# Ward-Takahashi identities at tree level in scalar QED

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1. Jun 3, 2017

### leo.

1. The problem statement, all variables and given/known data
Let $\Gamma^\mu$ be the three-point vertex in scalar QED and $\Gamma^{\mu\nu}$ be the four-point vertex. Use Feynman's rule at tree level and verify that the Ward-Takahashi identities are satisfied:
$$q^\mu \Gamma_\mu(p_1,p_2)=e[D_F^{-1}(p_1)-D_F^{-1}(p_2)],\\ q^{\mu}\Gamma_{\mu\nu}(q,k,p_1,p_2)=e[\Gamma_\nu(p_1+q,p_2)-\Gamma_\nu(p_1,p_2-k)]$$
where $D_F^{-1}(p)$ is the inverse of the scalar propagator and $q$ is the photon momentum.

Finaly show that $p_\mu \Pi^{\mu\nu}=0$ being $\Pi^{\mu\nu}$ the sum of the vacuum polarization diagrams.

2. Relevant equations
I believe the Feynman rules for the value of such vertices and the definition of those gammas, which I'm unsure. Based on the book I'm reading (Quantum Field Theory and the Standard Model by Matthew Schwartz) the three point vertex has 4 possible values: for particle scattering it is $ie(-p_1^\mu-p_2^\mu)$, for antiparticle scattering it is $ie(p_1^\mu+p_2^\mu)$, for pair annihilation it is $ie(-p_1^\mu+p_2^\mu)$ and for pair creation it is $ie(-p_1^\mu+p_2^\mu)$. The four-point vertex gives $2ie^2 g_{\mu\nu}$.

He summarizes this as "$-ie$ times the sum of the momentum of particles whose particle-flow arrows point to the right minus the momentum of the particles whose arrows point to the left". I believe this means that $\Gamma^\mu(p,q)=-ie(p^\mu+q^\mu)$ where $p^\mu$ is the total momentum of the particles entering the vertex and $q^\mu$ is the total momentum of the antiparticles going out of the vertex, but I'm unsure if this captures it all. For the four-point vertex I thought $\Gamma^{\mu\nu}=2ie^2g_{\mu\nu}$ but this is certainly wrong because the notation of the problem implies this $\Gamma$ depends on the momenta entering and exiting the vertex.

3. The attempt at a solution
My first issue is that I don't know what these $\Gamma$'s actually are. The problem says they are the vertices, and I thought this meant the value such vertex gives in a Feynman diagram. But we see that this isn't working properly here because the four-point vertex in diagrams gives a contribution that doesn't depend on momentum at all.

At first I thought I needed to consider the diagram with just the three external legs and the single vertex. But the Feynman rule for such diagram is trivial, it is just the value of the vertex times the polarization of the external photon. Plus it doesn't involve any scalar propagator.

My second guess was that I needed to use the s,t,u chanels diagrams because those have scalar propagators into them. But the issue is that if I calculate one such diagram, I'll end up with one equation involving the matrix element, which I have nothing else to eliminate. I thought on using Ward's identity, but I'm unsure.

Also when trying to build this up it got quite strange, because I've considered $\phi\gamma\to \phi\gamma$ and one internal particle. But since there is no antiparticle anywhere in my reasoning the definition I thought was true for $\Gamma^\mu$ doesn't help much because it will have zero in the second entry. So it is certainly wrong.

How can I get started with this?

2. Jun 9, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.