Correction to the scalar propagator - derivative coupling

Question

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. 
2. 

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?

Answer 1

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.