PHYS08200604017 Manimala Mitra - Homi Bhabha National Institute

In the above equations ... represents other terms which will not contribute to the neutral
component of the potential. With all these F k ’s, VF n is given by
VF n = |µH0 d |2 +|µHu 0 |2 + 1 2 |∑ Y Σi˜Σ0c R˜ν L i
| 2 + 1 ∑
|Y Σi H
2
u˜Σ 0 0c
R |2 + 1 2 |∑ Y Σi H u˜ν 0 L i
| (4.23)
2
i
i
i
−[µHd 0 (∑ Y
√
Σi
2˜Σ0c R˜ν L i
) ∗ +c.c]+|M| 2˜Σ0c ∗ 0c
R ˜Σ R +[∑ Y
√
Σi
H u˜ν 0 L i
(M˜Σ 0c
R
i
i 2 )∗ +c.c].(4.24)
As we have three generation of leptons hence the generation index i runs as i=1,2,3. The
D term contribution of V neutral is given as
V n D = 1 8 (g2 +g ′2 )(|H 0 d |2 −|H 0 u |2 + ∑ i
|˜ν Li | 2 ) 2 . (4.25)
˜Σ
0c
The component R which has Y = 0 and the third component of the isospin T 3 = 0 does
not contributes to VD n . The soft supersymmetry breaking contributions to the neutral
part of the potential is given by Vsoft n where
V n
soft = −(bH 0 u H0 d +c.c)+m2 H u
|H 0 u |2 +m 2 H d
|H 0 d |2 (4.26)
+m 2 Σ˜Σ 0c∗
R
+ ∑ i
˜Σ
0c
R +[˜m2˜Σ0c R
m 2˜ν i˜ν ∗ L i˜ν Li +[ ∑ i
˜Σ
0c
R +c.c]
A
√
Σi
Hu˜Σ 0 0c
R˜ν L i
+c.c]. 2
We represent the vacuum expectation values of Hu, 0 Hd 0, ˜ν L i
and R as 〈H0 u〉 = v 2 ,
〈Hd 0〉 = v 1, 〈˜ν Li 〉 = u i and 〈˜Σ 0c
R 〉 = ũ. We have considered a diagonal m2˜ν matrix. In terms
of these vacuum expectation values the neutral component of the potential is
〈V neutral 〉 = (|µ| 2 +m 2 H u
)|v 2 | 2 +(|µ| 2 +m 2 H d
)|v 1 | 2 −(bv 1 v 2 +c.c)
˜Σ
0c
+ 1 8 (g2 +g ′2 )(|v 1 | 2 −|v 2 | 2 + ∑ i
|u i | 2 ) 2 +(|M| 2 +m 2 Σ)|ũ| 2
+ ∑ i
m 2˜ν i
|u i | 2 +[˜m 2 ũ 2 +c.c]+ 1 2 |∑ i
Y Σi ũu i | 2 + 1 ∑
|Y Σi | 2 |ũv 2 | 2
2
i
+ 1 2 |∑ i
Y Σi u i v 2 | 2 +( ∑ i
A Σi
√
2
v 2 u i ũ+c.c)−(µv 1 ( ∑ i
Y Σi
√
2 ũu i ) ∗ +c.c)
+[ ∑ i
Y Σi
√
2
v 2 u i (Mũ) ∗ +c.c]. (4.27)
For simplicity we assume all the vacuum expectation values and all the parameters are
real. Hence 〈V neutral 〉 simplifies to
〈V neutral 〉 = (µ 2 +m 2 H u
)v 2 2 +(µ 2 +m 2 H d
)v 2 1 −2bv 1 v 2 + 1 8 (g2 +g ′2 )(v 2 1 −v 2 2 + ∑ i
u 2 i) 2
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