Graduate Lorentz invariance from Dirac spinor

[d8586]

I have a really naive question that I didn't manage to explain to myself. If I consider SUSY theory without R-parity conservation there exist an operator that mediates proton decay. This operator is

$$u^c d^c \tilde d^c $$

where $\tilde d$ is the scalar superpartner of down quark. Now, being a scalar, this field doesn't transform under Lorentz transformation. This means that the term $u^c d^c$ is Lorentz invariant. Being u and d 4-component Dirac spinor this has to be read as

$$(u^c)^T d^c$$

in order to proper contract rows and columns.

This means that also $u^T d$ should be Lorentz invariant...

However, Lorentz invariant are build with bar spinors, i.e.

$\bar \psi \psi$ is Lorentz invariant

while I don't see how

$\psi^T \psi$ can be Lorentz invariant.. Clearly I am missing something really basic here..

 

[haushofer]

If you want people to help you, I think it's a good idea to share some references. Naively, I don't see how you term can be Lorentz-invariant either, but without reference or context that's all I can say.

 

[d8586]

You are right, sorry. I am looking for example at the lower right box of Tab. 3 of https://arxiv.org/pdf/1008.4884.pdf
In the B-violating operator I have terms that goes (neglecting colour and SU(2) indices, and noting that Dirac indices are always contracted within the brakets) like

$$d^T C u$$

and similar. This is what I meant