Timeline for Group theoretical reason that Gluons carry color-charge and anti-colorcharge
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| Jun 1, 2022 at 19:31 | history | edited | Frederic Thomas | CC BY-SA 4.0 |
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Feb 23, 2016 at 20:41 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Oct 10, 2014 at 20:15 | comment | added | jak | Dear Lubos, thanks for your reading tip. I read today through the SU(3) related pages in Georgis book, but wasn't able to find an answer to my question. Nevertheless, I was able to come up with a "semi-satisfactory" answer to my question (see below). Despite my rusty group theory knowledge I would appreciate a technical correct answer or a reading tip where the charges of the gluons are derived explicitly using group theory. | |
| Oct 10, 2014 at 20:15 | answer | added | jak | timeline score: 4 | |
| Oct 9, 2014 at 16:25 | comment | added | Luboš Motl | Dear Jakob, I think that these are basic questions about group theory and group theory in physics. You find them on first pages of every introductory text about these matters. Take e.g. Howard Georgi, Lie algebras in particle physics, or anything simpler. It doesn't really make sense to answer your questions because you're effectively asking about all the basics of group theory and to answer, one would have to effectively reproduce a whole textbook on these matters because you seem to be starting from scratch. | |
| Oct 9, 2014 at 11:19 | comment | added | jak | What is the connection between these cartan-generator eigenvalues and the commonly used color labels? How exactly do we assign charge to the gauge fields? How does the $W$ quarks get Isospin $1,0,-1$? They are in the adjoint rep, therefore 2x2 matrices. Equally gluons are 3x3 matrices. In my understanding the objects transforming under the fundamental rep are labelled by the Cartan generator eigenvalues. Is this correct and if yes, how is it done for the adjoint rep objects? | |
| Oct 9, 2014 at 11:18 | comment | added | jak | Each quark is in addition an $SU(3)$ triplet $Q$, and transforming according to the fundamental rep $ e^{i a_A(x) \lambda_A /2} Q$. For $SU(3)$ there are two diagonal (Cartan) generators whose eigenvectors can be used as a basis for the corresponding vector space (onto which the group acts in the fundamental rep) and eigenvalues to label the fields (here the quarks). Unfortunately I don't know how to go on from here. | |
| Oct 9, 2014 at 11:17 | comment | added | jak | 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.. | |
| Oct 9, 2014 at 10:07 | comment | added | Luboš Motl | 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). | |
| Oct 9, 2014 at 10:05 | comment | added | Luboš Motl | 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. | |
| Oct 9, 2014 at 10:00 | comment | added | jak | 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. | |
| Oct 9, 2014 at 6:58 | comment | added | Luboš Motl | 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. | |
| Oct 9, 2014 at 6:20 | history | asked | jak | CC BY-SA 3.0 |