How Does the n Factor Arise in the QFT Generating Functional Equation?

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SUMMARY

The discussion focuses on the derivation of the n factor in the Quantum Field Theory (QFT) generating functional equation, specifically in the context of the equation's manipulation involving derivatives with respect to the source term J. The user attempts to differentiate the generating functional Z[J] and expresses confusion regarding the emergence of the n factor, particularly when considering the contributions from the action S'[\Phi] + J. The analysis reveals that the n factor arises from the interplay between the number of derivatives taken and the structure of the generating functional.

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Homework Statement



Hi

I am looking at the attached question part c)

generatingfunctional2.jpg


Homework Equations



below

The Attempt at a Solution



so if i take ##\frac{\partial^{(n-1)}}{\partial_{(n-1)}} ## of (2) it is clear I can get the ##\frac{i}{h} (\lambda_2 +\lambda_4 )## like-term, but I am unsure about the ##nG_{n-1}## .

There's obviously no other derivatives on the RHS so I will only yield a ##G_{n-1}## and that looks fine, I am a bit confused though, I can yield this from the ## Z[J] ## alone on the RHS, whereas the RHS is ##Z[J]## 'multiplied by' (it is already inside the integral) the extra term of ##S'[\Phi] + J## . So I suspect this extra term is the reason we get the ##n## factor but I am unsure how.

Looking at the LHS there is a single ##J## so it looks like this gives a factor of ##1## and then we take across ##(n-1)## from the RHS.

If I take a derivative wrt ##J##, on the LHS I can either act on the exponential or the single ##J## (but can only act on this ##J## once,) on the RHS it's the same story, with the difference that on the LHS the ##J## is outside the integral but on the RHS it is inside the integral, I'm trying to use this to deduce where the factor of ##n## comes from but I am struggling..
 

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