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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 :-)

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  • $\begingroup$ You may find section A. Linear state tomography in this paper useful. You've misunderstood the procedure. We don't require $\text{Tr}[\Pi_i\Delta_j] = \delta_{ij} $, instead we require $\sum_i{|\Delta _i\rangle \rangle \langle \langle \Pi _i|}=I$ where the notation $|\cdot \rangle \rangle $ is used to stand for matrix vectorization. $\endgroup$
    – narip
    Dec 2, 2023 at 11:00

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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 $$\rho = \sum_i \operatorname{tr}(\Pi_i\rho)\Delta_i =\sum_i \operatorname{tr}(\Delta_i\rho)\Pi_i.$$

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  • $\begingroup$ Thank you so much for your answer! I just had a quick follow-up question -- does this procedure work with any informationally complete basis? I.e. as long as we have an informationally complete set of matrices, this procedure should yield a dual basis that sums to the identity? $\endgroup$
    – junoswrld
    Dec 4, 2023 at 19:32
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    $\begingroup$ @junoswrld This procedure always works to get a dual frame, assuming the original POVM is informationally complete, yes. See eg the linked paper. The dual is not generally a POVM however. I don't think they generally need to sum to the identity. $\endgroup$
    – glS
    Dec 5, 2023 at 10:22
  • $\begingroup$ Is there any way to ensure completeness of the dual frame? I am referencing this paper: journals.aps.org/prxquantum/abstract/10.1103/…, Appendix A. Their procedure yields the same dual frame as you've described, but they say that they require the dual frame to sum to the identity... I'm unsure of how to ensure this condition. $\endgroup$
    – junoswrld
    Dec 5, 2023 at 15:58
  • $\begingroup$ @junoswrld I don't quite understand what they mean in that appendix tbh. They also say their dual matrices satisfy $\operatorname{tr}[\mathcal{B}_i\Delta_j]=\delta_{ij}$, but I don't think that's the case in general unless the original frame is a basis. Case in point, my example here, where you can see explicitly $\sum_i \Delta_i=I$ doesn't hold (though you do get a multiple of the identity), and nor does the above orthogonality condition. Another worked out example is Appendix H3 in arxiv.org/abs/2301.13229. I also don't see why they should care about having $\sum_i \Delta_i=I$ at all $\endgroup$
    – glS
    Dec 6, 2023 at 1:17
  • $\begingroup$ though one can notice that if the POVM elements are rank-1 operators of the form $\Pi_i=\alpha |\phi_i\rangle\!\langle\phi_i|$, then $\alpha=n/m$ with $n$ space dimension and $m$ number of POVM outcomes, $\operatorname{tr}(\Pi_i)=\alpha$, and any dual frame has to satisfy $\sum_i \Delta_i =m I/n$. You see it using $\rho=I$ in the last equation in the answer. That's why in my example you get $\sum_i \Delta_i=3I$ (as $m=6$ and $n=2$). It's not the same as summing to the identity, but I guess it's close enough? $\endgroup$
    – glS
    Dec 6, 2023 at 1:22

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