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Spontaneous symmetry breaking scalar field masses

  1. Dec 7, 2017 #1
    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. jcsd
  3. Dec 7, 2017 #2

    Orodruin

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    How would you usually make an expansion of any function ##f(\vec x)## about some point ##\vec x_0##?
     
  4. Dec 7, 2017 #3
    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 :)
     
  5. Dec 7, 2017 #4

    Orodruin

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    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.
     
  6. Dec 7, 2017 #5
    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?
     
  7. Dec 7, 2017 #6

    Orodruin

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    Correct. It does not need to be more difficult than that.
     
  8. Dec 7, 2017 #7
    Ah awesome, thanks! much clearer now :)
     
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