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I have a few questions related to the following interaction Lagrangian (no use of crossing symmetry in the following) involving the uncharged scalar $\chi$ and the charged scalar $\phi$:

$$\tag{1}\mathcal{L}_{\rm int} ~=~ \sqrt\lambda· g· \chi\phi^\dagger \phi - \frac{\lambda}{4}(\phi^\dagger\phi)^2.$$

Can we use (1) to say anything about the following reactions.

a) $\phi\phi\rightarrow\phi\phi$

b) $\phi\phi^\dagger\rightarrow \chi\chi$

c) $\chi\phi\rightarrow\chi\phi$

My thoughts: To say anything about a) shouldn't we have a term in $\mathcal{L}_{\rm int}$ not involving $\phi^\dagger~$?

If we use the "usual" rules of letting a dashed line represent the $\chi$-propagator and (properly directed) arrowed lines the charged $\phi$ then it seems to me that the first term in (1) involves three lines; one dashed, one particle $(\phi)$ and one anti-particle. The second term involves four lines with incoming antiparticle ($\phi^\dagger$)-line pointing away from the vertex and outgoing antiparticle pointing towards the vertex etc..

Using these rules it's easy to write down the rule say for c) but what is the vertex factor then since we don't have a term proportional to $\chi\phi^2$ (say) in $\mathcal{L}_{\rm int}$?

Please feel free to comment and explain in detail if you can or suggest examples of similar Lagrangians and reactions/interactions.

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Qualitatively, the first term looks like a Yukawa vertex, which is of the form $\overline{\psi}\psi \varphi$ for some fermion $\psi$ and scalar $\varphi$. So the details are different (different propagators etc.), but the diagrams look the same. The second term is a $\phi^4$ vertex. Both are described in every QFT textbook. – Vibert Dec 15 '12 at 14:36

"To say anything about a) shouldn't we have a term in $\mathcal{L}$ not involving $\phi^\dagger$?"

Nope. To observe an outgoing $\phi$-particle, you need the annihilation operators in $\phi^\dagger$. So, for example, the first term can be thought of as giving a diagram where an incoming $\phi$-particle and an incoming $\phi$-antiparticle annihilate and become a $\chi$-particle. But the same diagram can be thought of as a single $\phi$ evolving in time, and kicking off a $\chi$ as it goes.

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Oh, now I recall the keyword: "Topologically inequivalent diagrams". So for instance the diagram of $\phi\phi^\dagger\rightarrow\chi$ can be viewed as (if one rotates the diagram) $\phi\rightarrow\phi\chi$ (like you said?). – Maxwell's Demon Dec 15 '12 at 14:52
Yes, that's what I was getting at. – user1504 Dec 15 '12 at 14:57
OK, thanks for answering. – Maxwell's Demon Dec 15 '12 at 15:02

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