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Baker, Campbell, Hausdorff and all that

  1. Apr 29, 2008 #1
    [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.
     
  2. jcsd
  3. Apr 29, 2008 #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?
     
  4. Apr 29, 2008 #3
    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.
     
  5. Apr 30, 2008 #4
    I think I understand that. Thanks!
     
  6. Apr 30, 2008 #5
    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 =)
     
  7. May 1, 2008 #6
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