Homework Help: Spontaneous symmetry breaking scalar field masses

1. Dec 7, 2017

Milsomonk

1. The problem statement, all variables and given/known data
Determine the mass of the scalars and show that one remains zero in accordance with goldstones theorem.

2. Relevant 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
$$\phi_1 = \sqrt{\dfrac{-\mu^2}{\lambda}}, \phi_2 =0$$

3. 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 :)

Last edited: Dec 7, 2017
2. Dec 7, 2017

Orodruin

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

3. Dec 7, 2017

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 :)

4. Dec 7, 2017

Orodruin

Staff Emeritus
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.

5. Dec 7, 2017

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?

6. Dec 7, 2017

Orodruin

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

7. Dec 7, 2017

Milsomonk

Ah awesome, thanks! much clearer now :)