Consider a field redefinition $$ \phi \rightarrow \phi' = \phi+\lambda \phi^2 $$ Find the Feynman rules for this theory and work out the $2\rightarrow 2$ scattering amplitude at tree level (The result should be zero).

$$ \mathcal{L}_0 = -\frac{1}{2}(\partial^\mu \phi \partial_\mu \phi + m^2 \phi^2) \implies \text{EOM: } \Box\phi - m^2\phi=0 $$ Preforming the field redefinition on the Lagrangian I obtain: $$ \mathcal{L}_0 \rightarrow \mathcal{L}'= \mathcal{L}_0 -2\lambda\phi \partial^\mu \phi \partial_\mu \phi - 2 \lambda^2 \phi^2 \partial^\mu \phi \partial_\mu \phi -\lambda m^2 \phi^3 - \frac{\lambda^2}{2} m^2 \phi^4 $$ I then rewrote the two terms with partial derivatives as $$ -2\lambda\phi \partial^\mu \phi \partial_\mu \phi = -\lambda \partial^\mu (\phi^2 \partial_\mu \phi)+\lambda \phi^2 \Box \phi $$ which gets rid of the total derivative. Then I used the EOM to replace $\Box \phi$ with $m^2 \phi$ to get $$ \mathcal{L}' = \mathcal{L}_0 + \lambda (m^2 - 1)\phi^3 + \lambda^2 \left( \frac{m^2}{3}-\frac{1}{2} \right) \phi^4 $$ Now this looks like a normal scalar field theory with a cubic and quartic vertex. Now it wants a tree level diagram for two to two scattering, so I should put these two vertices in all the combinations that have no loops and result in two in, and two out? Then subsequently work out the scattering amplitude from the said diagrams using the Feynman rules I put together with the above Lagrangian?

  • $\begingroup$ Shouldn't you express $\phi$ via $\phi'$ by solving the defining square equation? $\endgroup$ – Vladimir Kalitvianski Dec 4 '12 at 14:06
  • $\begingroup$ Also, you cannot use EOM in Lagrangian transfromations because the corresponding variables are considered unknown. $\endgroup$ – Vladimir Kalitvianski Dec 4 '12 at 14:26
  • $\begingroup$ @VladimirKalitvianski If one is only interested in S matrix elements, one can use the lower order equation of motion. $\endgroup$ – Diego Mazón Dec 4 '12 at 17:36

I remember this as a nice problem from chapter 10 of Srenicki's QFT book. Your first expression for $\mathcal{L}'$ should have factors of $m^2$ in the last two terms. You should work with the lagrangian from here and derive the feynman rules and calculate the scattering amplitude. I will sketch this for you below. It is a very nice excercise to check whether you have understood everything up to that point.

You cannot use the EOM as you have done since that was for the original $\phi$ before the field redefinition. After the field redefinition your new $\phi$ (may help you to call it something else, e.g. $\phi \to \psi + \lambda \psi^2$) has a different equation of motion involving quadratic and cubic terms!

As I said above you should just work from your first expression for $\mathcal{L}'$ with the derivatives in and derive the Feynman rules from there. If you go through the whole procedure as is done in the previous chapter (being careful about the derivatives and symmetry factors) for each type of vertex (from each new term in the lagrangian) you will find a whole bunch of new vertex factors. Just calculate the 3 or 4 point function for each individual term and determine the feynman rules that way. Once you get used to how it works you can start to read them off directly from the lagrangian but it is good to go through the whole procedure a few times first. You should find for example, the cubic term with no derivatives gives a vertex joining three lines with a vertex factor of $$-i3! \lambda m^2.$$ The cubic vertex with derivatives has a vertex factor of $$4i\lambda(k_1k_2 + k_1k_3 + k_2k_3)$$ you see that the derivatives act to pull down powers of the momenta from the Fourier transforms of the propagators. It is sort of obvious if you just think about $\partial_\mu$ in momentum space.

So the total vertex joining three lines will just be the sum of these two. Do the same for the four point vertex and then as you say, draw all the feynman diagrams for $\phi\phi \to \phi\phi$ scattering, calculate the total amplitude and you should find if you do everything right that you get lots of nice cancellations and the answer is zero and you will have a nice warm feeling inside :-)

Derivative interactions such as these crop up in non-abelian gauge theories (e.g. the standard model) and gravity so they are important to understand. If you are still having trouble calculating the vertex factors I highly recommend this pdf which goes through in painstaking detail how to derive feynman rules. It follows the treatment by Srednicki too! Check section 5.3 and 5.4 for what you need.

