Taylor series expansion of functional

[vishal.ng]

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.

 

[Paul Colby]

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

 

[vishal.ng]

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

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?

 

[skynelson]

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

It is not helpful to quote an obvious result with no clarification on how to apply it to the given use case.

 

[MathematicalPhysicist]

You should give Mark Srednicki's textbook on QFT a try.