1
$\begingroup$

I'm having trouble with Electro-Weak Theory. I understand that one wants to develop a gauge theory with $SU(2)_L \otimes U(1)_Y$. What buggs me is that the coupling with different fermions has different hypercharge. I understand that up-type and down-type particles in the $SU(2)_L$ doublet have $\frac{1}{2}$ and $-\frac{1}{2}$ weak isospin, that is related to different eigenvalues of the $SU(2)_L$ generators. But $U(1)_Y$ has only one generator and it's proportional to the identity! So I think that different fermion hypercharges are related to:

1) There is a different generator of U(1) for each fermion. (all of them are proportional to the identity but differ in a multiplying factor) This could be rephrased as: Each fermion transforms in a different equivalent representation of $U(1)$.

2) There's only one generator but each fermion has a different coupling constant to the weak hypercharge field.

Neither of this sounds satisfactory...

$\endgroup$
  • $\begingroup$ How do you mean 2) is not satisfactory? EM is U(1), but several particles couple to its photon with couplings 1,2,3,... $\endgroup$ – Cosmas Zachos Apr 24 '17 at 18:46
3
$\begingroup$

Perhaps the right way for you to think about it is this: there is only one generator, and it is a big $N\times N$ matrix where $N$ is the number of all of the independent fermion degrees of freedom. To be exact, all of the generators of a symmetry are like this, but we are often able to ignore the full size of each matrix.

Any generator of a symmetry must specify an action of the symmetry on every degree of freedom. In general a symmetry can mix up all the different fermions. However, most of the symmetries of the Standard Model only mix up a few different groups of fermions. For example, if we write out all the fermions (just one generation) as $f = (u_L^r,u_L^g,u_L^b,d_L^r,d_L^g,d_L^b,\nu_L,e_L,u_R^r,u_R^g,u_R^b,d_R^r,d_R^g,d_R^b,e_R)^T$ then the generators of the color $SU(3)$ symmetries look like

$$T_{color}^a = \begin{pmatrix}\lambda^a_{3\times 3} &0 & \dots &\\ 0 & \lambda^a_{3\times 3} & 0 & \dots \\ 0 & \dots & 0_{1\times 1} &\dots \\ 0 &\dots & & 0_{1\times 1} & \dots \\ 0 & \dots & & &\lambda^a_{3\times 3} \\ 0 &\dots &&&& \lambda^a_{3\times 3}\\ 0 &\dots &&&&& 0_{1\times 1}\end{pmatrix} $$

where the $\lambda$'s are 3x3 Gell-Mann matrices that act on the subspaces $u_L^{\{r,g,b\}}$ etc. Likewise, the $SU_2(L)$ generators mix up the left-handed quarks and leptons in a specific way:

$$ T^i_{weak} = \begin{pmatrix}\Sigma^i_{6\times 6} & 0 &\dots \\ 0 &\sigma^i_{2\times 2} & \dots \\ 0 &\dots & 0_{7\times 7}\end{pmatrix} $$

where $\sigma^i$ are the Pauli matrices and $\Sigma^i$ are a simple extension of the Pauli matrices, e.g. $$\Sigma^2 = \begin{pmatrix}0_{3\times 3} & -iI_{3\times 3} \\ iI_{3\times 3} & 0_{3\times 3}\end{pmatrix}$$

Finally, the single $U(1)_Y$ generator is just a diagonal matrix: $$T_Y =\begin{pmatrix}\frac{1}{6}I_{6\times 6} & 0 &\dots \\ 0 & -\frac{1}{2}I_{2\times 2} & \dots \\ 0 &\dots & \frac{2}{3} I_{3\times 3} &\dots \\ 0 &\dots && -\frac{1}{3}I_{3\times 3} & \dots \\ 0 &\dots &&& -1\end{pmatrix}$$

If you look at these for a while you can see that the matrices break down into blocks, and the different blocks never mix with each other. The $(u_L^r,u_L^g,u_L^b,d_L^r,d_L^g,d_L^b)^T$ fermions forms a 6x6 block. The $(\nu_L,e_L)$ part forms a 2x2 block, and so on. These subspaces are called irreducible subrepresentations of the symmetry.

To finally answer your question, the full $U(1)_Y$ generator is not proportional to the identity, but you can see that its action on each irreducible subrepresentation is proportional to the identity. Since irreducible subrepresentations never mix, we often refer to them separately and act like they have seperate generator matrices. This is not really the case, but it is much more convenient that writing out the huge matrices every time, and almost never causes confusion once you understand it.

The reason we can always split fields into irreducible subrepresentations like this is explained by representation theory, which is a beautiful mathematical subject that it is well worth learning. (And perhaps you are already trying to!)

$\endgroup$
  • $\begingroup$ Thanks! So basically when one says that electro-weak is the gauge theory of SU(2)_L x U(1)_Y is not entirely correct. There's a SU(2) xU(1) transformation for each fermion and they can be different, adding up to a single bigger transformation in every fermion, like you pointed out (that is not SU(2) x U(1) ) $\endgroup$ – P. C. Spaniel Apr 24 '17 at 17:57
  • $\begingroup$ However, there's still one thing that I can't get. Every Doublet transforms in the same representation of SU(2)_L but in different representations of U(1)_Y. Is there a reason for this? Thanks again! $\endgroup$ – P. C. Spaniel Apr 24 '17 at 18:11
  • $\begingroup$ I wouldn't say that actually. There is only one U(1) transformation (and one set of SU(2) transformations), but every fermion behaves differently under the action of that single transformation. It's maybe a subtle distinction, but it's important. $\endgroup$ – Luke Pritchett Apr 24 '17 at 18:12
  • $\begingroup$ As for your followup, for any given symmetry group there are only certain ways that objects can transform consistently under the group. For SU(2) the options are singlet, doublet, triplet, etc. For U(1), you can make a consistent representation out of any single real number, so there's a lot more freedom. The question of why the U(1) are the ones they are is interesting, but a separate question. $\endgroup$ – Luke Pritchett Apr 24 '17 at 18:14
  • $\begingroup$ wow what an awesome answer $\endgroup$ – InertialObserver Jun 10 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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