I was wondering how it is possible to see from the $SU(3)$ Gauge Theory alone that Gluons carry two charges colors: $g\overline{b}$ etc.

Some background:

The W-Bosons (pre-symmetry breaking) form an $SU(2)$ triplet and carry the corresponding weak Isospin $1,0-1$. After SSB/Higgs the charged $W^\pm$-Bosons can be identified with complex linear combinations of the $W^{1,2}$, bosons, and therefore the corresponding term in the Lagrangian is $U(1)$ invariant, i.e. the $W^\pm$ carry electric charge, too.

For a local $SU(3)$ gauge theory 8 gauge fields, the gluon fields are needed. Exactly as it was the case for $SU(2)$, one for each generator $\lambda_a$ and one introduces consequently "matrix gauge fields"

$$ A^\mu = A^\mu_a \lambda_a$$

which can be seen as elements of the corresponding Lie algebra, because the $\lambda_a$ form a basis and the expression above can be seen as an expansion of $A^\mu$ in terms of this basis.

The transformation behaviour is the same for all $SU(N)$ theories

$$ A^\mu \rightarrow U A^\mu U^\dagger + \frac{i}{g} (\partial_\mu U) U^{-1} $$

As usual the fermions transform according to the fundamental representation, i.e. for $SU(3)$ are arranged in triplets. Each row represents a different color as explained in the answer here (What IS Color Charge? which recites from Griffith)

Therefore a red fermion, for example is

$$ c_{red} = \left(\begin{array}{c} f \\0\\0\end{array}\right) $$

where $f$ is the usual dirac spinor. An anti-red fermion would be

$$ c_{red} = \left(\begin{array}{c} \bar f 0 0\end{array}\right) $$

The red fermion transforms according to the fundamental rep $F$, the anti-red fermion according to the conjugated rep $F^\star$. Which is a difference to $SU(2)$, because $SU(2)$ has only real representations and therefore the normal and anti rep are equivalent (why is it enough, that they are equivalent? The conjugate rep for $SU(2)$ is different but considered equivalent because $r = U \bar r U^{-1}$, for some unitary matrix $U$. Any thoughts on this would be great, too), i.e. there is no anti-isospin. I guess this is the reason the $W$ do not carry anti-charge, simply because there isn't anti charge for $SU(2)$.

Now where is the point that we can see that the gluons carry anti-colorcharge and colorcharge? Is it because the matrix gluon fields defined above are part of the Lie algebra and transform therefore according to the adjoint rep of the group $A \rightarrow g A g^\dagger$, which could be seen as transforming according to the rep and anti-rep at the same time (or could be seen as completely non-sense idea from me ;) ) ?

Why does the gluon octed does not get charge assigned like the $SU(2)$ triplet, which would mean the gluons carry different values of one strong charge ? (Analogous to $1,0-1$ for weak isospin of the $W$ triplet.)

Any thoughts or ideas would be awesome!

