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Functional derivative

  • Thread starter parton
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  • #1
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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?
 
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  • #2
Cyosis
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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.
 
  • #3
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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.
 
  • #4
Cyosis
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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.
 
  • #5
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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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