Properties of boson in $\phi^4$ theory

[metroplex021]

I'm looking at the simplest example of an interacting theory, and this is the theory of a neutral scalar boson $\phi$ with $\lambda \phi^4$ interaction term. Can I ask: is there a physical interpretation of the `charge' through which this field interacts with itself? In particular, is \lambda a property of the field in something like the way that the electrical charge of the electron is both a property of it and a measure of the coupling strength?

(I am unsure of this latter analogy, as in the electrical case the charge property is understood as the invariant of the U(1) symmetry of the free electron Lagrangian; but the symmetry here is $Z_2$, so not continuous -- hence we can't use Noether's theorem.)

Any help much appreciated!

 

[The_Duck]

I guess we use "the charge of the electron" to mean two things. First, it can mean the coupling constant of the electromagnetic interaction. Second, it can mean the conserved quantity carried by the electron, associated with the U(1) symmetry of the action.

$\lambda$ is a coupling constant but it is not associated with any conserved quantity or symmetry.

If you want a classical intepretation of $\lambda$, I think that to lowest order the potential between two $\phi$ particles is a delta function $V(x, y) \propto \lambda \delta(x-y)$. Compare with the electric potential $V(x, y) \propto \frac{e^2}{|x-y|}$.

(PS: you can do inline latex like this: # # \lambda # #, but without the spaces between each pair of #'s )

 

[metroplex021]

Thank you, The_Duck. A final question: it seems strange to me that a particle can undergo an interaction without some interaction property to couple to. (In the case of real interacting theories, for example, we have properties like charge, color, and weak isospin in addition to their kinematic properties as the properties through which particles interact.) In QFT, is it somehow permitted that there can be coupling without properties to couple to? Or is this an aspect that makes $\phi^4$ theory merely a toy theory?

(Thanks too for the heads-up on tex!)

 

[The_Duck]

Gauge fields have to couple to conserved charges. That's why the electromagnetic field couples to the conserved electric charge, the gluon field couples to the conserved color charge, and so on. I think it can be proven that this is a requirement to build a consistent theory of massless spin-1 particles. But spin-0 particles have no such requirement. So scalar fields need not couple to conserved charges.

For example in the standard model the Higgs boson interacts with the leptons, quarks, W & Z bosons, and itself. It's not that these particles all carry some sort of "Higgs charge." It's just that there are terms in the Lagrangian that couple these fields to the Higgs.

 

[ChrisVer]

They somewhat did carry a "higgs" charge (it wasn't higgs it was the initial symmetry), that was eg the hypercharge.
The self-interactions is somewhat different? The mass term for example is such a self interaction (at least in the Dirac representation of your fermion fields).
There are also interactions like 3 or 4 gauge boson couplings (but they are associated with symmetry since they take the coupling constant coming from the gauge group).
For the scalars, that constant is an unknown free parameter of the theory, I am not sure if it is related to any symmetry because that would mean you would be able to determine it from the symmetry couplings instead (something that is not the case). You can just have the $\phi^4$ term and you put in front some coupling.

 

[vanhees71]

In effective hadronic field theories, the linear $\sigma$ model is a very early example for the linear realization of chiral symmetry in the strong interactions. And if you stick with the pions and its chiral partner, the socalled $\sigma$ (or in modern nomenclature $f_0$) meson, you have something only slightly more sophisticated $\phi^4$ theory. You have four real (self-adjoint), i.e., uncharged meson fields $\vec{\phi}$ and the Lagrangian with global O(4) symmetry (rotations in flavor space, which is the most simple representation of the $\mathrm{SU}_{2L} \otimes \mathrm{SU}_{2R}$ chiral symmetry of the light-quark sector of QCD:
$$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \vec{\varphi})(\partial_{\mu} \vec{\varphi}) - \frac{\mu^2}{2} \vec{\varphi}^2 -\frac{\lambda}{8} (\vec{\varphi}^2)^2.$$
Now you make $\mu^2<0$ (mass term with the wrong sign), which provides spontaneous symmetry breaking. The $\vec{\phi}$ field develops a vacuum-expectation value, i.e., you write $\vec{\phi}=(\vec{\pi},\sigma_0+\tilde{\sigma})$ with $\sigma_0=\text{const}$, where $\sigma_0$ is the value, minimizing the effective potential (i.e., at tree leavel $V(\vec{\phi})=\mu^2 \vec{\phi}^2/2+\lambda (\vec{\phi}^2)^2/8$. The ground state has only the SO(3) symmetry of the upper components left, which thus are the Goldstone modes of the spontaneously broken chiral symmetry. The residual SO(3) is the isospin symmetry. In the chiral limit the pions are thus massless and $\sigma_0$ is the pion-decay constant in this limit.

The interaction described by this model is the strong interaction among pions and sigmas. It's of course the residual interactions similar to the van der Waals interactions of neutral molecules for the electromagnetic interaction.