# Breaking/Tumbling gauge theory and composite fermions

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There is some statement learnt from this paper Tumbling gauge theories by [Raby, Dimopoulos, Susskind (1979)]:

Given 4d SU(5) gauge theory with fermions in the representation $$\bar 5\oplus 10$$

Add a scalar field in $$5$$ with a Yukawa coupling to two of fermions in $$10$$ and $$10$$. For an appropriate potential it condenses, Higgses $$SU(5) \to SU(4)$$ and gives masses to some of the fermions. We have left with $$SU(4)$$ with fermions in $$1 \oplus 4 \oplus \bar{4}$$ Standard 4d dynamics leaves a single massless fermion with the quantum numbers of a product of three microscopic fermions $$\bar 5 \cdot \bar 5 \cdot 10.$$ One can read the summary of statement in p.16 of this slide

questions:

1. Higgses $$SU(5) \to SU(4)$$ is the process of condensing $$5$$ by a Higgs potential, yes? I am not sure how does it give masses to some of the fermionsto be left with $$SU(4)$$ with fermions in $$1 \oplus 4 \oplus \bar{4}$$? This means out of 15 fermions there are only 9 fermions left to be massless? How come only 9 fermions left not 10 (or other number) fermions?

2. "Standard 4d dynamics leaves a single massless fermion with the quantum numbers of a product of three microscopic fermions $$\bar 5 \cdot \bar 5 \cdot 10.$$" But there were 9 fermions in $$1 \oplus 4 \oplus \bar{4}$$, is that true that both $$4 \oplus \bar{4}$$ are massive due to the $$SU(4)$$ gauge confinement dynamics? While the $$1$$ is the only remained massless fermion?

3. But the $$1$$ if it is the only remained massless fermion, should it be precisely from one of the 15 out of $$\bar 5\oplus 10$$? Then how come it is also a composite bound state out of $$\bar 5 \cdot \bar 5 \cdot 10 =(\bar 10+\bar 15) \cdot 10 = 1+ \dots?$$ It looks that the way to get the composite bound state still not clear...?

This post imported from StackExchange Physics at 2020-11-30 18:56 (UTC), posted by SE-user annie marie heart
asked Jul 19, 2020

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