y**t
发帖数: 50 | 1 assume e^A e^B=e^C
then for reasonable A and B,we have
C=B+\integral_0^1 h(e^{t adA} e^{ad B}) A dt
where h(z)=log(z)/(z-1) and adA M=[A, M]
you exapnd in adA and adB(to keep
to the right order,one could put an "x" before adA and adB and
collect x^n terms together) and then integrate
good luck,wonder how many terms you need,haha |
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