I'm trying to compute nucleon-nucleon scattering in scalar Yukawa theory. Here we view a nucleon as a complex scalar field $\psi$ and a meson as a real scalar field $\phi$. They interact through $H_I=g\int d^3x\psi^{\dagger}\psi\phi$.

Suppose we want to compute the amplitude for scattering from an initial state $|p_1,p_2\rangle$ to a final state $|p_1',p_2'\rangle$, which we assume are eigenstates of the free theory. At second order in perturbation theory we have the term

$$\frac{(-ig)^2}{2}\int d^4xd^4y\ T\left[\psi^{\dagger}(x)\psi(x)\phi(x)\psi^{\dagger}(y)\psi(y)\phi(y)\right]$$

using Dyson's formula, where all fields are in the interaction picture, so the Heisenberg picture of the free theory. Using Wick's theorem we may compute this amplitude explicity.

My notes claim that the only term giving a nonzero contribution is


I disagree with this however. Surely also we get contributions from the disconnected and unamputated diagrams? For example

$$:\psi^{\dagger}(x)\psi(x):\overbrace{\psi^{\dagger}(y)\psi(y)}\ \overbrace{\phi(x)\phi(y)}$$

and the various other permutations of such terms will produce extra terms (delta function divergences) won't they?

I know that if we are considering true scattering where our initial and final states are eigenstates of the interacting theory then there's a theorem which says we may ignore the disconnected and unamputated diagrams. Perhaps I'm meant to implicitly use this here, and the assumption in bold above is an error in the notes.

To summarise - am I right in thinking that disconnected and unamputated diagrams give a nonzero contribution to scattering for eigenstates of the free theory? And is the correct way to deal with this to assume we're working with eigenstates of the interacting theory? Are all my arguments and intuitions above correct?

Many thanks in advance!


1 Answer 1


The in and out states you use have two particles in them. The operator you mention with the double contraction only has a single annihilation and creation operator in it, so it only acts on a single particle from the two particle states. There is no momentum transfer due to this operator. Momentum conservation then constrains the outgoing momenta to equal exactly the incoming momenta and you have forward scattering, i.e. no scattering.

  • $\begingroup$ Thanks for your answer. I've drawn the Feynman diagrams and clearly there's no scattering. But the mathematical contribution from these diagrams seems to be $\propto ig\delta^{(3)}(0)$. Surely we must work with the true states of the interacting theory in order the cancel these contributions? If you happen to have Peskin and Schroeder I believe that's how they do it on p111, unless I'm missing something fundamental? $\endgroup$ Jan 2, 2013 at 14:32
  • 1
    $\begingroup$ I have P&S but not in front of me. I'll check it tomorrow. But yes, the Feynman diagram for the double contraction you wrote is a disconnected diagram with a tadpole attached to one of the legs. The delta function divergence comes from the short distance behaviour of the loop propagator, which needs regularisation. The tadpole contributes to the mass renormalisation of the $\psi$ field. When you sum up all the graphs with repeated (regularised) tadpole insertions you will partially "dress" the bare $\psi$ and get closer to the physical states. This is part of the renormalisation program. $\endgroup$
    – Michael
    Jan 2, 2013 at 15:33
  • $\begingroup$ I've check P&S. The point of the discussion on p.111 is not that you have to use the full interacting states, rather that disconnected diagrams don't contribute to the $T$ matrix, which is defined as the non-trivial part of the $S$ matrix. This is because the momentum transfer vanishes for these diagrams. An example of the mass renormalisation procedure for tadpole diagrams starts on p360 in the context of the linear sigma model. Your model is somewhat simpler, but the simplest model which gives an example of this is the $\phi^3$ model in Srednicki web.physics.ucsb.edu/~mark/qft.html $\endgroup$
    – Michael
    Jan 3, 2013 at 13:50
  • $\begingroup$ Don't be discouraged. All these things are hard to get your head around. You're asking good questions. You're well on your way to understanding QFT. :) $\endgroup$
    – Michael
    Jan 3, 2013 at 13:51
  • $\begingroup$ @EdwardHughes See chapter 9 of Srednicki, particularly the discussion following equation 9.16 if you want to jump straight to the discussion of tadpole diagrams. $\endgroup$
    – Michael
    Jan 3, 2013 at 14:02

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