Physics of Higgs Bosons Marco Aurelio D´ ıaz Departamento de F´ ısica, Pontificia Universidad Cat´ olica de Chile, Santiago 690441, Chile. (Dated: March 8, 2012) We briefly review the theory and phenomenology of Higgs boson physics, and include a presentation of the latest experimental results, concentrating with preference on the results from the ATLAS Experiment.
55
Embed
Physics of Higgs Bosons - University of …sheaff/PASI2012/lectures/Higgs.pdfPhysics of Higgs Bosons Marco Aurelio D´ıaz Departamento de F´ısica, ... The Branching Ratios of a
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Physics of Higgs Bosons
Marco Aurelio Dıaz
Departamento de Fısica, Pontificia Universidad Catolica de Chile, Santiago 690441, Chile.
(Dated: March 8, 2012)
We briefly review the theory and phenomenology of Higgs boson physics,and include a
presentation of the latest experimental results, concentrating with preference on the results
from the ATLAS Experiment.
2
I. HIGGS MECHANISM IN A U(1)U(1)U(1) GAUGE THEORY
From the Higgs field point of view, theU(1) gauge symmetry implies that the transformation
over theΦ field is,
Φ −→ Φ′ = e−iqθ(x)Φ (1)
whereΦ is a complex Higgs field,q is theU(1) charge of the Higgs field,θ(x) is the space-time
dependent transformation parameter, and the objectsexp[−iqθ(x)] form a representation of the
U(1) group.
The lagrangian is taken as,
L =[(∂µ − iqAµ) Φ†] [(∂µ + iqAµ) Φ] − 1
4FµνF
µν − V(Φ†Φ
)(2)
whereAµ is the gauge field, withFµν = ∂µAν − ∂νAµ and
Aµ −→ A′µ = Aµ + ∂µθ (3)
is theU(1) transformation of the gauge field. For the Higgs potential wetake,
V(Φ†Φ
)= µ2 Φ†Φ + λ
(Φ†Φ
)2(4)
whereµ2 has units of mass squared andλ is a dimensionless parameter that accounts for the
quartic Higgs self interaction. The Higgs potential is shown in Fig. 1 for two different values of
the sign ofµ2.
It is understood that the vacuum is the state with minimum energy, otherwise the extra en-
ergy could be used to create a particle, and the state would cease to be vacuum. The vacuum
expectation value of the Higgs field〈Φ〉 = v is the crucial quantity that gives mass to the gauge
bosonAµ, as we will see. Therefore, the situationµ2 > 0 at the left of Fig. 1 is not useful, and
we need a non zero value forv, thus
µ2 < 0 (5)
In fact, the Higgs field is complex, thus the minimum of the Higgs potential is a circle in this
complex plane, as indicated in Fig. 2. This vev though can always be chosen as real because
gauge invariance allow us to choose a gauge whereΦ′ = exp(−iqθ)Φ is real. Since the vev is
not gauge invariant, it is said that the gauge symmetry has been spontaneously broken.
3
FIG. 1: Higgs potential for two different signs ofµ2.
FIG. 2: Higgs potential in the complex plane of the Higgs field.
If we replace the Higgs field in eq. (4) by its vev, the potential becomesV = µ2v2 + λv4,
and looking for the minimum we find the following minimization condition,
∂V
∂v≡ t = 2µ2v + 4λv3 = 0 (6)
or tadpole equation. From here we confirm thatµ2 = −2λv2 is negative. The Higgs mass can
be directly obtained from the second derivative of the potential,
1
2
∂2V
∂v2≡ m2
H = µ2 + 6λv2 = 4λv2 (7)
as we will confirm below.
