- #1
Ravendark
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Homework Statement
Consider the fermionic part of the QCD Lagrangian: $$\mathcal{L} = \bar\psi (\mathrm{i} {\not{\!\partial}} - m) \psi \; ,$$ where I used a matrix notation to supress all the colour indices (i.e., ##\psi## is understood to be a three-component vector in colour space whilst each component is a Dirac four spinor).
To achieve local gauge invariance, we introduce a covariant derivative ##D_\mu## containing a gauge field with a transformation property that ensures local gauge invariance of ##\mathcal{L}##.
My question: Does the sign in the covariant derivative really matters? Or is it more like a convention and every author choose its own? I wrote down my thoughts in section 3.
Homework Equations
Covariant derivative: ##D_\mu = \partial_\mu \pm A_\mu##.
The Attempt at a Solution
##\mathcal{L}## is not invariant under local gauge transformations ##U = U(x) \in \mathrm{SU}(3)## of the fields since $$\partial_\mu \psi \to \partial_\mu \psi' = \partial_\mu U \psi = U \partial_\mu \psi + (\partial_\mu U) \psi \; .$$ Now we introduce a modified derivative ##D_\mu## and demand that it transforms like the fields, i.e., $$D_\mu \psi \stackrel{!}{\to} U D_\mu \psi \; .$$This would lead to a locally gauge invariant Lagrangian.
Now the sign of the covariant derivative comes into play:
- The choice ##D_\mu = \partial_\mu + A_\mu## implies that the gauge field has to transform as ##A_\mu \to U A_\mu U^{-1} - (\partial_\mu U)U^{-1}## to achieve local gauge invariance.
- The choice ##D_\mu = \partial_\mu - A_\mu## implies that the gauge field has to transform as ##A_\mu \to U A_\mu U^{-1} + (\partial_\mu U)U^{-1}## to achieve local gauge invariance.
Thus, from my point of view the sign in the covariant derivative does not really matters since we simply have to demand a slighty different transformation behaviour of the gauge field. Is this correct?