How Do You Derive Feynman Rules for Scalar QED Using Functional Methods?

[otaniyul]

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...

 

[babamarysol]

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

 

[vorcil]

man I've been trying to do this question

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

 

[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...

 

[paranormal]

Dear student,

Thank you for your detailed question. The Feynman rules for scalar QED are derived from the interaction Lagrangian, which contains the terms responsible for the interaction between the scalar and gauge fields. These terms are then used to construct the Feynman diagrams that represent the possible interactions between particles. The Feynman rules for scalar QED are as follows:

1. Propagator for scalar field (photon emission or absorption): This is given by the inverse of the Klein-Gordon operator, K = ∂^2 - m^2. In momentum space, it is represented by Δ(p) = 1/(p^2 - m^2 + iε).

2. Propagator for gauge field (photon emission or absorption): This is given by the inverse of the massless operator in Feynman gauge, P^μν = η^μν∂^2. In momentum space, it is represented by Dμν(p) = -ημν/(p^2 + iε).

3. Vertex factor: This is given by the coupling constant e and the momentum of the particle. For a scalar particle with momentum p, the vertex factor is ie. For a gauge boson with momentum p, the vertex factor is ie(p + k)^μ.

4. Momentum conservation: At each vertex, the total momentum must be conserved. This means that the incoming momenta must equal the outgoing momenta.

Using these rules, you can construct Feynman diagrams for scalar QED. The second order term in the interaction Lagrangian will give rise to a three-point vertex, which will have a vertex factor of ie and the product of the incoming and outgoing momenta. I hope this helps. Good luck with your calculations!