## 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?

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