Finding Feynman Rules for Reduced/Pseudo-QED I am trying to guess/compute the Feynman rules for the following theory.
$$ \mathcal{L} = \delta(x^d)\bar{\psi}\iota\gamma^\mu D_\mu \psi(x)-\frac{1}{4}F^{\hat{\mu}\hat{\nu}}F_{\hat{\mu}\hat{\nu}}(x) -\frac{1}{2}(\partial^{\hat{\mu}}A_\hat{\mu})^2(x) \, , $$
where for simplicity we assume that the electron is massless. The hatted Lorentz indices run from $0$ to $d,$ whereas unhatted ones run from $0$ to $d-1.$ That is to say, the electrons are confined in a hypersheet, but the photons are free to move in any direction.
Please let me know how I should go about solving it in $d+1$ spacetime dimensions.
 A: Let's write the action down and investigate each term individually.
$$ S = \int d^{d+1}x \Big[ \delta(x^d)\bar{\psi}\iota\gamma^\mu (\partial_\mu + ieA_\mu) \psi(x)-\frac{1}{4}F^{\hat{\mu}\hat{\nu}}F_{\hat{\mu}\hat{\nu}}(x) -\frac{1}{2}(\partial^{\hat{\mu}}A_\hat{\mu})^2(x)  \Big] $$
The first term is the kinetic term for the Fermionic field (massless electron) and therefore defines the propagator. First, we integrate out the delta function.
$$ \newcommand{\fsl}[1]{#1\kern-0.4em\raise0.22ex\hbox{/}} \int d^{d+1}x \ \delta(x^d)\bar{\psi}\iota \fsl{\partial} \psi(x) = \int d^dx \ \bar{\psi}\iota \fsl{\partial} \psi(x)$$
The Green's function is defined using
$$ \iota \fsl{\partial} S_F (x-y) = \iota \delta^{(d)}(x-y) \, ,$$
which in momentum space implies
$$ \fsl{k} \tilde{S}_F(k)=\iota$$
$$ \Rightarrow \boxed{\tilde{S}_F(k) = \frac{\iota}{\fsl{k}}}$$
Next term in the action gives the interaction between light and electrons. We will come to that shortly. The last two terms constitute the propagator for the photons. As usual, rewrite the electromagnetic field strengths in terms of the gauge field and then find the corresponding Green's function in momentum space.
\begin{align*}
\int  d^{d+1}x \ \Big[ -\frac{1}{4}F^{\hat{\mu}\hat{\nu}}F_{\hat{\mu}\hat{\nu}}-\frac{1}{2}(\partial^{\hat{\mu}}A_\hat{\mu})^2 \Big] &= \int  d^{d+1}x \ \Big[ -\frac{1}{2} A^\hat{\mu}(-\eta_{\hat{\mu} \hat{\nu}}\partial^2 + \partial_\hat{\mu}\partial_\hat{\nu}) A^\hat{\nu} + \frac{1}{2} A^\hat{\mu}\partial_\hat{\mu}\partial_\hat{\nu} A^\hat{\nu} \Big] \\
&= \int  d^{d+1}x \ \Big[  -\frac{1}{2} A^\hat{\mu} (-\eta_{\hat{\mu} \hat{\nu}}) \partial^2  A^\hat{\nu} \Big] \\
\end{align*}
Therefore, we get the Green's function as follows.
$$ \eta_{\hat{\mu} \hat{\lambda}} \partial^2 {D_F}^{\hat{\lambda}\hat{\nu}}(x-y) = \iota \ {\delta_\hat{\mu}}^\hat{\nu} \delta^{(d)}(x-y) $$
$$ \Rightarrow -k^2 \eta_{\hat{\mu} \hat{\lambda}}
 \ {\tilde{D}_F}^{\hat{\lambda}\hat{\nu}}(k) = \iota \ {\delta_\hat{\mu}}^\hat{\nu} $$
$$ \Rightarrow \boxed{ {\tilde{D}_F}^{\hat{\lambda}\hat{\nu}}(k) = \frac{-\eta_{\hat{\lambda}\hat{\nu}} } {k^2} } \, ,$$
where $k^2 = k_\hat{\mu} k^\hat{\mu}$.
Now, investigate the final term in the action. Note that the interaction happens with the photon field only in the d-brane (the polarisation vector of an interacting photon lies completely in the $d$-brane). We can safely integrate out the delta function.
$$ \int d^{d+1}x \ \delta(x^d)\bar{\psi}\iota \gamma^\mu (\iota e A_\mu) \psi = \int d^dx \ \bar{\psi}(-e) \fsl{A} \psi $$
Recall that the correlation function has an $e^{iS}$ inside the time-ordering which we expand perturbatively before carrying out the Wick contractions (alternatively, the generating integral is a functional integral of the above exponential). Therefore, the interaction vertex would contribute a term proportional to $ -\iota e \gamma^\mu \int d^dx $ which in momentum space would give us the following expression.

As a final remark, note that since the photon polarization aligned along the $x^d$-axis decouples from the $d$-brane interactions, we can safely ignore the $\hat{\mu}=d$ terms from the numerator of the photon propagator.
Please let me know if I made any conceptual mistakes.
