I have a Lagrangian with a two interacting complex scalar field $\phi$ and $\chi$.

$$ \mathcal{L} = |\partial_{\mu} \phi |^2 + |\partial_{\mu} \chi |^2 + \lambda_1 \left( |\phi|^2 - \frac{v_1^2}{2} \right)^2 + \lambda_2 \left( |\chi|^2 - \frac{v_2^2}{2} \right)^2 $$

First of all I have to find the spectrum of the particles. I'm not sure what a good answer would be. The two field $\phi$ and $\chi$ does not interact with each other, so I should have free particles and two kind of four particles composite systems with a mass gap of $$M_1 = 4 \lambda_1 v_1^2$$ and $$M_2 = 4 \lambda_2 v_2^2$$ for the $\phi$ and $\chi$ fields respectively.

Anything other to say? I feel like mine is a poor answer.

  • $\begingroup$ "free particles and two kind of four particles composite systems" - Even a single field has 1, 2, 3, 4, ... particle states. $\endgroup$ Commented Mar 22, 2020 at 23:20
  • $\begingroup$ @KeithMcClary A free field has only (free) particles with the mass given by the pole in the Spectral Representation. I'm convinced, as you said, that there could be be more than one particle, but they does not form a composite system. I'm not even sure of what "Find the spectrum of the particles" means. I have to find the mass? I have to find the energies? How can i compute the energies in an interacting theory? I need the Hamiltonian, but this is not an easy task if there are interactions. $\endgroup$
    – ACA
    Commented Mar 23, 2020 at 8:00
  • $\begingroup$ I know the mathematical definition of the spectrum of the Hamiltonian as an operator in Hilbert space. I don't know "spectrum of the particles". $\endgroup$ Commented Mar 23, 2020 at 16:32

1 Answer 1


In particle physics people refer to the masses and states of the physical particles as the spectrum. So when asked for the spectrum of this theory, they are asking to find the physical states and their masses.

I will proceed with only one field $\phi$ as the other one is analogous. Your parametrization suggests that your scalar fields have non-vanishing vacuum expectation values. I parametrize the complex scalar field as

$$ \phi = \frac{1}{\sqrt{2}}(v_1 + R + i I) $$

where $R$ and $I$ are real components. Note that $R$ is CP-even and $I$ is CP-odd. Expanding the potential terms we have the following mass term $$ \mathcal{L} \supset - \frac{1}{2} M_R^2 R^2 $$ with $M_R^2 = - 2 \lambda_1 v_1^2$. Note that $I$ remains massless. So the spectrum (only considering $\phi$) contains a massless CP-odd scalar field $I$ and a massive CP-even scalar $R$.

We could have expected this on the basis of the Goldstone theorem. Our Lagrangian for the complex scalar field has a global symmetry (rephasing with $U(1)$) that is spontaneously broken so we expect a massless Goldstone boson.

In your complete theory, you have therefore two massive CP-even states and two massless CP-odd Goldstone bosons.

  • $\begingroup$ So, when they ask me for the spectrum of the particles it is implicit that I have to study the theory around a vacuum expectation value? Should I specify the bound states of the theory ($R^4$, $R^3I$, $R^2I^2$ i guess...) and their (rest) masses? $\endgroup$
    – ACA
    Commented Mar 26, 2020 at 9:17
  • 1
    $\begingroup$ I was assuming the model is weakly coupled and perturbative, so there are no bound states. I would assume that was the idea of the exercise in your case also. What kind of course you are taking? $\endgroup$ Commented Mar 26, 2020 at 13:31
  • $\begingroup$ The fact that I never encountered bound states in QFT should fires my attention. Thanks to your comment I realize the things are quite complex and involve the so called Bethe-Salpeter equation. I undertaken a first course in QFT, in which we give a theoretical proof of Goldstone's theorem. I'm now preparing a PhD entrance exam so that I need to deepen some concepts and became more experienced with homework. $\endgroup$
    – ACA
    Commented Mar 26, 2020 at 14:00
  • 1
    $\begingroup$ Ok, yes strongly coupled QFT is very difficult. In your course, I guess the idea of the exercise was for you to show the understanding of the Goldstone theorem. Each scalar field in your Lagrangian is invariant under a global U(1), which spontaneously broken generates 1 Goldstone because of the broken generator. $\endgroup$ Commented Mar 26, 2020 at 14:18

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