## 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.

## I did it for you

11/03/2009

It is very easy to show, from Yang-Mills equations, how to obtain a scalar field equation through the Smilga’s choice. Let us write down Yang-Mills equations $\partial^\mu\partial_\mu A^a_\nu-\left(1-\frac{1}{\alpha}\right)\partial_\nu(\partial^\mu A^a_\mu)+gf^{abc}A^{b\mu}(\partial_\mu A^c_\nu-\partial_\nu A^c_\mu)+gf^{abc}\partial^\mu(A^b_\mu A^c_\nu)+g^2f^{abc}f^{cde}A^{b\mu}A^d_\mu A^e_\nu = 0$

using the choice $A_1^1=A_2^2=A_3^3=\phi$. This is really a great simplification. Smilga, in his book, already checked this for us but we give here the full computation. From above eqautions, the only critical term is the following $f^{abc}A^{b\mu}(\partial_\mu A^c_\nu-\partial_\nu A^c_\mu)$

as this term would produce terms deviating from the known form of the scalar theory. For SU(2) we have $f^{abc}=\epsilon^{abc}$ the fully-antisymmetric Levi-Civita tensor. This means that we will have $\epsilon^{a1c}A^{11}(\partial_1A_\nu^c-\partial_\nu A_1^c)+$ $\epsilon^{a2c}A^{22}(\partial_2A_\nu^c-\partial_\nu A_2^c)+$ $\epsilon^{a3c}A^{33}(\partial_3A_\nu^c-\partial_\nu A_3^c).$

Where we have used largely Smilga’s choice. Now do the following. Take the following components to evolve $\nu=1$ $a=1$, $\nu=2$ $a=2$ and $\nu=3$, $a=3$. It easy to see that the possible harmful term is zero with the Smilgaì’s choice. Now, for the cubic term you should use the useful relation $\epsilon^{abc}\epsilon^{cde}=\delta_{ad}\delta_{be}-\delta_{ae}\delta_{bd}$

and you will get back the quartic term.

The gauge fixing term can be easily disposed of through a rescaling of spatial variables while the kinematic term gives the right contribution. You will get three identical equations for the scalar field.

Of course, Smilga in his book already did this and I repeated his computations after the Editor of PLB asked for a revision having the referee already put out this problem. The Editorial work was done very well and two referees read the paper emphasizing errors where they were.

Finally, Tao’s critcism does not apply as I said. This does not mean that what he says is wrong. This means that does not apply to my case.

Update: As the question of the gauge fixing term appears so relevant, let me fix it once and for all. Firstly, I would like to point out that these solutions belong to a class of solutions in the Maximal Abelian Gauge (MAG). But let us forget about this and consider the question of gauge fixing. This term is arbitrarily introduced in the Lagrangian of the field in order to fix the gauge when a quantization procedure is applied. Due to gauge invariance and the fact that becomes an exact differential after partial integration, it useful to have it there for the above aims. The form that it  takes is $\frac{1}{\alpha}(\partial A)^2$

and is put directly into the Lagrangian. How does this term become with the Smilga’s choice? One has $\frac{1}{\alpha}\sum_{i=1}^3(\partial_iA_i^i)^2$

and the final effect is a pure rescaling into the space variables of the scalar field. In this way the argument is made consistent. One cannot take the other way around for the very nature of this term and claiming the result is wrong.

This particular class of solutions belongs to the subgroup of SU(N) given by the direct product of U(1). This is a property of MAG and all the matter is really consistent and works.

Finally, I invite people commenting this and other posts to limit herself to polite responses and in the realm of scientific discussion. Of course, doing something wrong happens and happened to anyone working in a scientifc endeavour for the simple reason that she is really doing things. People that only do useless criticisms boiling down to personal offenses are kindly invited to refrain from further interventions.

## A set of exact classical solutions of Yang-Mills equations

28/02/2009

Following the discussion with Rafael Frigori (see here) and returning to sane questions, I discuss here a class of exact classical solutions that must be considered in the class of Maximal Abelian Gauge.  As usual, we consider the following Yang-Mills action $S=\int d^4x\left[\frac{1}{2}\partial_\mu A^a_\nu\partial^\mu A^{a\nu}+\partial^\mu\bar c^a\partial_\mu c^a\right.$ $-gf^{abc}\partial_\mu A_\nu^aA^{b\mu}A^{c\nu}+\frac{g^2}{4}f^{abc}f^{ars}A^b_\mu A^c_\nu A^{r\mu}A^{s\nu}$ $\left.+gf^{abc}\partial_\mu\bar c^a A^{b\mu}c^c\right]$

being $c,\ \bar c$ the ghost field, $g$ the coupling constant and, for the moment we omit the gauge fixing term. Let us fix the gauge group being SU(2). We choose the following (Smilga’s choice, see the book): $A_1^1=A_2^2=A_3^3=\phi$

being $\phi$ a scalar field. The other components are taken to be zero. It easy to see that the action becomes $S=-6\int d^4x\left[\frac{1}{2}\partial_\mu\phi\partial^\mu\phi+\partial\bar c\partial c\right]+6\int d^4x\frac{2g^2}{4}\phi^4.$

This is a very nice result as if we have a solution of the scalar field theory we get immediately a classical solution of Yang-Mills equations while the ghost field decouples and behaves as that of a free particle. But such solutions do exist. We can solve exactly the  equation $\partial_t^2\phi-\partial_x^2\phi+2g^2\phi^3=0$

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

being sn Jacobi snoidal function, $\mu,\ \theta$ two arbitrary constants, if holds $p^2=\mu^2\left(\frac{2g^2}{2}\right)^\frac{1}{2}.$

We see that the field acquired a mass notwithstanding it was massless and the same happens to the Yang-Mills field. These are known as non-linear waves. These solutions do not represent a new theoretical view. A new theoretical view is given when they are used to build a quantum field theory. This is the core of the question.

What happens when we keep the gauge fixing term as $\frac{1}{\xi}(\partial\cdot A)^2?$

If you substitute Smilga’s choice in this term you will find a correction to the kinematic term implying a rescaling of space variables. This is harmless for the obtained solutions resulting in the end into a multiplicative factor for the action of the scalar field.

The set of Smilga’s choice is very large and increases with the choice of the gauge group. But such solutions always exist.

Yang-Mills propagators and QCD to appear in Nucl. Phys. B: Proc. Suppl.

Update: Together with Terry Tao, we agreed that these solutions hold in a perturbative sense, i.e. $A_\mu^a(x)=\eta_\mu^a\phi(x)+O(1/g)$

being $\eta_\mu^a$ a constant and $g$ the coupling taken to be very large. These become exact solutions when just time dependence is retained. So, the theorem contained in the above papers is correct for this latter case and approximate in the general case. As the case of interest is that of a large coupling, these results permit to say that all the conclusions drawn in the above papers are correct. This completed proof will appear shortly in Modern Physics Letters A (see http://arxiv.org/abs/0903.2357 ).

Thanks a lot to Terry for the very helpful criticism.

## 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?