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( I copied some text from a previous answer of mine) Defining the Choi and $\chi$ matrix The Choi matrix is a direct result of the Choi-Jamiolkowski isomorphism. Some intuition on what this is can be found in this previous answer. Consider the maximally entangled state $|\Omega \rangle = \sum_{\mathrm{i}}|\mathrm{i}\rangle \otimes |\mathrm{i}\rangle$, where $... 5 I am sure that since you are asking this question you probably already understand this, but for future & other's reference let me give a quick recap of what we are trying to achieve. Quantum channels Any process (in an open quantum system) is some map$\Lambda$from a space of density matrices to a space of density matrices. I write a, because these ... 5 The 2-norm difference typically isn't particularly physical. So no, this is most likely not the right distance. What you want from a physical point of view is a distance measure which measures the maximal distance between the two channels on any possible input (possibly on a larger system), such as the diamond norm. Note that the fact that "norms are ... 2 Long story short: if you want to characterize a circuit or operation, you have to perform Quantum process tomography (QPT), which is a generalization of quantum state tomography, which is used to characterize states. QPT is not very easy unfortunately, there is a little math involved. Please see this answer to a previous question for a general overview. A ... 2 The linear map$\mathcal{E}$is what characterizes a quantum process, $$\rho \rightarrow \mathcal{E}(\rho),$$ but$\mathcal{E}$can be determined by$\chi$. Using the operator-sum representation, $$\mathcal{E}(\rho) = \sum_i A_i \rho A^\dagger_i = \sum_{i}\sum_{m}\sum_{n}a_{im}\tilde{A}_m \rho a^*_{in}\tilde{A}^\dagger_n,$$ where the$a_{ij}are some set of ... 2 "Can gate set tomography be applied to a quantum channel" Yes, because gates are just unitary channels, if you wanted you could just let a qubit idle and undergo decay/dephasing processes instead of logic gates, and still perform whatever tomography you want. In general, any ideal unitary is actually a channel because gates aren't perfect (see ... 2 You want to calculate $$\rho_{out}=2\text{Tr}_0(\rho_{in}^T\otimes I\cdot\rho^{sys}_{choi}).$$ 2 Considering just the single qubit case, the four possible operators you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - ... 1 Let\Phi$be a channel acting on a state$\rho$(or more generally, a map acting on a linear operator; we don't actually need restrict to CPTP maps and states for these calculations). Let$J(\Phi)$be the Choi representation of$\Phi$, i.e. $$J(\Phi)\equiv \sum_{ij} \Phi(E_{ij})\otimes E_{ij},$$ where$E_{ij}\equiv|i\rangle\!\langle j|$. Denote with$J(\Phi)...

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For any map $\Lambda(\cdot)$ with Choi representation $\rho_{C}$ defined as you are doing it (i.e. the channel on the second biparition of the maximally entangled state), the output $\rho_{out} = \Lambda(\rho_{in})$ can be calculated as: $$\rho_{out} = \mathrm{tr}_{1}\big[\rho_{C}\big(\rho_{in}^{t} \otimes I\big)\big]$$ where the trace is over the ancilla ...

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We know what a measurement of a particular operator should yield: measuring $\sigma_A$ should yield the value $\langle\sigma_A\rangle$, and measuring $I$ should yield 1. Now, I assume the quoted protocol actually measures overlaps with the various eigenstates of the operators, and that $p_0$ and $p_1$ are the probabilities of being in the $+$ or $-$ ...

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Yes, $|i\rangle\langle j|$ is not a density matrix in general. But those matrices indeed form a basis of the space of all matrices. And if we know those $d^2$ values $\mathcal E(|i\rangle\langle j|)$ then we can fully reconstruct the action of $\mathcal E$. So, this is a sort of "mathematical" tomography. Note, however, that we can pick $d^2$ ...

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