Neutrino Masses in the MSSM - PDF Free Download

Neutrino Masses in the MSSM Steven Rimmer Supervisor: Dr. Athanasios Dedes Institute of Particle Physics Phenomenology, University of Durham

A supersymmetric standard model Find the most general Lagrangian which is invariant under Lorentz, SU(3) C SU(2) L U(1) Y and supersymmetry transformations; renormalizable minimal in particle content Neither lepton number (L) nor baryon number (B) are conserved Two options: constrain lagrangian parameters impose a further discrete symmetry when constructing the lagrangian

Superpotential The most general superpotential is given by W = Y E LH 1 E + Y D H 1 QD c + Y U QH 2 U c µh 1 H 2

Superpotential The most general superpotential is given by W = Y E LH 1 E + Y D H 1 QD c + Y U QH 2 U c µh 1 H 2 + 1 2 λllec + λ LQD c κlh 2

Superpotential The most general superpotential is given by W = Y E LH 1 E + Y D H 1 QD c + Y U QH 2 U c µh 1 H 2 + 1 2 λllec + λ LQD c κlh 2 + 1 2 λ U c D c D c Ibanez & Ross (1991) show that there are two preferred discrete symmetries, R-parity and a Z 3 allowing /L. L. E. Ibanez and G. G. Ross, Nucl. Phys. B 368, 3 (1992)

R-parity Under R-parity Standard model particles (including scalar Higgs bosons) are even corresonding superpartners are odd If R-parity is imposed when constructing the Lagrangian Terms in the Lagrangian which violate either L or B are excluded The lightest supersymmetric particle (LSP) is stable Sneutrino fields do not acquire a non-zero vacuum expectation values; R-parity is not violated spontaneously

R-parity conserving minimal supersymmetric standard model (Rpc-MSSM) Neutral scalar particle content neutral, complex scalar components of two higgs chiral supermultiplets, h 0 2 and h 0 1 neutral, complex scalar components of lepton supermultiplets, ν Li If R P conservation is imposed on the Lagrangian there are no bilinear terms which cause mixing between fields with different R P. No mixing between Higgs bosons and sneutrinos.

The neutral scalars decouple into CP-even and CP-odd real scalar eigenstates The Higgs bosons {h 0 2, h 0 1} {h 0, H 0, A 0, G 0 } and the sneutrinos ν Li { ν + i, ν i } The scalar potential can be minismised to obtain values for the vacuum expectation values (vevs) of h 0 2 and h 0 1.

L violating minimal supersymmetric standard model (/L-MSSM) Without imposing the conservation of L on the Lagrangian, bilinear terms exist between (and hence, mixing occurs between) charged gauginos, charged higgsinos and charged leptons neutral gauginos, neutral higgsinos and neutrinos Higgs bosons, sneutrinos... giving rise to some attractive features. B. C. Allanach, A. Dedes and H. K. Dreiner, Phys. Rev. D 69, 115002 (2004)

Neutrino Masses The mixing between neutrinos and neutral gauginos/higgsinos produces one tree-level, see-saw suppressed, neutrino mass. where 0 0 0 0 0 0 0 0 m tree ν m tree ν = v 2 d ( M1 g2 2 + M 2 g 2) ( κ1 2 + κ 4Det[M χ 0] 2 2 + κ 3 2)

The neutral scalar bosons all mix In general {h 0 2, h 0 1, ν Li } {h 0, H 0, A 0, G 0, ν + i, ν i } 10 10 mixing matrix of real, scalar fields. Solve minimisation conditions to find 5 complex vevs. Not clear which phases can be rotated away and which are physical; explicit CP-violation? Not clear if vevs are real; spontaneous CP-violation?