4
If in the gauge where the Higgs field is real we redefine the Higgs field as
Φ′(x) = v +1√2H(x) (8)
we find a lagrangian,
L =[(∂µ − iqAµ)
(v + H/
√2)] [
(∂µ + iqAµ)(v + H/
√2)]
− 1
4FµνF
µν
−µ2(v + H/
√2)2
− λ(v + H/
√2)4
(9)
We drop the constant, which is irrelevant in this context, and divide the lagrangian into two
pieces,
L = Lfree + Lint (10)
The free lagrangian contains the terms,
Lfree =1
2∂µH∂µH − m2
HH2 − 1
4FµνF
µν + q2v2AµAµ (11)
while the lagrangian that includes the interactions is,
Lint = q2AµAµ
(√2vH +
1
2H2
)− λ
(√2vH3 +
1
4H4
)(12)
From the free lagrangian in eq. (11) we see that the Higgs boson H has a mass proportional
to the quartic self couplingλ. In addition, a mass has been generated for the gauge boson
mA = 2q2v2, which is proportional to the Higgs vev. Notice that this mass cannot be included
by hand in the lagrangian since it is not gauge invariant.
Schematically, the interactions in eq. (12) are represented by the following Feynman rules,
A
A
H
A
A
H
H
H
H
H
H
H
H
H
of which only the first one is relevant for the present Higgs boson search.
5
II. HIGGS MECHANISM IN A SU(2) × U(1)SU(2) × U(1)SU(2) × U(1) GAUGE THEORY
Here we study the Higgs mechanism in the Standard Model. In addition to the (electroweak)
gauge groupSU(2) × U(1), it is assumed there is a unique Higgs doublet,
Φ =
φ1 + iφ2
φ3 + iφ4
(13)
with hyperchargeYΦ = 1. The relevant part of the lagrangian is,
L = (DµΦ)† (DµΦ) − V (Φ†Φ) (14)
with
DµΦ =
(∂µ +
i
2g′Bµ +
i
2gW k
µσk
)Φ (15)
The lagrangian is invariant under the gauge transformation,
Φ → Φ′ = e−iθUΦ (16)
whereU is aSU(2) 2 × 2 matrix. The Higgs potential is the same as in eq. (4), with theonly
difference that now the Higgs field is a complex doublet. Since Q = I3 + Y/2, the lower
(I3 = −1/2) Higgs field is electrically neutral. A gauge transformation can take us to the
unitary gauge whereφ1 = φ2 = φ4 = 0. In this gauge, the Higgs vev is
〈Φ〉 =
0
v/√
2
(17)
(note the change in normalization of the Higgs vev). If we do the analogous field redefinition
as in eq. (8)
Φ =
0
(v + H)/√
2
(18)
we find gauge boson masses,
m2γ = 0 , m2
Z =1
4(g2 + g′2)v2 , m2
W =1
4g2v2 (19)
6
and the measurement of the gauge boson masses and other couplings implies the valuev = 246
GeV. The Higgs kinetic terms leads to,
(DµΦ)† (DµΦ) =1
2∂µH∂µH + m2
W W+µ W−µ +
1
2g2W+
µ W−µ(vH + H2/2)
+1
2m2
ZZµZµ +
1
4(g2 + g′2)ZµZ
µ(vH + H2/2) (20)
generating the following Feynman rules,
W
W
H = igmW gµν
Z
Z
H = igmZ
cWgµν
which are very important for the production and decay of the Higgs particle. If the Higgs boson
had only couplings to gauge bosons, we would find the following set of branching ratios,
FIG. 3: Branching ratios of a fermiophobic Higgs boson.
7
As we will see in the next section, the SM Higgs boson also couples to fermions. But exten-
sions of the Higgs sector of the SM can contain Higgs bosons that do not couple to fermions:
fermiophobic Higgs bosons.
8
III. FERMION MASSES
The Higgs mechanism can be used also to give mass to the fermions. In this analysis we
exclude the neutrinos. Consider first the charged leptons. They can acquire a mass via the vev
of the Higgs doublet. We define the following lepton doublet and singlet,
L =
νeL
eL
, eR (21)
which transform as a doublet underSU(2). In addition, it transform underU(1) with an hy-
perchargeYeL= −1. The right handed electron is a singlet underSU(2) and transform under
U(1) with an hyperchargeYeR= −2. With this we can write the following Yukawa term,
LY uk = −fe
[(L†Φ
)eR + e†R
(Φ†L
)](22)
In the unitary gauge we find,
LY uk = −fe
(e†LeR + e†ReL
)(v + H) /
√2 (23)
Thus, a Dirac mass is generated for the charged lepton,
me =1√2fev (24)
which is proportional to the Higgs vev and the Yukawa coupling. In addition, a Yukawa inter-
action between the Higgs boson and the lepton is formed,
e+
e−
H = i√2he = i gme
2mW
As we can see, the Higgs boson interacts with the lepton with astrength given by the Yukawa
coupling: the heavier the lepton, the stronger the interaction. Analogous masses and interactions
are obtained for the heavier generations.
