# Color-charge conservation in proton decay

In some extensions of the Standard Model of particle physics (Supersymmetry with R-parity violation being a prominent example), the proton is allowed to decay, e.g. via $p\to e^+\pi^0$: While this decay is obviously violating the conservation of baryon and lepton number, I take it that color charge should still be conserved. My problem is that I don't see how this would work:

We have color-neutral objects in initial (the proton) and final state (the pion), but when I try to keep track of the color flow this fails, basically because I have to get rid of the third color charge somehow. Is the simple picture of assigning each (anti-)quark an (anti-)color too much of a simplification to be applied here? What would be the color assignment of the intermediate particle $X$ in the proton decay?

• well, the proton is color neutral, and d dbar cancel in color so the output is color neutral. What is the problem? Are you saying what are the color combinations of the proton to make it neutral and the color combinations of d dbar? – anna v Jul 9 '14 at 7:14
• – anna v Jul 9 '14 at 7:27
• @annav I think he is specifically trying to track down the colour charge flow along the diagram. Fuenfundachtzig, have you considered the mediator particle's colour charge? – PhotonBoom Jul 9 '14 at 7:30
• @PhotonicBoom: This is part of my question: What is the mediator particle's colour charge? It seems to have two colors from its production, but one anti-color at its annihilation vertex? – fuenfundachtzig Jul 9 '14 at 7:42
• @annav: Thanks for the link, but this is not about gluons? What are you hinting at? – fuenfundachtzig Jul 9 '14 at 7:43

## 3 Answers

It works if you assign colors like this: one red up, one green up, down is blue, $X$ takes red and green which are equivalent to antiblue ("yellow"), thus color is conserved. I didn't take into account the last fact which explains my confusion.

• I am happy to accept better answers which explain the same on more solid mathematical grounds. – fuenfundachtzig Jul 9 '14 at 8:52

An other way to see the argument of the answer of @fuenfundachtzig , is that, concerning $SU(3)$ representations, there is an equivalence between $(3*3)_\text{antisymmetrised}$ representation ("red * green") and $3^*$ representation ("antiblue"). Why ? Well, thanks to the completely anti-symmetric Levi_Civita symbol. Using objects upon which act the representations, you could write $\psi^{rg} = \epsilon ^{rgd} \bar \psi_{d}$. This means that the representations $(3*3)_\text{antisymmetrised}$ and $3^*$, which have both dimension $3$, correspond to the same structure.

Realize that the independent conservation of 3 colors r,g,b that we are familiar with from QCD, i.e., the possibility of drawing continuous red, green and blue lines throughout the Feynman diagrams, is a consequence of both $SU(3)_c$ gauge symmetry and $U(1)_B$ baryon number global symmetry at the same time. For a more detailed explanation, look, e.g. here.

As you have written, in the BSM theory under your consideration the baryon number is not conserved, and hence drawing lines of 3 independent colors does not make a good sense.

Instead on counting numbers of red, green and blue fields in the initial and final state and, you can only look at the mutually commuting generators of $SU(3)$, which are, e.g., $\lambda_3$ and $\lambda_8$. It can be easily seen that both proton and pion are simultaneous eigenstates of both $\lambda_3$ and $\lambda_8$ with both eigenvalues $0$ which is the precise meaning of being colorless. In this sense, color is conserved in the process on the picture.