  • 2
    $\begingroup$ It's charge-anticharge because the gluon is in the adjoint representation of a non-Abelian group - that's why there are nonzero charges. The adj rep isn't trivial (singlets), so it transforms under itself. The adj rep is a "matrix", and the entries of the matrix are specified by $ij$, the row and column. Matrix $U_{ij}$ is multiplied by a vector $v_i$ on the right side, so if $v_i$ is a quark, the $j$ in $U$ contracts with i.e. annihilates the $j$-th color quark, i.e. carries the charge of the $j$-th antiquark, but $U_{ij}$ creates the $i$-th quark's charge instead. $\endgroup$ Commented Oct 9, 2014 at 6:58
  • $\begingroup$ Thanks for your quick reply! I don't understand the difference to the $SU(2)$ case. The $W^\mu=W^\mu_a \sigma_a$ as matrix fields transform according to the adjoint rep, too. Wouldn't the same thought lead to the conclusion that the W-bosons carry two charges, for example $+1$ and $-1$ or so. As stated above the difference to $SU(3)$ is that the conjugate rep is really different for $SU(3)$, therefore anti-color exists. Nevertheless, I can't see how this leads to just one charge carried by the $W$ and two by the gluons. $\endgroup$
    – jak
    Commented Oct 9, 2014 at 10:00
  • $\begingroup$ Dear Jakobh, there is no real difference between $SU(2)$ and $SU(3)$. You write the elements of the $SU(2)$ algebra as a "vector", a combination of Pauli matrices, but this "vector" is really a composite, not-the-smallest, representation. The smallest representation of $SU(2)$ is the 2-component spinor (the "true vector" of $SU(2)$), and the 3-dimensional representation is built from two copies of the 2-component spinors by the same way as the 8-dimensional adjoint of $SU(3)$ is built from the 3-dimensional fundamental rep. $\endgroup$ Commented Oct 9, 2014 at 10:05
  • $\begingroup$ Take the polarizations $J_z$ of the $SU(2)=SO(3)$ generators. The 3-component vector has eigenvalues $-1,0,+1$ - that's for the combinations $L_x\pm i L_y$ and $L_z$ (the latter is the zero), respectively. But the smallest nontrivial representation has $J_z=\pm 1/2$ (this 2-colored "quark" is equivalent to its complex conjugate in this case). So you need to combine two of those to get $J_z=\pm 1$, so the W-bosons and the Z-boson also carry charges that may be obtained from the doublets (e.g. electron+neutrino). $\endgroup$ Commented Oct 9, 2014 at 10:07
  • $\begingroup$ I thought I understand, then I noticed things may be a bit more complicated: We have (left-handed) fermions as $SU(2)$ doublets $\Psi_L$, transforming according to the fundamental rep: $ e^{i a_i(x) \sigma_i /2} \Psi_L$ The objects in this doublet carry charge=isospin $I=\frac{1}{2}$. And $I_3$, the cartan generator, can be used to label them: The eigenvalues are $\pm \frac{1}{2}$ and we can speak of isospin $\pm \frac{1}{2}$ for the electron and the electron-neutrino etc.. $\endgroup$
    – jak
    Commented Oct 9, 2014 at 11:17

1 Answer 1


After reading through the corresponding chapters in several books I think I'm now able to give a "semi-satisfactory" answer to my own question (and to understand Lubos first comment ;) ).

I write semi-satisfactory, because I hope someone with a deeper understanding of these topics will give a better answer. My explanation is still a little bit heuristic, and I would love to see a more mathematical derivation of this curious fact of nature.

This answer is rather long, but figuring out these things took me quite some time, because I wasn't able to find a proper treatment of this topic. I'm almost certain that such a treatment exists somewhere, but after 20 books or so in libary of my university I just gave up.

Nevertheless, maybe this helps someone with similar problems.

As for Isospin $SU(2)$, we label the fields using the eigenvalues of the Cartan generators, which are the diagonal generators of the group. For $SU(2)$ there is just one, $I_3= \frac{\sigma_3}{2} $, with eigenvalues $\pm \frac{1}{2}$. The eigenvectors form a basis for the vector space of the fundamental representation and consequently we can write the fermion fields (which transform according to the fundamental rep) usign this basis and assign them charges, which correspond to the eigenvalues. Therefore we have $ \begin{pmatrix} v_e \\ e \end{pmatrix}$ and are able to assign the neutrino field $ \begin{pmatrix} v_e \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} v_e $, where $v_e$ is the usual spinor the Isospin charge $\frac{1}{2}$, because $$I_3 \begin{pmatrix} 1 \\ 0 \end{pmatrix} v_e = \frac{\sigma_3}{2} \begin{pmatrix} 1 \\ 0 \end{pmatrix} v_e = \frac{1}{2} \begin{pmatrix} 1 \\ 0 \end{pmatrix} v_e $$

Equally $- \frac{1}{2}$, for the electron field $\begin{pmatrix} 0 \\ 1 \end{pmatrix} e $

For $SU(3)$ things are a little bit more complicated, because we have two cartan generators $\frac{1}{2} \lambda_3$ and $\frac{1}{2} \lambda_8$ (With the usual Gell-Mann matrices $\lambda_i$). Consequently each field is labelled by two numbers.