Vanishing sneutrino vev basis Notice that if R P is not imposed, h 0 1 and ν Li carry the same quantum numbers, even before symmetry breaking. Define ν Lα = (h 0 1, ν Li ) µ α = (µ, κ i ) Solving the minimisation conditions defines a direction in this (h 0 1, ν Li ) space. In the interaction basis, we are free to define any direction in (h 0 1, ν Li ) to be the Higgs. Had the basis been rotated such that the Higgs points in this direction, a basis would have been chosen such that the sneutrino vevs are zero. A. Dedes, SR, J. Rosiek and M. Schmidt-Sommerfeld, Phys. Lett. B 627 (2005) 161

We have studied the scalar and fermion sector without making any assumptions concerning the CP or family structure of the theory, finding a basis suitable for calculation. A method for initialising parameters has been found, such that the correct values are generated for lepton/quark masses and CKM/PMNS matrices. FORTRAN code has been developed to produce the tree-level mass spectrum, mixing matrices and Feynman rules. The renormalisation of the theory has been studied, one-loop corrections to the neutralino-neutrino masses have been derived and implemented within the code. A. Dedes, SR, J. Rosiek and M. Schmidt-Sommerfeld, Phys. Lett. B 627 (2005) 161 A. Dedes, SR and J. Rosiek, in preparation

Vertex µ κ q W κ 0 p { ie ZN(4+α)p Z (2+α)q + ie } Z 2sW s N2pZ 1q γ µ P L W

Loop Diagrams H 0 s Z µ κ 0 q κ 0 p κ 0 q κ 0 p κ 0 r κ 0 r Other loops: Neutralino/CP-odd Higgs Chargino/charged Higgs Quark/squark Chargino/W-boson

Summary Our aim is to study the phenomenology of the most general /L-MSSM The full set of Feynman rules have been derived The one-loop corrections to the neutrino masses have been calculated This is interesting to compare limits on λ/λ with Tevatron/B-factory analyses Can now go on to study other processes connected with neutrino masses; neutrinoless double beta decay, lepton flavour violating events such as τ decays,... A. Dedes, SR and J. Rosiek, in preparation

Blank A. Dedes, SR and J. Rosiek, in preparation

Neutrino-neutralino mass matrix L 1 2 ( i B 0 i W 0 h 0 2 h 0 1 ν i ) M N i B 0 i W 0 h 0 2 h 0 1 ν i

Neutrino-neutralino mass matrix L 1 2 ( i B 0 i W 0 h 0 2 h 0 1 ν i ) M N i B 0 i W 0 h 0 2 h 0 1 ν i M N = M 1 0 gv u / 2 gv d / 2 gv j / 2 0 M 2 g 2 v u / 2 g 2 v d / 2 g 2 v j / 2 gv u / 2 g 2 v u / 2 0 µ κ j gv d / 2 g 2 v d / 2 µ 0 0 j gv i / 2 g 2 v i / 2 κ i 0 i 0 ij h 0 2 = v u h 0 1 = v d ν Li = v i

M N = M 1 0 gv u / 2 gv d / 2 gv j / 2 0 M 2 g 2 v u / 2 g 2 v d / 2 g 2 v j / 2 gv u / 2 g 2 v u / 2 0 µ κ j gv d / 2 g 2 v d / 2 µ 0 0 j gv i / 2 g 2 v i / 2 κ i 0 i 0 ij A. S. Joshipura and M. Nowakowski, Phys. Rev. D 51, 2421 (1995) M. Nowakowski and A. Pilaftsis, Nucl. Phys. B 461, 19 (1996)

M N = M 1 0 gv u / 2 gv d / 2 gv j / 2 0 M 2 g 2 v u / 2 g 2 v d / 2 g 2 v j / 2 gv u / 2 g 2 v u / 2 0 µ κ j gv d / 2 g 2 v d / 2 µ 0 0 j gv i / 2 g 2 v i / 2 κ i 0 i 0 ij M N = ( M χ 4 4 m 4 3 m T 3 4 0 3 3 ) Suggestive of the seesaw mechanism xxxm N M N xxxx 4 eigenvalues M χ 3 eigenvalues mmt M χ

