Spontaneous symmetry breaking scalar field masses

[Milsomonk]

Homework Statement

Determine the mass of the scalars and show that one remains zero in accordance with goldstones theorem.

Homework Equations

$$L=\dfrac {1}{2} (\partial_\mu \phi_a)(\partial^\mu \phi_a)-\dfrac{1}{2} \mu^2 (\phi_a \phi_a) - \dfrac{1}{4} \lambda (\phi_a \phi_a)^2+ i\bar{\psi} \gamma^\mu \partial_\mu \psi -g\bar{\psi} (\phi_1 +i\gamma^5 \phi_2)\psi$$
I have chosen a vacuum solution that breaks the symmetry
$$\phi_1 = \sqrt{\dfrac{-\mu^2}{\lambda}}, \phi_2 =0$$

The Attempt at a Solution

So I know that I need to expand the fields around the minimum and then write the new lagrangian, then I should be able to read the mass from the hyperbolic terms, but I'm not sure how to carry out the expansion. Any advice would be much appreciated :)

 

[Orodruin]

How would you usually make an expansion of any function $f(\vec x)$ about some point $\vec x_0$?

 

[Milsomonk]

I'd use a Mclaurin series ordinarily, I think the fact that there are two field is the part that is confusing me, but since I'm picking my vacuum solution such that Phi_2 is zero can I simply expand Phi_1 and discard all terms with Phi_2? also I'm not too sure if I need to rewrite the whole lagrangian once I've done the expansion or if just the potential would suffice. Thanks for your response :)

 

[Orodruin]

Just look at it as a function of two variables (that in turn happens to be functions of the space time coordinates, but that is besides the point) that you want to expand around a point which has one of the variables equal to zero. It is just a multidimensional Taylor series. In fact, you do not even need to take any derivatives, just express $\phi$ as a sum of the point you want to expand about and the deviation from that point.

 

[Milsomonk]

Ok so if I understand you correctly I can expand \phi as follows:

$$\phi_1=v+h$$
where v is my vacuum solution and h is my deviation.
then I can expand my second field but v is zero so I just get a deviation f.

$$\phi_2=f$$

Now I can substitute this into my potential and read the mass' from the terms that are squared in h and f?

 

[Orodruin]

Correct. It does not need to be more difficult than that.

 

[Milsomonk]

Ah awesome, thanks! much clearer now :)