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Differentiation of an exponential with operators (Peskin p.84)

  1. Jul 23, 2008 #1
    Does anyone know how to differentiate an exponential, which has an operator in its power? I found it quite a trouble in Peskin's QFT (page 84, formulas (4.17), (4.18)).
    Here we have these two formulas of Peskin:

    [tex]U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }e^{-iH\left( t-t_{0}\right) } [/tex];
    [tex]i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }\left( H-H_{0}\right) e^{-iH\left( t-t_{0}\right) } [/tex].

    I agree with this. However, if we write [tex]U\left( t,t_{0}\right)[/tex] as [tex]U\left( t,t_{0}\right)=e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }[/tex], then

    [tex]i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=\left( H-H_{0}\right)e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }[/tex]

    and we cannot transport [tex]e^{iH_{0}\left( t-t_{0}\right) }[/tex] to the left of [tex]\left( H-H_{0}\right)[/tex] so easily to obtain Peskin's result, since, according to my calculations, [tex]\left[ H,H_{0}\right]\neq0[/tex]. Do we have a rule, which explains where to put the operators from the exponential after differentiation, when we have several noncummuting operators in the power of exponential?
  2. jcsd
  3. Jul 23, 2008 #2


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    But you can't write this since H and H_0 don't commute. [tex] e^A e^B = e^{A+B} [/tex] only when A and B commute. Otherwise you have to use the Campbell-Hausdorf formula.
    You just differentiate as usual, making sure that you never pass an operator "through" another operator that does not commute with it.
  4. Jul 23, 2008 #3
    Thanks a lot! This has been tormenting me for ages!
  5. Jul 23, 2008 #4
    Thank you, nrqed, for pointing out my mistake. I appreciate it :)
  6. Jul 23, 2008 #5


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    :smile: You are very very welcome.

    And thank you for posting your question since this apparently helped Wasia too!!

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