1
$\begingroup$

I am reading the review on instantons. When I tried to derive formula (2.27) on page 17, I always get the different coefficient of $gF_{mn}$ term. My calculation is just directly expanding the first term $-\frac{1}{2} \int d^4x {\rm{tr_N}} F^2_{mn}$ in action to quadratic form, and leaving the second term $i\theta k$ invariant. Under the fluctuation $A_n \to A_n +\delta A_n $, the first term $-\frac{1}{2} \int d^4x {\rm{tr_N}} F^2_{mn}$ becomes $$ -\frac{1}{2} \int d^4x {\rm{tr_N}} F^2_{mn} \\-\frac{1}{2} \int d^4x {\rm{tr}_N }\left( 2(D_n\delta A_m)^2 -2D_n\delta A_mD_m\delta A_n +2g F_{mn} [\delta A_m,\delta A_n] \right) +...$$

Integrating by parts and disregarding the surface term, the quadratic term becomes $$ \int d^4x {\rm{tr}_N }\left( \delta A_m D^2 \delta_{mn}\delta A_n -\delta A_mD_nD_m\delta A_n+2g F_{mn} \delta A_n \delta A_m \right)$$ Using the equalities $D_n D_m=D_mD_n-gF_{mn}$, $D_n\delta A_n=0$, the quadratic form becomes $$ \int d^4x {\rm{tr}_N }\delta A_m\left( D^2 \delta_{mn} +g F_{mn}+2g F_{mn} \right)\delta A_n $$

So the coefficient of $gF_{mn}$ term is 3. What is wrong with the above calculation?

$\endgroup$
4
  • $\begingroup$ Why don't you write terms like $(\partial_m \delta A_n-\partial_n \delta A_m)g([\delta A^m, A^n] + [ A^m, \delta A^n])$ ? $\endgroup$ – Trimok Sep 20 '13 at 10:56
  • $\begingroup$ Using $D$ derivative is more convenient because of the final form and eom of $\delta A_n$. When I use integration by parts, I do use the $\partial$ derivative. $\endgroup$ – thone Sep 20 '13 at 12:13
  • $\begingroup$ Craig, I agree with Trimok that you should avoid the covariant derivatives in this calculation because you seem to be trapped into various traps. For example, it's very misleading to "use the equality" $D_nD_m-D_mD_n =- g F_{mn}$. The left-hand side is an operator and whether the "equality" holds depends on what the operator acts upon and how. It's OK when it acts on spinors in the fundamental rep but you want to act on $\delta A_m$ which is in adjoint, so the action is different, so all your coefficients are a bit wrong. Both coefficients are nonzero even for $U(1)$ which you also get wrong. $\endgroup$ – Luboš Motl Sep 20 '13 at 15:00
  • 1
    $\begingroup$ Follow thw advice of Trimok and Lubos, I abandoned the equality $D_m D_n -D_n D_m =gF_{mn} $ which I thought of as the operaor equation before (thanks for correction) and found that the coefficient is 4 which may be the right answer. $\endgroup$ – thone Sep 21 '13 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.