# Why the chi-matrix fidelity of the process is the fidelity of the chi-matrix noise map

I am following this paper, and I am stuggling with a derivation.

Basically, I consider an orthonormal basis $$\{B_i \}$$ with respect to Hilbert-Schmidt scalar product, on the density matrix space $$\mathcal{L}(H)$$.

I have a noisy process: $$\mathcal{E}=\mathcal{N} \circ \mathcal{U}$$ which tries to implement the unitary map $$\mathcal{U}$$. The noise map $$\mathcal{N}$$ is actually defined from this last equation.

We can decompose those process on the basis $$\{B_i\}$$ which define the so-called $$\chi$$ matrix. We have:

$$\mathcal{E}(\rho) = \sum_{ij} \chi_{\mathcal{E}}^{ij} B_i \rho B_j^{\dagger}$$ $$\mathcal{U}(\rho) = \sum_{ij} \chi_{\mathcal{U}}^{ij} B_i \rho B_j^{\dagger}$$ $$\mathcal{N}(\rho) = \sum_{ij} \chi_{\mathcal{N}}^{ij} B_i \rho B_j^{\dagger}$$

I am struggling to prove that the fidelity of the process which is defined as: $$F(\mathcal{E},\mathcal{U})=Tr(\chi_{\mathcal{E}}\chi_{\mathcal{U}})$$ verifies:

$$F(\mathcal{E},\mathcal{U})=F(\mathcal{N},\mathcal{I})$$

$$\mathcal{I}$$ being the identity map.

In the paper he work with the particular basis $$\{B_i\}$$ being the Pauli basis. I considered a general orthonormal basis in my derivation. I don't know if it is why I don't find the appropriate result. I would like to avoid using any external result such as the fidelity of the process is equal to the average fidelity of the output state with respect to the ideal one.

Is it possible to have a simple proof of that ? Does it indeed works even without considering the particular Pauli basis as the orthonormal basis ?

: I just realized that chi matrix associated to different basis of $$\mathcal{L}(H)$$ are related via unitary transformation. Thus prooving it on the Pauli basis is enough. So I just need to understand how to derive it on the Pauli basis.

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– JSdJ
Sep 9 at 13:47