Let the spacetime be 4-dimensional.

  • In the usual Maxwell theory of Abelian gauge fields $A$, where field strength $F=dA$ one considers the Maxwell action written as

$$ S_{Maxwell}\equiv\int -\frac{1}{2}(F \wedge \star F)=\int -\frac{1}{4}(F_{\mu \nu} F^{\mu \nu}) d^4x. $$ The normalization of $\frac{1}{4}$ has a precise physical meaning for the sake of maintaining the conventions of Maxwell equations. In the presence of source current 1-form $J$, solved from the equations of motions (through variational principles), from, $$ S_{Maxwell+source}\equiv \int-\frac{1}{2}(F \wedge \star F)+A \wedge \star J= \int(-\frac{1}{4}(F_{\mu \nu} F^{\mu \nu}) + A_\mu J^\mu )d^4x, $$ Maxwell equations becomes $$ d (\star F)= J, \quad dF=0, $$ namely there is no further numerical factor in front of the Maxwell equation. This gives a definite reason/answer behind the factor $1/4$.



enter image description here

In other words, we have $$ \boxed{S_{YM,1}\equiv\int \frac{1}{4 g^2}\operatorname{Tr}(F \wedge \star F)}= \int\frac{1}{8 g^2}\operatorname{Tr}(F_{\mu \nu} F^{\mu \nu})d^4x=\int \frac{1}{8 g^2} \operatorname{Tr}[T^a T^b] F^{a\mu \nu} F_{\mu \nu}^b d^4x $$ $$ =\int \frac{1}{8 g^2} \frac{1}{2} \delta^{ab} F^{a\mu \nu} F_{\mu \nu}^b d^4x =\int \frac{1}{16 g^2} \sum_a F^{a\mu \nu} F_{\mu \nu}^a d^4x =\int \frac{1}{16 g^2} F^{a\mu \nu} F_{\mu \nu}^a d^4x, $$ where Einstein summation notation is assumed in the end. Based on the SU(N) Lie algebra, $\operatorname{Tr}[T^a T^b] =\frac{1}{2} \delta^{ab} $. The $a,b$ are the indices for fundamental representation of SU(N).

While in Steven Weinberg's textbook on QFT Volume in (15.2.3), we see that $ S_{YM,2}\equiv -\int \frac{1}{4 } F^{a\mu \nu} F_{\mu \nu}^a d^4x, $ This means that Weinberg's and many other physicists' textbooks have: $$ \boxed{S_{YM,2}\equiv -\int \operatorname{Tr}(F \wedge \star F) =-\int \frac{1}{4 } F^{a\mu \nu} F_{\mu \nu}^a d^4x.} $$ It looks that M Atiyah also uses the similar convention as $S_{YM,2}$.

question: I wonder, whether there is a reason behind the convention factor $\frac{1}{4 g^2}$ in front of Clay Math Millennium Prizes Yang–Mills and Mass Gap note: $S_{YM,1}\equiv\int \frac{1}{4 g^2}\operatorname{Tr}(F \wedge \star F)$? Is there some physical or mathematical reason behind this convention in contrast to this $S_{YM,2}$ convention? Or other math / physics constraints that we should be aware of? (Note: This term is the kinetic term, not the topological term like $F \wedge F$.)

See also a related question on Yang-Mills normalization.

In principle, the partition function $Z$ of quantum Yang-Mills theory will be schematically written as the path integral form over the measure $[DA]$ of gauge connection $A$ $$ Z = \int [DA] \exp(i S_{YM}) $$

  • $\begingroup$ It's just a field redefinition. $\endgroup$ – Aaron Bergman Jun 23 '18 at 0:44
  • $\begingroup$ what is the benefit of which convention $S_{YM,1}$? $\endgroup$ – wonderich Jun 23 '18 at 0:45
  • $\begingroup$ If you redefine $A \mapsto g A$ to get rid of the $1/g^2$ in the kinetic term (the quadratic part of the action), this redefines $F \mapsto dA + g A \wedge A$, which shows how $g$ really acts as a coupling constant for the non-linear terms in the action. Reverse these steps to get the historical order. $\endgroup$ – Igor Khavkine Jun 23 '18 at 11:06
  • $\begingroup$ As Igor already remarked, $g$ usually has the interpretation of a coupling constant. This is usually only important if you have different "species" of fields whose Lagrangian is of the same Yang-Mills form but each with a different weighting. By the way, you can also absorb the constant in the bilinear form on the Lie algebra (which is implicitly used in the above action). $\endgroup$ – Tobias Diez Jun 24 '18 at 21:01
  • $\begingroup$ Is that still true if you have the topological term $$\frac{\theta}{8 \pi^2} \int\mathrm{Tr}_\mathfrak{g}(F\wedge F),$$ can you still rescale/redefine the field easily? Or if you have fermionic or bosonic matter fields? $\endgroup$ – wonderich Jun 25 '18 at 0:59

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.