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Functional Derivative fail

  1. Jan 16, 2012 #1
    Alright, so I feel kind of dumb...but:

    I have been working on some QFT material, specifically derivation of Feynman rules for some more simple models ([itex] \phi^{4} [/itex] for example), and I have been seriously failing with functional derivatives. Every time I try to use the definition I mess up somewhere. Usually, my best bet is to sort of treat it like a regular derivative and use some intuition, but that's not really legitimate.

    Oh, the Δ here is the Feynman propagator, I'm not sure what the standard is for notation so I suppose I should mention that.

    Take for example [itex] \frac{1}{i}\frac{\delta}{\delta J(z)} exp\left[-\frac{i}{2}\int J(x)\Delta_{F}(x-y)J(y)dxdy\right] [/itex]

    So what I'm supposed to get is
    [itex] - \int \Delta (z-x)_{F} J(x) dx \ exp \left[-\frac{i}{2} \int J(x) \Delta_{F}(x-y)J(y)dx dy \right] [/itex]

    And I can rationalize it as that we have an exponential and we treat exponential derivatives as we do traditionally, but then I get confused with where my factor of 1/2 went from the exponent...clearly, the i's cancel, but where did the two go? I have a feeling it has to do with the repeat of the J's in the J Δ J but I'm not sure.

    Ugh. . .this has wasted so very much of my time.

    Not to mention, I've yet to convince myself that we ARE allowed to treat an exponential the same way, though I think I might have an idea how to show it, I'm not sure how common identities like product rule would be proved with functionals.

    If someone could give me something to read over or something that would be awesome. I've looked for a while and not found much help. I found a paper by Feynman on Operator Calculus that had an Appendix relating to functionals but that didn't really help ^^;...
  2. jcsd
  3. Jan 17, 2012 #2


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    you could start with an example like

    [tex]\frac{\partial}{\partial x_i}\text{exp}\left[-\frac{1}{2}\,x_m\,A_{mn}\,x_n\right][/tex]

    with a symmetric matrix A and a sum over indices m,n
  4. Jan 17, 2012 #3
    Okay, so the 1/2 gets absorbed in quadratic terms for m=n and then absorbed in the symmetry otherwise.

    And for my case we treat the JΔJ term likewise. . .Wait that looks familiar .-. I lose.

    Like the form [itex] \int exp \left[ \frac{-1}{2} \int \phi (x) A \phi (x)dx\right] [/itex]

    with A being a differential operator we get (det A)-1/2 .-. I'm an idiot, I've seen this before and forgot. I thought that the discrete problem you typed looked familiar.

    Thank you!
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