Finding the "dual" basis of an overcomplete basis for Quantum State Tomography

Question

This question is related to this stack exchange post: What does the POVM corresponding to single-qubit state tomography look like?

From what I understand, when we are interested in reconstructing a state $\rho$ using the quantum state tomography protocol, we have a POVM set, $\mathcal{J} :=\{\Pi_i\}_{I=1}^L$. Then, we can gather (experimentally) the probabilities for each $\Pi_i$, and it should be in accordance with the Born rule, $p_i = \text {Tr}[\Pi_i\rho]$.

Using these probabilities, and the chosen POVM, the state can be expressed as

$$\rho = \sum_{i=1}^L p_i\Delta_i$$

Where $\mathcal{D}:=\{{\Delta_i}\}_{I=1}^L$ is defined as the dual set to the POVM. This set satisfies $\text{Tr}[\Pi_i\Delta_j] = \delta_{ij}$.

My question is, in practice we often use the POVM defined by

$$\mathcal{J} = \{ \dfrac{1}{3}|0\rangle \langle 0|,\dfrac{1}{3}|1\rangle \langle 1|,\dfrac{1}{3}|+\rangle \langle +|,\dfrac{1}{3}|-\rangle \langle -|,\dfrac{1}{3}|+i\rangle \langle +i|, \dfrac{1}{3}| -i\rangle\langle -i| \}$$

which is an overcomplete set, and it's not linearly independent. How do we go about constructing the dual set for this POVM? An explicit example on how to find the dual set would be very appreciated :-)

Answer 1

A general procedure to find the dual of any given frame $\{v_k\}_k$ is to compute the frame operator $S$, and then the (canonical) dual frame elements as $\tilde v_k\equiv S^{-1} v_k$. The dual frame is not unique (in general), but this procedure always works to give you a specific dual frame. In the case at hand, you're working with frames of operators, so the frame operator is a "superoperator" (think of it as a map) but the procedure is unchanged.

An explicit calculation for a case very similar to the one you're considering here is given in appendix H2 of https://arxiv.org/abs/2301.13229. You observe that the frame (super)operator equals (or rather, can be represented as) $S=\sum_i \operatorname{vec}(\Pi_i)\operatorname{vec}(\Pi_i)^\dagger$. For example if $\Pi_1=\frac13|0\rangle\!\langle 0|$ then $$\operatorname{vec}(\Pi_1)\operatorname{vec}(\Pi_1)^\dagger = \frac19\begin{pmatrix}1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}.$$ Carrying out similar calculations you get $$S = \frac19\begin{pmatrix}2&0&0&1\\ 0&1&0&0\\0&0&1&0 \\ 1&0&0&2\end{pmatrix},$$ and from this you can compute $\operatorname{vec}(\Delta_i)=S^{-1}\operatorname{vec}(\Pi_i)$, and finally obtain the dual frame elements: $$\Delta_1=\begin{pmatrix}2 & 0 \\ 0& -1\end{pmatrix}, \quad\Delta_2=\begin{pmatrix}-1 & 0 \\ 0& 2\end{pmatrix}, \\\Delta_3=\frac12\begin{pmatrix}1 & 3 \\ 3& 1\end{pmatrix}, \quad\Delta_4=\frac12\begin{pmatrix}1 & -3 \\ -3& 1\end{pmatrix}, \\ \Delta_5=\frac12\begin{pmatrix}1 & -3i \\ 3i& 1\end{pmatrix}, \quad\Delta_6=\frac12\begin{pmatrix}1 & 3i \\ -3i& 1\end{pmatrix}.$$ These are a set of operators that you can use to decompose any $\rho$ as a linear combination of the form $$\rho = \sum_i \operatorname{tr}(\Pi_i\rho)\Delta_i =\sum_i \operatorname{tr}(\Delta_i\rho)\Pi_i.$$