Understanding functional derivative

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Homework Help Overview

The discussion revolves around the evaluation of a functional derivative related to the functional \( J[f] = \int [f(y)]^p \phi(y) \, dy \). The original poster seeks to understand the transition from the limit definition of the functional derivative to the expression \( p[f(x)]^{p-1} \phi(x) \), as presented in a reference text on quantum field theory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to evaluate the limit by expanding the integral using the binomial theorem, questioning whether their approach is correct. Participants inquire about which terms in the expansion survive the limit as \( \varepsilon \to 0 \), and the original poster reflects on their understanding of the Dirac delta function.

Discussion Status

Participants are actively engaging with the original poster's reasoning, providing feedback on the use of the Dirac delta function and confirming the correctness of the steps taken. There is an ongoing exploration of the concepts involved, with no explicit consensus reached on the overall understanding yet.

Contextual Notes

The original poster expresses uncertainty regarding their grasp of the Dirac delta function and its application, indicating a potential gap in foundational knowledge that may affect their confidence in the current problem. Additionally, there is a mention of a desire to ask a separate question related to tensors, suggesting a broader context of inquiry.

Bishamonten
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Homework Statement



"The functional ## J[f] = \int [f(y)]^pφ(y)\, dy ## has a functional derivative with respect to ## f(x) ## given by:

$$ \frac {δJ[f]} {δf(x)} = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$

$$ = p[f(x)]^{p-1}φ(x) $$

Homework Equations



Limit definition of a functional derivative:

$$ \frac {δF} {δf(x)} = \lim_{ε \rightarrow 0} \frac {F[f(x') + εδ(x-x')] - F[f(x')]} ε$$

The Attempt at a Solution


From line 1:

$$ = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$


$$ = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int \sum_{k=0}^p
\begin{pmatrix}
p \\
k \\
\end{pmatrix}
[f(y)]^{p-k}[εδ(y-x)]^kφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$

$$ \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int \sum_{k=0}^p \frac {p!} {k!(p-k)!} [f(y)]^{p-k}[εδ(y-x)]^kφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$



This comes straight out of Lancaster and Blundell's "Quantum Field Theory for the Gifted Amateur", and is in the first chapter where they are introducing Lagrangians, Fermat's principle of least time, and Hamilton's principle of least action. This is simply them demonstrating functional derivatives, only, I wanted to see how they arrived from the limit definition of the functional ## J[f] ## to ## p[f(x)]^{p-1}φ(x) ##.

I wanted to evaluate the limit myself, through first evaluating the integral. I saw the power of the sum expression and thought ok, maybe I should introduce the binomial theorem, but I feel like I'm only making a mess of things and am not arriving any closer at ## p[f(x)]^{p-1}φ(x) ##, may someone let me know if I'm going at this in the right direction, and if not, any pointers, or if I'm simply making this more complicated than it should be?

 
Physics news on Phys.org
Which terms in your binomial expansion survive the limit ##\varepsilon \to 0##?
 
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Orodruin said:
Which terms in your binomial expansion survive the limit ##\varepsilon \to 0##?
Oooh, ok I don't know why I didn't put it in its expanded form earlier. Maybe it was my lack of confidence talking :DD Thank yo so much for responding!

So when I expand the term out, the 1st term in the series matches ## -\int [f(y)]^pφ(y)dy ##, except it's positive, thus cancelling. Every other term in the series besides the 2nd term ## \frac {p!} {(p-1)!} [f(y)]^{p-1}δ(y-x)φ(y) ## zeroes out when evaluating the limit because they will all continue to have the kth power of epsilon multiplying it, whereas the 2nd term has it's epsilon cancel the one distributed, and of course ## \frac {p!} {(p-1)!} ## is just p.

So here's what I got so far, I am a bit uncomfortable with the dirac delta function argument I make in this last step however, as you'll see:

$$ p\int[f(y)]^{p-1}φ(y)δ(y-x)dy $$

$$ [f(y)]^{p-1}φ(y) ≡ g(y) $$

$$ = p\int g(y)δ(y-x)dy = g(x) = p[f(x)]^{p-1}φ(x) $$

Is that a proper use of the indefinite integral of this delta function?
 
That is the definition of how the delta function works, so yes, it is perfectly fine.
 
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Orodruin said:
That is the definition of how the delta function works, so yes, it is perfectly fine.

Thank you very much for the help. We touched up on the dirac delta function during the last part of our differential equations class, but looks like I need to review how it works a bit. I remember that this was the definition of how it works, but the intuition for why it works that way didn't survive me unfortunately.

Also, I wish to ask another question while I'm at it, related to tensors, covariant and contravariant vectors. May I post it here, or should I start a new thread?
 
Bishamonten said:
I remember that this was the definition of how it works, but the intuition for why it works that way didn't survive me unfortunately.
It works like that because it is defined to work like that. For more in depth on what it really is you would need to study distributions (it really is a distribution and not a function), in particular in terms of derivatives and how it relates to differential equations.

Bishamonten said:
Also, I wish to ask another question while I'm at it, related to tensors, covariant and contravariant vectors. May I post it here, or should I start a new thread?
You should start a new thread. If it is a problem like this one, post it in the appropriate homework forum. If it is a more general question, you can post it in one of the technical forums.
 
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Once again, thank you for all your concise help Orodruin, I've no doubt I'll be needing it again soon enough :smile:
 
Sorry, post made by mistake
 

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