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I am trying to understand the meaning of the equation shown in the above image taken from this paper, but I am unfamiliar with the tensor network notation. My current strategy is trying to write down matrix representations for the LHS and RHS separately and see from there that they are the same, but I am having difficulty doing it. To me, the LHS looks like $I\otimes I$ and the RHS looks like $$\frac{I\otimes I+X\otimes X+Y\otimes Y+Z\otimes Z}{2},$$ but clearly they are not equal.

Could anybody shed more light on this?

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In your computation of the RHS of the equation, you treat the two $\sigma$s as if they were a tensor product of two 1-qubit operators, whereas you should consider the first as an effet (the dagger of a state), and the second as a state.

So this calculation should look like this in terms of matrices: $$\frac12\left( \begin{pmatrix}1\\0\\0\\1\end{pmatrix}^\dagger\otimes\begin{pmatrix}1\\0\\0\\1\end{pmatrix}+ \begin{pmatrix}0\\1\\1\\0\end{pmatrix}^\dagger\otimes\begin{pmatrix}0\\1\\1\\0\end{pmatrix}+ \begin{pmatrix}0\\-i\\i\\0\end{pmatrix}^\dagger\otimes\begin{pmatrix}0\\-i\\i\\0\end{pmatrix}+ \begin{pmatrix}1\\0\\0\\-1\end{pmatrix}^\dagger\otimes\begin{pmatrix}1\\0\\0\\-1\end{pmatrix} \right)$$

Aternatively, using the map/state duality, you could prove this equality by showing: $$\frac{I\otimes I^\dagger+X\otimes X^\dagger+Y\otimes Y^\dagger+ Z\otimes Z^\dagger}2 = \begin{pmatrix}1\\0\\0\\1\end{pmatrix}^\dagger\otimes\begin{pmatrix}1\\0\\0\\1\end{pmatrix}$$

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I would read the picture from left to right, as a quantum channel acting on density matrices -- here, the lower layer is ket, and the upper bra.

Then, the left side is the identity map $$ \mathcal E_L(\rho)=\rho\ , $$ and the right side is the map $$ \mathcal E_R(\rho) = \sum \sigma_i \mathrm{tr}(\sigma_i\rho)\ . $$ So the claim is that $E_R(\rho)$ is the identity channel.

This can be seen in various way, e.g. such as in Renaud's answer, or (from a more high-level perspective) understanding the space of density matrices as a Hilber space with scalar product $\mathrm{tr}[X^\dagger Y]$, and observing that the Paulis (including the identity) for a basis for that space, so the right hand side is just a resolution of the identity -- in a vector notation, this would amount to $$ |\rho\rangle = \sum_i |\sigma_i\rangle\langle\sigma_i|\rho\rangle $$ with an orthonormal basis $|\sigma_i\rangle$.

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