M N = M 1 0 gv u / 2 gv d / 2 gv j / 2 0 M 2 g 2 v u / 2 g 2 v d / 2 g 2 v j / 2 gv u / 2 g 2 v u / 2 0 µ κ j gv d / 2 g 2 v d / 2 µ 0 0 j gv i / 2 g 2 v i / 2 κ i 0 i 0 ij M N = ( ) N4 3 n 4 4 n 3 3 0 3 4 The mass 2 given by eigenvalues of M N M N xxxxxxxxxxxxxxx M χ xxx xxxxxxxxxxxxxxxx mmt M χ xx

n = gv d / 2 gv 1 / 2 gv 2 / 2 gv 3 / 2 g 2 v d / 2 g 2 v 1 / 2 g 2 v 2 / 2 g 2 v 3 / 2 µ κ 1 κ 2 κ 3 0 0 0 0 n has 3 linearly dependant rows, giving two zero eigenvalues with corresponding eigenvectors e 1,2 M N q i = ( N4 3 n 4 4 n 3 3 0 3 4 ) 0 0 0.. e i. =.. λ i e i. 0 0 0 q 1,2 are zero eigenvectors of the full matrix.

Vanishing sneutrino vev basis Notice that if R P is not imposed, h 0 1 and ν Li carry the same quantum numbers, even before symmetry breaking. Define ν Lα = (h 0 1, ν Li ) µ α = (µ, κ i ) Solving the minimisation conditions defines a direction in this (h 0 1, ν Li ) space. In the interaction basis, we are free to define any direction in (h 0 1, ν Li ) to be the Higgs. Had the basis been rotated such that the Higgs points in this direction, a basis would have been chosen such that the sneutrino vevs are zero. M. Bisset, O. C. W. Kong, C. Macesanu and L. H. Orr, Phys. Lett. B 430, 274 (1998) Y. Grossman and H. E. Haber, Phys. Rev. D 59, 093008 (1999)

Finding the vanishing sneutrino vev basis The neutral scalar potential takes the form V neutral = ( M 2 L )αβ ν Lα ν Lβ +m 2 2h 0 2 h 0 2 (b α ν Lα h 0 2+H.c)+ 1 8 (g2 +g2)[h 2 0 2 h 0 2 ν Lα ν Lα ] 2 where ( M 2 L) ) αβ ( m 2 L αβ + µ αµ β, m 2 2 m 2 H 2 + µ αµ α A. Dedes, SR, J. Rosiek and M. Schmidt-Sommerfeld, hep-ph/0506209.

Finding the vanishing sneutrino vev basis The neutral scalar potential takes the form V neutral = ( M 2 L )αβ ν Lα ν Lβ +m 2 2h 0 2 h 0 2 (b α ν Lα h 0 2+H.c)+ 1 8 (g2 +g2)[h 2 0 2 h 0 2 ν Lα ν Lα ] 2 Redefine the sneutrino fields ν Lα = U αβ ν Lβ, U = V diag(e iφ α ) Z, V unitary, Z orthogonal. V neutral = [ ( ) ] Z T ˆM 2 L Z ν Lα ν Lβ + m 2 2 h 0 [ αβ ] (b Z) α ν Lαh 0 2 + H.c 2 h 0 2 + 1 ( ) 2 8 (g2 + g2) 2 h 0 2 h 0 2 ν Lα ν Lα

Minimisation conditions [ ( ) ] Z T ˆM 2 L Z v β (b Z) α v u 1 αβ 8 (g2 + g2)(v 2 u 2 v γ v γ ) v α = 0

Minimisation conditions [ ( ) ] Z T ˆM 2 L Z v β (b Z) α v u 1 αβ 8 (g2 + g2)(v 2 u 2 v γ v γ ) v α = 0 [ ( ) ] [ ( ) ] Z T ˆM 2 L Z v 0 + Z T ˆM 2 L Z v j (b Z) 0 v u 1 00 0j 8 (g2 +g2)(v 2 u v 2 γ v γ )v 0 = 0

