Showing the equivalence of lagrangians

1. Jan 7, 2013

dingo_d

1. The problem statement, all variables and given/known data

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:

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

and

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

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 :)

2. Jan 9, 2013

fzero

You haven't specified the weak isospin $\mathbf{T}$ and hypercharge operator $Y$, but I think you will find that they are precisely defined so that this equivalence follows.