Construction of the Higgs Mechanism
and the Lee-Quigg-Thacker-bound
Franz Wilhelm
Supervisor: Tanumoy Mandal
Subject Reader: Rikard Enberg
Division of High Energy Physics,
Department of Physics and Astronomy,
Abstract
In this paper the higgs mechanism for the standard model is constructed in steps. First by considering spontaneous breaking of discrete and continuous global gauge invariance. Then spontaneous breaking of local gauge invariance. These results are then used to construct the electroweak part of the standard model through application of the higgs mechanism. Finally, the LQT -upper bound of 1 TeV for the higgs mass is calculated through unitarity constraints.
Contents
1 Introduction 2
2 Spontaneous Breaking of Discrete Symmetry 2
3 Spontaneous Breaking of Continuous Symmetry 5
4 Spontaneous Breaking of Gauge Symmetry 8 4.1 Abelian Case U(1) . . . 9 4.2 Non-Abelian Case SU(2) . . . 11
5 Higgs Mechanism in the Standard Model 14 6 Unitarity Constraints on the Higgs Mass 17
1
Introduction
Due to the many, experimentally known, conserved quantities in particle physics a suitable theory for it would most likely be a gauge theory. However, ordinary gauge theories lead to zero masses for gauge particles, while it is known exper-imentally that the masses of the weak gauge bosons (W±, Z0) are non-zero. To retain the symmetries in the theory, resulting in conserved quantities by Noether’s theorem, and still obtain non-zero masses for the gauge bosons spon-taneous symmetry breaking and the higgs mechanism can be applied. This solves the problem but also necessitates the existence of a new particle with unknown mass, known as the higgs particle.
This paper will derive a Lagrangian for the electroweak sector of the standard model by applying the higgs mechanism. It will be done over several sections with increasing complexity. Starting with spontaneous symmetry breaking with-out any gauge fields present for theories that are rotationally invariant (or in-variant under reflection in the 1-dimensional case) in sections 2 Spontaneous Breaking of Discrete Symmetry and 3 Spontaneous Breaking of Con-tinuous Symmetry.
Gauge fields will then be added in section 4 Spontaneous Breaking of Gauge Symmetry where the higgs mechanism will be introduced as the pro-cess giving gauge particles masses. This will be done first for the Abelian U(1) case and then for the non-Abelian SU(2) case.
Then all prerequisites for constructing the electroweak standard model have been given and section 5 Higgs Mechanism in the Standard Model is about the above mentioned construction.
Finally, the paper ends on deriving the Lee-Quigg-Thacker upper bound for the higgs mass in section 6 Unitarity Constraints on the Higgs Mass, before giving a Summary in section 7.
2
Spontaneous Breaking of Discrete Symmetry
Consider the following Lagrangian
L = −1 2∂µφ∂ µφ − 1 2µ 2φ2+λ 4φ 4 , (1)
which is invariant under reflection. This can be seen by making the substitution φ → −φ, in eqn. 1, which yields:
L = −1 2∂µ(−φ)∂ µ(−φ)− 1 2µ 2(−φ)2+λ 4(−φ) 4 = −1 2∂µφ∂ µφ− 1 2µ 2φ2+λ 4φ 4
The first term in the Lagrangian is recognized as the kinetic term and the expression in parenthesis as the potential V (φ). If µ2> 0, then the potential is
-10 -5 0 5 10 0 5000 10000 15000 V( )
Potential for the 2>0 case
Figure 1: In this graph V (φ) = φ4+ 50φ2.
If µ2< 0, then V (φ) is no longer positive semi-definite and the minima can be found by finding the zeroes of V0(φ):
0 = V0(φ) = µ2φ + λφ3⇔ ( φ1= 0 φ2,3= ± q |µ2| λ =: ±v (2)
-10 -5 0 5 10 -1000 0 1000 2000 3000 4000 5000 V( )
Potential for the 2<0 case
Figure 2: The graph shows that φ2,3 are the minima while φ1 is a local
maxi-mum. In this plot V (φ) = φ4− 50φ2.
