# Prove $\sum_{ij}(\mathcal{A_G})_{ij}(|\rho\rangle\!\rangle)_j|i\rangle\!\rangle={\cal A}_G|\rho\rangle\!\rangle$ in the Pauli-Liouville representation

Define the Pauli-Liouville representation of a (linear) map $$\mathcal{G}$$ as $$\mathcal{A_G}$$, which has components $$$$\label{2} (\mathcal{A_G})_{ij}:=\mathrm{tr}[P_i\mathcal{G}(P_j)]$$$$ $$P_i,P_j$$ are normalized Pauli matrices, then the post-state of $$\mathcal{G}$$ acting on a state $$\rho$$, written all in the Pauli basis, can be shown to be equal to a matrix multiplication: \begin{align}\label{Pauli-Liouville map} |\mathcal{G}(\rho)\rangle\!\rangle&=\sum_i\mathrm{tr}[\mathcal{G}(\rho)P_i] |i\rangle\!\rangle\nonumber\\ &= \sum_i\mathrm{tr}[\mathcal{G}(\sum_j\mathrm{tr}[\rho P_j]P_j)P_i] |i\rangle\!\rangle\nonumber \\ & = \sum_{ij}\mathrm{tr}[\rho P_j]\mathrm{tr}[\mathcal{G}(P_j)P_i]|i\rangle\!\rangle\nonumber \\ & =\sum_{ij}(\mathcal{A_G})_{ij}(|\rho\rangle\!\rangle)_j|i\rangle\!\rangle\nonumber \\ & =\mathcal{A_G}|\rho\rangle\!\rangle. \end{align} How comes the last equation?

• It's the definition of matrix multiplication: $y=Ax$ means that $y_i=\sum_j A_{ij}x_j$. BTW: the second scalar in the middle equation should be $\mathrm{tr}[\mathcal{G}(P_j)P_i]$. You should also define the symbols you're using (in particular, $P_i$ can't be a usual Pauli operator since you're assuming normalization w.r.t. the Hilbert-Schmidt inner product). Commented Oct 7, 2023 at 10:31
• Thanks. I revised the scalar. Commented Oct 7, 2023 at 11:21