Solving Dyson-Schwinger equations

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.

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5 Responses to Solving Dyson-Schwinger equations

  1. […] View original here: Solving Dyson-Schwinger equations […]

  2. Rafael says:

    Dear Marco,

    please, give a look at these cosmological implications of infrared QCD propagators:

    http://xxx.lanl.gov/abs/0906.2165v2
    http://xxx.lanl.gov/abs/0909.2684

    Are you able to formulate the gauge-to-scalar mapping also in curved backgrounds? What does this imply for ghosts (should they stay coupled as needed in aforementioned papers)?
    Best,

    Rafael.

    • mfrasca says:

      Dear Rafael,

      Your question is quite interesting as are the papers you put to my attention. Let me clarify what the mapping theorem implies. You can select a set of components of a Yang-Mills field and use them to solve the corresponding equations taking them to be all equal, space-independent and being also solution of a massless quartic scalar field equation. These are Smilga solutions. Otherwise, in the most general case, the mapping is only perturbative giving

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

      holding when the coupling g is taken large enough and being \phi(x) a solution of the equation for the massless quartic scalar field. As you can see, all amounts to a selection of components and so, I cannot expect that being or not in a general relativistic framework should change something. Rather, I would have serious troubles to solve exactly the corresponding equation for the scalar field. Indeed, as you know, cosmologists working with scalar fields in inflationary scenarios apply the so-called slow-roll approximation to avoid this problem. A posteriori they are able to show that this approximation is what one needs in this physical situation.

      About the ghost field, I should expect a decoupling also in the most general case because this happens for algebraical reasons. This arises from the antisymmetry of the structure constants of the gauge group and so I cannot see how this field can count here. But, as you may know, the running coupling of the Yang-Mills field has a quite peculiar behavior, as I also discussed in my paper, and this means that the scenario depicted by these authors could be correct provided dark matter is cold enough to be in a kind of intermediate regime where both asymptotic freedom and the low-energy scenario fail.

      Finally, let me say that a lattice analysis, using Robertson-Walker metric, could be very interesting here to see how cosmological effects enter into QCD.

      Best,

      Marco

  3. carlbrannen says:

    Congratulations on the new paper! (I can’t believe no one has spoken before me.)

    Regarding Smilga’s choice, I had this sudden flash that it might be related to a solution to a certain math problem I’ve been pursuing, which has to do with path integrals over finite configuration spaces and the like.

    Given a finite group, one defines the “group algebra over the complex numbers” by using the finite group elements as a basis for a vector space with multiplication defined by the finite group multiplication rule. This defines an algebra. Wolfram.com on “Group Algebra”.

    Now, given a group algebra, the stable long time propagators for individual particles correspond to the “primitive idempotents” of the algebra. Idempotent means uu = u. Primitive means that it’s not zero, and that it’s not the sum of two non zero idemopotent elements. Example, for spin-1/2, the projection operators for spin in various directions are primitive idempotents. The unit matrix is idempotent but not primitive. In a certain way, complete sets of annihilating group algebra primitive idempotents (i.e. sets where any pair multiply to zero and the sum of all of them is unity (where unity is 1 times the basis element corresponding to the identity of the finite group)) are a generalization of spinors.

    Finding the primitive idempotents of a group algebra can be difficult because, with a finite group of size N, the maximum possible number of idempotents is 2^ N.

    Anyway, the analogy to Smilga’s choice is that it is always possible to write down one primitive idempotent: The sum over n of g_n/N, where g_n is the basis element corresponding to the nth element of the finite group. It’s easy to show that this is a primitive idempotent.

    • mfrasca says:

      Hi Carl,

      Thank you. Indeed, this is a quite interesting result on a theory generally believed manageable only with perturbation techniques.

      Best,

      Marco

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