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I'm currently studying QFT from David Tong's lecture notes and video lectures. In meson to nucleon + antinucleon decay (section 3.2.1 in this ) in scalar Yukawa theory to order $g$, without using Feynman diagrams, he talks about the term in the mode expansion with a meson creation operator $a_\vec{p}^{\dagger}$ not contributing to the scattering amplitude because the two-meson state has no overlap with the final state.

I think I get what this means on a physical level, obviously we don't want any mesons in our final state, but I don't see how to get the contribution to disappear mathematically. What does "zero overlap" mean? I think I've seen that it means $\left\langle a|b \right\rangle = 0$ for whatever states you're looking at, but it seems to me that at order $g$ this term will look like $\left\langle f|\int d^4x\; \psi^{\dagger}(x)\psi (x) a^{\dagger} a^{\dagger}|0 \right\rangle$ (with some factors that I think should be irrelevant suppressed) and I don't see why this vanishes. Can anyone show me explicitly how this occurs?

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The general idea is that:

$$|f\rangle \propto b^{\dagger} c^{\dagger}|0\rangle$$

so that:

$$\langle f| \propto \langle 0| cb$$

The $\psi$ and $\psi^{\dagger}$ operators don't contain any $a$ operators, leaving those two $a^{\dagger}$ on the right to commutes right through the $cb$ on the left, annihilating the state and giving zero, as:

$$a|0\rangle = 0 \implies \langle 0|a^{\dagger}=0$$

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