6
$\begingroup$

The matching conditions for a breaking $G \rightarrow \prod_i G_i$ are

$$\omega_G-\frac{C_2(G)(\mu)}{12 \pi}=\omega_{G_i}-\frac{C_2(G_i)(\mu)}{12 \pi} ,$$

where $C_2(g)$ denotes the quadratic Casimir invariant for the adjoint representation of the group $g$. (See, for example, Eq. 7 in Implications of the CERN LEP results for SO(10) grand unification )

However, for example, for the braking chain $SU(4) \times SU(2)_L \times SU(2)_R \rightarrow SU(3_C) \times SU(2)_L \times U(1)_Y$ we have the matching condition

$$ \omega_{U(1)_Y} = \frac{3}{5}\left( \omega_{SU(2)_R} - \frac{2}{12\pi} \right) + \frac{2}{5}\left( \omega_{SU(4)_C} - \frac{4}{12\pi} \right)$$

(See, for example, Eq. 8 in Implications of the CERN LEP results for SO(10) grand unification )

Where does this matching condition for $U(1)$ subgroups come from?

|cite|improve this question
$\endgroup$
  • $\begingroup$ What's $\omega_i$? Also, note that the Casimir of $\mathrm{U}(1)$ vanishes on the adjoint since the adjoint is the trivial representation. $\endgroup$ – ACuriousMind♦ Apr 25 '16 at 10:39
  • $\begingroup$ @ACuriousMind 1.) $\omega_i$ corresponds to the subgroup $i$. I changed it in the question to make it clearer. 2.) Yes, that's why there is no term $ \propto \frac{1}{12\pi}$ on the left-hand side... $\endgroup$ – jak Apr 25 '16 at 10:43
  • $\begingroup$ It was clear to me that $\omega_i$ refers to the $\omega$ belonging to $G_i$ - I don't know what object you denote by $\omega$ in the first place! Are these coupling constants? Also, you don't mean the tensor product $\otimes$, you mean the direct product $\times$ of groups there (many physicists strangely use the tensor product there). $\endgroup$ – ACuriousMind♦ Apr 25 '16 at 10:46
  • $\begingroup$ @ACuriousMind $\omega_i := \alpha_i^{-1} := \frac{4\pi}{g_i^2}$ $\endgroup$ – jak Apr 25 '16 at 10:47
  • $\begingroup$ @ACuriousMind Presumably, the usage of $\otimes$ comes from the fact that one usually works with algebras in physics, instead of groups. $\endgroup$ – Danu Apr 25 '16 at 12:29
1
$\begingroup$

It results from the combination of two facts: $i$) the embedding of $U(1)_Y$ into $SU(2)_R \times U(1)_{B-L}$, and $ii$) the normalization of the $U(1)$ charges.

Let me take the simpler chain: $SU(2)_L \times SU(2)_R \times U(1)_{B-L} \to SU(2)_L \times U(1)_Y$.

$i$) At the scale of the left-right symmetry, the hypercharge gets merged into $SU(2)_R \times U(1)_{B-L}$, and we have the relation: \begin{equation} \frac{Y}{2} = T_{3R} + \frac{B-L}{2}\,. \end{equation} (the factors of 1/2 here are conventional). Here $T_{3R}$ is the generator of $SU(2)_R$.

$ii$) In order for the non-abelian generators to be normalized like their abelian counterparts, we define the new charges: $Y' = \sqrt{\frac{3}{5}} \left(\frac{Y}{2} \right)$ and $C' = \sqrt{\frac{3}{2}} \left(\frac{B-L}{2} \right)$. Therefore the equation above becomes:

\begin{equation} Y' = \sqrt{\frac{3}{5}} T_{3R}+ \sqrt{\frac{2}{5}} C' \,. \end{equation}

Finally, when you use this expression with the coupling constants (squared) and do the matching, you get the relation which appears in your question. (note that: $SU(4)_C \to SU(3)_C\times U(1)_{B-L}$ .)

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for your answer. I guess my problem is then, where does \begin{equation} \frac{Y}{2} = T_{3R} + \frac{B-L}{2}\,. \end{equation} come from? $\endgroup$ – jak Apr 26 '16 at 7:53
  • $\begingroup$ In general, we have: $Y/2= c_1 T_{3R} + c_2 {(B-L)/2}$, since the $U(1)_Y$ must be a linear combination of these generators. Then by matching with the (known) Standard Model hypercharges and the (known) $B-L$ charges which are fixed by the breaking, we obtain that $c_1=c_2=1$. $\endgroup$ – xi45 Apr 26 '16 at 11:34

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.