## Classical solutions of Yang-Mills equations

09/10/2009

So far, I have posted several posts in this blog about the question of classical solutions to Yang-Mills equations. This has produced some fuzz, mostly arisen from my published papers, as to such solutions may not be correct. Thanks to a wise intervention of Terry Tao, I was able to give a complete understanding of my solutions and a theorem was fully proved in a recent paper of mine to appear in Modern Physics Letters A (see here), agreed with Terry in a private communication. So, I think it is time to give a description of this result here as it appears really interesting showing how, already at a classical level, this theory can display massive solutions and a mass gap is already seen in this case. Then, it takes a really small step to get the corresponding proof in quantum field theory.

To understand how these solutions are obtained, let us consider the following equation for a scalar field

$\Box\phi+\lambda\phi^3=0.$

This is a massless self-interacting field. We can select a class of solutions by looking at the case of a rest reference frame. So, we put any dependence on spatial variables to zero and solve the equation

$\partial_{tt}\phi+\lambda\phi^3=0$

whose solutions are known and given by

$\phi(t,0)=\mu\left(\frac{2}{\lambda}\right)^{\frac{1}{4}}{\rm sn}\left[\left(\frac{\lambda}{2}\right)^{\frac{1}{4}}\mu t+\theta,i\right]$

being $\mu$ and $\theta$ two integration constants and sn a Jacobi elliptical function. Then, boosting this solution will produce an exact solution of the equation we started from given by

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

provided the following dispersion relation holds

$p^2=\left(\frac{\lambda}{2}\right)^{\frac{1}{2}}\mu^2$

and we see that, although we started with a massless field, self-interaction provided us massive solutions!

Now, the next question one should ask is if such a mechanism may be at work for classical Yang-Mills equations. These can be written down as

$\partial^\mu\partial_\mu A^a_\nu-\left(1-\frac{1}{\alpha}\right)\partial_\nu(\partial^\mu A^a_\mu)+gf^{abc}A^{b\mu}(\partial_\mu A^c_\nu-\partial_\nu A^c_\mu)$

$+gf^{abc}\partial^\mu(A^b_\mu A^c_\nu)+g^2f^{abc}f^{cde}A^{b\mu}A^d_\mu A^e_\nu = 0$

being $\alpha$ chosen depending on the gauge choice, $g$ the coupling and $f^{abc}$ the structure constants of the gauge group taken to be SU(N). The theorem I proved in my paper above states that the solution given for the scalar field theory is an exact solution of Yang-Mills equations, provided it will not depend on spatial coordinates, for a given choice of Yang-Mills components (Smilga’s choice) and $\lambda=Ng^2$, otherwise the following identity holds

$A_\mu^a(x)=\eta_\mu^a\phi(x)+O(1/g).$

Here $\eta_\mu^a$ is a set of constants arising with the Smilga’s choice. This theorem has the following implications: Firstly, when the coupling become increasingly large, a massless scalar field theory and Yang-Mills theory can be mapped each other. Secondly, already at the classical level, for a coupling large enough, a Yang-Mills theory gets massive solutions. We can see here that a mass gap arises already at a classical level for these theories. Finally, we emphasize that the above mapping appears to hold only in a strong coupling regime while, on the other side, these theories manifest different behaviors. Indeed, we know that Yang-Mills theory has asymptotic freedom while the scalar theory has not. The mapping theorem just mirrors this situation.

We note that these solutions are wave-like ones and describe free massive excitations. This means that these classical theories have to be considered trivial in some sense as these solutions seem to behave in the same way as the plane waves of a free theory.

One can build a quantum field theory on these classical solutions obtaining a theory manifesting a mass gap in some limit. This is has been done in several papers of mine and I will not repeat these arguments here.

## Solving Dyson-Schwinger equations

15/09/2009

Sunday I posted a paper of mine on arxiv (see here). I was interested on managing a simple interacting theory with the technique of Dyson-Schwinger equations. These are a set of exact equations that permit to compute all the n-point functions of a given theory. The critical point is that a lower order equation depends on higher order n-point functions making the solution of all set quite difficult. The most common approach is to try a truncation at some order relying on some physical insight. Of course, to have a control on such a truncation could be a difficult task and the results of a given computation should be carefully checked. The beauty of these equations relies on their non-perturbative nature to be contrasted with the severe difficulty in solving them.