In the case of quarks, we define anSU(2) left doublet and two right singlets,
Q =
uL
dL
, uR , dR (25)
9
with hyperchargesYQ = 1/3, YuR= 4/3, andYdR
= −2/3. The relevant Yukawa term for
down-type quarks in the lagrangian has the form,
LY uk = −fd
[(Q†Φ
)dR + d†
R
(Φ†Q
)](26)
With the same mechanism as before, the down-type quark receives a mass,
md =1√2fdv (27)
and a Yukawa interaction,
d
d
H = i√2hd = i gmd
2mW
which is also proportional to the mass of the quark. The case for the up-type quark is less
obvious. The previous terms are based on the fact that(Q†Φ) is anSU(2) invariant with com-
bined hypercharge2/3 that can be cancelled bydR. For the up-type quarks we use the fact that
(ΦT iσ2Q) is also anSU(2) invariant with combined hypercharge4/3 that can be cancelled by
u†R,
LY uk = fu
[(Q†iσ2Φ
∗) uR − u†R
(ΦT iσ2Q
)](28)
which leads to the mass,
mu =1√2fuv (29)
and a Yukawa interaction,
u
u
H = i√2hu = i gmu
2mW
10
FIG. 4: Branching ratios of a SM Higgs boson.
Notice that in the three cases, when including three generations, the Yukawa couplings can be
promoted to3 × 3 Yukawa matrices, which may or may not be diagonal.
The Branching Ratios of a SM Higgs boson, including decays into leptons, are shown in
Fig. 4. At low Higgs masses the decays intobb, cc, andτ+τ− dominates over theγγ decay. In
addition, thett decay apprears at the top quark threshold.
FIG. 5: Total width of the SM Higgs boson.
11
In Fig. 5 we can see the total width of the SM Higgs boson. It grows very fast with the mass
mH , from a few MeV at low masses to more than 100 GeV at large masses. For comparison we
have also supersymmetric Higgs bosons, which will be introduced later.
12
IV. UNITARITY OF WWW -WWW SCATTERING
The high energy behaviour of theW -W scattering allow us to show the role of the Higgs
boson. Consider the scattering,
W+(p1) + W−(p2) → W+(k1) + W−(k2) (30)
The relevant diagrams in the Standard Model involving only gauge bosons are,
Z, γ+ Z, γ +
and the ones involving the SM Higgs boson,
H+ H
The best strategy is to work in the Centre of Mass frame of reference, with the incomingW
bosons in thez axis. In this case, the incoming four momenta are,
p1 = (E, 0, 0, p) p2 = (E, 0, 0,−p) (31)
with E2 = p2 + m2W . We have the freedom to choose the orientation of thex andy axis, we
choose it such that the outgoingW bosons lie in theyz plane. In this case the outgoingW four
momenta are,
k1 = (E, 0, p sin θ, p cos θ) k2 = (E, 0,−p sin θ,−p cos θ) (32)
13
whereθ is the scattering angle. We are interested in the high energybehaviour of this scattering.
