NIKHEF Topical Lectures
Amsterdam, December 14 - 16, 2005.
Lectures on Supersymmetry
TUTORIALS
Exercise I: Basic calcultaional rules
1. Vertices and propagators
2. Dirac algebra
3. Cross sections and decay widths
4. Calculation of loop integrals.
Exercise II: Higgs boson production and decay mechanisms.
1. Higgs decays into fermions and gauge bosons
2. Higgs production in e+e− collisions
3. Higgs production in hadronic collisions
4. Higgs contributions to radiative corrections.
Exercise III: Divergences and symmetries.
1. Lagrangians and interactions
2. Electron and photon self-energies
3. Higgs boson self–energy: fermion loops
4. Higgs boson self–energy: scalar contributions.
Exercise IV: Higgs masses and coupling in the MSSM.
1. Derivation of the Higgs boson masses in the MSSM
2. Radiative corrections to the Higgs boson masses
3. The MSSM Higgs boson couplings
4. Phenomenology of the Higgs masses and couplings.
Exercise V: SUSY particle decay and production
1. Sparticle decays
2. Sfermion and gaugino production in e+e− collisions (LEP)
3. Squark/gluino production in pp collisions (Tevatron/LHC).
1
E1: Basic Calculational Rules
1. Vertices and propagators
Rules for fermions:
us (pi)
pi
p
i
p−m
6
−p̄i
v s(−pi)
6
= i pp+m
2 −m2
p̄f
−pf
us(pf )
vs(−pf )
Σs us(p) ūs(p) =6 p + m , Σs vs(p) v̄s(p) =6 p − m
Rules for gauge bosons:
µ(q)
−i
qµ
Σpol
gµν −qµ qν /(q 2 −MV2 )
q 2 −MV2
qν
∗ν (q)
•
•
q
= ∗ν µ = −(gµν − qµ qµ/MV2 )
For the photon, discard everything which is longitudinal (q µq ν )
above. Note that the trasversality of the photon implies: µ ·q µ = 0.
Vµ
f
f¯
−ieγµ(vf − af γ5)
Z : vf = (2If3 − 4ef s2W )/(4sW cW ) , af = 2If3 /(4sW cW )
√
W : vf = af = 1/(2 2sW )
Rules for Higgs bosons:
f
imf /v
f¯
γ : v f = ef , af = 0
H
i/(p2 − MH2 )
Vµ
−2iMV2 /v · gµν
Vν
2
2. Diracology: contractions and traces of γ matrices
Basic relations:
{γµ , γν } = γµγν + γν γµ = 2gµν and 6 p = pµγ µ
i
γ5 = µνρσ γ µγ ν γ ργ σ = iγ 0γ 1γ 2γ 3 and {γµ, γ5} = 0
4!
Tr(1) = 4 , Tr(γµ) = 0 , Tr(γ5) = 0
Tr(A1A2) = Tr(A2A1) , Tr(A1A2 · · · AN ) = Tr(A2 · · · AN A1)
Contractions of γ matrices
γµγ ν = 2gµν − γµγ ν ⇒ γ µγµ = δµµ = 4
γ µγν γµ = γ µ(2gµν − γµγν ) = 2γν − 4γν = −2γν
γ µγ ν γ ργµ = (2g µν − γ ν γ µ)(2gµρ − γµγ ρ)
= 4g νρ − 2γ ν γ ρ − 2γ ν γ ρ + 4γ ν γ ρ = 4g νρ
Traces of γ matrices:
Tr(γ µγ ν ) = Tr(2g µν − γ µγ ν ) = 2g µν Tr(1) − Tr(γ µγ ν ) ⇒ Tr(γ µγ ν ) = 4g µν
Trace of an odd number n of γ matrices (using γ52 = 1):
Tr(γ µ1 · γ µn ) = Tr(γ µ1 · · · γ µn γ 5γ 5) = (−1) Tr(γ µ1 · · · γ 5γ µn γ 5)
= (−1)n Tr(γ 5γ µ1 · · · γ µn γ 5) = − Tr(γ 5γ µ1 · · · γ µn γ 5)
⇒ Tr(γ µ1 · · · γ µn ) = 0
Tr(γ µγ5) = Tr(γ µγ ν γ σ γ5) = Tr(γ µ1 · · · γ µn γ 5) = 0
1
Tr(γ µγ ν γ5) = Tr(γ α γαγ µγ ν γ5) = (1/4) Tr(γα γ µγ ν γ5γ α)
4
= −(1/4) Tr(γα γ µγ ν γ α γ5) = −Tr(γ µγ ν γ5) = 0
Using the same tricks as above, proof the trace of 4 γ matrices:
Tr(γ µγ ν γ ργ σ ) = 4(g µν g ρσ + g µσ g νρ − g µρg νσ )
Tr(γ µγ ν γ ργ σ γ5) = −4iµνσρ
3
3. Cross sections and decay widths
The differential cross section for a 2 × n process i1i2 → f1 · · · fn is
|M (i1i2 → f1..fn )|2
d3 pf
dσ =
(2π)4δ 4(Σpi − Σpf ) S
Πn
2 2 1/2
3
2
(2π) 2ef
4[(p1 · p2) − m1m2]
• In the amplitude squared |M |2, one has to average (sum) on
degrees of freedom (polarisation, color) of initial (final) particles.
• There is a symmetry factor S = 1/n! for n identical particles.
• The flux factor is 2(p1 +p2)2 = 2s for 2 → n process with m1 = m2 = 0.
It is 2M for the decay of a particle with a mass M (1 → n process).
Calculation of phase–space for a two–body process a + b → f1 + f2:
1 d3 p1 d3 p2 4
δ (pa + pb − p1 − p2)
dPS2 =
16π 2 e1 e2
Z 3
d p2 4
1
δ (pa + pb − p1 − p2) = δ(ea + eb − e1 − e2)
e2
e2
with : |~p2| = |~pa + p~b − p~2| and e22 = |~p2|2 + m22
and
d3p1 = dΩ |p1|2 d|p1| with e21 = |~p1|2 + m21
1
1
In the c.m. frame: w = ea + eb, w0 = e1 + e2 = (m22 + p2) 2 (m21 + p2) 2 :
dw0
1
1
1
1
e1 + e 2
=p
+
⇒ dw0 = pdp
+
= e1de1
dp
e1 e2
e1 e2
e1 e2
dΩ
e1de1
dΩ
dw0
dΩ |p|
0
0
√
=
|p|
δ(w
−
w
)
=
|p|
δ(w
−
w
)
⇒
16π 2
e1 e2
16π 2
w0
16π 2 s
(for the last equality, the integral over dw 0 has been performed).
The differential cross section for a two body process is then:
dσ
1
1
|p|
√
=
× Σ|M (i1i2 → f1 f2)|2 ×
×S
dΩ 2s
16π 2
s
√
√
1
Note that |p| = 12 sλ = 12 s[1 − m21/s − m22/s)2 − 4m21m22/s2] 2 .