In my paper I consider a massless $\phi^4$ theory and I solve exactly all the set of Dyson-Schwinger equations. I am able to do this as I know a set of exact solutions of the classical equation of the theory and I am able to solve an apparently difficult equation for the two point function. At the end of the day,  one gets the exact propagator, the spectrum and the beta function. It is seen that this theory has only trivial fixed points. I was able to get these results on another paper of mine. So, it is surely comforting to get identical results with different approaches.

Finally,  I can apply  the mapping theorem with Yang-Mills theories, recently proved thanks also to Terry Tao intervention, to draw conclusions on them in the limit of a very large coupling. In the paper you can find a formulation of this theorem as agreed with Terry, a direct consequence of my latest accepted paper on this matter (see here).

I think this paper adds an important contribution to our understanding of Dyson-Schwinger equations presenting an exact non-trivial solution of them.

## Exact solutions on arxiv

24/07/2009

As promised in my preceding post (see here), I have posted yesterday a preprint on this matter on arxiv (see here). I have presented all the exact solutions I was able to obtain at a classical level and I have given a formulation of the quantum field theory for a massless quartic theory. The key point in this case is the solution of the equation for the propagator

$-\Box\Delta+3\lambda\phi_c^2(x)\Delta=\delta^D(x)$

being $\phi_c$ the given exact classical solution. As usual, I have used a gradient approximation and the solution of the equation

$\ddot\phi(t)+3\lambda\phi_c^2(t,0)\phi(t)=\delta(t)$

that I know when the phase in $\phi_c(t,0)$ is quantized as $(4n+1)K(i)$, being $n$ an integer and $K(i)$ an elliptic integral. This gives back a consistent result in the strong coupling limit, $\lambda\rightarrow\infty$, with my preceding paper on Physical Review D (see here).

The conclusion is rather interesting as quantum field theory, given from such subset of classical solutions, is trivial when the coupling becomes increasingly large as one has a Gaussian generating functional and the spectrum of a harmonic oscillator. This is in perfect agreement with common wisdom about this scalar theory. So, in some way, Jacobi elliptical functions that describe nonlinear waves behave as plane waves for a quantum field theory in a regime of a strong coupling.

## Is Terry wrong?

10/03/2009

I am a great estimator of Terry Tao and a reader of his blog. Tao is a Fields medalist and one of the greatest living mathematicians. Relying on such a giant authority may give someone the feeling of being a kind of dwarf trying to be listened around. Anyhow I will try. Terry come out with an intervention in Wikipedia here claiming:

“It may be relevant to point out that one of the references cited in the disputed section [3] has a significant error in it, despite being published. Namely, in the proof of Theorem 1, the author is assuming that an extremum A for the Yang-Mills action for a special class of connections (namely those in which $A^1_1=A^2_2=A^3_3$ and all other components vanish) is necessarily an extremum for the Yang-Mills action for all other connections also, but this is not the case (just because $YM(A) \geq YM(A')$, for instance, for A’ of this special form, does not imply that $YM(A) \geq YM(A')$ for general A’). Since one needs to be an extremiser (or critical point) in the space of all connections in order to be a solution to the Yang-Mills equations, the mapping provided in Theorem 1 has not been shown to actually produce solutions to the Yang-Mills equation (and I suspect that if one actually checks the Yang-Mills equation for this mapping, that one will not in fact get such a solution). Terry (talk) 20:32, 28 February 2009 (UTC)”

This claim of mistake by my side contains a misinterpretation of the mapping theorem. If the theorem would claim that this is true for all connections, as Terry says, it would be istantaneously false. I cannot map a scalar field on all the Y-M connections (think of chaotic solutions). The theorem simply states that there exists a class of solutions of the quartic scalar field that are also solution for the Yang-Mills equations and this can be easily proved by substitution (check Smilga’s book) and Tao is proved istantaneously wrong. So now, what is the point? I have a class of Yang-Mills solutions that Tao is claiming are not. But whoever can check by herself that I am right. So, is Terry wrong?

