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I'm reading Huang et al. (2020) (nature physics), where the authors present a version of Aaronson's *shadow tomography* scheme as follows (see page 11 in the arXiv version):

We want to estimate a state $\rho$. We apply a number of random unitary evolutions, $\rho\mapsto U\rho U^\dagger$, picking $U$ from an ensemble $\mathcal U$. For each choice of unitary $U$, we perform a measurement, observing a state $|b\rangle$. We then apply the inverse evolution to this state, obtaining $U^\dagger|b\rangle$. On average, this procedure leaves us with the state $$\mathcal M(\rho) =\mathbb E_{U\sim\mathcal U}\Bigg[ \sum_b \underbrace{\langle b|U\rho U^\dagger|b\rangle}_{\text{prob of observing $|b\rangle$}} \!\!\!\!\!\!\! \overbrace{(U^\dagger |b\rangle\!\langle b|U)}^{\text{post-measurement state}} \Bigg].$$ The claim is then that $\mathcal M$ can be inverted to obtain, on average, the original state: $$ \hat\rho=\mathcal M^{-1}(U^\dagger |\hat b\rangle\!\langle \hat b| U) \,\,\text{ is such that } \,\, \mathbb E[\hat \rho]=\rho. $$

Is there an easy way to see how one would go in performing such inversion? The authors mention that we are thinking here of $\mathcal M$ as a linear map, so I suppose we represent $\rho$ as a vector in some operatorial basis of Hermitians, which is fine, but to then perform the inversion of this linear map we would need to characterise it, which in this case I think would mean to know the values of $\mathcal M(\sigma_i)$ for some complete basis of Hermitians $\{\sigma_i\}_i$.

In practice, I think what you described is the best way. Just compute $\mathcal M(|i\rangle!\langle j|)$ and find the inverse by standard linear algebra methods. – Danylo Y – 2020-12-02T10:19:25.440