  • $\begingroup$ @knives You can use the equation of motion at $\lambda^n$ order in the terms of the Lagrangian that go with $\lambda^{n+1}$ as long as you just want to compute S-matrix elements. $\endgroup$ – Diego Mazón Dec 4 '12 at 17:41
  • $\begingroup$ I mean that one is allowed to do that because it does not change S matrix elements. $\endgroup$ – Diego Mazón Dec 4 '12 at 17:47
  • $\begingroup$ @MistakeInk Thank you. This is almost exactly what I ended up finding over the past week of working on this problem. Thanks for the link also. I will really try and get this. $\endgroup$ – kηives Dec 4 '12 at 19:03
  • $\begingroup$ @MistakeInk: Apart form non-abelian gauge theories, "derivative interactions" appear as counter-terms in abelian theories too. Cancellation of contrubutions from "derivative interactions" and "polynomial" ones means that polynomial interactions ("potential ones") contain implicitely some derivatives ("kinetic terms"). $\endgroup$ – Vladimir Kalitvianski Dec 5 '12 at 11:59

I guess that there are mistakes in your algebra. In any case:

Now this looks like a normal scalar field theory with a cubic and quartic vertex

Except for the fact that there are just two independent parameters $m$ and $\lambda$.

The cross section must be zero just because of the LSZ reduction formula and the last Lagrangian should be related to the first, which has no interactions, by a field redefinition. In deduction of the LSZ formula, it is clear that two Lagrangians related by a field redefinition give the same S-matrix elements regardless the off-shell n-point function are different in each case. The only condition is that both fields have the same matrix elements between the vacuum state and one particles states and the same vacuum expectation value. Matrix elements between the vacuum state and multiparticle states do not matter because of the Riemann-Lebesgue lemma and how these matrix elements enter in the LSZ formula.

  • $\begingroup$ Regardless the stupid downvoter who does not give an explanation for his downvote, this answer is essentially correct. Maybe there is some numerical mistake and the there are more interactions terms but the reason —as I wrote above— is LSZ reductio formula and field redefinitions. $\endgroup$ – Diego Mazón Dec 4 '12 at 17:49
  • $\begingroup$ The answer is unclear, 'just because the LSZ reduction' does not give much insight at all, how does that help the questioner? That is NOT the only term at first order in $\lambda$, what about the derivative term? Your statement that there is no tree level amplitude for $2 \to 2$ scattering from a $\phi^3$ theory is just wrong, you have $s$, $t$ and $u$ channel diagrams! Lots of erroneous and unclear information hence the down vote. $\endgroup$ – Mistake Ink Dec 4 '12 at 22:22
  • $\begingroup$ @MistakeInk 1) In my opinion, if one reads some wrong information or something totally wrong, first thing he should do is to point it out in the comments in order the reader does not confuse the reader and the writer may delete or correct it. And after he may or should downvote, which is less important. $\endgroup$ – Diego Mazón Dec 4 '12 at 22:37
  • $\begingroup$ Good point, I agree with you and should have written a comment first as you say. Sorry for not doing so. $\endgroup$ – Mistake Ink Dec 4 '12 at 22:44
  • $\begingroup$ @MistakeInk 2)The true reason why the amplitude is zero although the n-point functions do not correspond to a free theory is in the derivation os LSZ reduction formula. 3) You are probably right that there are more interaction terms and maybe even the coefficient os the cubic term is not correct, as I wrote, but this is not main part of the answer. 4) You are totally right about the tree-level contribution, I do not know what I was thinking about. 5) After reading your comment, I understand your downvote. Now, I can correct or delete my answer in order no one gets wrong information. Thank you. $\endgroup$ – Diego Mazón Dec 4 '12 at 22:44

See 't Hooft and Veltman's diagrammar lecture notes. There's a discussion of something called the 'equivalence theorem'. Essentially, it observes that since the S-matrix is obtained from the correlation function by finding the residue at a single-particle-pole, everything which would lead to multiple particle poles does not contribute. There's a loophole if the field transformation is not local, but leads to a one-particle pole.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.