0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.