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I'm studying a toy theory in quantum field theory. There are two free fields: a real massive scalar field $\phi$ with mass $M$ and a complex massive scalar field $\Psi$ with mass $m$.

They are coupled by $$ \mathcal{L} \subset g \Psi \Psi^\dagger \phi $$ I'm well aware that this interaction term results in a Lagrangian which is unbounded below, but it's just a toy model I'm using to try to get a grasp on the basics of quantum field theory.

Now, when I go to compute the tree-level scattering amplitude for $\Psi\Psi \to \Psi\Psi$ scattering I end up with a $T$-channel diagram and a $U$-channel diagram. (In both diagrams the two $\Psi$ particles come in, exchange an off-shell $\phi$ particle, and scatter into their final states).

The sum of these two diagrams gives me the total matrix element, which goes like $$ \frac1{(t-M^2)} + \frac1{(u-M^2)} $$ ignoring the factor's of $i$ and the factor of $g^2$. In computing this amplitude I worked in the center of mass/momentum frame, with the initial four momentum of the two particles being $(E(p),0,0,\pm p)$, with the final four momentum of the two particles being $(E(p),0, \pm \cos(θ)p, \pm \sin(θ)p)$. This ensures that four-momentum is conserved.

With this parametrization, $t$ becomes $-4(p\sin(θ/2))^2$, and $u$ becomes $-4(p\cos(θ/2))^2$. Things get interesting in the limit as $M \to 0$, i.e., the mass of the phi particle approaches zero. In this case the scattering amplitude becomes unbounded as $θ$ approaches zero or $\pi$. This is, to my mind, a highly unphysical result.

Is this divergence a result of a fundamental issue with this toy-model? How can we make sense of this result? What is the physical interpretation of a divergent scattering amplitude? Worse still is the fact that the integrated/total cross section also appears to diverge in all of these cases.

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    $\begingroup$ Won't e.g. $\phi \to \phi \phi$ give a width $\Gamma$ in the propagator for $\phi$, removing the singularity? That's an unusual, massless decay though, so I'm not sure. $\endgroup$ – innisfree May 1 '17 at 5:29
  • $\begingroup$ You'll also have $\Gamma / m \to \infty$, which suggests that the 'resonance' interpretation of the particle $\phi$ is a dubious. $\endgroup$ – innisfree May 1 '17 at 5:34
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    $\begingroup$ This has nothing to do with QFT. The same happens when we study Coulomb scattering in E&M, or classical scattering in a gravitational potential. The force has infinite range, so the cross section diverges at small angles $\endgroup$ – Thomas May 1 '17 at 13:01
  • $\begingroup$ In response to innisfree, I'm not sure if such a decay is possible. There's only a few three-point vertices in this theory, and none of them that I've found involve one phi particle coming in and two phi particles coming out. Keep in mind I'm completely new to QFT, so it's possible that I'm entirely mistaken. $\endgroup$ – chuckstables May 1 '17 at 21:15
  • $\begingroup$ @chuxley won't it happen at one-loop? E.g. draw a circle with a $\Psi$ and put 3 $\phi$ external lines on it. $\endgroup$ – innisfree May 2 '17 at 4:07
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What you found is a very simple example of an infra-red divergence, which plague all physical theories with massless particles.

This kind of divergences are already present in the classical case (see for example ref.1), and they usually signal that you are asking an unphysical question, not that the theory itself is unphysical. For example, if you have massless particles it becomes meaningless to ask about the total number of them in a certain physical configuration, while asking about the total energy is a well-posed question. This is reflected in the mathematics of the theory through divergences: the first question results in a divergent expression, while the second does not.

These divergences are inherited to the quantum mechanical case. For example, IR divergences are ubiquitous in QED, but it can be proven that these divergences always cancel out for "meaningful cross-sections", i.e., for predictions that are well-posed in the sense of the previous paragraph. This is sometimes known as the Bloch-Nordsieck cancellation, or in the more general case of the Standard model, as the Kinoshita-Lee-Nauenberg theorem. See for example refs.2,3,4.

In your particular case, as you are dealing with a scalar theory as opposed to a gauge boson, the analysis is slightly simpler (although one must in general abandon on-shell renormalisation schemes if we are to have massless particles). This is discussed in ref.5.

For a further discussion of infra-red divergences, see refs. 6-8.

References

  1. Itzykson C., Zuber J.-B. Quantum field theory, section 4-1-2.

  2. Peskin, Schroesder. An introduction To Quantum Field Theory, section 6.5.

  3. Itzykson C., Zuber J.-B. Quantum field theory, section 8-3-1.

  4. Schwartz M.D. Quantum Field Theory and the Standard Model, chapter 20.

  5. Srednicki M. Quantum Field Theory, chapters 26 and 27.

  6. Ticciati R. Quantum Field Theory for Mathematicians, section 19.9.

  7. Pokorski S. Gauge Field Theories, sections 5.5 and 8.7.

  8. Weinberg S. Quantum theory of fields, Vol.1. Foundations, chapter 13.

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  • $\begingroup$ Don't the IR divergences plague physical theories with $m=\Gamma =0$, e.g., photon? $\endgroup$ – innisfree May 1 '17 at 14:20
  • $\begingroup$ @innisfree hmm Im not sure what you mean. Yes, any theory with massless particles has potential IR divergences. QED, having a massless photon, is no exception. $\endgroup$ – AccidentalFourierTransform May 1 '17 at 16:33
  • $\begingroup$ Thanks for your reply. I'm going to be picking up an old copy of Peskin' and Shroesder's "An introduction to quantum field theory" from a friend tomorrow, so I'll definitely look into that. So just to confirm; these sorts of divergences (infra-red divergences) are ubiquitous in quantum field theories which involve mass-less particles? Also; thanks for changing the formatting of my question so that the mathematical formulas are readable. $\endgroup$ – chuckstables May 1 '17 at 21:08
  • $\begingroup$ @chuxley no problem, I'm glad I could help. And yes, infra-red divergences are those related to massless particles, and they are indeed ubiquitous in QFT. BTW: note that Srednicki's book is online in his webpage for free, so you can check that book too if you want. $\endgroup$ – AccidentalFourierTransform May 1 '17 at 21:29
  • $\begingroup$ I just looked that book up, thanks for letting me know that it's online. I don't know if I'm allowed to ask another question in the same thread, but it is related to my original question. In this toy theory, can you have a phi particle decay into two phi particles at one-loop? I'm trying to work out how it could happen (it came up in a comment above). Thanks :). $\endgroup$ – chuckstables May 2 '17 at 4:40
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This is a very crude answer. But this will help in your queries. You see the problem with the above method you implied is that at tree level $\psi \psi - > \psi \psi$ the internal propagator is $\phi$ particle. Now the propagator for massive and massless particles are very different. For massless particle the equation of motion in momentum space is not trivially invertible. This you cannot set M = 0 in the above equation. For massless $\phi$ you will need to find the propagator first.

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    $\begingroup$ Are you sure that propagators for massive versus massless scalars are non-trivially different? Can you write them explicitly? $\endgroup$ – innisfree May 1 '17 at 5:32

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