# S-matrix for $\phi^3$ theory

In the book Quantum Field Theory for the gifted amateur by Tom Lancaster & Stephen J Blunden, in the chapter about expanding the S matrix they give an example using the $\phi^4$ Langrangian, $$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^2-\frac{m^2}{2}\phi^2-\frac{\lambda}{4!}\phi^4$$ Where the $\phi^4$ term is the interacting part. In the example, they show the wick contractions for the first order term in the Dyson expansion and show the corresponding Feynman diagrams. In the exercises, they ask you to do the same thing for $\phi^3$ theory which has the langrangian: $$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^2-\frac{m^2}{2}\phi^2-\frac{\eta}{3!}\phi^3$$ Finding the interaction Hamiltonian $\mathcal{H}_I$ and expanding the s matrix is relatively straight-forward but when using the wick contractions on the terms of the S matrix, I ran into a problem. In $\phi^4$ theory, there was always an even number of terms in the string e.g. $\langle0|\hat{a}_q\hat\phi(z)\phi(z)\hat\phi(z)\hat\phi(z)\hat{a}^\dagger_p|0\rangle$. This means that in all permutations of contractions, all the terms will be contracted. But in $\phi^3$ theory, there is never an even number of terms so there will always be one term not contracted e.g.$$\langle0|\hat{a}_q\hat\phi(z)\hat\phi(z)\hat\phi(z)\hat{a}^\dagger_p|0\rangle \rightarrow \langle0|\hat{a}_q\hat{a}^\dagger_p|0\rangle \langle0|T[\hat\phi(z)\hat\phi(z)|0\rangle\mathbf{\langle0|\hat\phi(z)|0\rangle}$$ (This is one possible contraction). So when computing the amplitudes and drawing the Feynman diagrams, what do you do with the left over $\langle0|\hat\phi(z)|0\rangle$?

• Try using two creation and one annihilation operators, or try expanding interactions to the next order. The interaction vertex in $\phi^3$ takes two lines into one (or one into two). – Darkseid Oct 14 '17 at 7:23
• Regarding left-over $\phi$, it renders to zero. Expanding it in terms of creation/annihilation operators makes this obvious – Darkseid Oct 14 '17 at 10:00