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I want to calculate the amplitude for nucleon meson scattering $\psi \varphi \to \psi \phi$ in scalar Yukawa theory, with interaction term: $$H_{I} = g \int d^{3}x \psi^{\dagger} \psi \varphi.\tag{3.25}$$

This involves dealing with a time ordered string of operators $$T\{\psi^{\dagger}(x) \psi(x) \varphi(x)\psi^{\dagger}(y) \psi(y) \varphi(y)\}.\tag{3.46}$$ We can apply Wick's theorem to the above expression. Tong's lecture notes identify two relevant terms: $$:\psi^{\dagger}(x) \varphi(x)\psi(y) \varphi(y): \overbrace{\psi(x) \psi^{\dagger}(y)}\tag{3.53}$$ and $$:\psi(x) \varphi(x)\psi^{\dagger}(y) \varphi(y): \overbrace{\psi^{\dagger}(x) \psi(y)}.$$

My question is, what about the terms: $$:\psi^{\dagger}(x) \psi(x) \varphi(x) \varphi(y): \overbrace{\psi^{\dagger}(y) \psi(y)}$$ and $$:\varphi(x) \psi^{\dagger}(y) \psi(y) \varphi(y): \overbrace{\psi^{\dagger}(x) \psi(x)}.$$ They are not mentioned in the notes and I cannot come up with a Feynman diagram that would correspond to those. Is it even correct to contract two field operators defined at the same point?

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There is an implicitly written normal ordering symbol $:~:$ in the Hamiltonian (3.25), and therefore 2 implicitly written normal ordering symbols in the expression (3.46). The answer to OP's title question is that there are no contractions among operators within the same normal ordering symbol, i.e. belonging to the same spacetime point. This is explained in e.g. this related Phys.SE post.

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