Quantum corrections to massless fermionic field

[Mesmerized]

Hello,

in QED the corrections to electron propagator change the bare electron mass from [itex]m_0[/itex] to [itex]m=m_0+δm=m_0+∑({\not}p=m)[/itex] (Peskin, formula 7.27). This is the consequence of the result, that the propagator changes from [itex]i/({\not}p-m_0)[/itex] to [itex]i/({\not}p-m_0-∑({\not}p))[/itex]). This part is written very well in Peskin's book, formula 7.23. Then (after fomula 7.30) Peskin goes on to explain that if the fermion was massless, then the quantum corrections would never give it a mass, in other words, [itex]δm[/itex] is 0 when [itex]m_0[/itex] is 0.

The question is the following. When [itex]m_0[/itex] is 0, the quantum corrections change the massless fermion propagator from [itex]i/{\not}p[/itex] to [itex]i/({\not}p-∑({\not}p))[/itex], and to me this 'corrected' propagator looks like massive. Seems like the quantum corrections gave mass to the massless fermion. Am I wrong?

 

[andrien]

Fermions mass term couples left handed fermion to right handed fermion,which is because β matrix is off diagonal in weyl representation.In case,if you start with a massless fermion,No radiative correction will be able to make a left handed field a right handed one.You can only get a radiative correction to mass,when you start with a mass term and all radiative correction will give correction to it's initial value.It is a property atributed to fermions.Scalar particles however like Higgs boson,suffers a huge mass correction and are completely disastrous,you require fine tuning to give a huge cancellation so as to give a finite mass to it.

 

[Mesmerized]

Thanks for the reply andrien, I understand your point. But my question is a little different. When you draw the diagrams corresponding to fermion propagator corrections, the propagator itself becomes an infinite sum (like in formula 7.22 in Peskin's book), which in effects takes the propagator from [itex]i/{\not}p[/itex] to [itex]i/({\not}p-∑({\not}p))[/itex] (in the massless case). And the latter looks like a massive propagator.

Also, to renormalize that massless fermionic field, it seems like a field renormalization only wouldn't be enough, we will also have to make a mass renormalization (of a massless fermion field)

 

[andrien]

you can not define a renormalization constant like Z_m in this case,because Z_m connects the bare one to physical one by a proportionality constant and it will be useless because former is zero.The theory of quantum electrodynamics with massless fermions is trouble.

 

[Mesmerized]

what kind of trouble? I thought there should be no problem in passing from QED to limit where fermion mass is 0

 

[Avodyne]

When [itex]m_0[/itex] is 0, the quantum corrections change the massless fermion propagator from [itex]i/{\not}p[/itex] to [itex]i/({\not}p-∑({\not}p))[/itex], and to me this 'corrected' propagator looks like massive.

It's not massive, because if [itex]m_0=0[/itex], then [itex]\Sigma({\not}p)\propto{\not}p[/itex].

 

[Mesmerized]

thanks, that's right

 

[andrien]

what kind of trouble? I thought there should be no problem in passing from QED to limit where fermion mass is 0

In 4 dimensions,QED coupling constant is dimensionless(no mass dimension),if you set all fermions masses to zero also there will not be any mass scale except some cut-off,which will appear in description of physical quantities.Besides,the theory will contain a new infrared divergence in this limit,which according to Lee-Nauenberg theorem must be dealt by incorporating some probabilistic sum over initial states besides just summing over all final states.

 

[vanhees71]

Have a look at the following famous paper:

S. R. Coleman and E. J. Weinberg. Radiative Corrections as the Origin of Spontaneous Symmetry Breaking. Phys. Rev. D, 7:1888-1910, 1973.
http://dx.doi.org/10.1103/PhysRevD.7.1888

It's dealing with massless scalar QED. I'm not aware about similar works on the massless fermionic QED case.

 

[Avodyne]

In 4 dimensions,QED coupling constant is dimensionless(no mass dimension),if you set all fermions masses to zero also there will not be any mass scale except some cut-off,which will appear in description of physical quantities.

This is not quite correct. As in QCD, it would be necessary to introduce a renormalization scale μ, and the numerical value of the dimensionless coupling would depend on the choice of μ. However physical quantities (see below) would be independent of μ.

