1
$\begingroup$

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'.

$\endgroup$
1
$\begingroup$

"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, $U(1)$.)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – gautampk Apr 9 '17 at 15:42
  • $\begingroup$ 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? $\endgroup$ – gautampk Apr 9 '17 at 15:43
  • $\begingroup$ @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". $\endgroup$ – J.G. Apr 9 '17 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.