The longitudinal polarization of theW bosons are responsible for the high energy terms, and
the transverse polarizations can be neglected. The longitudinal polarizations are,
εL(p1,2) = (p, 0, 0,±E)/mW εL(k1,2) = (p, 0,±E sin θ,±E cos θ)/mW (33)
which are normalizedε2 = −1 and satisfy the Lorentz conditionε(q) · q = 0. The amplitude
for the graphs involving three gauge boson vertices is,
M3gb = g2 p4
m4W
(3 − 6 cos θ − cos2 θ
)+
g2
2
p2
m2W
(9 − 11 cos θ − 4 cos2 θ
)+ ... (34)
where we have neglected terms that do not grow with the momentump. Similarly, the diagram
with four gauge bosons gives the following amplitude,
M4gb = −g2 p4
m4W
(3 − 6 cos θ − cos2 θ
)+
g2
2
p2
m2W
(−8 + 12 cos θ + 4 cos2 θ
)+ ... (35)
Interestingly, the terms proportional top4 cancel among the gauge boson graphs, but not thep2
terms. Notice that these terms are unwanted because whenp is arbitrarily increased we do not
want the amplitude to arbitrarily increase also, since perturbation theory breaks down.
The graphs involving a Higgs boson contribute with,
MH = −g2
2
p2
m2W
(1 + cos θ) +g2
4
m2H
m2W
(s
s − m2H
+t
t − m2H
)+ ... (36)
with t = −2p2(1−cos θ). From here we see that the Higgs boson graphs are necessary tocancel
thep2 terms, otherwise the cross section could grow unacceptablylarge. The total amplitude
includes the also important terms proportional to the Higgsmass,
M =g2
4
m2H
m2W
(s
s − m2H
+t
t − m2H
)+ ... (37)
This term does not diverge at large momentump, but could be too large for a large Higgs boson
mass. The differential cross section is in terms of the amplitude,
dσ
dΩ=
1
64π2s|M|2 (38)
To study the effect of a large Higgs mass on this cross section, we do the partial wave decom-
position of the amplitude,
M = 8π∑
j
(2j + 1)AjPj(cos θ) (39)
14
where thePj(x) are the Legendre polynomials. Since these polynomials are orthogonal and
have a definite normalization,∫ 1
−1
dxPi(x)Pj(x) =2
2j + 1δij (40)
the total cross section becomes very simple,
σ =4π
s
∑
j
(2j + 1)|Aj|2 (41)
The optical theorem states that,
σ =1
2sImM(θ = 0) (42)
with each of these partial waves satisfying a ”unitarity bound” that in our language is expressed
as,
|Aj|2 = Im Aj =⇒ Aj = eiδj sin δj (43)
which means that any of the partial wave amplitudes cannot exceed unity. From eq. (37) we
find,
A0 = − m2H
64πm2W
[2 +
m2H
s − m2H
− m2H
sln
(1 +
s
m2H
)](44)
and if we take the limits ≫ m2H we get,
A0 −→ − m2H
32πm2W
(45)
Since this partial wave amplitude must have a magnitude smaller than unity we obtain the
following limit for the Higgs mass,
mH <√
32πmW ≈ 800 GeV (46)
with more precise calculations reaching the 1 TeV limit. So,the lesson is, unitarity needs a
Higgs boson, with a mass not larger than 1 TeV. The alternative is that perturbation theory
breaks down at these energies, thus, some strong interaction betweenW gauge bosons should
appear.
15
V. HIGGS BOSONS IN THE MSSM
In the Minimal Supersymmetric Standard Model we need two Higgs doubletsHd andHu
with hyperchargesYHd= −1 andYHu
= 1. The Higgs potential is,
V =1
8(g2 + g′2)
(|Hu|2 − |Hd|2
)2+
1
2g2|H†
dHu|2
+m21H |Hd|2 + m2
2H |Hu|2 − m212
(HT
d iσ2Hu + h.c)
(47)
where in the first line we have the supersymmetric quartic Higgs self interactions, in the second
line we havem21H andm2
2H mass terms that receive contributions from the supersymmetric
higgsino mass and soft supersymmetry breaking mass terms, and m212 which is a purely soft
mass term.
Notice that the quartic couplingλ in the SM is replaced by gauge couplings in the MSSM.
The fixed value of these couplings is the reason why the lightest Higgs boson in the MSSM
cannot be as heavy as the SM Higgs.
