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I'm given a lagrangian $$ \mathcal{L} = \partial_{\mu} \Phi^{\dagger} \partial^{\mu} \Phi + m^2 \Phi^{\dagger} \Phi - \lambda (\Phi^{\dagger} \Phi)^2 $$ where $m^2 > 0, \lambda > 0$. This Lagrangian is invariant under the following $SU(2) \times U(1)$ transformations $$ \Phi \mapsto \Phi^{'} = e^{i g \alpha^{a} t_a} e^{i g^{'} q_H \beta } \Phi. $$ Here $t_a$ are some generators of a $4$-dimensional representation of $SU(2)$. By letting $\alpha \mapsto \alpha(x)$ and $\beta \mapsto \beta(x)$ and replacing $\partial_{\mu}$ by $D_{\mu}$ (covariant derivative) one can make a gauge invariant lagrangian.

Now, I want to show that by choosing the ground state as $$ \Phi_0 = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 0 \\ 0 \\ v \end{pmatrix} $$ with $v^2 = m^2/ \lambda$ , one can always make a gauge choice (the unitary gauge) such that an arbitrary field is parametrized as $$ \Phi(x) = \begin{pmatrix} \phi_1 \\ \phi_2 \\ 0 \\ \frac{1}{\sqrt{2}} (v+ \sigma(x)) \end{pmatrix} $$ where $\sigma(x)$ is a real scalar field and $\phi_1, \phi_2$ are two arbitrary complex fields. How can I show this?

I think I may always write an arbitrary field as $$ \Phi(x) = \begin{pmatrix} \phi_1 \\ \phi_2 \\ \eta_1(x) + i \eta_2(x) \\ \frac{1}{\sqrt{2}} (v + \sigma(x) + i \eta_3(x)) \end{pmatrix} $$

How can I show explicitly that the $SU(2) \times U(1)$ gauge transformations can be used to eliminate $\eta_1, \eta_2$ and $\eta_3$?

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  • $\begingroup$ If $\Phi$ is a complex 4-plet, then you in general cannot bring the ground state to your form $\Phi_0$. Just think that all possible $\Phi$s with fixed $\Phi^\dagger\Phi$ define a 7-dimensional manifold, whereas $SU(2)\times U(1)$ has dimension only 4. Note also that your original Lagrangian (before gauging) has a global $SO(8)$ symmetry. This may of course be broken by coupling to gauge fields of the natural $SU(2)\times U(1)$ subgroup. $\endgroup$ – Tomáš Brauner Mar 8 at 21:13
  • $\begingroup$ @TomášBrauner is right. You will need two gauged SU(2)s to be able to rotate your vev into the desired form. $\endgroup$ – InertialObserver Mar 9 at 9:06

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