Question

In Section 15.7, we speculated that a grand unified group $G$ exists such that $$ G \supset S U(3) \times S U(2) \times U(1) \text {. } $$ Georgi and Glashow have shown that the smallest such group of gauge transformations is the group $S U(5)$. Of course, different GUT's with larger groups than $S U(5)$ can be constructed. Once a group is chosen for investigation, we have to assign the quarks and leptons to multiplets (i.e., irreducible representations) of the group. In the earlier chapters, we presented empirical evidence for distinct families (or generations) of fermions, (u,d; $\left.\nu_e, \mathrm{e}\right),\left(\mathrm{c}, \mathrm{s} ; \nu_\mu, \mu\right), \ldots$, where in the first family, for instance, we have $$ \begin{aligned} & \left.\left(\begin{array}{l} \mathrm{u} \\ \mathrm{d} \end{array}\right)_L, \mathrm{u}_R, \mathrm{~d}_R\right\} \quad \text { each with three colors, } \\ & \left(\begin{array}{c} \nu_e \\ \mathrm{e}^{-} \end{array}\right)_L, \mathrm{e}_R^{-}, \end{aligned} $$ together with their antiparticles. This grouping ensures that we can construct gauge theories free of anomalies, see Section 12.12 . In a family, there are thus 15 left-handed states, for example, those in (15.61) together with $\overline{\mathrm{u}}_L, \overline{\mathrm{d}}_L$, and $\mathrm{e}_L^{+}$. For the $S U(5)$ model, these can be accommodated in a fundamental $\overline{5}$ - and a 10-representation (the 10 is the antisymmetric part of the product of two fundamental 5-representations, compare (2.58)). Explicitly, we have for the lefthanded states, $$ \begin{aligned} \overline{5} & =(1,2)+(\overline{3}, 1)=\left(\nu_e, \mathrm{e}^{-}\right)_L+\overline{\mathrm{d}}_L, \\ 10 & =(1,1)+(\overline{3}, 1)+(3,2)=\mathrm{e}_L^{+}+\overline{\mathrm{u}}_L+(\mathrm{u}, \mathrm{d})_L, \end{aligned} $$ where we have shown the $\left(S U(3)_{\text {color }}, S U(2)_L\right)$ decomposition of the multiplets. What are the gauge bosons of $S U(5)$ ? An $S U(N)$ gauge theory has $N^2-1$ gauge bosons. For the $S U(5)$ model, these are $$ 24=(8,1)+\underbrace{(1,3)+(1,1)}_{\text {gluons }}+\underbrace{(3,2)+(\overline{3}, 2)}_{\mathrm{X}, \mathrm{Y}, \mathrm{Y} \text { bosens }} $$ So, we have a new pair of superheavy gauge bosons, $\mathrm{X}$ and $\mathrm{Y}$. They form a weak doublet and are colored. They mediate interactions which turn quarks into leptons: $$ (\mathrm{u}, \mathrm{d})_L \rightarrow \mathrm{e}_L^{+}+(\overline{\mathrm{Y}}, \overline{\mathrm{X}}), $$ or, in $S U(5)$ parlance, $$ (3,2) \rightarrow(1,1) \otimes(3,2) . $$ Is the appearance of such transitions really surprising? First, recall that at energies above $M_W$, the distinction between weak and electromagnetic interactions disappears. Similarly, at the GUT-scale $M_{X, \gamma}$, which we identify with (15.58), the strong color force merges with the electroweak force, and the sharp separation of particles into colored quarks and colorless leptons, which interact only through the electroweak force, disappears. This leads to lepton/baryon number-violating interactions such as (15.65).

   In Section 15.7, we speculated that a grand unified group $G$ exists such that
$$
G \supset S U(3) \times S U(2) \times U(1) \text {. }
$$

Georgi and Glashow have shown that the smallest such group of gauge transformations is the group $S U(5)$. Of course, different GUT's with larger groups than $S U(5)$ can be constructed.

Once a group is chosen for investigation, we have to assign the quarks and leptons to multiplets (i.e., irreducible representations) of the group. In the earlier chapters, we presented empirical evidence for distinct families (or generations) of fermions, (u,d; $\left.\nu_e, \mathrm{e}\right),\left(\mathrm{c}, \mathrm{s} ; \nu_\mu, \mu\right), \ldots$, where in the first family, for instance, we have
$$
\begin{aligned}
& \left.\left(\begin{array}{l}
\mathrm{u} \\
\mathrm{d}
\end{array}\right)_L, \mathrm{u}_R, \mathrm{~d}_R\right\} \quad \text { each with three colors, } \\
& \left(\begin{array}{c}
\nu_e \\
\mathrm{e}^{-}
\end{array}\right)_L, \mathrm{e}_R^{-},
\end{aligned}
$$
together with their antiparticles. This grouping ensures that we can construct gauge theories free of anomalies, see Section 12.12 . In a family, there are thus 15 left-handed states, for example, those in (15.61) together with $\overline{\mathrm{u}}_L, \overline{\mathrm{d}}_L$, and $\mathrm{e}_L^{+}$. For the $S U(5)$ model, these can be accommodated in a fundamental $\overline{5}$ - and a 10-representation (the 10 is the antisymmetric part of the product of two fundamental 5-representations, compare (2.58)). Explicitly, we have for the lefthanded states,
$$
\begin{aligned}
\overline{5} & =(1,2)+(\overline{3}, 1)=\left(\nu_e, \mathrm{e}^{-}\right)_L+\overline{\mathrm{d}}_L, \\
10 & =(1,1)+(\overline{3}, 1)+(3,2)=\mathrm{e}_L^{+}+\overline{\mathrm{u}}_L+(\mathrm{u}, \mathrm{d})_L,
\end{aligned}
$$
where we have shown the $\left(S U(3)_{\text {color }}, S U(2)_L\right)$ decomposition of the multiplets.
What are the gauge bosons of $S U(5)$ ? An $S U(N)$ gauge theory has $N^2-1$ gauge bosons. For the $S U(5)$ model, these are
$$
24=(8,1)+\underbrace{(1,3)+(1,1)}_{\text {gluons }}+\underbrace{(3,2)+(\overline{3}, 2)}_{\mathrm{X}, \mathrm{Y}, \mathrm{Y} \text { bosens }}
$$

