Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

191 submissions , 151 unreviewed
4,796 questions , 1,987 unanswered
5,288 answers , 22,472 comments
1,470 users with positive rep
773 active unimported users
More ...

  Elegant theoretical way(s) to prove the $U(1)$ electric charge of electron + proton $Q_e+Q_p=0$?

+ 3 like - 0 dislike
82 views

In Zee's QFT in Nutshell book p/410, he uses the following method to show $U(1)$ electric charge of proton + electron $$Q_e+Q_p=0$$ in Grand Unified Theory (GUT):

enter image description here

However, Zee only shows

  1. by the traceless properties of ${\bf 5}$ in $SU(5)$: $$3Q_{\bar{d}}+Q_e+Q_{\nu_e}=0$$

This means the sum of the charges of eq.11 in p.409 is zero:

enter image description here

  1. by the traceless properties of ${\bf 10}$ in$ SU(5)$, we still get the same relation: $$3Q_{\bar{d}}+Q_e+Q_{\nu_e}=0$$

enter image description here

If neutrino is neutral, $Q_{\nu_e}=0$,we get $$3Q_{\bar{d}}+Q_e =0$$ and follows the relation of $SU(2)$ weak doublet $$ Q_u+Q_{\bar{d}}=1 $$ Zee can only shows: $$ Q_p+Q_e= 2 Q_u-Q_{\bar{d}}+Q_e= 2-3Q_{\bar{d}}+Q_e=2+2Q_e $$

This does not directly show the $ Q_p+Q_e=0$ unless, $Q_e=-1$. This is not very satisfying...

question: What are some more elegant theoretical way(s) to prove the $U(1)$ electric charge of proton + electron $Q_e+Q_p=0$ in GUT or other TOE theories (of course, experiments tell us that is true...)? Perhpas, another way to ask, is there a different GUT which makes the $$ Q_p+Q_e= 2 Q_u-Q_{\bar{d}}+Q_e $$ as zero as a traceless condition or some condition?

This post imported from StackExchange Physics at 2020-11-30 18:54 (UTC), posted by SE-user annie marie heart
asked Jul 16, 2020 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

Not sure how helpful it will be, but this comes from my notes on GUT charge quantisation:

In the Standard Model, the electric charge is the quantum number related to the $U(1)$ of $SU(3)\times SU(2) \times U(1)$ symmetry. Strictly speaking it is a $U(1)$ subgroup of $SU(2)\times U(1)$, but that doesn't make a difference in this case. The Eigenvalues of $U(1)$ are continuous - it is just a phase - and so there is no reason why we should expect the electric charge to be quantised. However, if the overall symmetry is $SU(5)$ then we know that all the Eigenvalues of its generators are discrete, as this is the case for any simple non-Abelian group. The electric charge is then the Eigenvalue of one of the generators of $SU(5)$ and can hence only take on discrete values.

The electric charge is additive, i.e. the charge of a two-particle state is the sum of the charges of the two one-particle states. This is only possible if $Q$ is diagonal when acting on the one-particle states (otherwise they get mixed up). $SU(5)$ has four independent diagonal generators and we know that the electric charge generator commutes with the color generators, of which there are two, i.e. the two generators corresponding to the two independent Eigenvalues of $SU(3)$, i.e. the rank of $SU(3)$. There are thus two generators left in $SU(5)$ two construct the electric charge generator from and we arbitrarily chose to call these $T_3$ and $T_0$. These generators correspond to the diagonal generators of $SU(2)$ and $U(1)$ subgroups of $SU(5)$. We can thus write $$ Q = T_3+ c T_0 $$ This shows that in the $SU(5)$ GUT the formula for the electric charge is not enlarged to include any other terms than from the GWS theory.

We can estimate the value of $c$. First, we normalise the generators of $SU(5)$ in such a way that $T_a=\lambda_a/2$ with $\mathrm{tr}{\lambda_a \lambda_b} = 2 \delta_{ab}$. It then follows that $T_0$ and $T_3$ are given by \begin{align} T_0 = \frac{1}{2\sqrt{15}} \begin{pmatrix} 2 &0& 0 &0& 0 \\ 0 &2& 0 &0& 0 \\ 0 &0& 2 &0& 0 \\ 0 &0& 0 &-3& 0 \\ 0 &0& 0 &0& -3 \end{pmatrix} \quad \mathrm{and} \quad T_3 =\frac{1}{2} \begin{pmatrix} 0 &0& 0 &0& 0 \\ 0 &0& 0 &0& 0 \\ 0 &0& 0 &0& 0 \\ 0 &0& 0 &-1& 0 \\ 0 &0& 0 &0& +1 \end{pmatrix} \label{eq:smt0su5} \end{align} To find $c$ we e.g consider a left-handed d quark with electric charge $Q= -1/3$.

Acting on the $\bf{\bar 5}$ representation $\psi^i = \begin{pmatrix} d^\mathrm{r} & d^\mathrm{g} & d ^\mathrm{y} & \bar{u} & e^+ \end{pmatrix}^T$ we have $T_0 d = 1/\sqrt{15}d^\alpha$ and $T_3 d^\alpha =0$. Therefore $$ -\frac{1}{3} d^\alpha = 0+ \frac{1}{\sqrt{15}} c d^\alpha \quad \Rightarrow c=- \sqrt{\frac{5}{3}} $$ This is also valid for the other particles. E.g. for the antineutrino we have $$ Q \bar{\nu} = \Big( T_3-\sqrt{\frac{5}{3}} T_0 \Big) \bar{\nu} = \Big[-\frac{1}{2}-\sqrt{\frac{5}{3}} \Big( - \frac{3}{2\sqrt{15} }\Big)\Big] \bar{\nu} = 0 $$ corresponding to the electrically neutral antineutrino. Likewise, we recover that the positron has charge $+1$: $$ Q {e^+} = \Big( T_3-\sqrt{\frac{5}{3}} T_0 \Big) e^+ = \Big[+\frac{1}{2}-\sqrt{\frac{5}{3}} \Big( - \frac{3}{2\sqrt{15} }\Big)\Big] e^+ = e^+ $$ We have derived a relation between the electric charges of the quarks and the leptons. Indeed, the structure of $SU(5)$ automatically results in the fact that the charge of the d quark is $-1/3$ times the electric charge of the positron.

This post imported from StackExchange Physics at 2020-11-30 18:54 (UTC), posted by SE-user Oбжорoв
answered Jul 26, 2020 by Oбжорoв (10 points) [ no revision ]
many thanks --- any comment is welcome!!! voted up

This post imported from StackExchange Physics at 2020-11-30 18:54 (UTC), posted by SE-user annie marie heart

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...