# Is there such a thing as red - anti red gluon? Why such gluon posssibilities appear in equations?

Gluon properties have puzzled me for quite a while now.

I recently learned that there are 8 kinds of gluons. They are kind of represented by 8 linearly independent SU3 matrices. Matrices represent charge vertically and anticharge horizontally, so if we never have such a thing as white charged gluon, trace should be zero. I got this far. But that what bothers me that there still are blue-antiblue and red-antired gluons. They are white! If there is no such a thing as white charged gluon, why do we still have possibility of having it? I realize that we can't get equal probabilities of white charged gluons. But still, there is probability of getting green-antigreen. What does it mean physically?

## 2 Answers

Suppose you have a quark of color state $\lvert q\rangle$. Or equivalently, you could write the color state as a 3-component vector, but for now I'm abbreviating it with a smaller symbol. Anyway, if that quark interacts with a gluon whose color matrix is $T_g$, the outgoing quark has a color state of $T_g\lvert q\rangle$.

A red-antired gluon would be represented by the following color matrix, in the RGB basis:

$$T_{r\bar{r}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

If you have a red quark, with color state

$$\lvert r\rangle = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$$

and it interacts with that gluon, the resulting quark will have a color state of

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \lvert r\rangle$$

so, still red. However, if you naively apply this argument to the red-antired gluon and a green (or blue) quark, you get zero. In practice that means red-antired gluons don't interact with green or blue quarks at all. (The quantum mechanical amplitude for the interaction to happen is related to $\langle q_\text{final}\rvert T_g\lvert q_\text{init}\rangle$.)

You can then extrapolate this to what would happen if you let a red-antired gluon interact with a quark in the state $\frac{1}{\sqrt{2}}(\lvert r\rangle + \lvert g\rangle)$:

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \propto \lvert r\rangle$$

So the gluon actually changes the color state of some quarks. A white gluon wouldn't do that. White gluons don't change color states; that's what it means to be white.

Now, a true red-antired gluon doesn't exist because the corresponding matrix has a nonzero trace. In a sense, the hypothetical red-antired gluon is part white, but the white part doesn't exist. However you can run through the preceding arguments with a gluon whose color matrix is proportional to

$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{pmatrix}$$

which is physically valid, and the conclusions are all the same. This shows you why this gluon, despite containing colors and anticolors in equal combinations, is not white.

Note that the technical term for what you think of as "white" is "$SU(3)$ singlet".

I'm being a little loose with the notation in this post, but hopefully it makes sense.

• Great post, but one tiny nitpick. It is more conventional to associate the $\frac{1}{\sqrt{2}}(r\bar r- g\bar g)$ gluon with the third Gell-Mann matrix $\lambda_3=\operatorname{diag}(1,-1,0)$ rather than $\operatorname{diag}(1,0,-1)$. – Ryan Unger Feb 5 '15 at 2:19
• @0celo7 in the last equation I was going for the $r\bar{r} - b\bar{b}$ gluon, specifically because the blue component doesn't interact with the green component of the quark state. So I think this is consistent with what you're saying? – David Z Feb 5 '15 at 3:15
• I don't think you can have the $r\bar r-b\bar b$ gluon in the RGB basis. Griffiths has the $r\bar r-b\bar b$ gluon, but he uses RBG basis. After all, the gluon vertex is $\propto i\lambda^a_{ij}\gamma^\mu$. So when we use the standard Gell-Mann matrices, $\lambda^3$ is $r\bar r-g\bar g$. – Ryan Unger Feb 5 '15 at 3:20
• @0celo7 sure you can. If nothing else, QCD is perfectly color symmetric so if you can have $r\bar{r} - g\bar{g}$, a $g\leftrightarrow b$ rotation shows that you can have $r\bar{r} - b\bar{b}$. And besides, the point of it being a basis is that you can express any state (in this case, even unphysical ones like $r\bar{r}$). – David Z Feb 5 '15 at 3:29
• @0celo7 no, not at all. The Gell-Mann matrices are just one possible basis for the space of color matrices. Any color matrix can be expressed as a linear combination of basis elements (e.g. Gell-Mann matrices), but there is nothing special about the basis elements themselves. – David Z Feb 5 '15 at 3:37

