12.2 Problem Set 2 Solutions

Transcription

1 78 CHAPTER. PROBLEM SET SOLUTIONS. Problem Set Solutions. I will use a basis m, which ψ C = iγ ψ = Cγ ψ (.47) We can define left (light) handed Majorana fields as, so that ω = ψ L + (ψ L ) C (.48) χ = ψ R + (ψ R ) C (.49) Note that ω = ω C (.50) χ = χ C (.5) Then (ψ L ) C = (ψ C ) R (.5) (ψ R ) C = (ψ C ) L (.53) µ R ψψ = µ R (ψ R (ψ R ) C + ψr C ψ R ) (.54) µ L ωω = µ L (ψ L (ψ L ) C + ψl C ψ L ) (.55) are the right (left) handed mass terms for Majorana fields. Generalizing to N flavor, i,j =,,N, we have µ ij ψ Ci ψ j + h.c Using above definition for C and anti-symmetry of Grassmann variables are sees that µ ij can be taken as symmetric. For one flavor case general Dirac and Majorana mass is Since ( ) ( ) ( ) m µ ω ω χ L m (.56) µ χ R

2 .. PROBLEM SET SOLUTIONS 79 m (ωχ + ψω) = m(ψ L ψ R + ψ R ψ L ) = Dirac Mass (.57) One can diagnolize this by a unitary transformation to get the eigenvalues (in case µ L = 0 for instance) µ = µ R, m µ R This is an example of the see saw mechanism. Although one can not right down relevant (dimension 4) Majorana mess term in standard model, one can in some GUT s, e.g. SO(0). Therefore, the natural Majorana mass O(0 5 Gev), which is the GUT scale in typical theories.then above mechanism would give a LH fermion with a tiny mass m (00 Gev) 0 ev (.58) µ 0 5 Gev which is consistent with observation of scalar neutrinos. Note that for the typical Dirac mass µ, I take 00 Gev since they are obtained by Higgs much at weak scale. Same mechanism works for N > flavors in which case one diagonalizes the general mass make it with a unitary transformation.. (a) Let s begin with listing the matter content of Sµ indicating the hypercharges: Quarks: u R 3 d R 3 ρ L = q = q = ( ) ( ) ( ) u L c L t,, L (.59) d L s L b L 6 6 u R,c R,t R (.60) d R,s R,b R (.6) 6 Leptons: ( ) ( ) ( ) ν el ν µl ν l τl L =,, (.6) e L µ τ L L l R = l R,µ R,τ R (.63) Higgs:

3 80 CHAPTER. PROBLEM SET SOLUTIONS Φ = ( + ) (.64) We want to find a term with B = 0. Clearly this term should involve quarks. The restrictions are: SU(3) color, SU() weak, U() Y and Lorentz invariance. (l, ρ, τ separately are not fundamental). SU(3) requires three combinations: = (.65) = (.66) 3 3 = + 8 (.67) Last one can not violate B hence discarded. First two are ρρρ (or ρ c ρ c ρ c ) and ρρρ (or ρ c ρ c ρ c ) which are not Lorentz invariant unless we include another fermion which should be a lepton in order not to spoil SU(3) color. Hence the lowest dimensional operators which has B = 0 are ρρρl with dimension 6. (b) There are four types: ρρρl, ρ ρ ρ l, ρρρl, ρ ρ ρ l. First two does not violate B L, last two does. This problem amounts to see that last two are in violation of at last one of the after mentioned symmetries of Sµ. Consideration of ρρρl is sufficient: To have SU() invariance we need even number of left handed: ρ L ρ L ρ L ll, ρ R R L, ρ L ρ u ρ d L, ρ L ρ d ρ d L, ρ u ρ u ρ u R, ρ u ρ u ρ d R, ρ u ρ d ρ d R, ρ d ρ d ρ d l L ρ u ρ u l R Rl R Rl R R Rl R R Rl R R Rl R R R R. Note that ρ is either ρ or ρ c. l can only be a l. None of the above can be Lorentz invariant hence it is impossible to violate B L with a dimension 6 operator. Actually a generalization of above reasoning glons that B L can not be involved in Sµ neither perturbatively nor non-perturbatively. However B L violation would after a nice explanation for observed baryon asymmetry in the universe. One nice feature of GUT s is that there are consistent GUT s with relevant B L violating terms (e.g. SO(0)). (c) To violate L we need at least are lepton. If we insist to have only one lepton than we need to contract it with at least one quark. This would violate SU(3) hence we need three quarks, but this term (lρρρ) is dimension 6, no way. Consider two leptons, in order to violate L these should have same lepton number, hence Majorana type contraction: ll c l L or lr c l R. But these

4 .. PROBLEM SET SOLUTIONS 8 have total hypercharge and each. To cancel this we are only left with φ s to add. Therefore, we get ll c l L φ with dimension 5 or lr c l R φ 4 with dimension 7. As φ gets a VEV by Higgs l c l L φ l c l L v (.68) L L becomes a Majorana mess term. 3. (a) SU() in adjoint can be represented by τ 3 = 0 (.69) i τ = i i (.70) i τ = (.7) To have a cross product representation under SU() U() Y all elements in the triplet should carry same hypercharge Y. Then the covariant derivative takes the form Y where, Y = Y Y We want to give a VEV to triplet Higgs Δ µ = I µ + iga a µ τ a + ig B µ Y (.7) φ 3 = 3 (.73) such that only one of the linear combinations of generators τ, τ, τ 3, Y is unbroken. This will be the electric charge Q. This will be a diagonal U(), hence overall constant can be observed into charge Q = aτ 3 + by (.74)

