A quartic scalar field theory is trivial

About two years ago I have got published a paper of mine on PRD (see here). Generally this is a reason of great accomplishment for us working in the field of high-energy and nuclear physics as well. What I would like to show you in this post is that the referee, the editor and myself were all right and I will do this with a presentation at an undergraduate level, the one of American Journal of Physics if you want. The relevance of this paper is that I modified the Bender et al. method (see here) to make it work properly and finally obtain sensible results out of a strongly coupled quantum field theory.

A quantum field theory can be solved if we are able to solve the corresponding Heisenberg equations of motion. These equations, in the classical limit, are generally solvable in special cases (e.g. free particle) but the classical limit may be used to extract non perturbative results as happens for the WKB approximation. So, let us write down the equation for a massless quartic scalar field theory:

\frac{\partial^2\phi}{\partial t^2}-\Delta_2\phi+\lambda\phi^3=0.

As strange as may seem, this equation admits an exact solution that we can write down as

\phi(x) = \mu\left(\frac{2}{\lambda}\right)^{\frac{1}{4}}{\it sn}(p\cdot x,i)

when the following dispersion relation does hold

E^2=p^2+\mu^2\sqrt{\frac{\lambda}{2}}.

Here \mu is an integration constant and sn is the Jacobi snoidal function. We see that something strange is happening: We started with a massless theory but its solution is massive! And the mass depends on a free parameter being the theory massless. The Jacobi function describe a wave but this is not the same as a plane wave of a free particle. Rather, it is a linear combination of plane waves and so the theory has a solution proper to a spectrum of free particles! Let us see why. The following identity holds

{\it sn}(u,i) = \frac{2\pi}{K(i)}\sum_{n=0}^\infty\frac{(-1)^ne^{-(n+\frac{1}{2})\pi}}{1+e^{-(2n+1)\pi}}\sin\left[(2n+1)\frac{\pi u}{2K(i)}\right]

being K(i) an elliptic integral and then our solution can be written down as

\phi(x) = \sum_n A_n e^{i(2n+1)\frac{\pi}{2K(i)}p\cdot x} + c.c.

that is a superposition of plane waves, i.e. we are describing a spectrum of free particles. But what is the mass spectrum of these particles? From the above solution we have to put our particles to rest. We put {\bf p} = 0 and the spectrum is easily obtained using the above dispersion relation

E_n=(2n+1)\frac{\pi}{2K(i)}\left(\frac{\lambda}{2}\right)^{\frac{1}{4}}\mu

that is the one of an harmonic oscillator. The theory is trivial. Very easy, isn’t it?

Now, we put \sqrt{\sigma}=\left(\frac{\lambda}{2}\right)^{\frac{1}{4}}\mu and we compute the numbers

s_n=\frac{E_n}{\sqrt{\sigma}}=(2n+1)\frac{\pi}{2K(i)}

and we get the following table

to be compared with the one of Meyer for Yang-Mills in D=3+1

Still better are the Teper et al. values

This is really shocking. We have a satisfactory correspondence between our quartic scalar field theory and the spectrum of Yang-Mills theory in D=3+1. Indeed, we have already seen that lattice results about propagators seems to indicate that Yang-Mills theory is trivial. But this connection with the quartic scalar field turns out to be really unexpected. Indeed we will see that the solution we have found is also a solution to the classical Yang-Mills field.

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2 Responses to A quartic scalar field theory is trivial

  1. […] equations of motion of Yang-Mills theory. Indeed, as already seen for the quartic scalar field (see here), one can show, with undergraduate level arguments, that Yang-Mills theory has a massive exact […]

  2. […] I devised to solve Dirac equation using the massive solution of the massless scalar field given here and […]

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