The eigenvalues of $ \frac{1}{2} \lambda^3 = \frac{1}{2} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right) $ are $\pm \frac{1}{2}, 0$.

For $ \lambda^8 = \frac{1}{2 \sqrt{3}} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{array} \right)$ the eigenvalues are $\frac{1}{2 \sqrt{3}}, \frac{1}{2 \sqrt{3}} , \frac{-1}{\sqrt{3}}$

Therefore if we arrange the strong interacting fermions into triplets, according to the basis spanned by the eigenvectors of the cartan generators, we can assign them the following labels:

$$ (\frac{1}{2} , \frac{1}{2 \sqrt{3}}) \mathrm{ \ for \ } \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} f $$

where one usually defines red$:=(\frac{1}{2} , \frac{1}{2 \sqrt{3}})$

Analogous $$ (-\frac{1}{2} , \frac{1}{2 \sqrt{3}}) \mathrm{ \ for \ } \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} f $$

therefore blue$:=(\frac{-1}{2} , \frac{1}{2 \sqrt{3}})$ and equally green$:=(0 , \frac{-1}{ \sqrt{3}})$. The color idea comes from the fact that if we add the three colors, i.e.

$$ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} f $$, we get a state with charge zero (a colorless state), because

$$ \lambda_3 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0 $$ and $$ \lambda_8 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0$$

which is analogous to the sunlight, which contains all colors of light, but is colorless, nonetheless.

In contrast to $SU(2)$, for $SU(3)$ the representations are not real (the conjugate representation is not equivalent to the ordinary representation) and therefore we can speak of anti charge, here anti-color. The corresponding states are for example

$$ \begin{pmatrix} 1 & 0 &0 \end{pmatrix} f $$ for anti-red, with eigenvalue anti-red$:=(\frac{-1}{2} , \frac{-1}{2 \sqrt{3}})$. Where the minus sign comes from complex conjugation of the generators and can be a bit confusing because physicists and mathematicians define generators differently. Nevertheless, we may not forget that there is an $i$ in the exponent, which physicists factor out in order to work with hermitean generators (to get real eigenvalues, because the generators are identified with the Noether charges).

Now the gluons. The gluons are the gauge-bosons of $SU(3)$ and the covariant derivative reads $$ D_\mu = \partial_\mu - i g A_\mu $$, with $A_\mu = A_\mu^a \frac{ \lambda^a }{2}$. This way, the gluons are matrices, written in terms of the generators, which is necessary in order to make the Lagrangian local $SU(3)$ invariant. Therefore, the Gluons transform according to the adjoint representation. (The adjoint representation is the representation of the group on its own tangent space at the identity = the Lie algebra. $A_\mu = A_\mu^a \frac{ \lambda^a }{2}$ is a expansion in terms of the basis of the Lie algebra = the generators.)

What happens if a gluon field acts on a quark field? For example let the first gluon field act on a red quark,

$$ A_\mu^1 \frac{ \lambda^1 }{2} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} q = A_\mu^1 \frac{ 1 }{2} \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} q = A_\mu^1 \frac{ 1 }{2} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} q $$

Therefore, the gluon field transformed the red quark into a blue quark. From Noether's therorem we know that color is conserved, and we can conclude that the first gluon $A_\mu^1 \frac{ \lambda^1 }{2} $ must carry the color-charge anti-red|blue

  • 1
    $\begingroup$ Just a correction. The 2-dimensional rep of $SU(2)$ is not real. You need complex numbers but it is equivalent to its complex conjugate - we say that it is pseudoreal or quaternionic. $\endgroup$ Commented Oct 11, 2014 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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