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I am looking for the proof for the $e^{A+B} = \lim\limits_{n \to \infty} (e^{A/n} e^{B/n})^{n}$ and also the name of the proof. When $[A,B] \ne 0$, i want to understand how to expand the exponential of the matrices which helps me understand the Hamiltonian split-up in quantum computing . Please share the reference or links.

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    It is called the Trotter product formula https://en.m.wikipedia.org/wiki/Lie_product_formula – Mason Aug 29 '22 at 05:22
  • This is indeed known as the Trotter product formula in the case of unbounded operators, in the case of bounded operators it's commonly referred to as the Lie product formula, see https://math.stackexchange.com/questions/2030671/proving-the-lie-product-formula/2030685#2030685 . Are you looking for the proof in the unbounded case? – user159517 Aug 29 '22 at 05:37
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    Regarding your comment about the noncommutativity - the point of this statment, more or less, is that $e^{(A+B)/n}$ and $e^{A/n} e^{B/n}$ have matching first order terms - i.e., all terms where the lack of commutativity matters are $O(n^{-2}),$ and so can essentially be ignored. I don't think this statement can shed any light on how to expand exponentials in any convenient way. – stochasticboy321 Aug 29 '22 at 06:05
  • See here. – Jean Marie Aug 29 '22 at 09:41
  • Does this answer your question? Proving the Lie-Product formula – postmortes Sep 09 '22 at 06:08
  • @postmortes Thanks for the link. Not completely but it gives me an idea –  Sep 13 '22 at 10:30
  • @Mason Thanks for your time. now able to identify. Thank you –  Sep 13 '22 at 10:31
  • @stochasticboy321 Yeah it's true. That's the same reason QAOA is Quantum Approximation Algorithm –  Sep 13 '22 at 10:32

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I the case of matrices it is sometimes called the Lie product formulas. In the case of operators it is called the Trotter formula. You can find a proof in the book "Functional analysis" by Reed and Simon, in theorems VIII.29 and VIII.30.

LL 3.14
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