The Lagrangian can then be expanded around one of the minima, say +v, as follows: φ(x) = v + η(x), (3) resulting in: (4) L = −1 2∂µφ(x)∂ µφ(x) − 1 2µ 2φ2(x) + λ 4φ 4(x) = −1 2∂µ(v + η(x))∂ µ(v + η(x)) − 1 2µ 2(v + η(x))2+λ 4(v + η(x)) 4 = −1 2∂µη(x)∂ µη(x) − 1 2µ 2(v + η(x))2+λ 4(v + η(x)) 4 ,
Unlike the first Lagrangian this is not invariant under reflection, η → −η, which can be seen for instance from the term
−µ2vη
obtained by expanding the first term in the parenthesis. Upon making the substitution this term transforms into:
−µ2vη → µ2vη 6= −µ2vη
The mass corresponding to the field η is identified as the constant mη in the
term in the Lagrangian given on the following form:
−m
2 η
2 η
2
The first term in the Lagrangian does not yield any term of this form. The second gives:
−µ
2
2 η
2
and the third produces (keeping only terms of the right order):
−λ 4(v + η) 4= −λ 4(v 2+ 2vη + η2)2⇒ −λ 4(2v 2η2+ 4v2η2) = −3λv2 2 η Collecting these terms and simplifying with Eqn. (2) yields the mass term:
−µ 2+ 3λv2 2 η 2= −µ 2+ 3λ−µ2 λ 2 η 2= 2µ 2 2 η 2 (5)
The mass is then seen to be:
mη=p−2µ2, (6)
which is real and positive since µ2< 0.
Except for µ2 = 0, these are all possible cases, because µ2 must be real in
order for the action to be real. The last case, µ2= 0, does not yield anything
new and is similar to the µ2> 0 case.
3
Spontaneous Breaking of Continuous
Symme-try
In the following Lagrangian, that is symmetric under rotations in the φ1φ2
-plane, L = −1 2(∂µφ1) 2−1 2(∂µφ2) 2+1 2µ 2(φ2 1+ φ 2 2) − λ 4(φ 2 1+ φ 2 2) 2, (7)
the first two terms are recognized as the kinetic terms. Thus leaving
V (φ1, φ2) = λ 4(φ 2 1+ φ22)2− 1 2µ 2(φ2 1+ φ22), (8)
for the potential.
Extremizing the potential results in: ( 0 = ∂φ∂V 1 = λ(φ 2 1+ φ 2 2)φ1− µ2φ1 0 = ∂φ∂V 2 = λ(φ 2 1+ φ22)φ2− µ2φ2 ⇒ φ 1= 0 = φ2 (φ2 1+ φ22) = µ2 λ =: v 2 (9)
If µ2< 0, then V = 0 is the minimum. However, if µ2> 0, then the minimum
is equal to V = −µ42v2.
Henceforth it will be assumed that µ2 > 0. Then the potential will appear as in figure 3.
Figure 3: The graph shows that the circle φ2
1+ φ22= v2 is indeed the minima,
while φ1 = 0 = φ2 is a local maximum. VEV stands for vacuum expectation
value, which is the expectation value of the potential for the vacuum state. In this graph V (φ1, φ2) = −170(φ21+ φ 2 2) + (φ 2 1+ φ 2 2) 2.
