How would one show that the nonabelian ${F_{\mu\nu}}$ field strength tensor transforms as $${F_{\mu\nu}\to F_{\mu\nu}^{\prime}=UF_{\mu\nu}U^{-1}}$$ under a local gauge transformation? Rather than going through this in a very manual manner (i.e., gauge transform ${A_{\mu}\to A_{nu}^{\prime}}$ and use the explicit expression for ${F_{\mu\nu}}$), it was suggested to me to show first that $${D_{\mu}\to D_{\mu}^{\prime}=UD_{\mu}U^{-1}}$$ and then gauge transform the commutator relation ${\left[D_{\mu},D_{\nu}\right]}$ for ${F_{\mu\nu}}$: $${\left[D_{\mu},D_{\nu}\right]\psi\left(x\right)=gF_{\mu\nu}\psi\left(x\right)}.$$


2 Answers 2


The field strength tensor is proportional to the commutator of the covariant derivatives:

$$F_{\mu\nu}\propto[D_\mu,D_\nu]=D_\mu D_\nu-D_\nu D_\mu.$$

Transforming this expression according to $D_{\mu}\to D_{\mu}^{\prime}=UD_{\mu}U^{-1}$ gives

$$UD_{\mu}U^{-1}UD_{\nu}U^{-1}-UD_{\nu}U^{-1}UD_{\mu}U^{-1}=UD_\mu D_\nu U^{-1}-UD_\nu D_\mu U^{-1}=U[D_\mu,D_\nu]U^{-1},$$

where we have used $UU^{-1}=1$. Due to the proportionality of the field strength to the commutator, the desired transformation property

$${F_{\mu\nu}\to F_{\mu\nu}^{\prime}=UF_{\mu\nu}U^{-1}}$$

is evident.

The transformation property of the covariant derivative follows from both the transformation of the partial derivative and the gauge field itself.

  • This question is often asked typically because it is ambiguous from the notation how far the differential symbols act, cf. e.g. this Phys.SE post.

  • The short answer is that the set of operators $T$ that transform via a similarity transformation $T\mapsto UTU^{-1}$ form a subalgebra, i.e. the set is closed under

    1. addition,
    2. scalar multiplication, and
    3. operator composition.

    Since the commutator $[S,T]:=ST -TS$ is built from such elementary operations, the result follows.


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