Besides,the theory will contain a new infrared divergence in this limit,which according to Lee-Nauenberg theorem must be dealt by incorporating some probabilistic sum over initial states besides just summing over all final states.

This is the situation in QCD as well. Only "infrared safe" quantities such as jet-jet scattering cross sections can be computed. The values of infrared safe quantities are independent of the choice of μ.

 

[andrien]

This is not quite correct. As in QCD, it would be necessary to introduce a renormalization scale μ, and the numerical value of the dimensionless coupling would depend on the choice of μ. However physical quantities (see below) would be independent of μ.

No,in QED if all fermions are massless,then the massless limit is a singular point since there are no other scale present(photon is massless too).All the physical observables in this case must be measured by cut-off momentum.
However if one confines to QED in two dimensions,then the massless limit is not singular because coupling constant now has a mass dimension.All physical observables are measured by the coupling constant g now and which are continuous function of mass as m→0.

 

[vanhees71]

No,in QED if all fermions are massless,then the massless limit is a singular point since there are no other scale present(photon is massless too).All the physical observables in this case must be measured by cut-off momentum.
However if one confines to QED in two dimensions,then the massless limit is not singular because coupling constant now has a mass dimension.All physical observables are measured by the coupling constant g now and which are continuous function of mass as m→0.

As the Coleman-Weinberg paper shows, this argument is not so convincing. Even in a theory that has no dimensionful quantities due to the necessity to introduce a renormalization scale into the theory you get an energy scale. In QCD that's called [itex]\Lambda_{\text{QCD}}[/itex], and something similar should also happen in massless QED. As I said before, I've not seen this analyzed for fermionic QED but only for scalar QED in the Coleman-Weinberg paper.

 

[andrien]

As the Coleman-Weinberg paper shows, this argument is not so convincing. Even in a theory that has no dimensionful quantities due to the necessity to introduce a renormalization scale into the theory you get an energy scale. In QCD that's called [itex]\Lambda_{\text{QCD}}[/itex], and something similar should also happen in massless QED. As I said before, I've not seen this analyzed for fermionic QED but only for scalar QED in the Coleman-Weinberg paper.

Yes,I am aware of that paper and it deals with scalar QED for a reason, the reason is that scalar particles do get a contribution to their mass by radiative correction which is not the case for fermions.It does not work for Spinor QED.If you take massless spinor QED,then the problem is that there is no mass scale,apart from a cut off.For example One can try to calculate Lamb shift in massless QED,then one can treat this effect in non relativistic regime because the major contribution is there.Now when you try to do it as Bethe had done it for the first time,he cut off the frequency at ω=mc² by giving an argument that at frequency higher than this will cause integrand fall off more rapidly.But now we have problem in massless QED,at what value should we cut it off.We can cut it off at certain momentum which is arbitrary.This problem however is not very serious in relativistic version but you will again have a lnm (renormalized mass) type term which is problematic.

 

[vanhees71]

Do you know a reference, where massless QED with fermions is treated in a modern way? It's clear that the fermions do not become necessarily massive, because of chiral symmetry, but what's with anomalies, the IR problems etc.

Of course a cut-off is not a very good regularization scheme for gauge and chiral models. I'd prefer dim. reg. or BPHZ, etc.

 

[Avodyne]

Perturbative massless QED is considered to be a simple special case of massless nonabelian gauge theory (with massless quarks as well as massless gluons). A brief summary of the current understanding of perturbative IR divergences can be found in section 2.3 of http://arxiv.org/abs/1012.4001 by Radu Roiban. He writes, "The structure of soft and collinear singularities [that is, IR divergences] in a massless gauge theory in four dimensions has been extensively studied and understood [19–22]."

For more details on modern perturbative computations (which are done with dimensional regularization), see the lengthy (238 pages) review article by Henriette Elvang and Yu-tin Huang, http://arxiv.org/abs/1308.1697, which briefly discusses QED (or wait for their forthcoming book).

Nonperturbatively, nothing interesting is expected to happen in massless QED. It just runs to a free theory in the infrared. It is very hard to see how this could possibly be wrong.