Baker, Campbell, Hausdorff and all that

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[SOLVED] Baker, Campbell, Hausdorff and all that

I'm posting this here because, although it is a mathematics problem, it is related to perturbation theory and is the kind of problem physicists might be more skilled at answering.

Does anyone know an elegant proof of

[tex]e^{A+B} = \int_0^1 d\alpha_1 \,\delta(1-\alpha_1) e^{\alpha_1 A} + \int_0^1 d\alpha_1 d\alpha_2 \,\delta(1-\alpha_1 - \alpha_2) e^{\alpha_1 A} B e^{\alpha_2 A} + \frac{1}{2!}\int_0^1 d\alpha_1 d\alpha_2 d\alpha_3 \,\delta(1-\alpha_1 - \alpha_2 - \alpha_3) e^{\alpha_1 A} B e^{\alpha_2 A} B e^{\alpha_3 A} + \dots[/tex]

where of course [tex]A[/tex] and [tex]B[/tex] are matrices? I can prove it starting from the easy to prove identity
[tex]\frac{d}{ds} e^{A + s B} = \left( \int_0^1\!dt\, e^{t(A + s B)} B e^{-t(A + s B)} \right) e^{A + s B}[/tex]
but the proof gets a bit messy. I was hoping maybe someone recognizes the formula or knows a good references.
 
666
2
I can't help with your question but I'm curious. What does

[tex]\int_0^1 d\alpha_1 \,\delta(1-\alpha_1) e^{\alpha_1 A}[/tex]

mean?

That is, since

[tex]\int_0^{1+\epsilon} d\alpha_1 \,\delta(1-\alpha_1) e^{\alpha_1 A}=e^A[/tex] and [tex]\int_0^{1-\epsilon} d\alpha_1 \,\delta(1-\alpha_1) e^{\alpha_1 A}=0[/tex]

what happens when your limit of integration over a delta function is the singularity point?
 
410
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I'll give an example, based on (a more general version of) the second identity I gave. You can think of
[tex]\frac{d}{ds} \left( e^{A(s)} \right) = \int_0^1\int_0^1 \!d\alpha_1\,d\alpha_2 \delta(1-\alpha_1 - \alpha_2) e^{\alpha_1 A(s)} A'(s) e^{\alpha_2 A(s)} [/tex]
as short-hand for
[tex]\frac{d}{ds} \left( e^{A(s)} \right) = \int_0^1\!d\alpha\,e^{\alpha A(s)} A'(s) e^{(1-\alpha)A(s)} [/tex].
This the range of integration extends slightly beyond the singular point, in order that the short-hand works.
 
666
2
I think I understand that. Thanks!
 
410
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In case anyone is following the score, the correct identity turns out to be

[tex]e^{A+B} = \int_0^1 d\alpha_1 \,\delta(1-\alpha_1) e^{\alpha_1 A} + \int_0^1 d\alpha_1 d\alpha_2 \,\delta(1-\alpha_1 - \alpha_2) e^{\alpha_1 A} B e^{\alpha_2 A} + \int_0^1 d\alpha_1 d\alpha_2 d\alpha_3 \,\delta(1-\alpha_1 - \alpha_2 - \alpha_3) e^{\alpha_1 A} B e^{\alpha_2 A} B e^{\alpha_3 A} + \dots[/tex]

which is considerably easier to prove =)
 
410
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For anyone who cares for the solution, I've posted it here:
http://www.mathematics.thetangentbundle.net/wiki/Linear_algebra/Baker-Campbell-Hausdorff_formula/integral_expansion_proof [Broken]
Thanks to my friend Kory for his assistance.
 
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