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Suppose we want to express an N-qubit operator as

$$U=\sum_i \lambda_i \bigotimes_{k=1}^N W_{i,k}$$

where $W_{i,k}$ are each a two-by-two matrix. How can one find a minimal decomposition, that is one that has as few terms in the sum as possible?

I appreciate this may be difficult in general, so a partial solution or solution to a variant of the above statement could also be interesting if it is easier to tackle. The two-qubit case is addressed in an answer to this question (which inspired me to ask this question) using the Cartan decomposition, but it's not obvious how to generalize it to more than two qubits, nor that the machinery of the Cartan decomposition is even necessary.

user34722
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  • Assume $W$ are Pauli. Then there are $4^N$ possible N-fold tensor products of Paulis. Possibly multiple subsets of these $4^N$ possibilities will sum up to $U$. Finding the smallest such subset seems like a NP-complete problem. – Abdullah Khalid Mar 28 '23 at 05:42
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    Actually the Pauli decomposition of an operator is unique. Unique tensor products of Paulis form an orthonormal basis, so you can simply decompose $U$ using the inner product $\text{Tr}[UW]$. – user34722 Mar 28 '23 at 06:09
  • Of course, what was I thinking! Too late in the night. – Abdullah Khalid Mar 28 '23 at 06:12
  • In the $SU(4)$ case you can obtain the Cartan angles from the singular values of the matrix $\alpha_{ij} = \mathrm{Tr}(U \sigma_i \otimes \sigma_j)$. Maybe a higher order SVD for the tensors you get in the n-qubit case would be a starting point. – ChrisD Mar 29 '23 at 01:45

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