Now, let’s rewrite the Lagrangian in Eqn. (7) with the following substitution:
φ1(x) = π(x)
φ2(x) = v + σ(x)
(11)
where the minimum in the φ2-direction has been chosen to expand around. This
yields: L = −1 2(∂µφ1) 2−1 2(∂µφ2) 2+1 2µ 2(φ2 1+ φ 2 2) − λ 4(φ 2 1+ φ 2 2) 2= −1 2(∂µπ(x)) 2 −1 2(∂µ(v + σ(x))) 2+1 2µ 2(π2(x) + (v + σ(x))2) −λ 4(π 2(x) + (v + σ(x))2)2 = −1 2(∂µπ) 2 −1 2(∂µσ) 2+1 2µ 2(π2+ (v + σ)2) −λ 4(π 2+ (v + σ)2)2 (12) The field π while then have a mass term of the form:
−m
2 π
2 π
where mπ is the mass corresponding to the field.However, it turns out that this
term is zero. The first two terms in the Lagrangian do not contribute to the mass term. The second term gives:
µ2 2 π
2
and the last term yields:
−λ 4 · 2v 2π2= −λ 2 µ2 λπ 2= −µ2 2 π 2
Adding these two terms then gives a mass, mπ = 0. The result is as expected.
The π-field is just a different name for the φ1-field, which can be seen from the
transformation (Eqn. (11)). Since phi1 was massless it makes sense for π to be
massless as well.
The same process can be repeated to find the mass of the σ-field. The first two terms do not contribute in this case either, while the third gives:
µ2
2 σ
2
and the fourth:
−λv2σ2= −µ2σ2 which together give a mass of:
−m 2 σ 2 σ 2:=µ2 2 σ 2− µ2σ2= −µ2 2 σ 2⇒ m σ= µ
This can be further generalized to N -dimensions by replacing the Lagrangian in Eqn. (7) with: L = −1 2 N X i=1 (∂µφi)2+ 1 2µ 2 N X i=1 φ2i − λ 4 N X i=1 φ2i !2 , (13)
The potential is read off to be:
V (φ1, ..., φN) = λ 4 X i φ2i !2 −1 2µ 2X i φ2i, (14)
Extremizing the potential then yields:
(15) 0 = ∂V ∂φk = λ X i φ2i ! φk− µ2φk ⇔ φk=0 ∨ X i φ2i=µ 2 λ=:v 2
is easily seen that this is true now as well. Just set all but two of the fields equal to zero and the problem is reduced to the 2-dimensional one, but with different labels for the fields. This can be done for any pair of fields in the Lagrangian.
Furthermore, the potential can be rewritten as a function ofP φ2 i, which
means that the minimum of the N -dimensional potential can be written as a sum of the 2-dimensional minima. ThereforeP φ2
i must equal v2to obtain the
minimum.
Then the transformation: φ1(x) = π(x) + v φk(x) = σk(x), 2 ≤ k ≤ N (16)
yields the following Lagrangian:
(17) L = −1 2(∂µ(π + v)) 2 −1 2 N X k=2 (∂µσk)2+ 1 2µ 2(π + v)2 +1 2µ 2 N X k=2 σk2− λ 4 (π + v) 2+ N X k=2 σk2 !2 = −1 2(∂µπ) 2 −1 2 N X k=2 (∂µσk)2+ 1 2µ 2(π + v)2 +1 2µ 2 N X k=2 σk2−λ 4 (π + v) 2+ N X k=2 σk2 !2
The first three terms do not contribute to the mass term for the fileds σk. The
fourth yields a term:
µ2
2 σ
2 k
The final term gives (keeping only terms of the right order at each step and changing the summation index to avoid confusion):
−λ 4 (π + v)2+ N X j=2 σj2 2 ⇒ −λ 4 v 2+ σ2 k 2 ⇒ −λ 2v 2σ2 k = − λ 2 µ2 λσ 2 k = − µ2 2 σ 2 k
Adding these two terms yields a mass of zero for all σk-fields.
4
Spontaneous Breaking of Gauge Symmetry
A theory is gauge invariant/symmetric if the action is invariant under a trans-formation of the quantum fields. If that transtrans-formation also depends on the coordinates of the system, then it is called a local gauge transformation. Quite often, the Lagrangian will also be gauge invariant or nearly gauge invariant if the action is.
