^3 \Psi(x)}{\partial x_a \partial x_b \partial x_c} = \frac{1}{3} \Big\{\text{tr}(E_a E_b E_c) + \text{tr}(E_a E_c E_b) + \text{tr}(E_b E_a E_c) + \text{tr}(E_c E_a E_b) + \text{tr}(E_b E_c E_a) + \text{tr}(E_c E_b E_a) \Big\} + \dots.\end{aligned}\]

Now the six terms in the brackets divide into two groups of three: the three in each group are all cyclically permuted versions of each other. We are left with \[\tag{3.13}\begin{aligned}\frac{\partial^3 \Psi(x)}{\partial x_a \partial x_b \partial x_c}\Big|_{x=0} = \text{tr}(E_a E_b E_c) + \text{tr}(E_a E_c E_b) = 2 \Re[ \text{tr}(E_a E_b E_c)].\end{aligned}\]

In other words, \(\Psi(x)\) acts as a generating function for symmetrized traces of \(n\)-products of POVM elements: we extract them by taking ever higher order derivatives and evaluating at \(x=0\). For an IC-POVM, the \(2\)-product gives a Gram matrix from which a dual frame can be constructed. The symmetrized \(3\)-product teaches one how to take the Jordan product, from which the geometry of quantum state space flows [16].

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