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.

  • $\begingroup$ In principle, you're right -- such a term is of the form $\left(\text{barred spinor}\right)\left(\text{spinor}\right)\, \left(\text{scalar}\right)$. How that is written out may depend on subtleties such as Dirac/Majorana/Wey spinors, conventions, or whether the author even bothered (they might have wanted to focus on the colour structure and proto decay). Maybe you can show where you found this specific expression, and then one could say more about it. $\endgroup$ – Toffomat Nov 1 '18 at 12:10
  • $\begingroup$ I am looking for example at the lower right box of Tab. 3 of 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 $\endgroup$ – d8586 Nov 6 '18 at 8:17

Ah: The $C$ is key. The spinor $d^TC$ is the charge-conugate spinor (often denoted $d^c$), which behaves like a barred spinor as far as Lorentz transformations are concerned.

You can look up charge conjugation, Majorana spinors and similar things in most QFT textbook, e.g. in Sohnius' introduction to SUSY (the Appendix): https://iktp.tu-dresden.de/Lehre/SS2009/SUSY/literatur/sohnius_article.pdf

(Note that are several conventions, so you have to make sure who uses what.)

  • $\begingroup$ Hi, I need digest that a little bitt! It seems that this explains why $d^T C u$ is a Lorentz scalar, but I'm not sure how an operator such as $(u^c)^T d^c$ can then be (this operators are for example listed here en.wikipedia.org/wiki/…) I need to work it out explicitly I guess, it seems I am missing something $\endgroup$ – d8586 Nov 6 '18 at 12:54
  • $\begingroup$ It seems that the operators on the wikipedia site you linked are somewhat schematic and leave out the Lorentz structure. (So, they are not Lorentz invariant the way they're written.) $\endgroup$ – Toffomat Nov 6 '18 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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