# Complicated Lagrangian - checking Feynman rules

I have the following example Lagrangian:

\begin{equation} \begin{split} \mathcal{L} &= \left[ \bar{\psi} (i \require{cancel}\cancel{D} -M)\psi - \frac14 F_{\mu \nu} F^{\mu \nu} \right] + \\ & +\sum_{i={1,2}}\bigg[ (D_\mu\phi_i) (D^\mu\phi_i)^\dagger -m_i^2 |\phi_i|^2 \bigg] + \\ & +\frac12 \bigg[ \chi^\dagger \gamma^0(i \cancel{\partial}-m_\chi)\chi \bigg] + \\ & +\sqrt{2} \lambda \bigg[ \phi_1\chi^\dagger\gamma^0 P_L \psi - \phi_2\chi^\dagger\gamma^0 P_R \psi +\mbox{h.c.}\bigg] \end{split} \end{equation}

With $D_\mu=\partial_\mu-i\lambda A_\mu$, $P_{R/L} = (1 \pm \gamma^5)/2$. The field $\psi$ is a fermion, the $\phi_i$ are complex scalar fields and $\chi$ is a Majorana fermion. I want to write down the Feynman rules.

Now, the first bracket is just a regular QED Lagrangian, so it includes the $\psi$ and photon propagators and the usual interaction vertex. Similarly, the second bracket contains the two complex scalar fielda coupled to the electromagnetic field, so basically it's just two Scalar QED Lagrangians.

I'm not sure about the third bracket, because I don't know how to treat Majorana fermions at all. It's the kinetic term, and the only difference is the presence of $\gamma^0$, but I'm not sure how that should would appear in the propagator? Does it just get multiplied, like:

$$\gamma^0 \frac{i}{\cancel{p}-m_\chi + i\epsilon} \quad ?$$

If so, how do I know which side to put the gamma matrix on?

As for the fourth bracket, the interactions are between one of the complex scalar fields, the Majorana spinor and a Dirac spinor. I don't know how to include the projection and gamma into this. Are the vertices just gonna be:

$$\sqrt{2} \lambda \gamma^0 P_{R/L} \quad ?$$

What will the vertices look like?

In general, my trouble is figuring out how to incorporate objects like the gamma matrices or projection operators into the appropriate Feynman rules for strange interactions like this.

About the Majorana spinors, think of them as 'real' spinors, while Dirac spinors are 'complex'. For a real scalar field, $\phi = \phi^*$, while for a complex scalar field $\phi$ and $\phi^*$ are independent degrees of freedom. Similarly, for a Dirac spinor $\psi$ and $\psi^*$ are independent degrees of freedom, while for a Majorana spinor $\psi^* = \psi$. The anti-commutation relations for the gamma matrices will be the same as before.

To find the Green's function or propagator of the Majorana spinor $\chi$, you have to - identical to the case for a Dirac spinor - find the inverse of this operator $$\bigg[ \gamma^0(i \cancel{\partial}-m_\chi) \bigg]G(x-y) = i\delta(x-y)$$ which will give you $$G(x-y) = \int \frac{d^4 p}{(2\pi)^4} \frac{i(\cancel{p} + m_\chi)\gamma^0}{p^2 - m^2 + i\epsilon} e^{-ip\cdot(x-y)}$$ Note that $\gamma^0\cancel{p}\gamma^0\cancel{k} = \gamma^0 \gamma^\mu \partial_\mu \gamma^0 \gamma^\nu \partial_\nu = - (\gamma^0)^2 \gamma^\mu \partial_\mu \gamma^\nu \partial_\nu$. You therefore have to put the $\gamma^0$ in the above propagator on the right, to ensure the minus sign will cancel (as every time you pull the $\gamma^0$ 'through' a $\cancel{p}$ it obtains a minus sign).

Because the Majorana spinor is 'real', $\chi$ and $\chi^*$ are the same. $\psi$ is a Dirac spinor and therefore has different chiral components (that are obtained from the projection operators $P_L$ and $P_R$). The interaction terms tell you that this Majorana spinor $\chi$ interacts with the difference chiral parts (the left and right handed components) of the Dirac spinor $\psi$ in different ways $$\sqrt{2} \lambda \chi^\dagger\gamma^0 \bigg[ \phi_1 P_L - \phi_2 P_R \bigg] \psi+\mbox{h.c.}$$ If you want to do Feynman diagrams this means that we will have two distinguish everywhere between the diagrams for $\phi_1$ and $\phi_2$, and $P_L \psi$ and $P_R \psi$, since these different components of the fields will interact in different ways.

For the first interaction term (involving $\chi^\dagger$, $\phi_1$ and $P_L \psi$) the interaction term will indeed be what you said it would be: $\sqrt{2}\lambda\gamma^0$.

• I'm not sure how to write out the h.c. part. Is the projection operator going to switch from R to L? I want to write out the vertices for $\chi, \phi_1, \psi$. Also, since $\chi$ is supposed to be a 'real' fermion, is $\chi ^\dagger = \chi$? – Spine Feast Jun 11 '17 at 18:58
• So for example, what will be the difference between the vertices for $(\bar{\psi}, \chi, \phi_1)$ and $(\psi, \chi^\dagger, \phi_1^\dagger)$ ? $P_L$ anticommutes with $\gamma^0$, so I'll get a $P_R$ out of that right? – Spine Feast Jun 11 '17 at 19:05
• Remember the projection operators can be written as $P_{R,L} = (1 \pm \gamma_5)$ and $\gamma_5^\dagger = \gamma_5$. – JgL Jun 11 '17 at 19:17
• right, so when I conjugate $\gamma^0 P_L$ I'll get $P_L \gamma^0 = \gamma^0 P_R$ since $\{ \gamma^0 , \gamma^5 \} = 1$ – Spine Feast Jun 11 '17 at 19:21
• Yes. (In my previous comment it should say $P_{R,L} = \frac 12 (1 \pm \gamma_5)$, but it's not very relevant) Indeed, $(\gamma^0 P_L)^\dagger = \gamma^0 P_R$ as you point out. – JgL Jun 11 '17 at 19:24