4
4. Calculation of loop integrals
p+k
p
= −iΠ(p2)
k
R
• Measure of loop integral over internal momentum: d4k/(2π)4.
(For fermion loops: take trace and factor (−1) for Fermi stats).
Z
i
i
d4 k
2
−iΓ = (ig)
(2π)4 (p + k)2 − m2 k 2 − m2
Z
d4 k
1
1
2
⇒ Γ = ig
(2π)4 (p + k)2 − m2 k 2 − m2
R1
• Symmetrize the integrand using: 1/ab = 0 dx/[a + (b − a)x]2
Z
Z 1
4
d
k
1
dx
Γ = ig 2
(2π)4 0
(k 2 + 2pkx + p2x − m2)2
• Shift variable k → k 0 = k + px (integrand becomes k 2 symmetric)
Z 1
Z
4
1
d
k
dx
Γ = ig 2
(2π)4 0
(k 2 + p2x(1 − x) − m2)2
• Wick rotation k0 → ik0 to go to Euclidean space (k 2 → −k 2)
Z 1
Z
4
d
k
1
Γ = −g 2
dx
(2π)4 0
(k 2 − p2x(1 − x) + m2)2
R
R +∞ 4
2
2 ∞
4
• Polar coordinates for d k: −∞ d k F (k ) = π 0 dk 2 k 2 F (k 2)
Z 1 Z ∞
g2
1
Γ=−
dx
ydy
16π 2 0
(y − p2x(1 − x) + m2)2
0
• Perform the integrals over the variables y and x:
R Λ2
– If integral divergent: cut-off at the energy Λ ( 0 dk 2 ).
– Eventually, use the on–shell mass relation p2 = m2.
5
E2: Higgs production and decay mechanisms
1. Higgs bosons decays
1.1 Decays into fermions: H → f f¯
q
>
X
s1 ,s2
MM†
f (p1)
−iM = ūs1 (p1)(imf /v)v s2 (−p2)
†
s2
s1
f¯(−p2) +iM = v̄ (−p2)(−imf /v)u (p1)
m 2 X
f
= Nc
v̄ s2 (−p2) us1 (p1)ūs1 (p1)v s2 (−p2)
v
s ,s
1 2
with Nc = 3(1) for quarks (leptons). Only one polarisation for H.
(v/mf )2/Nc × Σ|M |2 =
=
=
=
Tr(6 p1 + m)(− 6 p2 − m)
Tr(γµpµ1 + m)(−γν pν2 − m)
−pµ1 pν2 Tr(γµγν ) − m2Tr(1)
−4p1.p2 − 4m2
Using q 2 = (p1 − p2)2 = 2m2f − 2p1 · p2 = MH2 and defining the velocity
of the final fermions βf = 2|pf |/MH = (1 − 4m2f /MH2 )1/2
⇒ Σ|M |2 = Nc (mf /v)2 2(MH2 − 4m2f ) = 2Nc (m2f /v 2) MH2 βf2
The differential decay width is then simply given by:
dΓ
1
1
2|pf |
=
× Σ|M |2 ×
×
dΩ 2MH
32π 2
MH
Integrating over
R dΩ = dφd cos θ (and since there is no angular
dependence, dΩ = 4π), one obtains the partial decay width:
m2f MH 3
Γ(H → f f¯) = Nc 2
β
v 8π f
H decays dominantly into heaviest fermion and width ∝ MH .
6
1.2 Decays into massive gauge bosons
q
>
Vµ(p1)
−iM = ∗µ(p1) (−2iMV2 /v g µν ) ∗ν (−p2)
0 0
†
2
µν
Vν (−p2) +iM = µ0 (p1) (−2iMV /v g ) ν 0 (−p2)
X
X
4MV4 µν µ0ν 0 X ∗
2
∗
0 (p1 )
|M | =
g
g
(p
)
1
µ
µ
ν (−p2 )ν 0 (−p2 )
v2
pol
pol
pol
0 0
(v 2/4MV4 )Σ = g µν g µ ν (gµµ0 − p1µp1µ0 /MV2 )(gνν 0 − p2ν p2ν 0 /MV2 )
0
0
= (gµµ0 − p1µp1µ0 /MV2 )(g µµ − pµ2 pµ2 /MV2 )
= 4 − p21/MV2 − p22/MV2 + (p1 · p2)2/MV4
= (MH4 /4MV4 ) [1 − 4MV2 /MH2 + 12MV4 /MH4 ]
dΓ
1 2|pV |
The differential decay width, dΩ
= 2M1 × |M |2 × 32π
× S, with
2 M
H
H
SR = δV = 21 for two identical final Z bosons. This finally gives
( dΩ = 4π):
1/2 MV2
δV MH3
4MV2
MV4
Γ(H → V V ) =
1−
1 − 4 2 + 12 4
16πv 2
MH2
MH
MH
The dependence on MV is hidden, since v ≡ 2MW /g2 = 2MZ cW /g2.
For large enough MH [recall that H → f f¯ ∝ MH ], one has:
Γ(H → V V ) ' δV MH3 /(8πv 2) ⇒ Γ(H → W W ) ' 2Γ(H → ZZ)
The decay widths grows like MH3 i.e. is very large for MH MV .
For small MH , one (two) V bosons can be off–shell, the width is
Z 2
Z M 2 −q2
1
H
dq12 MV ΓV
Γ 0 MH
dq22MV ΓV
Γ= 2
π 0
(q12 − MV2 )2 + MV2 Γ2V 0
(q22 − MV2 )2 + MV2 Γ2V
2
2
2
2 2
3
q1
q1
12q1 q2
4q12q22
δV MH 1/2
, λ= 1− 2 − 2
λ
λ−
−
Γ0 =
8πv 2
MH4
MH MH
MH4
7
1.3 Decays into photons and gluons: H → γγ, gg
H does not couple to massless particles at tree-level: loop induced.
We have vertex diagrams with fermion (top) and W exchange
for H → γγ(Zγ); only top for H → gg: calculation complicated.
