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otaniyul
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Homework Statement
Hello all, thanks for reading...
I was assigned to calculate feynman rules for the scalar QED theory via functional methods.
The fields are a scalar complex field \phi, \phi^* and gauge field A_\mu, and the lagrangean is
\mathfrak{L} = (D_\mu \phi)^* (D^\mu \phi) - m^2 \phi \phi^* - (1/4)F_{\mu,\nu}F^{\mu\nu}
where D_\mu = \partial_\mu + ieA_\mu is the covariant derivative, and F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu is the maxwell tensor.
For the fonts J (complex scalar) and j (4-vector), the generating function will then be
Z[J,j] = \int \mathfrak{D}\phi\mathfrak{D}\phi^*\mathfrak{D}A \exp(i\int\mathfrak{L} + J^*\phi + J\phi^* + j^\mu A_\mu)
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Homework Equations
to simplify notation, we will use
the klein gordon operator: K = \partial^2 - m^2,
its propagator: \Delta(x) = \int \frac{d^4k}{(2\pi)^4} \frac{e^{-ikx}}{k^2-m^2+i\epsilon}
the massless operator (feynman gauge): P^{\mu\nu} = \eta^{\mu\nu} \partial^2,
and its propagator: D_{\mu\nu}(x) = \int \frac{d^4k}{(2\pi)^4} \frac{-\eta_{\mu\nu}e^{-ikx}}{k^2}
We have
K\Delta(x) = \delta(x)
P^{\mu\nu} D_{\nu\rho}(x) = \delta^\mu_\rho \delta(x)
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The Attempt at a Solution
Ok, I have then split the Lagrangean in the free-theory part and interaction part:
\mathfrak{L}_0 = (\partial_\mu \phi)^* (\partial^\mu \phi) - m^2 \phi \phi^* - (1/4)F_{\mu,\nu}F^{\mu\nu},
\mathfrak{L}_{int} = eA^\mu (\phi i\partial_\mu \phi* - \phi* i\partial_\mu \phi) + e^2 A^\mu A_\mu \phi^* \phi
First we derive the free theory generating function. We add the Gauge Fixing term to the Lagrangean, with Feynman gauge ((-1/2)(\partial_\mu A^\mu)^2. The lagrangean integral can be made to the form
\int dx \mathfrak{L}_0 = -\int \phi K \phi^* + (1/2) A\mu P^{\mu\nu} A_\nu
and the generating functional will then be
Z_0 = N \exp \big( -i \int dx dy J^*(x) \Delta(x-y) J(y) + (1/2) j^\mu(x) D_{\mu\nu}(x-y) j^\nu(y) \big) = N e^{-i \int J^* \Delta J + (1/2) j D j^}
soo, to my question. I have the scalar and photon propagatos, those i / k^2 - m^2 from the propagators, but how do i get the vertex terms?
The generating function for the interaction theory will be
Z[J,j] = N \exp( -i \int \mathfrak{L}_{int} ) Z_0[J,j]
where in L_int we substitute the fields for functional derivatives of the sources. so in first order, i did this:
\int dx (ie\eta^{\mu\nu}) A_\mu(x) (\phi(x) \partial_\nu\phi^*(x) - \phi^*(x) \partial_\nu\phi(x))
goes to
\int dx dz \delta(x-z) (ie\eta^{\mu\nu}) A_\mu(x) \frac{\partial}{\partial z^\nu}(\phi(x) \phi^*(z) - \phi^*(x) \phi(z))
where i used a dirac delta to let the derivative act only where it is needed. then, passing to functional derivatives, we should get
\int dx dz \delta(x-z) (ie \eta^{\mu\nu}) \frac{\delta}{i\delta j^\mu(x)} \frac{\partial}{\partial z^\nu} \left( \frac{\delta}{i\delta J^*(x)}\frac{\delta}{i\delta J(z)} - \frac{\delta}{i\delta J(x)} \frac{\delta}{i\delta J^*(z)} \right)
then aplying to the free generating function, using the explicit form of the propagators, after some work... we get the first order term:
e \int dx dy dz \frac{dp}{(2\pi)^4}\frac{dk}{(2\pi)^4}\frac{dkp}{(2\pi)^4} (2\pi)^4\delta(p+k+k') j^\mu(x) \left( -\frac{e^{ipx}\eta_{\mu\nu}} {p^2} \right) (k+k')^\nu J(y)\frac{e^{iky}}{k^2-m^2} J^*(z)\frac{e^{ik'z}}{k'^2-m^2}
so... after all calculations, how do i read out the feynman rule? I mean, from the interaction lagrangean i know it must be something like
ie(k+k')^\mu
but why, formally? the generation funtion is mixed up with positions and momentum spaces, so its hard for me to see... where are the momenta goint, entering or exiting the vertex and all those stuff... any help!?
and gosh, there is also the second order!
thank you a lot, i write too much...
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