Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 4 / c / i

Table of contents
- Solution i

Solution

For the Grassmann-odd ghost

c

, the BRST transformations are

Q A_{μ} = D_{μ} c = \partial_{μ} c - i g [A_{μ}, c], Qϕ = i g [c, ϕ], Q c = i g c^{2},

up to a simultaneous convention-dependent sign for

Q

and

c

. The ghost transformation makes the BRST charge nilpotent:

Q^{2} A_{μ} = Q^{2} ϕ = 0

.

 Ancestors (11)

 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