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Given a Lagrangian coupling a complex scalar field $\psi$ to a real scalar field $\phi$:

$$\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi + \partial_{\mu}\psi\partial^{\mu}\psi^*+ \frac{1}{2}m^2 \phi^2 + \frac{1}{2}M^2 \psi^2 + g \phi \psi \psi^*$$

I'm struggling to see how there can be any non-zero Feynman diagrams for the $\psi \psi \rightarrow \psi \psi$ scattering. (As in figure 9 of Tong's notes)

If we have two internal vertices I understand that we need to calculate the quantity: $$<0| T\{\psi(x_1)\psi(x_2)\psi(y_1)\psi(y_2)\ \int \phi \psi \psi^* \int \phi \psi \psi^*\} |0>$$

Now By Wick's theorem this should be given by the sum of all possible contractions. However the contraction of $\psi$ with $\psi$ is zero so I can't see anyway to completely contract this?

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    $\begingroup$ If you have an incoming particle $\psi$, you have to include a factor $\psi(x)$ in the correlator. On the other hand, if you have an outgoing particle $\psi$, you have to include a field $\psi^*(x)$ in the correlator. Thus, you want to calculate $\langle\psi(x_1)\psi(x_2)\psi^*(x_3)\psi^*(x_4) (\phi\psi\psi^*)(\phi\psi\psi^*)\rangle$, which is non-zero in general. $\endgroup$ – AccidentalFourierTransform Dec 31 '15 at 19:24
  • $\begingroup$ Thanks, I didn't realise that - what is the reason for that? So we are actually calculating the four point function $G^{(4)}(\psi(x_1), \psi(x_2), \psi(y_1)^*, \psi(y_2)^*)$ $\endgroup$ – Wooster Dec 31 '15 at 19:28
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    $\begingroup$ In a nutshell, the reason is that the scattering amplitude is given by $\langle \text{out}|\text{in}\rangle$, so that whatever is outgoing gets conjugated compared to if it were incoming. $\endgroup$ – AccidentalFourierTransform Dec 31 '15 at 20:10

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