A general (local) gauge transformation can be written on the form:
where δij is the Kroenecker delta. In this thesis we will be interested in the
SU(N ) and SO(N ) transformations, i.e. the Special Unitary and the Special Orthogonal transformations. Special means that the determinant of the trans-formation matrix, θ, is +1, while unitary means that U U† = 1, which implies that θ must be hermitian. Finally, orthogonal means that U x · U y = x · y, i.e. that the transformation preserves the dot product. This requires θ to be an orthogonal matrix (θTθ = 1). N stands for the number of dimensions.
The above imposes restrictions on θij. In the SO(N ) case θ must be real
and antisymmetric, due to orthogonality. While in the SU(N ) case θ must be hermitian, as stated previously, and traceless, for the determinant to be +1. In both cases it must be an N × N matrix.
For an arbitrary N × N matrix there are N2parameters. In the SO(N ) case the restrictions leaves 1
2N (N − 1) free parameters in θ, because antisymmetry
requires all N -elements on the diagonal to be zero and fixing the parameters on one side of the diagonal also fixes the N22−N parameters on the other side of the diagonal. Thus reducing the number of free parameters to:
N2− N −N
2− N
2 = 1
2N (N − 1)
as claimed. For the SU case the traceless condition only reduces the number of free parameters with 1, since N − 1 elements in the diagonal can be chosen arbitrarily as long as the last element is chosen so that the trace vanishes. This gives N2− 1 free parameters for the SU case.
Using this Eqn. (18) can be reexpressed as:
Uij = δij− iθa(x)(Ta)ij+ O(θ2) (19)
where θa are real scalar valued functions and Ta are matrices s.t. they satisfy
the same conditions put on θ as well as:
T r(TaTb) = 2δab (20)
The Ta:s are called generators and there number equals the number of free parameters.[1]
4.1
Abelian Case U(1)
In general the generators of a gauge group do not commute. However, in the special case of groups with only one generator, the generators are commuting trivially. Such is the case for U(1) and SO(2). If the generators commute the group is called Abelian.
The following Lagrangian:
L = −(Dµφ)†(Dµφ) + µ2φ†φ − λ 2(φ †φ)2−1 2F µνF µν (21)
where Fµν is the electromagnetic field tensor (Fµν = ∂µAν− ∂νAµ) and D µ=
∂µ− ieAµ, is invariant under a local U(1) gauge transformation:
φ(x) → eiα(x)φ(x), Aµ(x) → Aµ(x) −
1
Here eiα corresponds to the 1 × 1 matrix θ, but unlike Eqn. (18) this is not
restricted to first order in θ.
The potential of the Lagrangian in Eqn. (21) is made up of the middle terms:
V (φ, φ†) = λ 2(φ
†φ)2− µ2φ†φ (23)
and its extrema can be found in the usual way: 0 = ∂V ∂φ = λφ †φφ†− µ2φ† 0 = ∂φ∂V† = λφ†φφ − µ 2φ ⇔ φ = 0 = φ† or φ†φ = µλ2 =: v2 (24)
The second derivative then shows that the first alternative is a local maximum and the second a local minimum, assuming µ2> 0.