5 8 CHAPTER. PROBLEM SET SOLUTIONS b is arbitrary but for convenience we take as Q = τ 3 + by (.75) Y Q = τ 3 + Y = 0 Y Y (.76) This should have a zero eigenvalue, hence Y {+,, 0}. However 0 in case Y = 0 only τ and τ are broken by Y, hence we get SU() 0 U() U() U(). We should choose Y {+, }. (b) Consider both a doublet φ with the covariant derivative and a triplet with σ a D µ φ = (I µ + ig A µ + ig B µ )φ (.77) D µ φ 3 = (I µ + iga τ a + ig Y B µ )φ 3 (.78) µ Expanding out D µφ + D µ φ 3 for φ = v, φ 3 = v3 we get for Y = ±: m w± = g (v + v 3 ) (.79) 4 m = (g + ρ z )(v + 4v 3 ) (.80) 4 m γ = 0 (.8) If we keep φ we have the option Y = 0 in contrast to above since φ already breaks to U(). For this case, Y = 0: w± 4 m = (v + 4v 3 ) (.8) m = 4 (g z + ρ )v (.83) m γ = 0 (.84) Note that m is entirely coming from usual doublet Higgs. z

6 .. PROBLEM SET SOLUTIONS 83 v 3 can be constrained as follows: See H. E. Haber, Minimal and Nonv minimal Higgs Bosons, in Phenomenology of Sµ and Beyond, D.P. Rey and P. Rey world scientific, 989 (this is in library: QC793.W66 989). An experimental fact that is very close to : ρ µ µ w z cos θw (.85) ρ = ɛ, 0 < ɛ (.86) On the other hand for a general Higgs content one can express ρ in term so f the casming of SU() and U() as: T,Y (T(T + ) Y ) < φ T,Y > ρ = (.87) T,Y Y < φ T,Y > where T,Y denote the representation, < φ T,Y > is the VEV of particular Higgs in the sum. Note that for T =, Y = ± one naturally gets ρ = (for any number of Higgs fields with T =, Y = ± ). For our problem we get + ( v 3 v ) + v 3 v ρ = = = ɛ (.88) 3 + 4( v 3 + v ) v v Therefore, 3 should be very small. v v 3 v 3 ρ ( ) ɛ = ɛ = (.89) v v This shows that adding a new type of Higgs field to the usual doublet is highly constrained by experiments. However, one can clearly add any number of doublets without violating ρ = ɛ constraint. This possibly is explored in the next problem. (c) Initially we have = 0 real D.O.F. 3 is eaten and we have left with 7 real scalar D.O.F. One of them is usual Higgs with Q = 0, isospin. The rest for Y = ± are two neutral scalars, two scalars of charge ±, two scalars of charge ± as clear from above Q matrix. (d) We now have the possibility of a Higgs field with Y = ±. Recall from Problem that the biggest constraint for a L violating term was imposed c by preserving hypercharge. Now we can write down l L l L φ 3 +, which is dimension 4 hence marginal. Note however, that lepton number violating v

7 84 CHAPTER. PROBLEM SET SOLUTIONS processes are quick constrained by experiments hence v 3 should again be v very small in accord with our discussion in previous part. 4. (a) We have two Higgs doublets φ and φ with condensation Then the masses of the gauge fields are ( ) 0 < φ >= (.90) v ( ) 0 < φ >= (.9) v w± m = m z = m γ = g (v + v ) (.9) 4 (ρ + ρ )(v + v 3 ) (.93) 4 0 (.94) Therefore, m w m z g = g + g (.95) is same as in the case of single doublet. Furthermore, using the general formula for ρ (from previous problem) we saw that m ρ = = w (.96) m z cos θ w Therefore, θ w is also the same as before. Started with 8 real scalar D.O.F. 3 absorbed into congitiduval modes of gauge mesons w± and z. ( Therefore, ) 5 left out of which ( one is the ) 0 α + iβ usual neutral Higgs: φ =, 4 others are φ v + h =, γ + iδ α and β are charge +(α + iβ), (α iβ), γ and δ are both charge zero. (b) See H. E. Haber in QC793.W (c) General Yukawa coupling to quarks reads α α d Lφ qr + λ ɛ αβ qlφ α β u q d + λ 3 qlφ α α d qr + λ 4 ɛ αβ qlφ α β u λ q qr+ (.97)

8 .. PROBLEM SET SOLUTIONS 85 d However, under new U(), φ has charge, φ has +, q R, qr u has + then nd and 3 rd terms not allowed. u d This means φ cannot couple to q d, φ cannot couple to q R as in part above. Extra restriction on the potential in the previous part is that (φ + φ )(φ φ + ) term is not allowed. From continuous U() breaking we get an additional Goldstone boson, the axion. It is proportional to Tv, where T is the U() generator that generates φ φ e iλ φ e iλ, φ. Therefore, its coupling to quarks are α u ɛ αβ q L q R φ β (u L u R λ u + c L c R λ c + t L t R λ t ) }{{} ax (.98) axion