Mastering Functional Derivatives in Quantum Field Theory

In summary, the speaker has been struggling with functional derivatives in their work on QFT material. They often have to resort to using intuition rather than the proper definition. They have also been unsure about how to treat exponentials in functional derivatives and have not been able to find much help in their research. However, they have come to a realization that they have seen similar problems before and are grateful for the help.
  • #1
Elwin.Martin
207
0
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 ^^;...
 
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  • #2
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
 
  • #3
tom.stoer said:
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

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!
 

1. What is a functional derivative?

A functional derivative is a mathematical concept used in the field of functional analysis to describe the change in a functional as a result of a small change in its input. It is similar to a regular derivative, but instead of a function with a finite number of variables, it operates on a functional with an infinite number of variables.

2. How is a functional derivative calculated?

The functional derivative is calculated by taking the limit of the difference quotient as the change in the input approaches zero. This results in an expression involving the original functional and its modified input, which is then evaluated at the point of interest.

3. What is the purpose of a functional derivative?

The functional derivative is used in many areas of mathematics and physics, including variational calculus, optimization problems, and quantum field theory. It allows for the characterization of the sensitivity of a functional to small changes in its input, which is useful in solving many complex problems.

4. Can functional derivatives fail?

Yes, functional derivatives can fail in certain cases where the functional is not well-defined or does not have the necessary properties for the derivative to exist. This can happen when the functional is not smooth enough, or when the input is not in the appropriate function space.

5. How can functional derivatives be used in scientific research?

Functional derivatives are a powerful tool in scientific research, particularly in fields such as physics, mathematics, and engineering. They can be used to optimize systems, solve difficult differential equations, and model complex physical phenomena. They are also used in data analysis and machine learning to extract useful information from large datasets.

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