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Consider a Hermitian operator $H_\mathcal{AB}$ acting on the Hilbert space $ \mathcal{A}\otimes\mathcal{B}$. What is the smallest commuting set of separable operators in the form $A_k\otimes B_k$, such that $$ H_\mathcal{AB}=\sum_k A_k\otimes B_k? $$ As pointed out in the comments, a set like this might not even exist.

It would also be interesting to constrain $A_k$ and $B_k$ to be Hermitian themselves. Note that if we didn't require commutativity, we could use the operator singular value decomposition.

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    $\begingroup$ In fact, I'm not convinced that it's necessarily possible to do this with any number of commuting operators of that form. $\endgroup$ – Omnomnomnom May 1 '17 at 3:46
  • $\begingroup$ What is a separable operator ? $\endgroup$ – user42761 May 1 '17 at 9:53
  • $\begingroup$ @AndréS. In this context $O$ is separable if it can be written $O=A\otimes B$, with $A$ and $B$ acting on $\mathcal A$ and $\mathcal B$. $\endgroup$ – Ziofil May 1 '17 at 15:32
  • $\begingroup$ @Omnomnomnom, are there explicit examples of operators that cannot be decomposed using a commuting set? $\endgroup$ – Ziofil May 8 '17 at 18:25
  • $\begingroup$ @Ziofil I don't know whether the statement is true, let alone whether there are explicit examples. $\endgroup$ – Omnomnomnom May 8 '17 at 18:28

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