# How to expand $(D_\mu\Phi)^\dagger(D^\mu\Phi)$ in $SU(2)$?

I would like to calculate the following expression: $$(D_\mu\Phi)^\dagger(D^\mu\Phi)$$ where $$D_\mu\Phi = (\partial_\mu-\frac{ig}{2}\tau^aA_\mu^a)\Phi$$ and $$A_\mu^a$$ are the components of a real $$SU(2)$$ gauge Field and $$\Phi$$ is a complex two component spinor.

Unfortunately I have not yet come to a result. Here is what I have tried, $$(\tau^a)^\dagger = \tau^a .$$

Inserting the definition of $$D_\mu$$ yields
$$\partial_\mu\Phi^\dagger \partial^\mu \Phi - \partial^\mu(\Phi^\dagger)\frac{ig}{2}\tau^a A_\mu^a\Phi+\frac{ig}{2} \Phi^\dagger \tau^aA^{a}_{\mu}\partial^\mu \Phi+\frac{g^2}{4}\Phi^\dagger\tau^aA^a_\mu \tau^bA^{b\mu}\Phi.$$

can I simplify this expression even further?

• Antidistribute the two middle terms and condense the evident anticommutator of the Pauli matrices in the last term. Commented Jul 12 at 13:11
• For the two middle terms I get $\frac{ig}{2}[\Phi^\dagger\tau^a A^a_\mu,\partial^\mu]\Phi$ and for the last term using $\tau^a\tau^b = \delta^{ab} 1 + i \epsilon^{abc} \tau^c$, $\frac{g^2}{4}[\Phi^\dagger A^a_\mu A^{a\mu}\Phi + i \Phi^\dagger \epsilon^{abc} \tau^c A^a_\mu A^{b\mu} \Phi$] is this korrekt? @CosmasZachos Commented Jul 12 at 15:11

You appear to not appreciate your expression as a row vector dotted on a column vector (possibly sandwiching operators). I corrected your expression to $$\partial_\mu\Phi^\dagger \partial^\mu \Phi - \partial^\mu(\Phi^\dagger)\frac{ig}{2}\tau^a A_\mu^a\Phi+\frac{ig}{2} \Phi^\dagger \tau^aA^{a}_{\mu}\partial^\mu \Phi+\frac{g^2}{4}\Phi^\dagger\tau^aA^a_\mu \tau^bA^{b\mu}\Phi\\ = \partial_\mu\Phi^\dagger \partial^\mu \Phi +\frac{ig}{2} A^{a~\mu}~~\Phi^\dagger \tau^a\overset{\leftrightarrow}\partial_\mu \Phi + \frac{g^2}{8}A^a_\mu A^{b\mu} \Phi^\dagger\{\tau^a, \tau^b\} \Phi\\ =\partial_\mu\Phi^\dagger \partial^\mu \Phi +\frac{ig}{2} A^{a~\mu}~~\Phi^\dagger \tau^a\overset{\leftrightarrow}\partial_\mu \Phi + \frac{g^2}{4}A^a_\mu A^{a\mu} \Phi^\dagger \Phi .$$