# How do $\phi^4$ terms result in an interacting (classical) field?

It was my understanding that a scalar field, $\phi(t;\vec{x})$, varying in time and parametrised by position, was 'non-interacting' if the field evaluated at one point affected the field evaluated at another. That is, the set of equations of motion (one for each $\vec{x}$) for the field were decoupled in the position degrees of freedom.

However, a field with a Lagrangian that has a term like

$$\mathcal{L}_{\mathrm{int}} = -\lambda\phi^{4}$$

is still said to be 'interacting', even though the equations of motion are still decoupled.

I would have thought non-local terms like

$$\mathcal{L}_{\mathrm{int}} = -\lambda\phi^{*}(t;\vec{x})\phi(t;\vec{x}')$$

would be needed, though I get that they're not allowed to keep the theory local.

I understand that a $\phi^4$ term results in non-linear equations of motion, but my question is what is 'interacting' in the these fields?

Additionally, I can see how if there are two fields then a Yukawa term like:

$$\mathcal{L}_{\mathrm{int}} = -g\psi^{*}\phi\psi$$

can be interpreted as 'interaction', but I'm not clear on why a second order coupling term like $-g\psi^{*}\phi$ isn't an 'interaction'.

"Interacting" doesn't mean an interaction between values of $\phi$ at space-like separated points; that violates causality. Let me try to explain what it does mean. I'll focus on real fields for simplicity; I'll leave the complex case to you. I'll work with $c=\hbar=1$.

A Lagrangian density $\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi$ gives $\square\phi=0$, which has plane-wave solutions in Minkowski space. This is analogous to the classical-mechanics choice $L=\frac{1}{2}m\dot{q}^2$, giving $m\ddot{q}=0$. Clearly, this is non-interacting because there's no "force" acting on $\phi$ or $q$.

Classically any non-uniform $V$ implies $L=\frac{1}{2}m\dot{q}^2-V(q)$ incudes an interaction, $m\ddot{q}=-V'(q)$. So for an "interacting" field, it'd be enough to take $L=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-V(\phi)$ so that $\square\phi=-V'(\phi)$, right?

Well, not quite. Perhaps the simplest choice for $V$ is one that is a quadratic with a global minimum (note this precludes a linear term), since expansion around such a turning point gives a quadratic at lowest order. This gives us the familiar Hooke's law, viz. $V=\frac{1}{2}m\omega^2q^2$ so $\ddot{q}=-\omega^2 q$. The field-theoretic equivalent is $V=\frac{1}{2}m^2\phi^2$ (in this context we use $m$ instead of $\omega$), giving $\square\phi =-m^2\phi$. That still has plane-wave solutions, though; they just have an $m$-dependent change to the relativistic dispersion relation. We interpret this not as an "interaction", but just as the particle having a mass.

So if we don't think of quadratic terms in the potential as "interacting" in field theory, what would deserve that name? Well, just go to the next order, viz. $\square\phi=-m^2\phi+O(\phi^3)$. Even a massive field won't obey this equation unless it's also subject to an "interaction's" potential.

Note: why do we use a quartic potential and cubic force rather than a cubic potential and a quadratic force? Because the parity of the exponents is crucial to whether the theory has a desirable symmetry, $\mathbb{Z}_2$ (or, in the complex case, $U(1)$.)

• So the $\phi^2$ term gives rise to a self-interaction which is interpreted as the rest mass for quanta of the field? That does make sense, and also makes the Higgs mechanism more clear. Instead of a self-interaction which isn't gauge invariant (?) it's replaced by an interaction with the Higgs field. – gautampk Apr 9 '17 at 15:42
• One follow-up: why are higher order 'self-interactions' like $\phi^4$ not interpreted as mass terms? Is it just because they prevent plane wave 'free' solutions? – gautampk Apr 9 '17 at 15:43
• @gautampk: This has nothing to do with the Higgs field, which provides a mechanism for giving vector (not scalar) fields a mass besides just including a term quadratic in the vector (which would break the symmetry anyway). The only way to interpret higher-order terms as due to a mass would be if $\phi$ had a $\phi$-dependent "mass". – J.G. Apr 9 '17 at 15:51