However it is simple if H momentum is small (i.e. MH Mloop):
q=0
p
p
−m
∂
× v
≡ ∂m
i
i
i
im
p/−m
Hγγ
=
−iMµν
v
p/−m
H
p/−m
γ
≡
p+k
im
v
∂
∂m
p
p
= −i
−m
v
∂
γγ
∂m Πµν
k
γ
Let’s calculate the derivative of the fermionic photon self-energy:
Z
d4 k
i
i
γγ 2
(−1)Tr(−iee
γ
)
−iΠµν (p ) = Nc
(−iee
γ
)
f
µ
f
ν
(2π)4
6k − m
6 p+ 6 k − m
Πγγ
µν (p)
=
−i Nc e2e2f
Z
d4k Trγµ(6 k + m)γν (6 p+ 6 k + m)
(2π)4 [(p + k)2 − m2](k 2 − m2)
Using the rules for Diracology and loop integral calculations:
Z 1
Z 1
dx
dx
=
D=
2
2
2 2
2
2
2 2
0 [(k + px) + p x(1 − x) − m ]
0 (k + 2pkx + p x − m )
N = Tr[γµ 6 k γν (6 p+ 6 k ) + m2γµγν ] = k ρ(k + p)σ Tr[γµγρ γν γσ + m2γµγν ]
= 4[2kµkν + (m2 − k 2 − p.k)gµν ]
Shift k → k + px, Wick rotation k0 →
ik0 for EuclideanR space ⇒
R
+∞
∞
k 2 → −k 2 and sym. integrand with −∞ d4k F (k 2) = π 2 0 dyyF (y)
8
R
d4k(kµkν ) = 14 gµν d4k(k 2)]:
Z 1 Z ∞
i
2
2 2
×
dx
ydy
Πγγ
µν (p) = −iNc ef e × 4 × π ×
16π 4
0
0
1 2
2
2
[ 2 k + m − x(1 − x)p ]gµν + 2x(1 − x)[gµν p2 − pµpν ]
[y + m2 − p2x(1 − x)]2
[also use of symmetry relation
R
Because of gauge invariance, photon is transverse (∝ gµν p2 − pµpν ):
the first term (∝ gµν should vanish* and we are left with:
Z 1 Z ∞
2 2
N
e
e
2x(1 − x)
c
f
2
ydy
(p)
=
dx
Πγγ
(g
p
−
p
p
)
µν
µ ν
µν
4π 2
[y + m2 − p2x(1 − x)]2
0
0
We can now calculate the Hγγ vertex [photons to symmetrize
→ 2; they are on–shell and p1,2 6= p but p2 = p1 · p2 = 12 MH2 ]
4m2 ∂ γγ
m ∂ γγ
Π (p1, p2) = −
Π (p , p )
= −2
2 µν 1 2
v ∂m µν
v
∂m
Z 1 Z ∞
−2x(1 − x)ydy
2m2 Nc e2f e2
=−
dx
(g
p
.p
−
p
p
)
µν 1 2
1µ 2ν
v
π2
[y + m2 − p2x(1 − x)]3
0
0
Hγγ
Mµν
2
2
2
Inside the integral,
we
can
suppose
m
p
(M
) and integrate
H
R
R
over x and y [ x(1 − x)dx = 1/6 and y/(y + m2)3dy = 1/2m2]
α
2
Hγγ
Nc e2f (gµν p1.p2 − p1µp2ν )
Mµν
=
3v
π
Now we use the same machinery as for decays into gauge bosons:
4 2 4 α2 MH4
2MH4 2 4 α2
µν ∗
∗
2
|M| = 2 Nc ef 2
Σ|g µ(p1)ν (p2)| =
N e
9v
π 2
9v 2 c f π 2
Integrating over phase space (with factor 12 for identical photons):
2
MH3 2 4 α2
Γ(H → γγ) = 2 Nc ef
9v
16π 3
*
This statement is not trivial to prove and we will come back to this discussion later on.
9
Several remarks to be made:
• The amplitude was of course finite (no tree level contribution)!
• The approximation mf MH is in practice good up to MH ∼ 2mf !
• Only tops contribute, other f have negligible Yukawa coupling.
• Infinitely heavy fermions do not decouple from the amplitude:
a way to count the number of heavy particles coupling to the H!
• There are also contributions from W bosons. Also in the
< 140 GeV), one has:
limit MH MW (valid for MH ∼
21 2
MH3 α2 2
Σ f N c ef −
Γ(H → γγ) = 2
9v 16π 3
4
• The W contribution is larger (∼ 4) than the t quark contribution
and the interference of the two is destructive.
• With the same calculation, one can get the amplitude for H →
Zγ. Only difference, Zf f, ZW W couplings and MZ in phase space.
> 10) larger than that of top.
Here again, the W contr. is much ( ∼
• The calculation holds also for gluons if we make the changes:
Qe e → gs Ta which means α → αs and Nc2 → |Tr(TaTa )|2 = | 12 δab |2 = 2:
MH3 αs2
Γ(H → gg) = 2 3
9v 8π
Decay width and branching ratios:
The total decay width of the Higgs is the sum of partial widths:
Γtot(H → all) = Σf Γ(H → f f¯) + ΣV Γ(H → V V )
and the branching ratio for Higgs decay into a given final state is:
BR(H → X) = Γ(H → X)/Γtot(H → all)
10
2. Higgs bosons production in e+e− Collisions
2.1 The Higgs–strahlung process:
us(l1)
p1
l1
q
v s(−l2)
−l2
−i(gµν −qµ qν /MZ2 )
q 2 −MZ2
∗ν 0 (p1)
−p2
∗
−i(g µν − q µ q ν /MZ2 ) −2iMZ2
0
g
−
−iM =
νν ν 0
q 2 − MZ2
v
First thing to use for simplification is Dirac equation 6 l u(l) = me ∼ 0:
v s(−l2)(−ie)γµ(ve
aeγ5)us(l1)
v s(−l2 )γµq µ us(l1) = v s(−l2) [− 6 l2+ 6 l1]us(l1) = 2me ∼ 0 ⇒ qµqν → 0
p
√
where me is supposed to be much smaller than s = q 2. Then:
4e2MZ4 v −2 ν ∗µ s
s
s
s
|M | = 2
2 2 (p1 ) (p1 ) v (−l2 ) γµ (ve − ae γ5 )u(l1 ) u(l1 ) γν (ve − ae γ5 )v(−l2 )
(q − MZ )
2
Average over polarizations of e± and sum on those of photon:
k
pµ1 pν1 1
2
µν
Σ|M | = Tr 6 l1(ve − aeγ5)γµ(− 6 l2)(ve − aeγ5)γν − g +
4
4
MZ2
−Tr = (ve2 + a2e )Tr 6 l1γµ 6 l2γν − 2aeveTr 6 l1γµ 6 l2γν γ5
= 4(ve2 + a2e )[l1µl2ν + l2µl1ν − l1.l2gµν ] − 8iaeve l1α l2β αµµβν
2
1
(l
.p
)(l
.p
)
p
1
1
2
1
Σ|M |2 = k(ve2 + a2e ) 2(l1.l2) − 2
− 4(l1.l2) + (l1.l2) 12
2
4
MZ
MZ
= k(ve2 + a2e ) −(l1.l2) − 2(l1.p1)(l2.p1)/MZ2
where we have used the fact that (αµβν ) g µν −pµ1 pν1 is (anti)symmetric.
√
2
2
In the c.m. frame, one has (with E1,2
= MZ,H
+|p|2) and |p| = s/2λ):
√
s
l1,2 =
(1, 0, 0, ±1) and p1,2 = (EZ,H , 0, ±|p| sin θ, ±|p| cos θ)
2
11
2
2 2
2
2
2
2
−
|p|
cos
θ)
s(E
λ
sin
θ
s
M
s
2
2
Z
Z
+
+
=
k(v
+
a
)
⇒ k(ve2 + a2e )
e
e
2
2MZ2
MZ2 s
8
The differential cross section is given by:
dσ
1 4e2MZ4 (ve2 + a2e )s MZ2 1 2 2
λ
=
+
λ
sin
θ
d cos θdφ 2s v 2(s − MZ2 )2MZ2
s
8
32π 2
R
R
with dφ = 2π and sin2 θd cos θ = 4/3 one gets the cross section
ve2 + a2e
αMZ2
2
2
σ(e e → HZ) =
λ(λ
+
12M
/s)
Z
12v 2 s(1 − MZ2 /s)2
+ −
A few remarks:
• The cross section drops like 1/s at high–energies (typical√of an
√
s–channel process). The maximum is reached at s = MZ + 2MH .
