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I have a lagrangian written as:

$$\mathcal{L}_H = \text{Tr}\left[\,(D_\mu \Phi)^\dagger D^\mu \Phi\right] - \mu^2 \text{Tr}\left[\,\Phi^\dagger \Phi\right] - \lambda (\text{Tr}\left[\,\Phi^\dagger \Phi\right])^2 $$

Where the field is:

$$\Phi \equiv \frac{1}{\sqrt{2}}(i \sigma_{2} \phi^*, \phi) = \frac{1}{\sqrt{2}} \begin{pmatrix}\phi^{0*} & \phi^+ \\-\phi^- & \phi^0\end{pmatrix} \;,$$


$$D_\mu \Phi \equiv \partial_\mu \Phi + i g \frac{\sigma^{i}}{2} W^{i}_\mu \Phi - i \frac{g'}{2} B_\mu \Phi \sigma_3 \;.$$

And I need to show that that is the same as:

$$\mathcal{L}_H = \left[(\partial_\mu+ig{\bf{T}\cdot W_\mu}+ig'\frac{Y}{2}B_\mu)\phi\right]^\dagger\left[(\partial_\mu+ig{\bf{T}\cdot W_\mu}+ig'\frac{Y}{2}B_\mu)\phi\right]-\mu^2(\phi^\dagger\phi)-\lambda(\phi^\dagger\phi)^2 $$.

Now, I have easily shown that the potential part really is the same, by explicitly doing matrix multiplication, and showing that the expression with the trace is the same as the one without, but how to show the kinetic part? What to do here?

Any help is welcome :)

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