Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 1 / d

Table of contents
- Solution d

Solution

Change variables in the defining integral from the fluctuation to the total field,

φ = ϕ_{0} + η

. Then

e^{- W (J; ϕ_{0}) /ℏ} = \int d φ e^{- [S (φ) + J (φ - ϕ_{0})] /ℏ} = e^{J ϕ_{0} /ℏ} e^{- W (J; 0) /ℏ},

and hence

W (J; ϕ_{0}) = W (J; 0) - J ϕ_{0} .

It follows nonperturbatively that

χ = \partial_{J} W (J; ϕ_{0}) = \partial_{J} W (J; 0) - ϕ_{0} .

Thus the same source

J_{χ}

corresponds at zero background to the mean field

Φ = χ + ϕ_{0}

. Using the source-sign-compatible Legendre transform

Γ (χ; ϕ_{0}) = W (J_{χ}; ϕ_{0}) - J_{χ} χ

,

Γ (χ; ϕ_{0}) = W (J_{χ}; 0) - J_{χ} (χ + ϕ_{0}) = Γ (χ + ϕ_{0}; 0) .

Renaming

χ

η

gives the requested identity

Γ (η; ϕ_{0}) = Γ (ϕ_{0} + η; 0) .

 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