√
• At the maximum LEP2 energy, s = 209 GeV, the cross section for MH = (100) 115 GeV is given by (using the fact that σ0 =
4πα2(0)/3 = 86.8 nb with α(0) = 1/137, α(s) ' 1/128 and sin2 θW = 0.232):
σ = 0.42 (0.16) pb for MH = 100 (115) GeV
R
−1
If we have an integrated
luminosity
of
L
∼
100
pb
, this means
R
that we have N = σ × L ∼ 42(16) Higgs boson events.
• Since for MH ∼ 100 GeV, BR(H → bb̄) ∼ 90%, the signal is
e+ e− → ZH → Zbb̄ and the main background is e+e− → ZZ → Z b̄b.
• At high energies s MZ2 , one has a differential cross section
dσ
ve2 + a2e
3
αMZ2
2
'
sin θ with σ '
λ3
2
2
2
d cos θ
4σ
12v s(1 − MZ /s)
the behaviour in sin2 θ of the angular distribution and in λ3 of the
total cross section is typical for the production of two spin–zero
particles (here, the Z boson is almost a Goldstone boson).
12
2.2 The vector boson fusion mechanism:
us(p1)
s
u(l1)
p1
q1
p
l1
−l2
q
v s(l2)
2
−p2 v s
(p2 )
0
µ
µ
0
q1 q1
ūs(p1)γµ(v − aγ5)us(l1) g µµ − M
2
2
2
2
i(−ie) (−i) (−2iMV /v)
V
gµ 0 ν 0 ×
M=
ν qν0
0
q
(q12 − MV2 )(q22 − MV2 )
s
νν
s
2 2
γ
(v
−
aγ
)v̄
v̄(−p
g
−
ν
5
2
(l
)
)
M
2
2
V
0
0
Using the relations q1µ γµ = q6 =6 l1− 6 p1 ∝ me ∼ 0 and g νν g µµ gµ0ν 0 = g µν :
s
s
ūs(p1)γµ(v − aγ5)us(l1).v̄(−l
γ
(v
−
aγ
)v
4e4MV4 /v 2
ν
5
)
(−p
2
2)
|M | =
× s µ
ν
s
s
s
2
2
u(l1)γ (v − aγ5)u(p1).v̄(−p2)γ (v − aγ5)v(−l2)
D1 D2
2
4e4MV4 /v 2
Tr 6 l1γν (v − aγ5) 6 p1γµ(v − aγ5)
=
×
Tr 6 l2γ ν (v − aγ5) 6 p2γ µ(v − aγ5)
D12D22
4e4MV4 /v 2
(v 2 + a2)Tr 6 l1γν 6 p1γµ − 2vaTr 6 l1γν 6 p1γµγ5
× 2
=
(v + a2)Tr 6 l2γ ν 6 p2γ µ − 2vaTr 6 l2γ ν 6 p2γ µγ5
D12D22
Performing the trace and product using µναβ µνα0β 0 = δαα0 δββ0 − δβα0 δαβ0
32e4MV4 /v 2
1
2
× [gS (l1.p2)(l2.p1) + gA (l1.l2)(p1.p2)]
|M | =
4
D12D22
with gS = (v 2 + a2)2 + 4a2v 2 and gA = (v 2 + a2)2 − 4a2v 2
Let’s write the momenta of the particles in a convenient way:
q
l1 = (E, 0, 0, E) , p1 = ( x21E 2 + p2T 1, pT 1 sin θ1 , pT 1 cos θ1 , x1E)
q
l2 = (E, 0, 0, −E) , p2 = ( x22E 2 + p2T 2, pT 2 sin θ1, pT 1 cos θ1, −x2E)
13
and assume high energies s MV2 so that pT 1,T 2/E are rather small:
li.pi ∼ p2T i/2xi , l1.p2 ∼ 2E 2x2 , l2.p1 ∼ 2E 2x1 , p1.p2 ∼ 2E 2x1x2
hold, together with 2l1.l2 = s and the Higgs momentum squared:
MH2 = (q − p1 − p2)2 = s − 2q.p1 − 2q.p2 + 2p1.p2 = s(1 − x1)(1 − x2)
Using these products, one has then for the amplitude squared:
1
32e4MV4
4E 4(gS + gA )x1x2
2
|M | =
× 2
2 )2 (p2 /x + M 2 )2
4
v2
(pT 1/x1 + MW
2
W
T2
4
4
2 3 3
8e MV
(gS + gA )s x1x2
=
×
2 2 2
2 2
v2
(p2T 1 + x1MW
) (pT 2 + x2MW
)
Let us now deal with the three body phase space:
1 d3 p1 d3 p2 d3 p 4
δ (q − p1 − p2 − p)
dPS3 =
(2π)5 2E1 2E2 2EH
Defining τH = MH2 /s and using the known relation for δ functions:
Z
Z
Z 3
dp
= d4p δ(p2 − MH2 ) = d4p δ[s(1 − x1)(1 − x2) − sτH ]
2EH
and decomposing the momenta along the 3 directions, one obtains:
1 d(x1E) 2 d(x2E) 2
d pT 1
d pT 2 δ[s(1 − x1)(1 − x2) − sτH ]
(2π)5 2x1E
2x2E
R 2
R∞ 2 2
2
2 2
Noting that dpT i/(pT i + xiMV ) = π 0 dp /(p + xiMV2 )2 = π 2/(xiMV2 )
and using MW = ev/(2sW ), the differential cross section is given by:
6 4 8e MV
1 dx1 dx2 π 2
1
2 3 3
×
(gS + gA )s x1x2 ×
δ
dσ =
2 2
2s
4MW
sW
(2π)5 2x1 2x2 x1x2MV4
Z
Z
α3
dx1 dx2x1x2 sδ[s(1 − x1)(1 − x2) − sτH ]
σ =
2 s2 (gS + gA )
2MW
W
dPS3 =
14
R
Now perform the integrals using δ[f (x)] = |f 0 (x)|−1
x=x0 with f (x0 ) = 0
Z
Z
Z 1−τH
τH
1
s
dx1 dx2 · · · =
dx1x1 1 −
1 − x1
s(1 − x1)
0
Z 1−τH 1
τH
1 + τH
=
(1
+
τ
)log
+
− 2(1 − τH )
=
dx1 −1 +
H
2
1
−
x
(1
−
x
)
τ
1
1
H
0
where the boundary conditions are obtained by requiring that
p1Z = p2Z = x1,2E = 0 ⇒ x1 = 0 and x2 = 1−τH /(1−x1) = 0 → x1 = 1−τH .
