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Given the scalar field Lagrangian $$\mathscr{L}=\frac{1}{2}e^{-\lambda\phi}\partial_\mu\phi\partial^\mu\phi,$$ evaluate the order $\lambda^2$ correction to the propagator.

At that order in $\lambda$, the Lagrangian is $$\mathscr{L}=\frac{1}{2}\left(\partial_\mu\phi\right)^2 - \frac{\lambda}{2}\phi\left(\partial_\mu\phi\right)^2 + \frac{\lambda^2}{4}\phi^2 \left(\partial_\mu\phi\right)^2 + \mathcal{O}\left(\lambda^3\right).$$

The vertices are:

  1. (a)
  2. (b)

Since $\phi$s are indistinguishable and because of the derivative coupling, Feynman rules for the vertices should be:

  1. $$-i\lambda\left(k_1 k_2 + k_1 k_3 + k_2 k_3\right)$$
  2. $$i\lambda^2 \left(k_1 k_2 + k_1 k_3 + k_1 k_4 + k_2 k_3 + k_2 k_4 + k_3 k_4\right)$$

At order $\lambda$, there's nothing.

At order $\lambda^2$ there are contributions from the tadpole diagram with a $\phi^2 \left(\partial_\mu\phi\right)^2$ vertex and from the diagram with two $\phi\left(\partial_\mu\phi\right)^2$ vertices.

Is it right? Or am I missing something? Are the Feynman rules for the vertices correct?

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It seems to me that your Lagrangian is just a free Lagrangian in disguise.

Start from

$$\mathcal L =\frac{1}{2} (\partial_\mu \phi)^2 $$

and do a field redefinition

$$\phi(x) \to \frac{2}{\lambda} e^{-\frac{\lambda \phi(x)}{2}}$$

With this you find back your Lagrangian. Field redefinitions don't change correlation functions, so whatever you are going to compute with your Lagrangian will be identical to a free Lagrangian and thus there is no correction to the propagator.

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  • $\begingroup$ This is an exam problem, so I get confused knowing it's trivial. How can I see that field redefinitions don't change correlation functions and propagators? Do I need path integral? $\endgroup$ – Vincenzo Ventriglia Sep 27 '18 at 7:35
  • $\begingroup$ From path integrals it"s really trivial to see that correlations functions are invariant, since the field $\phi$ is just an integrated variable. Without path integrals I don't know if there is a quick way to see it. $\endgroup$ – FrodCube Sep 27 '18 at 10:15
  • $\begingroup$ I might say something wrong, but there are many subtleties here. The correlation function will likely change under a field redefinition. What could remain invariant are the scattering amplitudes (with the LSZ prescription). However in that case (I think) one has to assume that the redefinition has a linear term with coefficient $1$, otherwise the asymptotic states change and you can't relate the things that you are computing before and after the redefinition. Again, I'm not sure, I'd be happy if someone could prove me wrong (or right). $\endgroup$ – MannyC Mar 8 '20 at 0:33

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