Under certain reasonable conditions on the soft masses, theelectroweak symmetry is sponta-
neously broken when the two Higgs doublets acquire a non trivial vev. We do the replacement,
Hd =
1√2(vd + χd + iφd)
H−d
, Hu =
H+u
1√2(vu + χu + iφu)
(48)
Gauge boson masses are generated in a similar way as in the SM,obtaining
m2W =
1
4g2(v2
u + v2d) , m2
Z =1
4(g2 + g′2)(v2
u + v2d) (49)
These masses come from the Higgs kinetic terms, and there is acontribution from both Higgs
FIG. 21: Observed and expected (median, for the background-only hypothesis) 95% C.L. upper limits
on the ratios to the SM cross section, as functions of the Higgs boson mass for the combined CDF and
D0 analyses.
They see a small1σ “excess” of data with respect to background in the mass range125 <
mH < 155 GeV. The production mechanisms considered are Higgs-strahlung qq → W/ZH,
gluon-gluon fusiongg → H, and vector boson fusionqq → q′q′H. The decay modes studied
areH → bb, H → W+W−, H → ZZ, H → τ+τ−, andH → γγ.
40
XIV. HIGGS SEARCHES WITH ATLAS AND CMS AT THE LHC
FIG. 22: ATLAS Detector.
FIG. 23: CMS Detector.
41
A. H → WW ∗ → ℓ+νℓ−νH → WW ∗ → ℓ+νℓ−νH → WW ∗ → ℓ+νℓ−ν channel.
A search for the SM Higgs boson was done with the ATLAS detector in the channelH →WW ∗ → ℓ+νℓ−ν where the lepton can be electron or muon. This is done inpp collisions
at√
s = 7 TeV, with an integrated luminosity of2.05 fb−1. The decay modeH → WW ∗ is
dominant formH<∼ 135 GeV, as we can see in Fig. 4.
Electron candidates are selected from clustered energy deposits in the electromagnetic
calorimeter, with an associated track reconstructed in theinner detector. They are identified
with an efficiency of71% for electrons with transverse energyET > 20 GeV and|η| < 2.47.
Muons are reconstructed by combining tracks from the inner detector and muon spectrometer,
with an efficiency of92% for pT > 20 GeV and|η| < 2.4. Electrons and muons should be
produced at the primary vertex, which should have more than 3tracks. Both leptons should be
isolated within a cone
∆R =√
∆φ2 + ∆η2 < 0.2 (72)
One important parameter is the invariant mass of the pair of leptonsmℓℓ, defined as,
m2ℓℓ = (pℓ1 + pℓ2)
2 (73)
If the two leptons have different flavours their invariant mass is required to bemℓℓ > 10 GeV,
otherwisemℓℓ > 15. This is to suppress the background fromΥ. In addition, to suppress the
background fromZ production it is required|mℓℓ − mZ | > 15 GeV.
A second important variable is the azimuthal angle between the two leptons∆φℓℓ, calculated
simply with the product of the two 3-vectors,
~p1T · ~p2T = |~p1T ||~p2T | cos ∆φℓℓ (74)
This angle is used to exploit differences in spin correlations between signal and background:
∆φℓℓ < 1.3 for mH < 170 GeV, and∆φℓℓ < 1.8 for mH < 120 GeV.
42
FIG. 24: mT distribution shown after all cuts formH = 150 GeV, except for themT cut itself. The
top graph shows the selection for theH + (0 jet) channel and the bottom for theH + (1 jet) channel.
The background distributions are stacked, so that the top of the diboson background coincides with the
Standard Model (SM) line which includes the statistical and systematic uncertainties on the expectation
in the absence of a signal. The expected signal formH = 150 GeV is shown as a separate thicker line,
and the final bin includes the overflow.
For the transverse massmT we require0.75mH < mT < mH if mH < 220 GeV, and
0.6mH < mT < mH otherwise. These requirements reduce theWW and top backgrounds.
43
TABLE III: The expected numbers of signal (mH = 150 GeV) and background events after the require-
ments listed in the first column, as well as the observed numbers of events in data. All numbers are
summed over lepton flavor.
H + 0-jet Channel Signal WW W + jets Z/γ∗ + jets tt tW/tb/tqb WZ/ZZ/Wγ Total Bkg. Observed