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

Table of contents
- Solution d

Solution

For

n = 2

, the operator formula becomes

Z_{2} [J] = [1 + \frac{i g}{2} \int d^{d} x \frac{δ ^{2}}{δ J ( x ) ^{2}} + O (g^{2})] Z_{0} [J] .

Two source derivatives of the free Gaussian produce a

J

-independent coincident propagator and a source-dependent insertion between two free propagators. The former is a vacuum bubble and disappears when

Z_{2} [0] = 1

. Thus

\frac{Z _{2} [ J ]}{Z _{2} [ 0 ]} = Z_{0} [J] [1 - \frac{i g}{2} \int \frac{d ^{d} p}{( 2 π ) ^{d}} \frac{J ( - p ) J ( p )}{( p ^{2} + i 0 ) ^{2}} + O (g^{2})]

Expanding the exact denominator from part (b),

\frac{1}{p ^{2} - g + i 0} = \frac{1}{p ^{2} + i 0} + \frac{g}{( p ^{2} + i 0 ) ^{2}} + O (g^{2}),

gives the same source-dependent correction.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version