Collecting all results, one obtains then the total cross section*:
α3
1
σ=
2 s2 (gS + gA ) (1 + τH )log τ − 2(1 − τH )
2MW
H
W
Let us now make a few remarks:
√
• The cross section rises as log(s/MH2 ): small at low s and large
√
at high s. Dominant Higgs production process for s MH2 .
• This approximation is good only within a factor of 2 and works
better at higher energies. It can be obtained in an easier way
using the effective longitudinal vector boson approximation.
√ 4
• In the case of W W fusion, gs = 8/(2 2) = 1/8 and gA = 0, one has:
3
α
s
σ(e+e− → H ν̄ν) =
(1
+
τ
)log
H
2 s2
2 − 2(1 − τH )
16MW
M
H
W
√
• At LEP2 energies, s ∼ 200 GeV, the cross section is σ ∼R 5(2)·10−3
pb for MH = 100(115) GeV, i.e. less than one event for L = 100
pb−1. This process is not very useful for Higgs searches at LEP2.
• For ZZ fusion with s2W ∼ 1/4, gs ∼ gA ∼ a4e ∼ 1/(16 × 9): the cross
section σ(e+e− → e+e− H) is ∼ 9 times smaller than for W W fusion.
*
This calculation, including details is done in: G. Altarelli, B. Mele and F. Pitolli, Nucl. Phys. B287 (1987) 205.
15
3. Higgs bosons production in hadronic Collisions
3.1 The gluon–gluon fusion process*
X
P
g
g
H
P
X
The cross section of the subprocess, gg → H, is given by:
d3 pH
1
1
1
4 4
2
(2π
)δ (q − pH )
×
×
|MHgg |
dσ̂ =
2ŝ 2 · 8 2 · 8
(2π)32EH
R 3
R 4
Using the fact that d pH /(2EH ) = d pH δ(p2H − MH2 ) and that
|MHgg |2 = 32πMH Γ(H → gg) calculated before, one obtains for σ̂:
π 2 MH
Γ(H → gg) δ(ŝ − MH2 )
σ̂ =
8ŝ
Convolute with gluon densities to obtain the total cross section
Z 1
Z 1
π 2 MH
σ=
Γ(H → gg)g(x1)g(x2)δ(ŝ − MH2 )
dx1
dx2
8ŝ
0
0
with ŝ = sx1x2, implying ŝ − MH2 = s(x1x2 − τH ) with τH = MH2 /s:
Z 1
Z 1
π2
Γ(H → gg) g(x1)g(x2)δ[s(x1x2 − τH )]
dx2
σ=
dx1
8M
H
0
0
R
We perform the integral on x2 [ δ[f (x)] = |f 0 (x)|−1
x=x0 with f (x0 ) = 0]
Z 1
π2
1 αs2 dLgg
dx
σ=
Γ(H → gg) τH
g(x)g(x/τH ) =
τH
8MH3
576v 2 π
dτH
τH x
where the integration bounds are xmax
= 1, xmin
= x1(for x2 = 1) = τ .
1
1
At LHC, gg luminosity is large and gg → H dominant process!
*
Calculation to be checked!!!
16
3.2 The Higgs strahlung and vector boson fusion process
q
V
q
V∗
q
H
0
q
q̄ 0
H
q0
The cross sections for these processes are the same as in e+e−
collisions, provided that the following changes are performed:
√
– The total energy s is replaced by the subprocess energy ŝ.
– The average over the quark colors is made: factor 13 · 13 .
– In the bremsstrahlung process, possibility of q q¯0 → W ∗ → W H.
– The couplings of the electrons are replaced by those of quarks:
in q q̄ → V H: a2e + ve2 → a2q + vq2.
in qq → Hqq: gS,A → [(v 2 + a2)(v 02 + a02) ± 4(av)(a0v 0).
The cross sections for a given initial state, are given by:
vq2 + a2q
1 αMV2
λ̂(λ̂2 + 12MV2 /ŝ)
σ(q q̄ → HV ) =
2
2
2
9 12v ŝ(1 − MV /ŝ)
1
1 α3
− 2(1 − τ̂H )
σ(qq → qqH) =
2 2 (gS + gA ) (1 + τ̂H )log
9 2MW
sW
τ̂H
0
Summing over all possibilities for quark/antiquark initial states
and folding with the proper densities, the total cross sections are:
XZ 1 Z 1
dx1 dx2fq (x1)fq0 (x2) σ̂[qq 0 → H + X]
σ[pp → H + X] =
q,q 0
0
0
Remarks:
• At LHC, qq → Hqq is the dominant process but not as gg → H.
• The cross section for q q̄ → HV is OK for low MH ; σ(HW ) ∼ 2σ(HZ).
• At Tevatron, Higgs–strahlung (esp. q q¯0 → HW ) more important.
17
E3: Divergences and Symmetries
1. Lagrangians and interactions
• Take QED Lagrangian for a fermion of charge e and mass m:
1
LQED = − Fµν F µν + iψγ µ∂µψ − mψψ + eψγ µAµψ
4
with Aµ and Fµν = ∂µAν −∂ν Aµ, the electromagnetic field and tensor.
U(1) gauge invariance: no AµAµ term so that photon is massless.
• Let’s add a scalar field φ:
Lφ = |∂µφ|2 − m2S |φ|2 + λ(φ+φ)2
which leads to a spontaneously broken symmetry (SSB), hφi = √v2 ,
√
and we write φ = (H +v)/ 2 with H being the physical Higgs boson.
• And couple this field to a fermion f (à la Yukawa):
Lf = −λf . ψψφ
√
After SSB, the fermion acquires a mass mf = λf v/ 2.
• Let us introduce two scalar fields φ1 and φ2:
Lkin = |∂µφ1|2 + |∂µφ2|2 − m21|φ1|2 − m22|φ2|2
they will have a coupling to the scalar field φ after SSB:
LS = λS |φ|2(|φ1|2 + |φ2|2) + 2vλS φ(|φ1|2 + |φ2|2)
Plus, eventually, terms in φ1φ2 that we take zero for simplicity.
18
2. Self–energy of the electron and photon
γ
Electron self–energy
e−
Z
≡ −iΣe(6 p)
i
−ig µν
d4 k
(−ieγµ)
(−ieγν )
−iΣe(p) =
(2π)4
6 p+ 6 k − m
k2
– Perform the Dirac Algebra:
R 1 γµγν + γν γµ =2 2gµν , etc...
– Symmetrisation: 1/ab = 0 dx/[a + (b − a)x] +change of variable.
– Switch to Euclidean space with Wick rotation k0 → ik0, k 2 → −k 2.
– Integrate over momentum (symmetric integrand and regul.):
R +∞ 4
R 2 2 2
2
2
2 Λ
d
k
F
(k
)
=
π
0 dk k F (k )
−∞
Definition of the correction to the mass: δme = Σe(6 p)|p=m:
Z
Z Λ2
m e e2 1
dyy[y + m2e x2]−2
dx(1 + x)
⇒ δme =
2
8π 0
0
2
2
me e 3 Λ
Λ2
3α
=
log
melog 2 + · · ·
+ ··· =
8π 2 m2e
4π
me
UV divergence (at large k 2) logarithmic.
