Gauge covariant derivative

Consider a generic (possibly non-Abelian) Gauge transformation, defined by a symmetry operator ${\displaystyle U(x)=e^{i\alpha (x)}}$, acting on a field ${\displaystyle \phi (x)}$, such that

${\displaystyle \phi (x)\rightarrow \phi '(x)=U(x)\phi (x)\equiv e^{i\alpha (x)}\phi (x),}$

${\displaystyle \phi ^{\dagger }(x)\rightarrow \phi {'}^{\dagger }=\phi ^{\dagger }(x)U^{\dagger }(x)\equiv \phi ^{\dagger }(x)e^{-i\alpha (x)},\qquad U^{\dagger }=U^{-1}.}$

where ${\displaystyle \alpha (x)}$ is an element of the Lie algebra associated with the Lie group of symmetry transformations, and can be expressed in terms of the generators of the group, ${\displaystyle \{t^{a}\}_{a}}$, as ${\displaystyle \alpha (x)=\alpha ^{a}(x)t^{a}}$.

The partial derivative ${\displaystyle \partial _{\mu }}$ transforms, accordingly, as

${\displaystyle \partial _{\mu }\phi (x)\rightarrow \partial _{\mu }\phi '(x)=U(x)\partial _{\mu }\phi (x)+(\partial _{\mu }U)\phi (x)\equiv e^{i\alpha (x)}\partial _{\mu }\phi (x)+i(\partial _{\mu }\alpha )e^{i\alpha (x)}\phi (x)}$

and a kinetic term of the form ${\displaystyle \phi ^{\dagger }\partial _{\mu }\phi }$ is thus not invariant under this transformation.

We can introduce the covariant derivative ${\displaystyle D_{\mu }}$ in this context as a generalization of the partial derivative ${\displaystyle \partial _{\mu }}$ which transforms covariantly under the Gauge transformation, i.e. an object satisfying

${\displaystyle D_{\mu }\phi (x)\rightarrow D'_{\mu }\phi '(x)=U(x)D_{\mu }\phi (x),}$

which in operatorial form takes the form

${\displaystyle D'_{\mu }=U(x)D_{\mu }U^{\dagger }(x).}$

We thus compute (omitting the explicit ${\displaystyle x}$ dependencies for brevity)

${\displaystyle D_{\mu }\phi \rightarrow D'_{\mu }U\phi =UD_{\mu }\phi +(\delta D_{\mu }U+[D_{\mu },U])\phi }$,

where

${\displaystyle D_{\mu }\rightarrow D'_{\mu }\equiv D_{\mu }+\delta D_{\mu }}$.

The requirement for ${\displaystyle D_{\mu }}$ to transform covariantly is now translated in the condition

${\displaystyle (\delta D_{\mu }U+[D_{\mu },U])\phi =0.}$

To obtain an explicit expression, we follow QED and make the Ansatz

${\displaystyle D_{\mu }=\partial _{\mu }-igA_{\mu },}$

where the vector field ${\displaystyle A_{\mu }}$ satisfies,

${\displaystyle A_{\mu }\rightarrow A'_{\mu }=A_{\mu }+\delta A_{\mu },}$

from which it follows that

${\displaystyle \delta D_{\mu }\equiv -ig\delta A_{\mu }}$

and

${\displaystyle \delta A_{\mu }=[U,A_{\mu }]U^{\dagger }-{\frac {i}{g}}[\partial _{\mu },U]U^{\dagger }}$

which, using ${\displaystyle U(x)=1+i\alpha (x)+{\mathcal {O}}(\alpha ^{2})}$, takes the form

${\displaystyle \delta A_{\mu }={\frac {1}{g}}([\partial _{\mu },\alpha ]-ig[A_{\mu },\alpha ])+{\mathcal {O}}(\alpha ^{2})={\frac {1}{g}}[D_{\mu },\alpha ]+{\mathcal {O}}(\alpha ^{2})}$

