# (Basic) confusion about value of scalar-fermion vertex $\phi\psi_i\psi_j$ for Majorana fermions

If I have a toy model with $$N$$ Majorana fermions $$\lbrace\psi_i\rbrace_{i=1,\ldots,N}$$ and a scalar field $$\phi$$ where the interaction among the fields is

$$\mathcal L_\text{int}= \sum_{i,j=1}^N a_{ij}\phi\bar\psi_i\psi_j$$

For example for $$N=2$$ it would be

$$\mathcal L_\text{int}= a_{11}\phi \bar\psi_1\psi_1 +a_{12}\phi \bar\psi_1\psi_2 +a_{21}\phi \bar\psi_2\psi_1 +a_{22}\phi \bar\psi_2\psi_2$$ But for Majorana fermions one can demonstrate that $$\bar\psi_i\psi_j=\bar\psi_j\psi_i$$, so the term including the interaction $$\phi\psi_1\psi_2$$ is $$\mathcal L_\text{12}= (a_{12}+a_{21})\phi \bar\psi_1\psi_2$$ I would think the respective vertices of the different interactions are $$\Gamma(\phi\psi_1\psi_1)=a_{11}$$ $$\Gamma(\phi\psi_1\psi_2)=a_{12}+a_{21}$$ $$\Gamma(\phi\psi_2\psi_2)=a_{22}$$ But what I generally see is the vertex being $$\Gamma(\phi\psi_i\psi_j)=a_{ij}$$ Which is different to my conclusion for $$i\neq j$$. For example, I see this in Complete set of Feynman rules for the MSSM in the interaction neutral Higgs-Neutralinos ($$H^0\chi^0_i\chi^0_j$$).

Whether the way I try to see the value of the vertex is wrong or it is wrong the reference about the vertex, I don't know which it is.

I was wrong about what I initially thought, the vertex is in fact

$$\Gamma(\phi\psi_i\psi_j)=a_{ij}+a_{ji}$$

for all $$i,j$$.

The way I saw it is with an example, consider the process $$\phi+\psi_a\to\psi_b$$. The way to deduce the vertex is calculating

$$\Gamma(\phi\psi_a\psi_b)\to\sum_{i,j}\langle0| b_b \text{:}(a_{ij}\phi\bar\psi_i\psi_j)\text{:} b_a^\dagger|0\rangle$$

where $$\text{:}(\cdots)\text{:}$$ is normal ordering, $$b_i^\dagger$$ and $$b_i$$ are the creation and anhilition operators of the fermions respectively (ignoring dependencies like momentum or spin), they have anticommutation properties

$$\lbrace b_i,b_j \rbrace= \lbrace b_i^\dagger,b_j^\dagger \rbrace=0 ,\quad \lbrace b_i,b_j^\dagger \rbrace\sim\delta_{ij}$$

For the sake of the argument we are going to ignore the integral, sum of spins and constants in the decomposition of the fermion fields in $$b,b^\dagger$$

$$\psi_i\sim b_iu_i+b_i^\dagger v_i$$ $$\bar\psi_i\sim b_i\bar v_i+b_i^\dagger\bar u_i$$ where $$u,v$$ are 4-component spinors representing the particle and antiparticle state. Also we are going to ignore the $$\phi$$ field, then we have the vertex is

$$\Gamma(\phi\psi_a\psi_b) \to\sum_{i,j} a_{ij} \langle0| b_b \text{:}(b_i\bar v_i +b_i^\dagger\bar u_i ) (b_j u_j +b_j^\dagger v_j ) \text{:} b_a^\dagger|0\rangle$$

Using $$b|0\rangle=0$$, $$\langle0|b^\dagger=0$$ and the anticommutation relations it is straightforward to see that $$\Gamma(\phi\psi_a\psi_b) \to\sum_{i,j} a_{ij} \langle0| b_b \text{:} (b_i^\dagger b_j\bar u_i u_j +b_ib_j^\dagger\bar v_i v_j) \text{:} b_a^\dagger|0\rangle$$

Applying normal ordering

$$\Gamma(\phi\psi_a\psi_b) \to\sum_{i,j} a_{ij} \langle0| b_b (b_i^\dagger b_j\bar u_i u_j -b_j^\dagger b_i\bar v_i v_j) b_a^\dagger|0\rangle$$

Applying the anticonmutation rules and $$\langle 0|0\rangle=1$$

$$\Gamma(\phi\psi_a\psi_b) \to\sum_{i,j} a_{ij} ( \delta_{aj}\delta_{bi}\bar u_i u_j -\delta_{ai}\delta_{bj}\bar v_i v_j )$$

$$\Gamma(\phi\psi_a\psi_b) \to a_{ba}\bar u_b u_a -a_{ab}\bar v_a v_b$$

Using certain properties one can prove, at least in certain representations, that $$\bar v_av_b=-\bar u_bu_a$$ and then

$$\Gamma(\phi\psi_a\psi_b) \to (a_{ab}+a_{ba})\bar u_b u_a$$

For simplicity one can define $$a_{ij}$$ such that $$a_{ij}=a_{ji}$$, that implies directly that

$$\Gamma(\phi\psi_a\psi_b) = 2a_{ab}$$

We see that the result is the same, regardless $$a=b$$ or $$a\neq b$$, also there's a number 2 present, which can be avoided redefining $$a\to\frac12a$$. I based this partially on the appendix D of The search for supersymmetry: Probing physics beyond the standard model, where they explain things about the vertices in Majorana fermions.