## Smilga’s choice and the mapping theorem

25/10/2008

After the acceptance of my paper (see here) I wondered what should have been Smilga’s choice for the SU(3) given its existence. Let me explain what a Smilga’s choice is. Firstly I point out his beatiful book about QCD that was my starting point for all this matter (see here). You will find in this book, in the chapter about classical solutions, that when one chooses a set of values different from zero of the components of the Yang-Mills potential and take these non-null components all equal one gets for space-homogeneous Yang-Mills equations the motion equation of a massless anharmonic oscillator. Smilga asked in the book what the physical meaning of such homogeneous solutions should be. I answered to this smart question with a mapping theorem (take a look at my paper here): Anytime a Smilga’s choice is done one has mapped a solution of the classical Yang-Mills equations onto a solution of a quartic massless scalar field. This result is extremely important as we obtain in this way both the physical meaning of such solutions and a set of classical solutions to do perturbation theory without recurring to some small parameter. Indeed, we know how to manage a quantum scalar field in the strong coupling limit (see here and here).

Now, for SU(2) it is very easy to do a choice that reaches our aim: One takes

$A=((0,\phi,0,0),(0,0,\phi,0),(0,0,0,\phi))$

and you are done. I have not exploited all the phase space in this case as one should consider that Smilga’s choice is not unique and things get worse as the gauge group is taken more complex. For SU(3) things are really horrible as one has to explore a very large phase space and the product of the structure constants of the group does not admit a simple expression. So, I reduced myself to write down a few lines of code both in C and matlab working out such a product of structure constants. My PC worked fine for me and obtained a lot of results. As said above Smilga’s choice is not unique and one can have a huge number of choices increasing the number of structure constants of the group. So, e.g. the following Smilga’s choice is good for SU(3) leaving you with the right ‘t Hooft coupling in the mapped scalar field

$A=((\phi,0,0,0),(0,\phi,0,0),(\phi,0,0,0),$

$(\phi,0,0,0),(\phi,0,0,0),(0,0,0,0),(0,0,\phi,0),(0,0,0,0)).$

This Smilga’s choice gives a multiplicative overall factor 2 to the scalar field action. Smilga’s choice for SU(2) will leave a factor -3. Of course, these factors will depend on the gauge group but one can ask a couple of mathematical questions that are worth exploring. Firstly, whatever Smilga’s choice one takes that grants the correct ‘t Hooft coupling in the Lagrangian of the scalar field, is the overall factor always the same? Better, does a Smilga’s choice exist that grants for any SU(N) group the same overall factor equal in absolute value to the number $N^2-1$ as happens to SU(2)?  These results would extend the understanding of the existence of the mapping to a stunning level taking into account that already for SU(3) the number of configurations is really overwhelming.

Concluding, we just remark the essential points to be taken into account for a choice to be a proper Smilga’s choice: 1) The proper Lagrangian of a quartic massless scalar field should be reproduced multiplicated with an overall factor (negative or positive is not important). 2) The coupling $\lambda$ of the scalar field must be the ‘t Hooft coupling $Ng^2$ for a Yang-Mills theory with a SU(N) gauge group and coupling constant $g$.

Update: Found! After I have extended the search space with my C program, I was able to obtain a set of proper Smilga’s choices for SU(3). These behave exactly as for SU(2). Here is an example

$A=((0,0,0,0),(0,\phi,0,0),(0,\phi,0,0),(0,0,\phi,0),$

$(0,\phi,\phi,0),(0,0,\phi,0),(0,0,0,\phi),(0,0,0,\phi))$

This gives an overall factor $N^2-1=8$ and the proper ‘t Hooft coupling $Ng^2=3g^2$ with the same signs in the Lagrangian of the scalar field as seen for the SU(2) case, that is one has in the end the following mapped Lagrangian

$L=-8\int d^4x {1\over 2}(\partial\phi)^2+8\int d^4x {{3g^2}\over 4}\phi^4$

and things are done! We are left we an interesting mathematical question: As the gauge group is changed the number of proper Smilga’s choices increases vastly. What should be the meaning of such a large number? What kind of symmetry is hidden behind this?