In summary, the individual is studying QFT in the path integral formalism and is having trouble deriving the Schwinger Dyson equation for a real free scalar field. They are particularly stuck on the Taylor series expansion of the functional exponential and are unsure about the correct formula for the differential. They are seeking clarification and resources, such as Mark Srednicki's textbook on QFT, to help them understand these concepts better.
  • #1
vishal.ng
2
1
I'm studying QFT in the path integral formalism, and got stuck in deriving the Schwinger Dyson equation for a real free scalar field,
L=½(∂φ)^2 - m^2 φ^2
in the equation,
S[φ]=∫ d4x L[φ]
∫ Dφ e^{i S[φ]} φ(x1) φ(x2) = ∫ Dφ e^{i S[φ']} φ'(x1) φ'(x2)
Particularly, it is in the Taylor series expansion of the functional exponential
e^{i S[φ']}=e^{i S[φ+iα]} . Can anybody please tell me about the expansion? I have searched and haven't found anything quite helpful on the net. Thank you.
 
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  • #2
##e^x = \sum_{n=0}^\infty \frac{x^n}{n!}##
 
  • #3
Paul Colby said:
##e^x = \sum_{n=0}^\infty \frac{x^n}{n!}##
I believe that is a power series expansion. The final answer should contain the exponential still since, we have to relate it to the path integral. I have just started learning the functional formalism and I wanted to know whether,
##F[\phi']=F[\phi+\epsilon]=F[\phi]+\epsilon\left.\frac{dF}{d\phi'}\right|_{\phi'=\phi}+O(\epsilon^2)##
Which I believe is the functional analog of the Taylor series expansion, is correct and if the differential is indeed given by,
##\frac{dF}{d\phi'}=\int d^4y \:\varepsilon(y) \frac{\delta F[\phi(x)]}{\delta\phi(y)}##
Should ##\epsilon## and ##\varepsilon## be the same or should one of them be omitted. Or is the formula incorrect?
 
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  • #4
Paul Colby said:
##e^x = \sum_{n=0}^\infty \frac{x^n}{n!}##
It is not helpful to quote an obvious result with no clarification on how to apply it to the given use case.
 
  • #5
You should give Mark Srednicki's textbook on QFT a try.
 
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Related to Taylor series expansion of functional

1. What is the Taylor series expansion of a functional?

The Taylor series expansion of a functional is a way of representing a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point. This can provide a more accurate approximation of the function's behavior compared to a single polynomial or linear approximation.

2. How is the Taylor series expansion different from a regular Taylor series?

A regular Taylor series is used to approximate a function at a single point, while a Taylor series expansion of a functional takes into account the entire behavior of the function. This allows for a more precise representation of the function's behavior, especially for non-polynomial functions.

3. What is the purpose of using a Taylor series expansion of a functional?

The purpose of using a Taylor series expansion of a functional is to approximate a function at a certain point or over a certain interval. This can be useful in many areas of mathematics and physics, such as optimization, numerical analysis, and differential equations.

4. How is a Taylor series expansion of a functional calculated?

A Taylor series expansion of a functional is calculated by taking the derivatives of the function at a specific point and plugging them into the Taylor series formula. The formula involves a summation of terms, each of which is a derivative of the function evaluated at that point.

5. Can a Taylor series expansion of a functional be used for any function?

No, a Taylor series expansion of a functional can only be used for functions that are infinitely differentiable, meaning that they have an infinite number of derivatives at every point. This means that some functions, such as absolute value or square root, cannot be represented by a Taylor series expansion of a functional.

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