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Let $A$ and $B$ be two anti-Hermitian operators on a finite-dimensional Hilbert space. BCH formula gives an explicit expression for $e^A e^B$ as $e^C=e^A e^B$, for $C$ in the Lie algebra generated by $A$ and $B$. However, this formula holds only for sufficiently small $A$ and $B$.

My question is for general anti-Hermitian operators $A$ and $B$, which are not necessarily small, is it still true that there exists an operator $C$ in the Lie algebra generated by $A$ and $B$ such that $e^C=e^A e^B$?

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    $\begingroup$ The Lie algebra of anti-Hermitian operators is $\mathfrak{u}(n)$, and the corresponding Lie group is $U(n)$ which is compact and connected, so the exponential map is surjective. $\endgroup$ Feb 4, 2020 at 23:34
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    $\begingroup$ I agree that there exists an anti-Hermitian $C$ with this property. The question is if $C$ is an element of the Lie algebra generated by two operators $A$ and $B$. $\endgroup$
    – user149918
    Feb 4, 2020 at 23:41
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    $\begingroup$ Oh, I see. I thought it seemed too simple. $\endgroup$ Feb 4, 2020 at 23:49
  • $\begingroup$ For anyone interested: The Lie algebra generated by finite dimensional anti-Hermitian operators (i.e. the linear combinations of all nested commutators - which is necessarily finite dimensional as a subalgebra of $\frak{u}(n)$) is still compact, and the exponential map will still be surjective. $\endgroup$ Feb 29 at 14:27

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