Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 1 / c

Solution

ϕ (x) e^{i \int J ϕ} = \frac{1}{i} \frac{δ}{δ J ( x )} e^{i \int J ϕ} = - i \frac{δ}{δ J ( x )} e^{i \int J ϕ} .

Expanding the interaction exponential, making this replacement in every term, and resumming gives

Z_{n} [J] = exp [- i g \frac{( - i ) ^{n}}{n !} \int d^{d} x \frac{δ ^{n}}{δ J ( x ) ^{n}}] Z_{0} [J] .

This formal identity assumes a common regulator, a source-independent normalization, and permission to interchange the path integral, power series, and functional derivatives. A normalized functional with

Z_{n} [0] = 1

requires division by the same expression evaluated at

J = 0

, which removes connected Vacuum Feynman diagrams.

There are no discussions about this article yet.

There are currently no matching articles.