Minimisation conditions [ ( ) ] Z T ˆM 2 L Z v β (b Z) α v u 1 αβ 8 (g2 + g2)(v 2 u 2 v γ v γ ) v α = 0 [ ( ) ] [ ( ) ] Z T ˆM 2 L Z v 0 + Z T ˆM 2 L Z v j (b Z) 0 v u 1 00 0j 8 (g2 +g2)(v 2 u v 2 γ v γ )v 0 = 0 [ ( ) ] [ ( ) ] Z T ˆM 2 L Z v 0 + Z T ˆM 2 L Z v j (b Z) i v u 1 i0 ij 8 (g2 +g2)(v 2 u v 2 γ v γ )v i = 0

Z α0 = ( ˆM 2 L) αα b α tan β 1 2 M 2 Z tan 2 β 1 tan 2 β+1. However, Z is an orthogonal matrix and orthoginality must be imposed 3 Z α0 Z α0 = 1 α=0 giving multiple solutions for tan β = v u v0 Each tan β corresponds to a different extrema. By considering all solutions for tan β can find the solution which corresponds to the deepest minima. This is the correct vanishing sneutrino vev basis. The first column of Z is now fully defined (3 parameters)

Have the freedom to define 3 3 orthogonal sub-matrix of Z such that [ ( ) ] Z T ˆM 2 L Z

Scalar potential in vanishing sneutrino vev basis V neutral = (Me 2 L ) αβ ν Lα ν Lβ + m 2 2 h 0 2 h 0 2 [ ] B α ν Lα h 0 2 + H.c + 1 8 (g2 + g2) 2 ( h 0 2 h 0 2 ν Lα ν Lα ) 2 where [ ( ) ] (Me 2 L ) αβ Z T ˆM 2 L Z αβ Have now moved to a basis where and B α (b Z) α sneutrino vevs are zero sneutrino masses diagonal, ( ˆM 2 ν ) i all the parameters of the scalar potential real; can split into CP-even and CP-odd eigenstates

Parametrizing the neutral scalar mass matrices M 2 A = 2 B 0 sin 2β and tan β where M 2 A is the mass of the lightest CP-odd Higgs boson in the Rpc limit. The Lagrangian contains the terms L ( x 2 r γ ) M 2 EVEN ( x2 r δ ) ( y 2 t γ ) M 2 ODD ( y2 t δ ) where h 0 2 = x 2 + iy 2 ν Lα = r α + it α

CP-even Higgs boson mass matrix M 2 EVEN = cos 2 βm 2 A + sin2 βm 2 Z 1 2 sin 2β(M 2 A + M 2 Z ) B j 1 2 sin 2β(M A 2 + M Z 2) sin2 βma 2 + cos2 βmz 2 B j tan β ( B i B i tan β ( ˆM ) 2 ν ) i + 1 2 cos 2βM Z 2 δ ij

CP-even Higgs boson mass matrix M 2 EVEN = cos 2 βm 2 A + sin2 βm 2 Z 1 2 sin 2β(M 2 A + M 2 Z ) B j 1 2 sin 2β(M A 2 + M Z 2) sin2 βma 2 + cos2 βmz 2 B j tan β ( B i B i tan β ( ˆM ) 2 ν ) i + 1 2 cos 2βM Z 2 Noticing that the top-left 2 2 sub-matrix is identical to the Rpc case, for which the Higgs masses are given by M 2 h,h = 1 2 ( M 2 Z + M 2 A ± ) (MZ 2 + M A 2)2 4MA 2M Z 2 cos2 2β δ ij Applying Courant-Fischer theorem, it can be seen that in Rpv case one eigenvalue which is smaller that M 2 Z exists

CP-odd Higgs boson mass matrix The CP-odd mass matrix reads M 2 ODD = cos 2 βma 2 + ξ sin2 βmz 2 1 2 sin 2β(M A 2 ξm Z 2) B j 1 2 sin 2β(M A 2 ξm Z 2) sin2 βma 2 + ξ cos2 βmz 2 B j tan β B i B i tan β ( ˆM 2 ν ) iδ ij ξ is the gauge fixing parameter. we can project out without approximation the gauge dependent part, the would-be Goldstone mode, of the CP-odd scalar matrix as