- #1
d8586
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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..
$$u^c d^c \tilde d^c $$
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..