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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 $p\to e^+\pi^0$:

Feynman diagram of proton decay in a GUT

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?

fuenfundachtzig
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3 Answers3

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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.

fuenfundachtzig
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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.

rob
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Trimok
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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.

Community
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Hoody
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