1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Showing the equivalence of lagrangians

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a lagrangian written as:

    [itex]\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 [/itex]

    Where the field is:

    [itex]\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} \;,[/itex]

    and

    [itex]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 \;.[/itex]

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

    [itex]$\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 [/itex].

    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. jcsd
  3. Jan 9, 2013 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You haven't specified the weak isospin [itex]\mathbf{T}[/itex] and hypercharge operator [itex]Y[/itex], but I think you will find that they are precisely defined so that this equivalence follows.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Showing the equivalence of lagrangians
Loading...