In principle Λ = ∞ ⇒ renormalization: mphys
= mnu
e
e + δme .
But QED valid up to GUT (MP ) scale, i.e. Λ = MGUT(MP ).
Correction logarithmic AND proportional to me, therefore small:
δme ∼ 0.2me for Λ ∼ MP ∼ 1019 GeV.
More fundamental: correction small due to chiral symmetry:
if me = 0, LQED is invariant under the chiral transformation:
ψL → eiθL ψL and ψR → eiθR ψR with ψL,R = 1/2(1 ∓ γ5)ψ.
But me breaks the chiral symmetry → correction ∝ to the mass.
⇒ Symmetry ≡ protection for the mass.
19
Photon self–energy
e+
γγ 2
γ ≡ −iΠµν (p )
γ
e−
i
i
d4 k
(−ieγ
)
(−1)Tr(−ieγ
)
=
ν
µ
(2π)4
6k − m
6 p+ 6 k − m
We already made the calculation and we reached the level where:
Z 1 Z ∞
i
2
2
2
×
Πγγ
dx
ydy
µν (p ) = −ie × 4 × 2π ×
16π 4
0
0
[ 12 y 2 + m2 − x(1 − x)p2]gµν + 2x(1 − x)[gµν p2 − pµpν ]
[y + m2 − p2x(1 − x)]2
2
−iΠγγ
µν (p )
Z
using usual tricks and a cut–off Λ for integral on k 2, one gets
Z 1 Z Λ2
α 2
e2
1 µν γγ
y 2 + 2m2y
∼
Λ !!!
δmγ = g Πµν (0) =
dx
dy
4
16π 2 0
(y + m2)2
4π
0
But we must have mγ ≡ 0 at all orders because of gauge invariance.
Problem: the cut–off Λ2 violates the QED gauge invariance.
Solution: dimensional regularization, preserves gauge invariance!
We work in a space–time of n = 4R− dimensions:
– Internal momentum in n dim: dnk/(2π)n etc...
– Dirac algebra in n dim: Tr(I) = n, γµγ µ = nI, gµµ = n, etc..
– UV divergence: poles in 1/(n − 4) = 1/ with → 0.
In this case,
one would have for the integral:
R
A(m2) = [k 2 − m2]−1 ∼ am2/ + · · · ⇒ m2∂A/∂2m2 ∼ am2/.
With this regularization, δmγ = 0 at all orders: ⇒ massless photon.
Another example of a protection for a mass by a symmetry...
20
3. Higgs boson self–energy
Fermionic contributions:
f¯
H
H
f
2
−iΣH (p ) = Nf
Z
d4 k
iλf
i
iλf
i
√
√
(−1)Tr
(−
(−
)
)
(2π)4
2 6k − m
2 6 p+ 6 k − m
Usual calculation. Simpler: p2 = MH2 = 0 (fermion heavy compared
to MH ). Using a cut–off Λ for the integral on k 2, one obtains:
Z 1 Z Λ2
2
y(−y
+
m
λ
1
2
f
f)
ΣH (p2 = 0) = 4Nf √
dy
dx
(y + m2f )2
2 16π 2 0
0
After the trivial integral on x and the one on y, one gets:
∆MH2
λ2f Λ
2
2
2
= = Nf 2 − Λ + 6mf log
− 2mf + O(1/Λ2)
8π
mf
We have thus a quadratic divergence, δMH2 ∼ Λ2.
Divergence is independent of MH , and does not disappear if MH =0:
The choice MH =0 does not increase the symmetry of LSM .
Here, the cut–off does not break any symmetry and the problem is
not solved with dimensional regularization (though we have only
poles in 1/ and the quadratic divergence is not apparent).
If we fix the cut–off Λ to MGUT or MP : ⇒ MH ∼ 1014 to 1017 GeV!
The Higgs boson mass prefers to be close to the very high scale:
This is the hierarchy problem.
21
< 1 TeV) for unitarity etc... reasons
But we want a light Higgs ( ∼
We need thus to make: MH2 |Physical = MH2 |0 + ∆MH2 + countreterm
And adjust this counterterm with a precision of 1030 (30 digits)!
This is the naturalness problem.
In a complete theory, no problem formally: we adjust the bare
MH and the counterterm which are infinite, to have the physical
finite mass. This is the case of the log divergence of me in QED.
However, we want to give a physical meaning to the cut–off Λ and
the logarithmic and quadratic divergences are of different nature.
In the Standard Model:
besides the fermions, there are also contributions to MH from the
massive gauge bosons and from the Higgs boson itself:
W, Z, H
W, Z, H
H
H
H
W, Z, H
Total contributions of fermions and bosons in the SM at one–loop:
X
2
2
2
2
2
∆MH ∝ [3(MW + MZ + MH )/4 −
m2f ](Λ2/MW
)
We can adjust the unknown MH so that the quadratic divergence
disappears (would be a prediction for Higgs mass, MH ∼ 200 GeV).
However: does not work at two–loop level or at higher orders....
Summary: the problem of the quadratic divergences to MH is there.
There is no symmetry which protects MH in the SM.
22
The contributions of (2) scalars
f˜
f˜
φ
φ
f˜
i
i
dk
+
∆MH2 = ΣH (p2) = (i)NS
(iλ
)
S
(2π)4
k 2 − m21 k 2 − m22
Z
4
i
i
d
k
+ m1 ↔ m2
+NS (iλS v)2
(2π)4 k 2 − m21 (k + p)2 − m21
Z
4
And here, really simple calculation... We get after integration:
Λ
Λ λS NS 2
2
2
2
− 2Λ + 2m1log
+ 2m2log
∆MH =
16π 2
m1
m2 1
Λ
Λ λ2S v 2NS −
2
+
2log
+
2log
+
O
−
16π 2
m1
m2
Λ2
Again, quadratic divergences. But let us now assume:
• Scalar couplings related to fermion couplings: λ2f = −λS (!).
• Multiplicative factors are the same: NS = Nf (nb: 2 scalars!).
• To simplify, the scalars have the same mass: m1 = m2 = mS .
Let us now add the fermionic and scalar contributions:
∆MH2 |tot
λ2f Nf 2
Λ mS 2
2
=
(mf − mS )log
+ 3mf log
4π 2
mS
mf
The quadratic divergences have disappeared in the sum!!
Logarithmic divergence still there, but even with Λ = MP , contribution small. It disappears also if in addition we have mS = mf !
⇒ Symmetry fermions–scalars → no divergence in Λ2
“Supersymmetry” no divergences at all: MH is protected!
