# Feynman rules for scalar qed

1. Oct 26, 2009

### otaniyul

1. The problem statement, all variables and given/known data
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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2. Relevant 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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3. 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 fucntion. 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 fucntional 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 fucntion, 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...

Last edited: Oct 26, 2009
2. May 31, 2010

### babamarysol

Feynman's rules are better understood writing down the perturbation expansion for correlation's function...

3. Jun 4, 2010

### vorcil

man i've been trying to do this question

I can't do it, can someone do it?

4. Jun 8, 2010

### babamarysol

hamiltonian density (interaction part)

$$\mathscr{H}_{int}=eA^\mu j_\mu$$

two point function

$$<0| T(\phi_1\phi_2) |0> =<0| \phi_1 \phi_2 \text{exp} \left[-i \int d^4 x \mathscr{H}_{int} \right] |0>_{connected}$$

four point function

$$<0|T(\phi_1\phi_2 \phi_3 \phi_4) |0>=<0| \phi_1 \phi_2 \phi_3 \phi_4 exp[-i \int d^4 x \mathscr{H}_{int} ] |0>_connected$$

expanding the exponent you find what you want order by order...

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