In the red-blue-green basis, the purported red-antired gluon: $$T_{r\bar{r}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ is a linear combination of $$T_{r\bar{r}} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} =\frac{2}{6}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \frac{3}{6}\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} + \frac{1}{6}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix}.$$ The astute readers may recognize the second and the third traceless matrices as the standard Gell-Mann matrices $$T_3 = T_{r\bar{r}-b\bar{b}}$$ and $$T_8 = T_{r\bar{r}+b\bar{b}-2g\bar{g}}$$, part of the $$SU(3)_C$$ 8-gluon gang.

What about the first (identity) matrix in the above linear combination? $$I= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ It has the characteristics:

• It is NOT traceless
• It commutes with all the traceless Gell-Mann matrices, thus it's NOT part of the Gell-Mannian 8-gluon irreducible representation.

Therefor, OP's red-antired gluon $$T_{r\bar{r}}$$ is a combination of two Gell-Mann matrices and something outside the Gell-Mannian universe.

Now the million dollar question: can $$T_{r\bar{r}}$$ be part of mother nature? According to the received wisdom, the answer is NO. The rational is that the identity matrix, which is portion of the decomposed $$T_{r\bar{r}}$$, is white (commutes with $$SU(3)_C$$) and thus it is NOT confined as the other 8 $$SU(3)_C$$ gluons. If the red-antired gluon $$T_{r\bar{r}}$$ existed in nature, it would have manifested as a long-ranged strong force, which is hitherto not observed.

And now the punchline: the red-antired gluon $$T_{r\bar{r}}$$ is actually materialized in the physical world as part of the electromagnetic force.

The full story:

The identity matrix $$I$$ mentioned earlier forms a $$U(1)$$ group with gauge transformation $$e^{\theta (B-L) i I},$$ where $$iI$$ is the gauge generator of the 9th "gluon". The charge of this 9th "gluon" is baryon number B (1/3 for quarks) minus lepton number L (1 for leptons). Hence the gauge group is more appropriately denoted as $$U(1)_{B−L}$$. It worth noting that $$U(1)_{B−L}*SU(3)_C = U(3)_C \subset SU(4),$$ where leptons are deemed as the 4th color in addition to the 3 colors of quarks.

In the context of beyond standard model, the hypercharge $$U(1)_Y$$ is replaced with the combined gauge group $$U(1)_{B−L}∗U(1)_R,$$ where the right-handed isospin $$U(1)_R$$ gauge (right-handed counterpart of $$T_3$$ component of weak $$SU(2)$$ on left-handed fermion doublets) is only applicable to the right-handed fermions.

The combined gauge group $$U(1)_{B−L}∗U(1)_R$$ is spontaneously broken at the GUT (Majonara mass) energy scale. There are two leftovers from the GUT-scale spontaneous symmetry breaking:

• One combination ($$Z′$$ boson) of the $$U(1)_{B−L}$$ and $$U(1)_R$$ gauge fields acquires GUT scale mass, thus short-ranged.
• The other combination of the $$U(1)_{B−L}$$ and $$U(1)_R$$ constitutes the hypercharge $$U(1)_Y$$ gauge field, which in turn combines with the $$W_0$$ component of the electroweak $$SU(2)$$ gauge field to form the electromagnetic gauge field via the electroweak Higgs mechanism. In other words, some relic of the "9th gluon" is long-ranged, manifested as being a portion of the electromagnetic force.

This color-neutral $$U(1)$$ 9th gluon is also mentioned in the answer of @Luboš Motl to a similar question:Do color-neutral gluons exist?