Well, standard model has $SU(3)\times SU(2)\times U(1)$ gauge group, so this is a direct product of multiple groups embedded in larger set. So how quarks that must interact with every gauge field and leptons that must interact only with $SU(2)$ and $U(1)$ fields would interact with this embedded group?
$\begingroup$
$\endgroup$
6
-
3$\begingroup$ WP. Can you expand on what this embedding you are talking about might possibly be? You already know quarks couple to all 3 factor groups but leptons only to the latter 2. What is the question? $\endgroup$– Cosmas ZachosCommented May 22, 2022 at 13:35
-
$\begingroup$ Gauge fields are usually NxN traceless matrices, and dirac fields are N vectors. Well, in the fermionic action, the Dirac fields are multiplied with the gauge field. How do you do this if the number of rows of the vector does not match the number of columns of the matrix? That's why I ask, fermions( both quarks and leptons) should be embedded in larger set to couple with gauge field $\endgroup$– PeterCommented May 22, 2022 at 13:52
-
$\begingroup$ "Usually"? In the fundamental representation of SU(N) they are NxN matrices. In the M-dim representation of the same group, they are MxM matrices. Is this the question? $\endgroup$– Cosmas ZachosCommented May 22, 2022 at 13:59
-
$\begingroup$ No. Combining all groups of the standard model into one large group, we get one large matrix that has elements of each group. So are we acting with this one big matrix on one Dirac vector right? Or does each gauge field have its own matrix of its dimension (SU(3) is 3x3, SU(2) is 2x2, U(1) is just number) and acts separately on different dirac vectors. $\endgroup$– PeterCommented May 22, 2022 at 14:09
-
$\begingroup$ Can you answer me please? I want to know where I'm wrong. $\endgroup$– PeterCommented May 22, 2022 at 14:49
|
Show 1 more comment
2 Answers
$\begingroup$
$\endgroup$
1
I'm not sure what it is you wish to know, per your comment, but take a quark in the fundamental of these groups, $q^i_\alpha$. You may think of it as a 2×3 matrix, the rows of which represent weak isospin, and the columns color. Not as a 5-vector!
So the weak isospin 2×2 matrices (dotted onto the SU(2) gauge fields, if you like) act on the left, transforming the Latin indices, resulting in a 2×3 matrix; while the color 3×3 matrices (gauge fields) act on the right, acting on the color Greek indices, resulting in a 2×3 matrix as well.$^\natural$ The weak hypercharge amounts to a multiplication of the entire array by a complex number.
If you had umpteen gauge groups, you'd have an umpteen-dimensional-hypersolid array, with umpteen different indices, each one operated above by the suitable group matrices of each group Cartesian factor.
You thus see that leptons only have a weak isospin and a hypercharge specification, so they look like $l^i$ 2-vectors, with the suitable hypercharge specifying the power of the complex number their hypercharge transformation dictates.
A field in the isospin 7 representation of SU(2) with indices J,... and the 10-dim representation of color SU(3), with indices A,..., so $q^J_A$, will transform by 15×15 weak isospin (2$\cdot$7+1=15) matrices on the left and 10×10 color matrices on the right, while again multiplied by a suitable phase for hypercharge. Is this the question?
$^\natural$ Ignoring hypercharge, the action of a particular rotation of the full group would thus look like $$ q^i_\alpha \mapsto (e^{i\theta \tau^1})^i_{~j} (e^{i\phi \lambda^1})_{\alpha}^{~\beta} ~ q^j_\beta ~. $$
answered May 22, 2022 at 15:54
-
1
$\begingroup$
$\endgroup$
1
The quarks interact with all three gauge fields, while the leptons interact only with the gauge fields of the weak and electromagnetic interactions.
-
$\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$– Community BotCommented May 22, 2022 at 14:35