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First we should take a step back. Is there any machine learning done a quantum computer that cannot be efficiently simulated on a classical computer? The answer currently (2020) is no. In this respect quantum machine learning (which has many variants) is at the fundamental research phase. None of this is at a stage where it is at all considered something ...


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If I understand what you are asking for (I don't know much about tensor networks), both equations are singular value decompositions of $U$, just with respect to different indices (and in the latter case highlighting the singular values, which in the first equation is "hidden" in $A,B$). There are two things to notice here. First of all, given an arbitrary ...


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A tensor is a multi-dimensional array. A vector is a 1d tensor, a matrix is a 2d tensor, and so forth. Tensor contraction is a generalization of matrix multiplication. In matrix multiplication, you pair up the second axis of the first matrix with the first axis of the second matrix and sum over their pointwise product to produce new entries. A tensor ...


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A Different Way Of Looking At Linear Algebra Tensor Networks provide a different way of looking at linear algebra particularly within the context of tensor space systems. Quantum Circuits Are Just Products of Vectors and Operators A quantum circuit is inherently a tensor space system as when we have multiple qubits we must think of the whole circuit with ...


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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 ...


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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}^\...


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By treating a quantum circuit as a network model, the network model can be optimized the order of the calculation by making a contract between tensor and tensor. I think it is definitely important to learn TN for future qc.


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