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How to cancel infinite mass corrections for quantities without counterterms?

+ 5 like - 0 dislike
8881 views

I'm trying to understand how infinite mass corrections are cancelled for a particle that is massless at tree level. As a simple example consider a theory with three charged Weyl fermions, $ \chi _{ ++ }, \psi _- , \psi _+ $, as well a complex scalar, $ \phi _- $.

 Furthermore, assume that charge is approximately broken in another sector such that one of the fermions, $ \psi _- $, gets a small Majorana mass (this is very similar to the situation I'm actually interested in, a R symmetric SUSY model with R breaking through anomaly mediated SUSY breaking, so its not as far fetched as it may sound). The Lagrangian takes the form,
\begin{equation} 
{\cal L} = {\cal L} _{ kin} - M ( \psi _- \psi _+ + h.c. ) - g ( \phi _-  \chi _{ + + } \psi _- + h.c. )  - V ( \phi ) - m ( \psi _- \psi _- + h.c. )  
\end{equation} 

where $ V ( \phi )  $ is the scalar potential.

Due to the symmetry breaking from $ \psi _- $ we can't also get a symmetry breaking Majorana mass for the $ \chi _{++} $ under loop corrections. However, we don't have a counterterm for it! For example to first order we have,

$\hspace{1.5cm}$

Usually the $ 1 / \epsilon $ is harmless as we hide it in the counterterm. But in this case since we don't have a tree-level contribution to the Majorana mass for $ \chi _{ + + } $, we also don't have a counterterm for it. How is this issue resolved?

Edit:  

  1. I've found a related topic in the context of the weak interaction discussed in the appendix of arXiv:1106.3587. Here, if I understand correctly, they use the $Z$ boson to cancel the infinity. However, I don't understand how that would work here since this is not even a gauge theory.
  2. Weinbeg also discusses a similar topic in the context of the weak interaction in "Perturbative Calculations of Symmetry Breaking", Phys Rev D Vol 7 Num 10.
asked Jun 19, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
edited Jun 23, 2014 by JeffDror

1 Answer

+ 3 like - 0 dislike

You simply have to include the Majorana mass term at a tree level because the masslessness of $\chi_{++}$ isn't guaranteed by any symmetry that is valid at the quantum level.

Then the counterterm for this Majorana mass term will swallow the divergent contribution, too. Note that in many other cases, the Weyl fermions have the chiral symmetry

$$ \chi \to \chi \times \exp(i\phi\gamma_5) $$

where $\gamma_5$ is omitted (set to one) in the two-component formalism, a symmetry that protects the masslessness even at the quantum level (the reason why it's natural for the Higgsinos to assume that they stay light, for example, so why SUSY makes the light Higgs - the bosonic partner of the higgsinos - natural).

But the cubic term involving one factor of the $\chi_{++}$ field explicitly violates this chiral symmetry in your example. Note that the violation of the chiral symmetry may already be seen classically.

answered Jun 23, 2014 by Luboš Motl (10,248 points) [ revision history ]

Thanks for your insight. Saying that $\chi_{++}$ must have a tree level majorana mass if it acquires a loop induced mass is equivalent to claiming that you can't build a hidden sector such that $\psi_- $ will get a majorana mass but not $\chi_{++}$. I disagree on this point. This is in fact common when giving neutrinos loop level masses. As an example I point out this paper that my above toy model was based on where R charge is the analog to the $U(1)$ above. In the paper the neutrinos (analog of $\chi_{++}$) are massless at tree level. However, the gauginos (analog of $\psi_- $) gets a mass through anomaly mediated SUSY breaking (Eq 44). Then the second diagram in Figure 3 of the linked paper is allowed ($\chi_i^0$ are just mixtures of the gauginos), which gives the neutrinos masses at loop level.

Thanks. Right, it's a frequent practice of phenomenologists to treat some approximate symmetries as real etc. Without the hidden sector, one could identify the exact chiral symmetry etc. which protects the masslessness, so in this sense, it's OK to assume the massless particles as the visible sector. But if you talk about the whole theory including the hidden sector producing the masses, then the symmetry isn't there and the kosher rules of QFT dictate that one should consider the most general tree-level Lagrangian that is compatible with the symmetries - i.e. a Lagrangian with general Majorana masses, too. A special justification of the mixed treatment could be if the chiral symmetry breaking were spontaneous in some way but in your example it is explicit.

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