113 group contains an element analogous to a 2 \u03c0 rotation in the case of SU 2

# 113 group contains an element analogous to a 2 π

This preview shows page 114 - 117 out of 126 pages.

113
group contains an element (analogous to a 2 π rotation in the case of SU (2)), which reverses the sign of spinors, but not tensors. Calling this element r , we have r : ψ → − ψ and H H. (14.8) This is not the same as R parity because it lacks θ → − θ , which amounts to reversing the signs of space-time spinors. This operation – which is just 2 π rotation – is sometimes denoted ( 1) F , where F is fermion number. It is an exact symmetry, since fermion number is conserved modulo two. Thus R parity is just r ( 1) F , which means that it is equivalent to the gauge group element r . This element r is related to baryon number and lepton number by r = ( 1) 3( B L ) . (14.9) Now for the key point: This discrete gauge symmetry is unbroken if and only if no Spin(10) spinor fields acquire vevs as part of a Higgs mechanism. In other words, only r - invariant fields condense and acquire vevs. In this case, 3( B L ) would be conserved mod 2, which could be possible in a realistic model. In that case we would have related R parity to an unbroken gauge symmetry, so that it could be exactly conserved. This would ensure that the LSP, the lightest particle with negative R parity, is absolutely stable. Phenomenologically, to control baryon number and lepton number violation, this seems to be quite desirable. Nonetheless, there is a considerable literature on models with broken R parity symmetry in which the LSP is not absolutely stable. Some of them may be viable. 14.5 Susy unification The fact that there is a single gauge coupling constant for a simple GUT group G , such as SU (5) or SO (10), means that for any properly normalized generator T in the standard model subalgebra, the quantity g 2 i tr ( T 2 ) , (14.10) should be the same, when evaluated at the unification scale m X . Here, the index i = 1 , 2 , 3 labels the standard model groups U (1), SU (2), SU (3). The trace runs over any representation of G . Let’s illustrate this for a few choices of T : Q : electric charge T 3 : third component of weak isospin T c : a generator of color SU (3) e Y = Y/ 2 : weak hypercharge (suitably normalized) (14.11) Unification then requires that g 2 1 tr ( e Y 2 ) = g 2 2 tr ( T 2 3 ) = g 2 3 tr ( T 2 c ) . (14.12) 114
These are also the same as e 2 tr Q 2 . To check this substitute Q = T 3 + e Y and e = g 1 g 2 g 2 1 + g 2 2 and use tr ( T 3 e Y ) = 0, to get tr Q 2 = tr T 2 3 + tr e Y 2 = ( g 2 1 g 2 2 + 1 ) tr e Y 2 = g 2 1 e 2 tr e Y 2 = g 2 2 e 2 tr T 2 3 , (14.13) Note that e Y = Y/ 2 is the “properly” normalized generator. As in the standard model, the weak mixing angle (or Weinberg angle) is sin 2 θ W = e 2 /g 2 2 . Using the formulas given above, this can be evaluated at the unification scale m X . sin 2 θ W = tr T 2 3 tr Q 2 = g 2 1 g 2 1 + g 2 2 . (14.14) Let’s evaluate this for the ¯ 5 representation of SU (5) which contains ¯ d and l . Any other representation gives the same answer.

#### You've reached the end of your free preview.

Want to read all 126 pages?

Stuck? We have tutors online 24/7 who can help you get unstuck.
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes