How to reconstruct the density matrix $\rho$ from the overlap matrix $T_{a,a'}={\rm Tr}(M^{(a)}M^{(a')})$?

Question

Suppose we have $N$-qubit POVM $${\bf M} = \{M^{(a_1)} \otimes M^{(a_2)} \otimes \cdots \otimes M^{(a_N)}\}_{a_1, \ldots, a_N}.$$ Given an $N$-qubit state $\rho$, the measurement outcome ${\bf a} = (a_1, a_2, \ldots, a_N)$ occures with probability $P({\bf a})$ such that $\sum_{{\bf a}}P({\bf a})=1$. Equivalently, we can write $$P({\bf a}) = Tr(M^{({\bf a})} \rho).$$

Provided that the measurement is informationally complete, the density matrix can be unambiguously inferred from the probability distribution of measurement outcomes: $$\tag{1} \rho = \sum_{{\bf a, a'}} P({\bf a}) T^{-1}_{{\bf a, a'}} M^{({\bf a'})},$$ where the matrix $T$ is called an overlap matrix defined as follows: $$T_{{\bf a, a'}} := Tr(M^{({\bf a})}M^{({\bf a'})}).$$

I would like to know how (1) can be understood from intuitive and mathematical points of view. My main point of confusion is the matrix $T^{-1}_{{\bf a,a'}}$.

Answer 1

Let $\rho$ be an arbitrary state, and $\{\mu_b\}_b$ some POVM, so that the outcome probabilities are $p_b(\rho)=\operatorname{tr}(\mu_b \rho)$. You're asking what are the (Hermitian) operators $M_b$ such that you get the decomposition $\rho=\sum_b p_b(\rho) M_b$ for all $\rho$.

This is completely analogous to the following question: given a finite-dimensional vector space $V$ and a finite subset $\{v_k\}\subset V$ that spans $V$, is there some set $\{w_k\}$ such that for all $v\in V$ we have $v=\sum_k \langle v_k,v\rangle w_k$? One way to answer this question is to observe this relation is equivalent to $v=WV^\dagger v$, with $V$ and $W$ matrices whose columns are the vectors $\{v_k\}$ and $\{w_k\}$, respectively. If we want this for all $v\in V$, that means we're asking for vectors $w_k$ such that $WV^\dagger =I$. This is an inhomogeneous linear system, a solution of which can always be written $$W=(V^\dagger)^+\equiv (V V^\dagger)^{-1} V,$$ assuming $VV^\dagger$ is invertible, that is, $V$ is surjective (which is true iff $\{v_k\}$ span $V$). You might now observe that $VV^\dagger=\sum_k v_kv_k^\dagger$. Defining $S\equiv VV^\dagger$, we can restate this result as saying that a viable set of "dual" vectors $w_k$ is given by $w_k = S^{-1}v_k$, and therefore $$v = \sum_k \langle v_k,v\rangle S^{-1}(v_k). %= \sum_k \langle S^{-1}(v_k),v\rangle v_k.$$

All of this can be applied to decompose states in terms of POVM elements, just replacing $v\to \rho$ and $v_k\to \mu_b$, and using the trace inner product: $\langle\mu_b,\rho\rangle\equiv \operatorname{tr}(\mu_b\rho)$. You'll then end up with the decomposition $$\rho = \sum_b \langle\mu_b,\rho\rangle S^{-1}(\mu_b), \qquad S \equiv \sum_b |\mu_b\rangle\!\rangle\langle\!\langle\mu_b|,$$ where $S$ is the superoperator defined as $S(X)\equiv \sum_b \mu_b \langle\mu_b ,X\rangle$ for all $X$. I think this is essentially the expression you mentioned, though your $T$ should not be the "overlap matrix" but rather the $S$ operator above. Unless you're making some additional assumption (the expressions might be identical for minimal IC-POVMs, though I haven't really checked).

I should mention that this is a direct application of the formalism of frame theory to measurements and state reconstruction. See e.g. (Scott 2006) and (Zhu 2004).