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Homework Help: Functional derivative

  1. Apr 1, 2010 #1
    In the literature (Ryder, path-integrals) I have found the following relation for the functional derivative with respect to a real scalar field [tex] \phi(x) [/tex]:

    [tex] i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) ( \square + m^2 ) \phi(x)} = ( \square + m^2 ) \phi(x) e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) ( \square + m^2 ) \phi(x)}[/tex]

    But how do I compute this? I am just confused about this d'Alembert operator [tex] \square [/tex] and I never end up with the right solution as above.

    Could anybody explain how to obtain this solution, please?
    Last edited: Apr 1, 2010
  2. jcsd
  3. Apr 2, 2010 #2


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    Homework Helper

    It would help if you show us your steps until the point where the d'Alembert operator is causing problems.

    Using the Minkowski metric (+---) (c=1) the d'Alembert operator is given by [itex]\partial_{tt}-\partial_{xx}-\partial_{yy}-\partial_{zz}[/itex]. With this you can evaluate the integral [itex]\int d^4 x \; \phi(x) \square \delta \phi(x)[/itex] using partial integration. What do you know about the volume integrals that appear due to partial integration? Try it for [itex]\int d^4x \; \phi(x) \partial_{xx} \phi(x)[/itex]. The other terms work exactly the same.
  4. Apr 2, 2010 #3
    Ok, my problem is the following. If I consider

    [tex] i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) m^2 \phi(x)} [/tex]

    I could write the exponent as [tex] -i \int \mathrm{d}^{4} x \frac{1}{2} m \phi^{2} [/tex] and use some kind of product rule and obtain:

    [tex] i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) m^2 \phi(x)} = m^2 \phi(x) e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) m^2 \phi(x)} [/tex]

    Ok, that seems to be correct, but now I if want to compute

    [tex] i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) \square \phi(x)} [/tex]

    I can not rewrite the exponent in such a form like above and otherwise I don't really know how to build the derivative.

    I just differentiated the first [tex] \phi(x) [/tex] and obtained

    [tex] = \dfrac{1}{2} \square \phi(x) e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) \square \phi(x)} [/tex]

    where I have a factor 1/2 which is wrong, but I don't know how to derive this correctly.
  5. Apr 2, 2010 #4


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    You have treated [itex] \square \phi(x) [/itex] as a constant now, but it is not a constant.

    Take the variation of the argument of the exponent.

    [tex] \delta \int d^4x \; \phi(x) \square \phi(x) = \int d^4x \; \left[ \delta \phi(x) \square \phi(x) +\phi(x) \square \delta \phi(x) \right] [/tex]

    Now integrate the second term by parts.
  6. Apr 2, 2010 #5
    Ok, I integrated the second term by parts (2 times) and finally obtained the correct result.

    Thanks a lot for your help :smile:
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