23
E4: Higgs masses and couplings in the MSSM
1. The MSSM Higgs boson masses
To obtain the physical Higgs fields and their masses from VH ,
VH = m21(|H01|2 + |H1+|2) + m22(|H20|2 + |H2− |2) − m23(H1+ H2− − H10 H20 + hc)
g22 + g12
g22 +∗ 0
0 2
+ 2
0 2
− 2 2
+
(|H1 | + |H1 | − |H2 | − |H2 | ) + |H1 H1 + H20∗H2− |2
8
2
develop H1 = (H10, H1− ) and H2 = (H2+, H20) into real (corresponding
to CP–even and charged Higgses) and imaginary (CP–odd Higgs
and Goldstones) parts and diagonalize the mass matrices:
1
M2ij = ∂ 2VH /∂Hi ∂Hj |hRe(H 0 )i=v1,2,hIm(H 0 )i=0,hH ± i=0
1,2
1,2
1,2
2
To obtain the masses and mixing angles, two useful relations are:
Tr(M2) = M12 + M22 , Det(M2) = M12M22
2M12
M11 − M22
p
sin 2θ = p
,
cos
2θ
=
(M11 − M22)2 + 4M212
(M11 − M22)2 + 4M212
where M1 and M2 are the physical masses and θ the mixing angle.
The procedure in the case of the CP–even Higgs bosons:
The neutral part of the scalar potential is (drop subscripts ...):
VH = m21|H1|2 + m22|H2|2 + m23(H1H2 + hc) + (MZ2 /4v 2)(|H1|2 − |H2|2)2
First perform the first derivative of the scalar potential:
∂VH /∂H10 = 2m̄21H1 + 2m̄23H2 + MZ2 /v 2 H1 (H12 − H22)
∂VH /∂H20 = 2m̄22H2 + 2m̄23H1 + MZ2 /v 2 H2 (H22 − H12)
24
At the minimum, ∂VH /∂H1,2 = 0, leading to the two relations:
1
1
m̄21 = −m̄23 tan β − MZ2 cos(2β) , m̄22 = −m̄23cotβ + MZ2 cos(2β)
2
2
Then make the second derivative with respect to H1 and H2:
∂ 2 VH
MZ2 2
2
(H1 − H22 + 2H1H1) = 2m̄21 + MZ2 (3c2β − s2β )
0
0 = 2m̄1 +
2
∂H1 ∂H1
v
MZ2 2
∂ 2 VH
2
= 2m̄2 + 2 (H2 − H12 + 2H2H2) = 2m̄22 + MZ2 (3s2β − c2β )
0
0
∂H2 ∂H2
v
MZ2
∂ 2 VH
2
= 2m̄3 + 2 (−2H1H2 ) = 2m̄23 − MZ2 sin 2β
0
0
∂H1 ∂H2
v
Using the previous relations for m̄1 and m̄2 in terms of m̄3 and
MZ , we then obtain the mass matrix for the CP even Higgs bosons:
2
2
2
2
2
−
m̄
tan
β
+
M
cos
β
m̄
−
M
sin
β
cos
β
3
3
Z
Z
M2R =
2
2
2
−m̄3cotβ + MZ2 sin2 β
m̄3MZ sin β cos β
In the case of the CP–odd Higgs boson: use the same expressions
for ∂ 2V /∂Hi0∂Hj0 as above but set the fields to zero at the minimum:
2
2
−m̄3 tan β
m̄3
M2I =
−m̄23cotβ
m̄23
Since DetM2I = 0, one eigenvalue is zero (the Goldstone) while the
other one corresponds to the mass of the pseudoscalar Higgs:
MA2 = −m̄23(tan β + cotβ) = −2m̄23/ sin 2β
The mixing angle θ is, in fact, just the angle β:
#− 21
"
#− 12
"
2
2
c2β
sβ c β
+4
=2 2
= s2β
−
+4
sin 2θ = 2
c β sβ
s2β /4
−1/2
sβ c β c2β cos 2θ = − +
······
s2β = c2β
=
c β sβ
s2β
25
In the case of the charged Higgs boson, same excercice:
±
±
H1
G
cos β sin β
=
H2±
− sin β cos β
H±
with a massless charged Goldstone and a charged Higgs with mass:
2
MH2 ± = MA2 + MW
Back to the CP–even Higgsses: inject expression of MA2 into M2R :
2
2
2
2
2
2
)
sin
β
cos
β
+
M
cos
β
−(M
sin
β
+
M
M
Z
A
Z
A
M2R =
2
2
2
2
−(MA + MZ ) sin β cos β MA cos β + MZ2 sin2 β
Calculating determinant and trace, one obtains for the masses:
DetM2R = (MA2 s2 +MZ2 c2 )(MA2 c2 +MZ2 s2)−(MA2 +MZ2 )s2c2 = MA2 MZ2 c22β ≡ Mh2MH2
TrM2R = MA2 s2 + MZ2 c2 + MA2 c2 + MZ2 s2 = MA2 + MZ2 ≡ Mh2 + MH2
To obtain the CP–even Higgs masses, solve the equation:
Mh2(MA2 +MZ2 −Mh2) = MA2 MZ2 c22β ⇒ Mh4 −Mh2(MA2 +MZ2 )+MA2 MZ2 c22β = 0
2
2
2
2 2
2
with discriminant ∆ = (M√
A + MZ ) −4MA MZ cos 2β, the two solutions
2
are: Mh,H
= 12 (MA2 + MZ2 ∓ ∆) giving (h is the lightest Higgs):
q
1
2
Mh,H
=
MA2 + MZ2 ∓ (MA2 + MZ2 )2 − 4MA2 MZ2 cos2 2β
2
The mixing angle α which rotates the fields is (− π2 ≤ α ≤ 0)
MA2 + MZ2
2M12
−(MA2 + MZ2 ) sin 2β
= tan2β 2
tan2α =
=
M11 − M22
(MZ2 − MA2 ) cos 2β
MA − MZ2
We see that we have an important constraint on the lightest h
boson mass in the MSSM:
Mh ≤ min(MA, MZ ) · | cos 2β| ≤ MZ
26
besides some other (alos important) relations:
MH > max(MA, MZ ) and MH ± > MW
If we send MA to infinity, we will have for the Higgs masses and α:
π
Mh ∼ MZ | cos 2β| , MH ∼ MH ± ∼ MA ,
α ∼ −β
2
This is the decoupling regime: all Higgses are heavy except for h.
The h boson is lighter than MZ and should have been seen at LEP2
√
(we have sLEP2 ∼ 200 GeV > Mh + MZ ∼ 180 GeV).
So what happened in this case? Maybe the MSSM is ruled out?
No! This relation holds only at first order (tree–level): there are
strong couplings involved, in particular the htt and ht̃t̃ couplings.
⇒ Calculation of radiative corrections to Mh necessary.
We have to include the important corrections due to top (s)quarks.
2. Calculation of radiative corrections to Mh
Let us do the calculation, but with some simplifications:
– take the (decoupling) limit MA → 0 and use tan β 1 (Mhmax)
– assume no stop mixing and same masses, mt̃1 = mt̃2 = mt̃
√
∗
2
– simple couplings: ht̄t ∼ ht̃t̃ ∼ λt , hht̃ t̃ ∼ λt with λt = 2mt/v
– work in the limit Mh mt, mt̃.
