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.


1 Answer 1


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.


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