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The existence of the inverse of a linear map is independent of the way the map affects the trace. Moreover, if an invertible map preserves a property then its inverse necessarily also preserves the property. Since depolarizing channel preserves the trace, so does its inverse. Inverse of depolarizing channel We can derive the formula for the inverse $\mathcal{... 6 Let's say you have a magical machine that gives you$\langle P_{p} \rangle$(which are expectation values and therefor, well, numbers) and only the$\langle P_{p} \rangle$. It does this for all the$3$- (n)-fold tensor products of the traceless Paulis. That is:$p \in \{X,Y,Z\}^{\otimes 3 (n)}$, for a total of$3^{3 (n)}$Paulis, where$n$is the general ... 5 Without really checking your arguments, there is a fundamental reason why the scaling could be fine, but it is still not strange at all. The point is that you estimate with additive precision, but your fidelity could be exponentially small. E.g. if the fidelity is$10^{-8}$, you need a much smaller error to resolve that compare to a fidelity of, say, 1/2. ... 4 Under the assumption that the ensemble$\mathcal{U}$faithfully produces the Haar expectations at least to the second moment, the inversion can be performed as suggested in the last paragraph of the question: Define: $$\theta_b = U^{\dagger}|b\rangle\langle b| U$$ and replace the averaging over the ensemble by Haar averaging. (This step is is done only to ... 3 This is not a complete answer but a couple of comments that should help clarify these details: (i)$\mathcal{M}$is not invertible as a quantum channel but as a linear map; namely, even though the inverse exists,$\mathcal{M}^{-1}$is not CP, even though it is a linear map. Assuming the input-output dimensions are the same, the only CP maps that are ... 3 There's a few bugs in your code as well as a slight misunderstanding about the guarantees of the protocol. First to clarify some details: The protocol you implement samples$U \in \text{Cl}(2)^{\otimes 2}$(not the same as$U\in \text{Cl}(2^2)$!) and then applies these operators pre-measurement. Up to global phase this is equivalent to performing local ... 3 Let$\mathcal X$be an$N$-dimensional (complex) vector space space. The (real) vector space of Hermitian operators defined on it,$\mathrm{Herm}(\mathcal X)$, has dimension$N^2$. An easy way to see this is to realise that generic complex matrices are characterised by$2N^2$real parameters, and$A^\dagger=A$imposes$N^2$independent constraints. ... 2 The method stim.Tableau.random(n) generates a uniformly random n-qubit Clifford operation, using the algorithm from "Hadamard-free circuits expose the structure of the Clifford group", and returns its stabilizer tableau. example import stim t = stim.Tableau.random(10) print(repr(t)) sample output stim.Tableau.from_conjugated_generators( xs=[ ... 2 Ah, the channel is trace preserving so its straightforward to invert in this case. Let$Y = \mathcal{D}_p (X)\$ so that \begin{align} \text{Tr}(Y) &= p\text{Tr}(X) + (1-p) \frac{\text{Tr}(X)}{2^n} \text{Tr}\left(\mathbb{I}_{2^n}\right) \\&= \text{Tr}(X) \end{align} So that \begin{align} Y &= p X + (1-p) \frac{\text{Tr}(Y)}{2^n} \mathbb{I}_{2^n} \...