PHYS08200604017 Manimala Mitra - Homi Bhabha National Institute
PHYS08200604017 Manimala Mitra - Homi Bhabha National Institute
PHYS08200604017 Manimala Mitra - Homi Bhabha National Institute
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Appendix<br />
A: The Scalar Potential and Higgs Spectrum<br />
Our model has two SU(2) complex Higgs doublets Φ 1 and Φ 2 , with hypercharge Y = 1.<br />
The scalar potential can then be written as<br />
( ) 2 ( ) 2 ) 2<br />
V = λ 1 Φ † 1Φ 1 −v 2 +λ2 Φ † 2Φ 2 −v ′2 +λ3<br />
((Φ † 1Φ 1 −v 2 )+(Φ † 2Φ 2 −v ′2 )<br />
( ) 2<br />
+λ 4<br />
((Φ † 1Φ 1 )(Φ † 2Φ 2 )−(Φ † 1Φ 2 )(Φ † 2Φ 1 )<br />
)+λ 5 Re(Φ † 1Φ 2 )−vv ′ cosξ<br />
where<br />
+λ 6<br />
(Im(Φ † 1Φ 2 )−vv ′ sinξ) 2,<br />
(A1)<br />
〈Φ 1 〉 =<br />
( ( )<br />
0 0<br />
, 〈Φ<br />
v)<br />
2 〉 =<br />
v ′ e iξ , and tanβ = v′<br />
v .<br />
(A2)<br />
According to our Z 2 charge assignment, Φ 1 carries charge +1, while Φ 2 has −1 charge.<br />
Therefore, the λ 5 term is zero when the symmetry is exact. We will discuss the phenomenological<br />
consequences of this and argue in favor of a mild breaking of this Z 2 symmetry.<br />
With the scalar potential Eq. (A1) it is straightforward to obtain the Higgs mass<br />
matrix and obtain the corresponding mass spectrum. The physical degrees of freedoms<br />
contain the charged Higgs H ± and the neutral Higgs H 0 , h 0 , and A 0 . While H 0 and h 0<br />
are CP even, A 0 is CP odd. If we work in a simplified scenario where ξ is taken as zero,<br />
then it is is quite straightforward to derive the mass of the charged Higgs H ± and the<br />
CP-odd Higgs A 0 . The masses are given as<br />
M 2 H ± = λ 4(v 2 +v ′2 ), and M 2 A 0 = λ 6(v 2 +v ′2 ),<br />
(A3)<br />
respectively. The mass matrix for the neutral CP-even Higgs is<br />
( )<br />
4v<br />
M ′ =<br />
2 (λ 1 +λ 3 )+v ′2 λ 5 (4λ 3 +λ 5 )vv ′<br />
(4λ 3 +λ 5 )vv ′ 4v ′2 (λ 2 +λ 3 )+v 2 . (A4)<br />
λ 5<br />
The mixing angle, obtained from diagonalizing the above matrix is given by<br />
and the corresponding masses are<br />
tan2α = 2M 12<br />
M 11 −M 22<br />
,<br />
M 2 H 0 ,h 0 = 1 2 {M 11 +M 22 ±<br />
(A5)<br />
√<br />
(M 11 −M 22 ) 2 +4M 2 12. (A6)<br />
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