Given two non-commuting matrices $X$ and $Y$, consider the following quantity $$ e^{X+\lambda Y} $$ What is the Taylor series expansion of the above quantity about $\lambda=0$? Note that this is non-trivial since $X$ and $Y$ do not commute and is not simply $e^X(1+\lambda Y+ \lambda^2 Y/2 + \cdots)$.

One way to proceed is perhaps the Zassenhaus formula, which expresses the above quantity as an infinite product of matrix exponentials $$e^{(X+\lambda Y)}= e^{X}~ e^{\lambda Y} ~e^{-\frac{1}{2} \lambda[X,Y]} ~ e^{\frac{1}{6}(2\lambda^2[Y,[X,Y]]+\lambda [X,[X,Y]] )} ~ e^{\frac{-t^4}{24}(\lambda[[[X,Y],X],X] + 3\lambda^2[[[X,Y],X],Y] + 3\lambda^3[[[X,Y],Y],Y]) }\cdots$$ It can be seen that even at the first order in $\lambda$ in the Taylor expansion, there are contributions from each of the exponentials. The question is : can these terms in the Taylor series be written in closed form, order-by-order?

PS : I would eventually be interested in the trace, Tr[$e^{X+\lambda Y}$].


1 Answer 1


Your ultimate question about the trace is much easier to answer.

Let $A = (X+\lambda Y)$, then find the differential and gradient of the trace of its exponential. $$\eqalign{ \phi &= {\rm Tr}(\exp(A)) \\ d\phi &= \exp(A):dA^T \\ &= \exp(A):Y^T\,d\lambda \\ &= {\rm Tr}\big(Y\exp(A)\big)\,d\lambda \\ \frac{d\phi}{d\lambda} &= {\rm Tr}\big(Y\exp(A)\big) \\ &= {\rm Tr}\Big(Y\exp(X+\lambda Y)\Big) \\ }$$ At $\lambda=0$ this reduces to $$\frac{d\phi}{d\lambda} ={\rm Tr}\Big(Y\exp(X)\Big)$$ NB: In some of the steps above, a colon is used to denote the trace/Frobenius product, i.e. $$A:B = {\rm Tr}(A^TB)$$

  • $\begingroup$ Thanks! Can you also write something similar at the second order or for the second derivative? $\endgroup$ Commented Sep 26, 2019 at 23:20
  • $\begingroup$ Unless your matrices have some kind of special structure that you can take advantage of (e.g. $3\times 3$ skew), you're back to some variant of Baker-Campbell-Hausdorff for the higher order derivatives. $\endgroup$
    – greg
    Commented Sep 27, 2019 at 13:51

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