1
$\begingroup$

During a course I came across a formula for the quantum effective action of a Yang-Mills theory in euclidean space and it appears like this (some indices may be dropped but I hope that won't be a problem):

$$\Gamma(A) = \frac{1}{2g^2}S(A) + \frac{1}{2}\log\det(-\nabla^2_A\cdot\delta_{\mu\nu}-2i\mathrm{ad}(F_{\mu\nu})\cdot)-\log\det(-\nabla^2_A\cdot)$$

where

$$\nabla^2_A\cdot = D^\mu_A\cdot D^\mu_A\cdot,\quad D^\mu_A\cdot = \partial^\mu+i[A^\mu, \cdot], \quad \mathrm{ad}(F_{\mu\nu})\cdot = [F_{\mu\nu}, \cdot].$$

Do you know a book/article where I can find such an evaluation? I think that could be useful to learn about this topic.

$\endgroup$

1 Answer 1

0
$\begingroup$

Your action seems identical to the effective action in section 2.4.2 of David Tong's notes on gauge theory (modulo the sign in front of $[F_{\mu \nu}, \cdotp]$).

It can be obtained using the background field method to integrate out quantum fluctuations. The first determinant corresponds to the gauge field, and the second to the ghost field. The covariant derivatives $D^\mu_A$ are with respect to the (classical) background field.

His notes are excellent, I suggest you take a look. I believe the action, and ultimately also the Yang-Mills beta function, are also derived in (the slightly more convoluted) section 16.6 of Peskin & Schroeder, but your notation more closely resembles that of Tong.


I may be wrong, and am unable to post comments, but I hope this is of some help. Would be useful if you provided the source of your equation.

$\endgroup$
1
  • $\begingroup$ Thank you very much, I think you gave me what I was looking for, I'll check for the sign. I believe my problem was that the calculation is better known as background method $\endgroup$ Commented Mar 5 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.