I Feynman diagram for scalar - vector interaction

[gerald V]

TL;DR

How does the respective diagram look? How can the current conservation be seen?

The term for the electromagnetic interaction of a Fermion is $g \bar{\Psi} \gamma_\mu \Psi A^\mu$, where $g$ is a dimensionless coupling constant, $\Psi$ is the wave function of the Fermion, $\gamma$ are the gamma matrices and $A$ is the electromagnetic field. One can quite simply see that in lowest order, one Fermion line goes in, one Fermion line goes out, and one electromagnetic line goes out or in.

In contrast, for charged scalars with complex wave function $\varphi$, the interaction term reads $g \varphi^* A^\mu \partial_\mu \varphi - cc.$. How does the interaction diagram look in this case, in particular the part encoding $\partial_\mu \varphi$?

The electromagnetic current produced by the scalar field is conserved. How can this be seen from the diagram?I apologize if this is a stupid question. Thank you in advance.

 

[Paul Colby]

Complete nonexpert here but I would guess the $\partial_\mu$ contributes a factor of $ik_\mu$ in the integral.

 

[vanhees71]

In the usual fermionic QED you have only one tree-level vertex with one incoming and one outgoing fermion line, where the arrows indicate charge flow and a wiggly photon line carrying a Lorentz index $\mu$. The vertex reads $-\mathrm{i} \gamma^{\mu}$ (in the $\gamma$ matrix you have of course also two Dirac-spinor indices, but they are usually not written out but you use the bispinor formalism with four-component bi-spinors $u$ and $v$ and the Dirac matrices as building blocks).

The current conservation holds for "on-shell" diagrams only (as in classical field theory, the Noether currents obey the continuity equation for the solution of the field equations only). The current is $\propto \bar{\psi} \gamma^{\mu} \psi$. In the Feynman diagrams you usually work in energy-momentum space. The external fermion legs in a diagram have, e.g., the meaning of $u(p_1,\sigma_1)$ (leg with arrow pointing towards the vertex, meaning an incoming particle) and an $\bar{u}(p_2,\sigma_2)$ (leg with arrow pointing away from the vertex, meaning an outgoing particle). The spinors fullfil the "on-shell conditions" $(p_{\mu} \gamma^{\mu}-m) u(p,\sigma)=0$ and $\bar{u}(p,\sigma) (p_{\mu} \gamma^{\mu}-m)$. At each vertex energy-momentum conservation holds, i.e., the photon momentum is $k=p_1-p_2$. Current conservation thus is reflected for this example by
$$k_{\mu} \bar{u}(p_2,\sigma_2)\gamma^{\mu} u(p_1,\sigma_2)=\bar{u}(p_2,\sigma_2)\gamma^{\mu} (p_{1\mu}-p_{2 \mu}) u(p_1,\sigma_2)=\bar{u}(p_2,\sigma_2) (m-m)u(p_1,\sigma_2)=0.$$
Note that it is important that both spinors belong to the same particle, i.e., $m$ is the same for both particles in the vertex.

In "scalar QED" it is important to note that you have two vertices at the tree level. One looking as before, i.e., an incoming and an outgoing scalar-boson leg and a photon line but also another one with an incoming and an outgoing scalar-boson leg and two photon lines. The first one has the meaning $-\mathrm{i} g (p_1^{\mu} + p_2^{\mu})$ (in this notation the photon line carries the momentum $p_1^{\mu}-p_2^{\mu}$, the 2nd one $\mathrm{i} g^2 g^{\mu \nu}$. To see the current conservation again you only need the first diagram, and there the current conservation for on-shell external legs, implying $p_1^2=p_2^2=m^2$ the current conservation comes simply from $(p_1-p_2) \cdot (p_1+p_2)=p_1^2-p_2^2=m^2-m^2-0$.