# Generalisation of Baker–Campbell–Hausdorff (Matrix exponential simplification)

Let there be some matrix exponential function for matrices of the form $$M=\sum_{i=1}^p c_i A_i$$ where $$M,A_i\in\mathbb{R}^{n\times n}$$ are given, real square matrices and $$c_i \in \mathbb{R}$$. In other words, we are interested in the matrix exponential of the following form: $$f(M)=e^M=e^{\sum c_i A_i}.$$ The matrix exponent is thus expressible in some finite real basis for the subspace we are interested in, as $$M \in \mbox{Span}(A_i)$$. Does an approximation (or even identity) exist for the case in which this basis of matrices $$A_i$$ and their commutators is known?

Concretely, I currently have a basis of less than 10 matrices (i.e., $$p<10$$), if this makes the problem easier. This problem arises as part of an optimization problem in which the $$A_i$$ are fixed (some chosen basis) and the $$c_i$$ are variable.

• Have you tried the case where $p=2$? Commented Mar 6, 2023 at 13:36
• That seems to boil down to the Campbell-Baker-Hausdorf formula. In a certain sense, I guess I'm asking for whether a generalization for $p > 2$ is known. Commented Mar 6, 2023 at 16:48
• "Generalisation of Baker–Campbell–Hausdorff" would be a good title for this question Commented Mar 6, 2023 at 16:51
• I think it's more like a generalization of the Lie product formula than the Baker-Campbell-Hausdorff formula. It also reminds me of the Magnus expansion but it's not immediately clear to me how to solve your particular problem. Commented Mar 6, 2023 at 17:08
• OP ls possibly asking for a generalized Zassenhaus formula. Commented Mar 9, 2023 at 0:21

In your comment, you seem to consider that a generalization for $$n > 2$$ indeterminates of the Baker-Campbell-Hausdorff formula would answer your question. Such a generalization is indeed known. You can for example read it in Section 2 of the following paper.
The formula is the following: for noncommuting indeterminates $$x_1, \dotsc, x_n$$, one has $$e^{x_1} \dotsb e^{x_n} = e^z,$$ where $$z = \sum_{m=1}^{+\infty} \sum_{p_{j,k}} \frac{(-1)^{m-1}}{m} \frac{(\mathrm{ad}~{x_n})^{p_{m,n}} \dotsb (\mathrm{ad}~{x_1})^{p_{m,1}} \dotsb (\mathrm{ad}~{x_n})^{p_{1,n}} \dotsb (\mathrm{ad}~{x_1})^{p_{1,1}}}{ \left(\sum_{j=1}^m \sum_{k=1}^n p_{j,k}\right) \Pi_{j=1}^m \Pi_{k=1}^n (p_{j,k}!)} ,$$ where the $$p_{j,k}$$ range over all nonnegative integers such that, for each $$j = 1, \dotsc, m$$, one has $$\sum_{k=1}^n p_{j,k} > 0$$. Strichartz uses the notation $$(\mathrm{ad}~x) y$$ to denote the commutator $$[x,y]$$. Similarly $$(\mathrm{ad}~x)^2 y = [x,[x,y]]$$ and so on, with the convention that, when it is not followed by an argument $$(\mathrm{ad}~x) = x$$.
In particular, with these notations, the formula for $$n = 2$$ is $$e^{x} e^y = e^z$$ where $$z = \sum_{m=1}^{+\infty} \sum_{p_j,q_j} \frac{(-1)^{m-1}}{m} \frac{(\mathrm{ad}~{y})^{q_m} (\mathrm{ad}~x)^{p_m} \dotsb (\mathrm{ad}~{y})^{q_1} (\mathrm{ad}~{x})^{p_1}}{ \left(\sum_{j=1}^m p_j + q_j \right) \Pi_{j=1}^m (p_j! q_j!)},$$ where the sum ranges over nonnegative integers $$p_j, q_j$$ such that, for each $$j = 1, \dotsc, m$$, one has $$p_j+q_j > 0$$.
Complementing the comments of @kc9jud and @Qmechanic: as OP asked for an identity or an approximation another approach which, admittedly, may be too "brute force" (as it does not use any information regarding commutators of the $$A_i$$) could be the Lie product formula $$e^{\sum_i c_iA_i}=\lim_{n\to\infty} \big( \prod_i e^{\frac{c_iA_i}n} \big)^n\,,$$ i.e. once $$n$$ is "large enough" one has $$e^{\sum_i c_iA_i}\approx e^{\frac{c_1A_1}n}e^{\frac{c_2A_2}n}\ldots e^{\frac{c_kA_k}n}e^{\frac{c_1A_1}n}e^{\frac{c_2A_2}n}\ldots \text{ (n times)} \ldots e^{\frac{c_{k-1}A_{k-1}}n}e^{\frac{c_kA_k}n}$$ Note that, while this formula is usually stated for just two matrices the standard proof readily generalizes to an arbitrary (finite) sum in the exponent.
• Unfortunately, this doesn't really help because it still requires calculating exponentials and boils down to the same approximation one would use naively, namely $(\mathbb{1}+\frac{cA}{n})^n$ Commented Mar 9, 2023 at 16:15
• That's fair, I only suggested this because it may have been that your $A_i$'s are of such convenient form that $e^{c{\bf A_i}}$ is very simple to calculate for arbitrary $c$ (which would've distinguished this approach from $({\bf 1}+\frac{cA}n)^n$), so this would've reduced the problem to simple matrix multiplication. Commented Mar 9, 2023 at 16:21