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The commutator formula states that for $A,B$ elements of a Lie algebra,

$$ \lim_{n\to \infty}\left\{ \exp\left(-A\tfrac{t}{n}\right)\exp\left(-B\tfrac{t}{n}\right)\exp\left(A\tfrac{t}{n}\right)\exp\left(B\tfrac{t}{n}\right)\right\}^{n^2}=\exp\left(t^2[A,B]\right)$$

I am interested in the case where $A=iH_1$ and $B=iH_2$ with $H_i$ self-adjoint. For finite dimensions the above certainly holds, but what happens in infinite dimensions? Under which conditions? Bounded/unbouded operators? I know that Trotter's formula has some complications in infinite dimensions, I'd be very thankful for any hints here.

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up vote -1 down vote accepted

This is answered in "Topics in Dynamics 1: Flows" by Edward Nelson (Theorem 7, Chapter 8): if $A$ and $B$ are skew-adjoint operators on a Hilbert space $\mathcal{H}$ and the restriction of $[A,B]$ to $\mathcal{D}=\mathcal{D}(AB)\cap\mathcal{D}(BA)\cap\mathcal{D}(A^2)\cap\mathcal{D}(B^{2})$ is essentially skew-adjoint, then

$\lim_{n\rightarrow\infty}[e^{-\sqrt{\frac{t}{n}}A}e^{-\sqrt{\frac{t}{n}}B}e^{\sqrt{\frac{t}{n}}A}e^{\sqrt{\frac{t}{n}}B}]^nu=e^{\overline{[A,B]}t}u$ for all $u\in\mathcal{H}$, uniformly for $t$ in any compact subset in $[0,\infty)$.

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – TravisJ Jul 9 at 20:07
    
Others also may need to know more of what you know while making it known to OP. – Narasimham Jul 9 at 20:15

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