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We know that for a $\phi^4$ interacting theory with $$ \mathcal{L}_{int} = -\frac{\lambda}{4!}\phi^4 $$ the interaction $\phi \phi \rightarrow \phi \phi$ gives a vertex with a factor $-\frac{i\lambda}{4!}$ which stems from Wicks theorem.

If we now use the same procedure for an interacting lagrangian of the form $$ \mathcal{L}_{int} = -\frac{g}{6}\phi^3-g\Phi^*\Phi\phi $$ for example, what would be the vertex? When we go through Wicks theorem, do we sum the terms or just treat them seperately?

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Your second Lagrangian would generate of two vertices:

A cubic self interaction for $\phi$ (in your notation $\phi \phi \rightarrow \phi$, say) and an inter-field interaction between $\Phi$ and $\phi$ (in your notation $\Phi^{*}\Phi \rightarrow \phi$).

Both vertices have three legs; in the first they're all $\phi$ fields and in the second there is a $\Phi$ anti-particle, a $\Phi$ particle and a $\phi$. In your example, both couplings are equal, $g$.

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  • $\begingroup$ Thanks! So that would mean that when calculating the matrix elements the vertices would correspond to $-\frac{ig}{6}\int{d^4x}$ for $\phi\phi \rightarrow \phi$ and $-ig\int{d^4x} $ for $\phi \rightarrow \Phi^* \Phi$ ? Also, do both of them(and all triple field terms for that matter) interpret themselves as a decaying particle or a pair which annihilates? $\endgroup$ – fielder Nov 21 '19 at 16:51
  • $\begingroup$ Yes, in position space that's fine (except you forget the combinatorics of Wick's theorem which will modify the factor of $1 / 3!$). Those are the tree level vertices, which are then used at higher loop order to create more complicated processes. The interpretation of the vertex you ask about depends upon the way you draw the Feynman diagrams and assign momenta. It could be decay, or it could be particle splitting (see crossing symmetry). $\endgroup$ – lux Nov 21 '19 at 16:57
  • $\begingroup$ Oh of course, because we can interchange $\phi_1 \phi_2 \phi_3$ without changing the diagram. $\endgroup$ – fielder Nov 21 '19 at 17:00
  • $\begingroup$ Correct! Of course you also need the kinetic terms in the lagrangian to generate the specific form of the field propagators but this is another question $\endgroup$ – lux Nov 21 '19 at 17:03

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