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  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


  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 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]

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