# Unitary gauge for spontaneous symmetry breaking

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 $$\eta_3$$?

• 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. – Tomáš Brauner Mar 8 '19 at 21:13
• @TomášBrauner is right. You will need two gauged SU(2)s to be able to rotate your vev into the desired form. – InertialObserver Mar 9 '19 at 9:06