The fields can then be expanded around the minima in the following way: φ(x) = v +√1 2(ρ(x) + iχ(x)) φ†(x) = v +√1 2(ρ(x) − iχ(x)) (25)
where ρ and χ are real fields. Making this change in the Lagrangian (Eqn. (21)) yields: (26) L = −(Dµ(v2+ 1 √ 2(ρ(x) + iχ(x)))) †(Dµ(v +√1 2(ρ(x) + iχ(x)))) + µ2(v +√1 2(ρ(x) − iχ(x)))(v + 1 √ 2(ρ(x) + iχ(x))) −λ 2((v + 1 √ 2(ρ(x) − iχ(x)))(v + 1 √ 2(ρ(x) + iχ(x)))) 2−1 2F µνF µν
The mass terms for ρ and χ get contributions from the two middle terms. The first term does not contribute since the terms in it contain either derivatives or crossterms. The last term does not contain ρ or χ at all. The second term gives the following contributions:
µ2
2 ρ
2, µ2
2 χ
2
while the third term yields (keeping only terms of the right order in each step):
(27) −λ 2((v + 1 √ 2(ρ − iχ))(v + 1 √ 2(ρ + iχ))) 2 = −λ 2(v 2+√2vρ +1 2(ρ 2+ χ2))2⇒ −λ 2((2v 2+ v2)ρ2+ v2χ2) = −3λv 2 2 ρ 2−λv2 2 χ 2= −3µ2 2 ρ 2−µ2 2 χ 2
So the mass terms for the two fields are:
The masses are then seen to be√2µ, for ρ, and 0, for χ.
Finally, there is the mass term for the gauge field Aµ. The last three terms
in Eqn. (26) either do not contain Aµ or only derivatives of Aµ, hence they do
not contribute to the mass term. The first term on the other hand yields (in each step only keeping terms of the right order):
(28) −(Dµ(v2+ 1 √ 2(ρ(x) + iχ(x)))) †(Dµ(v +√1 2(ρ(x) + iχ(x)))) ⇒ −(Dµv)†(Dµv)= − ((∂µ− ieAµ)v)†((∂µ− ieAµ)v) = −e2v2A2 = −2(ev) 2 2 A 2
Thus the gauge field, Aµ, has obtained a mass of
√ 2ev.
SO(2) corresponds to a rotation in a plane, while U(1) corresponds to a change of phase in the complex plane. Since the complex plane can be viewed as a bijection from R2, these two cases are equivalent.
4.2
Non-Abelian Case SU(2)
The SU(2) group has three traceless hermitian matrices as generators. These could be for instance the three Pauli matrices, σa, that also satisfy Eqn. (20).
The Lagrangian in Eqn. (21) can be generalized to higher dimensions in the following way: L = N X i=1 −((Dµφ)i)†((Dµφ)i) + µ2φ†iφi− λ 2(φ † iφi)2 −1 2 N2−1 X a=1 (Fa)µν(Fa)µν (29) where (Dµφ)i= ∂µφi− igP N2−1 a=1 Aaµ(Ta) j iφj, (Fa)µν = ∂µ(Aa)ν− ∂ν(Aa)µ and
g is a coupling constant replacing e from the U(1) case. In the SU(2) case this reads: L = −((Dµφ)1)†((Dµφ)1) − ((Dµφ)2)†((Dµφ)2) + µ2φ†1φ1+ µ2φ†2φ2− λ 2(φ † 1φ1)2 −λ 2(φ † 2φ2)2− 1 2(F 1)µν(F1) µν− 1 2(F 2)µν(F2) µν− 1 2(F 3)µν(F3) µν (30)
which is invariant under the gauge transformation:
φi(x) → eiα(x)(σ a)j i φj(x), Aµ(x) → Aa µ(x) − 1 gσ a∂ µα(x) (31)
The potential of this Lagrangian is then identified as:
V (φ1, φ†1, φ2, φ†2) = λ 2(φ † 1φ1)2+ λ 2(φ † 2φ2)2− µ2φ†1φ1− µ2φ†2φ2 (32)
same index. Regarding terms with different indices separately, it is seen that they are the same as in Eqn. (23). These two facts give that the minima are:
φ†2φ2= φ†1φ1=
µ2
λ =: v
2 (33)
The fields can be expanded around the minima. φ1(x) = v +√12(ρ1(x) + iχ1(x)) φ2(x) = v +√12(ρ2(x) + iχ2(x)) (34)
where the ρ:s and χ:s are real valued. Inserting this into the Lagrangian from Eqn. (30) results in:
(35) L = −((Dµφ)1)†((Dµφ)1) − ((Dµφ)2)†((Dµφ)2) + µ2(v +√1 2(ρ1− iχ1))(v + 1 √ 2(ρ1+ iχ1)) + µ2(v +√1 2(ρ2− iχ2))(v + 1 √ 2(ρ2+ iχ2)) −λ 2((v + 1 √ 2(ρ1− iχ1))(v + 1 √ 2(ρ1+ iχ1))) 2 −λ 2((v + 1 √ 2(ρ2− iχ2))(v + 1 √ 2(ρ2+ iχ2))) 2 −1 2(F 1)µν(F1) µν− 1 2(F 2)µν(F2) µν− 1 2(F 3)µν(F3) µν
where the (Dµφ)i:s have been kept to reduce messiness.
