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I've been given a small project to ease me into life as a PhD student and part of it involves working out Feynman rules for scalar bosons coupling to gauge bosons.
I've been reading through Peskin and Schroder Chapter 9 and want to clarify if I've got my understanding right of how to do it.
The QED Lagrangian is L = \int d^{4}x \,\mathcal{L} = \int d^{4}x \, \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - e\psi\gamma^{\mu}\psi A_{\mu}
The electron propogator's Feynman diagram rule is obtained by finding the Green's function of the Dirac operator i\gamma^{\mu}\partial_{\mu}-m because it just describes an electron moving without interaction.
Similarly, the photon propogator is found by rearranging -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} into the form A_{\mu}\Delta^{\mu\nu}A_{\nu} and finding the Green's function of \Delta^{\mu\nu}.
One thing I'm a bit hazy on (it seems to work but I want to be sure) is getting the vertex Feynman rules.
Suppose \mathcal{L} = \mathcal{L}_{0}+\mathcal{L}_{I}. Suppose the interaction term involves 3 fields (like \psi, \bar{\psi} and A_{\mu}. Is the vertex rule, in momentum space, then simply
\frac{\delta}{\delta \psi}\frac{\delta}{\delta \bar{\psi}}\frac{\delta}{\delta A_{\mu}}\int d^{4}x \mathcal{L}_{I}
That would give the QED one as -ie\gamma^{\mu}\int d^{4}x which is right. Is this true for all interaction Lagrangians? I follow most of the "Computing the correlation function" bits and the Z = e^{iS...} bits involving functional derivatives, but when it comes to then computing the Feynman rules I just kind of noticed that relation between the Lagrangian and the rules rather than find any step by step guide in P&S.
Am I going about it the right way or have I just hit on the right answer by (dumb) luck? Thanks :)
I've been reading through Peskin and Schroder Chapter 9 and want to clarify if I've got my understanding right of how to do it.
The QED Lagrangian is L = \int d^{4}x \,\mathcal{L} = \int d^{4}x \, \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - e\psi\gamma^{\mu}\psi A_{\mu}
The electron propogator's Feynman diagram rule is obtained by finding the Green's function of the Dirac operator i\gamma^{\mu}\partial_{\mu}-m because it just describes an electron moving without interaction.
Similarly, the photon propogator is found by rearranging -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} into the form A_{\mu}\Delta^{\mu\nu}A_{\nu} and finding the Green's function of \Delta^{\mu\nu}.
One thing I'm a bit hazy on (it seems to work but I want to be sure) is getting the vertex Feynman rules.
Suppose \mathcal{L} = \mathcal{L}_{0}+\mathcal{L}_{I}. Suppose the interaction term involves 3 fields (like \psi, \bar{\psi} and A_{\mu}. Is the vertex rule, in momentum space, then simply
\frac{\delta}{\delta \psi}\frac{\delta}{\delta \bar{\psi}}\frac{\delta}{\delta A_{\mu}}\int d^{4}x \mathcal{L}_{I}
That would give the QED one as -ie\gamma^{\mu}\int d^{4}x which is right. Is this true for all interaction Lagrangians? I follow most of the "Computing the correlation function" bits and the Z = e^{iS...} bits involving functional derivatives, but when it comes to then computing the Feynman rules I just kind of noticed that relation between the Lagrangian and the rules rather than find any step by step guide in P&S.
Am I going about it the right way or have I just hit on the right answer by (dumb) luck? Thanks :)