In addition to two–point functions including fermion/scalar loops,
we have also tadpole contributions (counterterm corrections):
27
t
h
•
h
•
t̃i
t̃i
h
h
•
t
h
•
•
h
t̃i
•
•
h •
h •
• The calculation is almost already done: for two–point function:
∆Mh2|2
3λ2t 2
mt̃ Λ
2
2
=
+ 3mt log
(mt − mt̃ )log
4π 2
mt̃
mt
• For the tadpole contributions, the calculation is very simple:
Z
2
i
λ
1
−iM
d4 k
f
2
H
√
−i
∆Mh |1 = iNf
(−4mi)
v
−Mh2
(2π)4 k 2 − m2f
2
Z
−iMH2
i
d4 k
1
+ iNS
(ivλ
)i
S
v
−Mh2
(2π)4 k 2 − m2S
Z 2
Z 2
y
4Nf mf λf Λ
NS λS Λ
y
dy
= √
+
dy
y + m2f
16π 2 0
y + m2S
2v16π 2 0
R
Using λS =-λ2f =-2m2f /v and ... = Λ2 − m2log(Λ2/m2), one obtains
∆Mh2|1
Λ
Λ 3λ2t 2
− m2t log
= − 2 2 mt̃ log
4π v
mt̃
mt̃
• The total correction to the h boson mass is then given by:
∆Mh2
m2t̃
mt̃ 3g 2 m4t
3m4t
log
=
=
2 log m2
2π 2v 2 mt 2π 2 MW
t
Its depends on m4t and log(m2t̃ /m2t ), and is large: Mhmax → MZ + 40
GeV! This explains why the h boson has not been seen at LEP2.
28
3. The Higgs boson couplings
Trilinear and Quartic scalar couplings
The trilinear (3 fields) et quartic (4 fields) couplings among Higgs
bosons can be obtained from the scalar potential VH by making:
λ2ijk
λ2ijkl
∂ 3 VH
0
=
±
0
∂Hi ∂Hj ∂Hk hH1 i=v1,hH2 i=v2,hH1,2i=0
∂ 4 VH
0
=
±
0
∂Hi ∂Hj ∂Hk ∂Hl hH1 i=v1,hH2 i=v2,hH1,2i=0
with the Hi expressed in terms of the fields h, H, A, H ± and G0, G±
with rotations of angles β et α. Examples (unit: λ0 = −iMZ2 /v):
λhhh = 3 cos 2α sin(β + α) + rad. corr.
λhhhh = 3 cos2 α/MZ2 + rad. corr.
(in units of λ20)
In the decoupling limit, MA MZ we have α → β − π/2:
λhhh → 3 cos2(2β) = 3Mh2/MZ2 = λ3|MS
λhhhh → 3 cos2(2β)/MZ2 = 3Mh2/MZ4 = λ4|MS
⇒ In the decoupling limit, MA MZ , the Higgs potential of the
MSSM becomes like the one of the SM: only one light Higgs with a
< 130 GeV and with standard interactions. All other Higmass Mh ∼
gsses are heavy and decouple (but self–couplings are non–zero).
Couplings to gauge bosons:
Higgs couplings to massive gauge bosons are obtained from the
kinetic terms of H1 et H2 in the SU(2)L × U(1)Y Lagrangian:
Lkin. = (DµH1)† (DµH1) + (D µH2 )†(DµH2)
29
Develop Dµ and make the usual transformations on the fields:
YH
~τa
Dµ = i∂µ − g Wµa − g 0 i Bµ
2
2
±
±
W1,2,3, B → W , Z, γ ; H1,2 → h, H, A, H , G0, G± via rotations β, α
ghiV V ≡ coefficients de hiVµVµ (gµν )
ghihj V ≡ coefficients de hihj Vµ (∂µ → pµ)
ghi hj V V ≡ coefficients de hihj VµVµ (gµν )
Some very important couplings for Higgs phenomenology:
igMZ
igMZ
sin(β − α)g µν , Z µ Z ν H :
cos(β − α)g µν
cos θW
cos θW
µ
ν
µν
µ
ν
W W h : igMW sin(β − α)g , W W H : igMW cos(β − α)g µν
g sin(β − α)
g cos(β − α)
(p + p0)µ , Z µ HA : −
(p + p0)µ
Z µhA :
2 cos θW
2 cos θW
Z µZ ν h :
– γ massless: no coupling with the neutral Higgsses at tree–level.
– CP invariance: no ZZA and Zhh, ZHh, ZHH couplings e.g.
2
2
2
!
= gMS
+ gHZZ
– Couplings of h and H complementary: ghZZ
– Decoupling limit (MA → ∞, α → β− π2 ): sin(β−α) → 1, cos(β−α) → 0:
⇒ ghV V = gHMSV V , gHV V = 0 (= gAV V )
.
Yukawa couplings to fermions:
The Higgs couplings to fermions come from Superpotential W :
X
W =
Yiju ûiR Ĥ2.Q̂j + Yijd dˆiR Ĥ1 .Q̂j + Yijl lˆRi Ĥ1.L̂j + µĤ1.Ĥ2
i,j=gen
with LYuk = − 12
P ∂2W
ψ
ψ
+
h.c.
evaluated in terms of H1, H2 .
iL ∂zi ∂zj j
ij
30
Take bilinears out, digagonal Y matrices Y with relations to masses,
and expressings H1, H2 in terms of the physical fields, we get:
gmu
LYuk = −
[ūu(H sin α + h cos α) − iūγ5u A cos β]
2MW sin β
gmd
¯
¯
dd(H cos α − h sin α) − idγ5d A sin β
−
2MW cos β
g
m
u
+ √
(1 + γ5)]d + hc
H + ū[md tan β(1 + γ5) +
tan β
2 2MW
Couplings in termsof those of HSM [factor −(i)gmf /2MW = −imf /v]
and their values in the decoupling limit [cos α → sin β, sin α → − cos β]:
f
gf f h
gf f H
gf f A
u cos α/ sin β → 1 sin α/ sin β → − tan β cot β
d − sin α/ cos β → 1 cos α/ cos β → tan β tan β
–
–
–
–
The couplings of H ± have the same intensity as those of A.
For tan β > 1: Cplgs to d enhanced, cplgs to u suppressed.
For tan β 1: couplings to b quarks b (mb tan β) very strong.
For MA MZ : h couples like the SM Higgs boson and H like A.
4. Summary of masses and couplings including RC
Radiative corrections very important in the MSSM Higgs sector
A large activity for the RC calculation in the last 15 years
• The full one–loop corrections have been calculated:
– the parameters µ, At and Ab appear at the subleading level.
– the h boson mass is maximal (minimal) for At ∼ 2MQ̃(0).
• Approximate calculation for the dominant two–loop QCD RC:
– dominant QCD RC large but absorbed by mt |pole → mt|MS .
31
– Yukawa corrections rather small in the limit Mh = 0.
• Recently, complete corrections in effective potential approach:
– O(αt αS ): including squark mixing and gluino loops.
– O(αt2): including mixing and O(αb αS ), O(ατ αS ).
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