# Does the quadratic mass term $\frac{1}{2}m^2\phi^2$ belong to the free Lagrangian or is it an interaction term?

$$L = -\frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2.$$ Why is the $$\phi^2$$ term in the scalar Lagrangian not considered a self-interaction?

• You can treat it as either, though treating it as part of the kinetic term is more traditional and definitely infinitely simpler. Jun 13 '20 at 6:42

Conventionally, bilinear terms in the Lagrangian are called kinetic terms; the rest (terms with three or more fields) is considered interactions. The reason is simple: if you solve the theory with just the kinetic terms, the solution describes the behavior of free (non-interacting) particles. For the example you provided, the solution is that of a free scalar particle and the $$m$$ in front of the $$\phi^2$$ term happens to be the mass of the particle. That is why the term is called a mass term, part of the kinetic terms.

Nonetheless, it is just a convention. If you insist, you can treat the term as a self-interaction. But then when you start solving the theory, you would discover that the self-interaction term give the particle, which you thought was massless, a mass.

• What bothers me is that in the case of classical mechanics (no fields), if we add an $x^2$ then we will be dealing with a harmonic oscillator instead of a free particle. Does the convention changes when moving to fields? Apr 22 '20 at 15:28
• @Pinkman98 Well the convention is the same, we pick out the terms that describe free particles and the rest would be interactions. It just happens that a Lagrangian of that particular form with a mass term is what we need to describe free massive particles. If you really want to find out exactly why that must be the case, Weinberg in his QFT book vol.1 has a very logical development going from free particle states to quantum fields to the required forms for the Hamiltonian and Lagrangian. Apr 22 '20 at 18:20
• Ah, I see. Thank you very much. Also I would say that maybe the difference is that in classical mechanics, the mass appears already in the $\frac{p^2}{2m}$ term which is not the case in the scalar Lagrangian. Apr 23 '20 at 8:22

You definitely can treat it as an interaction. If you do, the "free" propagator is that of a massless theory: $$G_0(k^2) = {1 \over k^2 + i \epsilon}$$

The "interacting" propagator can be summed geometrically:

$$G_i(k^2) = G_0 + G_0 {m^2} G_0 + ...$$ $$G_i(k^2) = {G_0 \over 1 - G_0 m^2} = {1 \over k^2 - m^2 + i \epsilon}$$

However the main reason why we treat the "free" propagator as anything quadratic in the field is because anything that is bilinear in the fields can be evaluated exactly. It is just a gaussian integral:

$$Z(J) = \int \mathcal{D} \phi e^{i \phi M \phi + J \phi}$$ $$Z(J) \propto {1 \over \sqrt{\det M}} e^{\frac12 J M^{-1} J}$$

Because it can be solved exactly, we consider it "free" solution, from which we compute perturbatively interactions.

Because the solutions are diagonalized in k-space, the propagation of the eigenstates do not scatter each other.

• I'm sorry to post a comment here months after the question was answered, but how would you deal with the 1/2 factor appearing in the "mass interaction"? Mar 18 at 15:35
• The $M$ operator here is defined below: $M \equiv -\frac12 (\partial^2 + m^2)$. You can use integration by part of the original lagrangian ($\frac12 (\partial \phi)^2 - m^2 \phi^2$) to show the 2 integrals are the same. Sorry for confusion. May 9 at 7:24