1

Consider the free scalar field theory $$ \mathcal{L}=\partial _\mu \phi \partial ^\mu \phi^* - V(|\phi|) $$ where $V(|\phi|)=-m^2|\phi|^2$.

There are several answers on this site about why the quadratic term in the Lagrangian corresponds to the mass (see here and here for instance). But most of these explanations assume there are no higher-order terms in the scalar potential. What happens if we have a more complicated scalar potential such as $V(|\phi|)=-a^2|\phi|^2+b^2|\phi|^4$? Is the mass still just $a$ or would it be some combination of $a$ and $b$? The difference here is that now the potential has minima at $V(|\phi|)\neq 0$.

Do the locations of the minima of the potential have any affect on what we consider the mass of the particle in the theory? i.e. in an interacting theory would a field configuration be considered more or less massive than the free theory?

Superbee
  • 701

2 Answers2

2

As far as I understand the mass $m_P$ of an asymptotic (i.e. observable) particle,e.g. corresponding to some scalar field $\phi$, is the location of the pole of the full two-point function, i.e. $$ \int d^4x~e^{ip(x-y)}\langle \Omega | T \phi(x) \phi(y) |\Omega \rangle = \frac{i Z}{p^2 - m_P^2 + i0} + \text{terms regular at }p^2 = m_P^2. $$ Note that $m_P$ is not the bare mass $m$ appearing in the Lagrangian, in fact the bare mass is formally divergent. $m_P$ receives perturbative corrections, so it does indeed depend on the interaction terms. The bare mass is the mass of the free theory, where all the interactions terms, i.e. all non-quadratic terms, are zero, but it is by no means a physical quantity, just a parameter.

I should also mention that the minima business that you mention is related to the Spontaneously broken symmetry Lagrangians. In such cases in order to do perturbation theory, one has to expand the fields around the minimum which changes the terms in the Lagrangian. It is important to understand that the actual observable particle corresponds to excitations around the vacuum, i.e. it corresponds to the field after you did the expansion. As said above, the mass of this particle is then the location of the pole of its full 2-pt function.

jkb1603
  • 1,249
  • 6
  • 15
1

So the simplest explanation of why the quadratic term is the mass term is because it generates the correct mass term in the equations of motion, \begin{equation} \mathcal{L} = \partial_\mu \phi \partial^\mu \phi^* - m^2 |\phi|^2 \quad \longrightarrow (\square + m^2 ) \phi = 0. \end{equation} Under normal circumstances, the quadratic part is not included in the potential, simply because it is the mass term. This is true as long as $a^2 < 0$ in \begin{equation} \mathcal{L} = \partial_\mu \phi \partial^\mu \phi^* - V(|\phi|), \quad V(|\phi|) = -a^2 |\phi|^2 + b^2 |\phi|^4. \end{equation} In this case you have a minimum at $\phi = 0$ and thus no spontaneous symmetry breaking (SSB). If $a^2 >0$, then you are right and the vacuum state is no longer at $\phi =0$ hence we have SSB. However then $a^2$ is no longer the mass-term because $\phi$ is no longer a physical field since physical fields require a vanishing vacuum expectation value (VEV). The physical fields are obtained by expanding around the VEV. A quick calculation shows you that you'll get one massless scalar (the Goldstone boson) and one boson with mass $\sqrt{2} a$.

tomtom1-4
  • 1,249