So, we have a new pair of superheavy gauge bosons, $\mathrm{X}$ and $\mathrm{Y}$. They form a weak doublet and are colored. They mediate interactions which turn quarks into leptons:
$$
(\mathrm{u}, \mathrm{d})_L \rightarrow \mathrm{e}_L^{+}+(\overline{\mathrm{Y}}, \overline{\mathrm{X}}),
$$
or, in $S U(5)$ parlance,
$$
(3,2) \rightarrow(1,1) \otimes(3,2) .
$$

Is the appearance of such transitions really surprising? First, recall that at energies above $M_W$, the distinction between weak and electromagnetic interactions disappears. Similarly, at the GUT-scale $M_{X, \gamma}$, which we identify with (15.58), the strong color force merges with the electroweak force, and the sharp separation of particles into colored quarks and colorless leptons, which interact only through the electroweak force, disappears. This leads to lepton/baryon number-violating interactions such as (15.65).
Show more…
Quarks and leptons: introductory course in modern particle physics
Quarks and leptons: introductory course in modern particle physics
Francis Halzen, Alan… 1st Edition
Chapter 15, Problem 7 ↓

Instant Answer

verified

Step 1

The group $SU(5)$ has $5^2 - 1 = 24$ generators, corresponding to 24 gauge bosons. These gauge bosons facilitate the interactions among the fermions.  Show more…

Show all steps

lock
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
In Section 15.7, we speculated that a grand unified group $G$ exists such that $$ G \supset S U(3) \times S U(2) \times U(1) \text {. } $$ Georgi and Glashow have shown that the smallest such group of gauge transformations is the group $S U(5)$. Of course, different GUT's with larger groups than $S U(5)$ can be constructed. Once a group is chosen for investigation, we have to assign the quarks and leptons to multiplets (i.e., irreducible representations) of the group. In the earlier chapters, we presented empirical evidence for distinct families (or generations) of fermions, (u,d; $\left.\nu_e, \mathrm{e}\right),\left(\mathrm{c}, \mathrm{s} ; \nu_\mu, \mu\right), \ldots$, where in the first family, for instance, we have $$ \begin{aligned} & \left.\left(\begin{array}{l} \mathrm{u} \\ \mathrm{d} \end{array}\right)_L, \mathrm{u}_R, \mathrm{~d}_R\right\} \quad \text { each with three colors, } \\ & \left(\begin{array}{c} \nu_e \\ \mathrm{e}^{-} \end{array}\right)_L, \mathrm{e}_R^{-}, \end{aligned} $$ together with their antiparticles. This grouping ensures that we can construct gauge theories free of anomalies, see Section 12.12 . In a family, there are thus 15 left-handed states, for example, those in (15.61) together with $\overline{\mathrm{u}}_L, \overline{\mathrm{d}}_L$, and $\mathrm{e}_L^{+}$. For the $S U(5)$ model, these can be accommodated in a fundamental $\overline{5}$ - and a 10-representation (the 10 is the antisymmetric part of the product of two fundamental 5-representations, compare (2.58)). Explicitly, we have for the lefthanded states, $$ \begin{aligned} \overline{5} & =(1,2)+(\overline{3}, 1)=\left(\nu_e, \mathrm{e}^{-}\right)_L+\overline{\mathrm{d}}_L, \\ 10 & =(1,1)+(\overline{3}, 1)+(3,2)=\mathrm{e}_L^{+}+\overline{\mathrm{u}}_L+(\mathrm{u}, \mathrm{d})_L, \end{aligned} $$ where we have shown the $\left(S U(3)_{\text {color }}, S U(2)_L\right)$ decomposition of the multiplets. What are the gauge bosons of $S U(5)$ ? An $S U(N)$ gauge theory has $N^2-1$ gauge bosons. For the $S U(5)$ model, these are $$ 24=(8,1)+\underbrace{(1,3)+(1,1)}_{\text {gluons }}+\underbrace{(3,2)+(\overline{3}, 2)}_{\mathrm{X}, \mathrm{Y}, \mathrm{Y} \text { bosens }} $$ So, we have a new pair of superheavy gauge bosons, $\mathrm{X}$ and $\mathrm{Y}$. They form a weak doublet and are colored. They mediate interactions which turn quarks into leptons: $$ (\mathrm{u}, \mathrm{d})_L \rightarrow \mathrm{e}_L^{+}+(\overline{\mathrm{Y}}, \overline{\mathrm{X}}), $$ or, in $S U(5)$ parlance, $$ (3,2) \rightarrow(1,1) \otimes(3,2) . $$ Is the appearance of such transitions really surprising? First, recall that at energies above $M_W$, the distinction between weak and electromagnetic interactions disappears. Similarly, at the GUT-scale $M_{X, \gamma}$, which we identify with (15.58), the strong color force merges with the electroweak force, and the sharp separation of particles into colored quarks and colorless leptons, which interact only through the electroweak force, disappears. This leads to lepton/baryon number-violating interactions such as (15.65).
Close icon
Play audio
Feedback
Powered by NumerAI
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever