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
$$:\psi^{\dagger}(x)\psi(x)\psi^{\dagger}(y)\psi(y):\overbrace{\phi(x)\phi(y)}$$
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!
Best Answer
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.