We have thus found an object ${\displaystyle D_{\mu }}$ such that

${\displaystyle \phi ^{\dagger }(x)D_{\mu }\phi (x)\rightarrow \phi '^{\dagger }(x)D'_{\mu }\phi '(x)=\phi ^{\dagger }(x)D_{\mu }\phi (x).}$

If a gauge transformation is given by

${\displaystyle \psi \mapsto e^{i\Lambda }\psi }$

and for the gauge potential

${\displaystyle A_{\mu }\mapsto A_{\mu }+{1 \over e}(\partial _{\mu }\Lambda )}$

then ${\displaystyle D_{\mu }}$ transforms as

${\displaystyle D_{\mu }\mapsto \partial _{\mu }-ieA_{\mu }-i(\partial _{\mu }\Lambda )}$,

and ${\displaystyle D_{\mu }\psi }$ transforms as

${\displaystyle D_{\mu }\psi \mapsto e^{i\Lambda }D_{\mu }\psi }$

and ${\displaystyle {\bar {\psi }}:=\psi ^{\dagger }\gamma ^{0}}$ transforms as

${\displaystyle {\bar {\psi }}\mapsto {\bar {\psi }}e^{-i\Lambda }}$

so that

${\displaystyle {\bar {\psi }}D_{\mu }\psi \mapsto {\bar {\psi }}D_{\mu }\psi }$

and ${\displaystyle {\bar {\psi }}D_{\mu }\psi }$ in the QED Lagrangian is therefore gauge invariant, and the gauge covariant derivative is thus named aptly.

On the other hand, the non-covariant derivative ${\displaystyle \partial _{\mu }}$ would not preserve the Lagrangian's gauge symmetry, since

${\displaystyle {\bar {\psi }}\partial _{\mu }\psi \mapsto {\bar {\psi }}\partial _{\mu }\psi +i{\bar {\psi }}(\partial _{\mu }\Lambda )\psi }$.

In quantum chromodynamics, the gauge covariant derivative is[11]

${\displaystyle D_{\mu }:=\partial _{\mu }-ig_{s}\,G_{\mu }^{\alpha }\,\lambda _{\alpha }/2}$

where ${\displaystyle g_{s}}$ is the coupling constant of the strong interaction, ${\displaystyle G}$ is the gluon gauge field, for eight different gluons ${\displaystyle \alpha =1\dots 8}$, and where ${\displaystyle \lambda _{\alpha }}$ is one of the eight Gell-Mann matrices. The Gell-Mann matrices give a representation of the color symmetry group SU(3). For quarks, the representation is the fundamental representation, for gluons, the representation is the adjoint representation.

The covariant derivative in the Standard Model combines the electromagnetic, the weak and the strong interactions. It can be expressed in the following form:[12]

${\displaystyle D_{\mu }:=\partial _{\mu }-i{\frac {g'}{2}}Y\,B_{\mu }-i{\frac {g}{2}}\sigma _{j}\,W_{\mu }^{j}-i{\frac {g_{s}}{2}}\lambda _{\alpha }\,G_{\mu }^{\alpha }}$

The gauge fields here belong to the fundamental representations of the electroweak Lie group ${\displaystyle U(1)\otimes SU(2)}$ times the color symmetry Lie group SU(3). The coupling constant ${\displaystyle g'}$ provides the coupling of the hypercharge ${\displaystyle Y}$ to the ${\displaystyle B}$ boson and ${\displaystyle g}$ the coupling via the three vector bosons ${\displaystyle W^{j}}$ ${\displaystyle (j=1,2,3)}$ to the weak isospin, whose components are written here as the Pauli matrices ${\displaystyle \sigma _{j}}$. Via the Higgs mechanism, these boson fields combine into the massless electromagnetic field ${\displaystyle A_{\mu }}$ and the fields for the three massive vector bosons ${\displaystyle W^{\pm }}$ and ${\displaystyle Z}$.