# Two component field theory and unitary transformation

1. Jan 5, 2012

### zakk87

I have a theory described by a 2-component field $\psi_i$ (i'm working with BCS in Nambu-Gorkov representation, but any other field theory would be ok, that's why I'm posting in this subforum), and the lagrangian it's defined in the following way:

$$\mathscr{L}=\psi^\dagger \Gamma \psi$$

where $\Gamma$ is an opportune 2x2 matrix. Now, I know from theory that applying a unitary transformation won't change the physics of the system, but I can't always see how it's so. The simplest counter-example I came up with is this one:

$$\Gamma=\left( \begin{array}{cc} \partial_\mu & 0 \\ 0 & \partial_\mu \end{array} \right)$$

and by imposing $\det \left( \Gamma \right) = 0$ and taking the Fourier transform I can find the dispersion relation: $k_\mu k^\mu = 0 \Longrightarrow \omega = \left| \mathbf{k} \right|$. But if I apply the following unitary transformation $U=\left( \begin{array}{cc} e^{- \mathrm{i} \alpha \left( x \right)} & 0 \\ 0 & e^{\mathrm{i} \alpha \left( x \right)} \end{array} \right)$ i get:

$$\Gamma' \equiv U^{-1} \Gamma U =\left( \begin{array}{cc} \partial_\mu - \mathrm{i} \partial_\mu \alpha \left( x \right) & 0 \\ 0 & \partial_\mu + \mathrm{i} \partial_\mu \alpha \left( x \right) \end{array} \right)$$

and now the dispersion relation reads: $k_\mu k^\mu + k^2 \alpha \left( x \right)^2 = 0$ which is quite different. Now, the dispersion relation contains lots of physical information about the system and it shouldn't have changed.

My guess is that those two dispersion relations are essentially the same one. The change is related to $U$ being the infinitesimal generator of translation in momenta space, and that the second dispersion relation is just the first one "Doppler shifted", i.e. written in another intertial frame of reference. Am I right? And if so, is there anyway I can show that the second dispersion relation is just the first one written in a different frame of reference?