First off, a pre-question: if I got this wrong, then probably the whole reasoning is wrong as well.

Studying the lagrangian for a two-particle scalar field with a quartic interaction in the context of SSB and Higgs mechanism, one can find many different cases:

  • the only fields are the scalars, and the quadratic term is positive: there is no SSB, the fields have the same mass.
  • the only fields are the scalars, but the quadratic term is negative: there is SSB, one field is massless (Goldstone mode), the other one gains more mass than the previous case.
  • there is another, vector, field, and the quadratic term is positive: there is no SSB, I have now three particles (two massive scalars, one massless vector)
  • there is another, vector, field, but the quadratic term is negative: there is SSB, and the Higgs mechanism makes the Goldstone mode disappear while the giving mass to the vector, so I have one massive scalar and one massive vector.

Is this correct?

Now, for the actual question: I want to find the masses of the particles in the various different cases. From what I understand, this means that I want to find the mass matrix elements $$m^2_{ab}=\frac{\partial^2V}{\partial\phi_a\partial\phi_b}\Big|_{\min}$$ and when there is no SSB, I actually only have to take the second derivative and evaluate it to zero. When there is SSB, I firstly have to shift my fields so that they both have a $VEV=0$, and then do the same thing.

In my lecture notes, the first two cases are done with a $O(2)$ symmetry, with a lagrangian $$\mathcal{L}=\frac{1}{2}\partial_\mu\vec{\phi}\partial^\mu\vec{\phi}-\frac{\mu^2}{2}\vec{\phi}\cdot\vec{\phi}-\frac{\lambda}{4}(\vec{\phi}\cdot\vec{\phi})^2$$ and the calculations of $$m^2_{11}=\frac{\partial^2V}{\partial\phi_1^2}\Big|_{\min}\quad m^2_{12}=\frac{\partial^2V}{\partial\phi_1\partial\phi_2}\Big|_{\min}$$ etc lead to the masses I mentioned.

However, to treat the "scalar+vector" cases, in my notes it says that it's preferable to use the $U(1)$ symmetry with a lagrangian $$\mathcal{L}=\partial_\mu\phi^\dagger\partial^\mu\phi-\mu^2\phi^\dagger\phi-\lambda(\phi^\dagger\phi)^2$$ but in this case if I try to calculate the masses with $$m^2=\frac{\partial^2V}{\partial\phi^\dagger\partial\phi}\Big|_{\min}$$I find that they don't change at all in the SSB case!

Is there something wrong with the reasoning, or should I just check my calculations better?

  • 1
    $\begingroup$ Check your calculation. For SSB, at the bottom of the potential (point of contact of sombrero with the table) there is a massless Goldstone scalar and a massive one. $\endgroup$ Jul 8, 2020 at 14:49
  • $\begingroup$ @CosmasZachos I know this, and it is the reason why I asked this question. Checking my calculation I find no errors, so I'm asking: should I just check better, or is there anything you can see in the question (definitions, formulas) that I got wrong? $\endgroup$ Jul 27, 2020 at 21:04
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    $\begingroup$ ? I am not sure what you mean by "vector". You are describing the same situation in two different parameterizations. Of course you should check your calculations. $\endgroup$ Jul 27, 2020 at 21:12

1 Answer 1


I'm a little rusty on this (since 2008, it's 2022) and I am not an expert, but a vector boson should have a field $A^\mu$ or $G_a^\mu$ depending whether it's a photon or a gauge boson, where the $\mu$ refers to the component of the vector and the $a$ refers to the basis matrix of QCD. The Lagrangian should have an interaction component of $\phi$ with $A^\mu$ or $G^\mu$ for a scalar field coupled with a vector field. I'm not sure, it may take the form ($\phi^\dagger\phi+\phi\phi^\dagger)(A+A^\dagger$). This should give you a different minimum, I believe, when the entire Lagrangian's minimum is considered.


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