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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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2 Answers 2

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Let us start with the Campbell-Hausdorff formula for selfadjoint operators: let $H_j$ be bounded selfadjoint operators on a Hilbert space. Then $$ e^{i\tau H_1}e^{i\tau H_2}=e^{i\tau (H_1+H_2)-\frac{\tau ^2}2[H_1,H_2]+\tau^3 R_\tau}, $$ where $ \sup_{\vert\tau\vert\le \alpha_0}\Vert R_\tau\Vert_{\mathcal B(H)}<+\infty $ for some positive value of $\alpha_0$. Then $$ e^{-i\tau H_1}e^{-i\tau H_2}e^{i\tau H_1}e^{i\tau H_2}=e^{-i\tau (H_1+H_2)-\frac{\tau ^2}2[H_1,H_2]-\tau^3 R_{-\tau}} e^{i\tau (H_1+H_2)-\frac{\tau ^2}2[H_1,H_2]+\tau^3 R_{\tau}} $$ so that applying the formula again, you get with $S_\tau$ bounded in operator-norm near the origin $$ e^{-i\tau H_1}e^{-i\tau H_2}e^{i\tau H_1}e^{i\tau H_2}=e^{-\tau^2[H_1,H_2]+\tau^3S_\tau}. $$ Replacing $\tau$ by $t/n$ and waiting for $t/n$ to get smaller than $\alpha_0$, you find the sought formula $$ \lim_{n\rightarrow+\infty}\bigl( e^{-i\frac tn H_1}e^{-i\frac tn H_2}e^{i\frac tn H_1}e^{i\frac tn H_2}\bigr)^{n^2}=\lim_{n\rightarrow+\infty}e^{- {t^2}[H_1,H_2]+O(n^{-1})}. $$ When the operators are unbounded, there are complications with the domains and the size of the remainders. Note also that the operator $[H_1,H_2]$ is skew-adjoint as the commutator of two selfadjoint operators so the limit is indeed unitary.

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For the unbounded case, see

Kato, Tosio Trotter's product formula for an arbitrary pair of self-adjoint contraction semigroups. Topics in functional analysis (essays dedicated to M. G. Kreĭn on the occasion of his 70th birthday), pp. 185–195, Adv. in Math. Suppl. Stud., 3, Academic Press, New York-London, 1978.

The domain condition is that the square roots of $A$ and $B$ should have a dense intersection, then the convergence holds, to the exponential of the form sum of these operators.

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  • $\begingroup$ The Trotter sum formulae are also listed in Reed-Simon Vol.1 ("Functional Analysis", Theorem VIII.31): "For all $A,B$ self-adjoint with $A+B$ essentially self-adjoint on the intersection one has $\operatorname{s-lim} (e^{itA/n}e^{itB/n})^n=e^{i(A+B)t}$; if $A,B$ additionally are bounded from below, then $\operatorname{s-lim}(e^{-tA/n}e^{-tB/n})^n=e^{-t(A+B)}$". References are listed in the notes section following the chapter. $\endgroup$ Jun 30, 2022 at 15:20

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