Following the same argument that lead to the minima it can be seen that, due to no cross-terms and similarities to the transformed Lagrangian in the U(1) case, the χ-fields are massless and the masses of the ρ-fields are√2µ.
As for the gauge masses, only the terms containing (Dµφ)i contribute to
the mass terms, for the usual reasons. Expanding the first term and identifying Ta= σa yields (keeping only terms of the right order in each step and omitting the minus sign):
The second term yields: (37) ((Dµφ)2)†((Dµφ)2) = (∂µφ2− ig N2−1 X a=1 Aaµ(σa)2jφj)†(∂µφ2 − ig N2−1 X a=1 (Aa)µ(σa)2jφj) ⇒ ig 3 X a=1 Aaµ(σa)2jφj) !† −ig 3 X a=1 (Aa)µ(σa)2jφj ! = g2 (A1)µφ1+ i(A2)µφ1 − (A3) µφ2 † (A1)µφ1+ i(A2)µφ1− (A3)µφ2 ⇒ g2 (A1) µv + i(A2)µv − (A3)µv † (A1)µv + i(A2)µv − (A3)µv ⇒ g2v2(A1)2+g2v2(A2)2+g2v2(A3)2
Thus it is seen that all three gauge fields have obtained a mass of 2gv. SU(2) is similar to SO(3). Consider the bijection between su(2) and R3:
su(2) 3 x = xiσi↔ (x1, x2, x3) = ~x ∈ R3 (38)
This shows that su(2) is isomorphic to R3. Furthermore, consider the most general form of a matrix U ∈ SU(2):
U (a, b) = a b −¯b ¯a , a, b ∈ C (39)
Now create the linear map x → U xU−1. Since:
Tr(U xU−1) = 0 (U xU−1)†= U xU−1 (40) U xU−1 is traceless and Hermitian, i.e. U xU−1 ∈ su(2) and according to Eqn. (38) it can be identified with R3. Lastly, notice that:
x · y = 1 2Tr(xy), x, y ∈ su(2) (41) and therefore (42) x · y → Tr(U xU−1U yU−1) = Tr(U xyU−1) = Tr(U xiyiU−1) = Tr(xiyiU U−1) = Tr(xiyi) =1 2Tr(xy) = x · y
where the following relations have been used to get to the third line:
Eqn. (42) shows that this transformation preserves the dot product, which means that the transformation is orthogonal in three dimensions. Hence it be-longs to SO(3).Thus for every U ∈ SU (2), ∃(x → U xU−1) ∈ SO(3) and the mapping from SU(2) to SO(3) is surjective. It is not however isomorphic, since the map is surjective but not injective. In fact each element, except the zero element, in SO(3) gets mapped twice. This means that while SU(2) and SO(3) are not exactly the same, they are similar enough that several problems can be worked out in either language.[2]
What has been shown by spontaneous symmetry breaking in both the Abelian and non-Abelian cases are examples of Goldstone’s theorem:[3]
If a quantum field has a non-zero vacuum expectation value (VEV) and the VEV of the field is invariant under the same transformation that leaves the Lagrangian invariant. Then, the vacuum states are degenerate and there exists massless fields.
5
Higgs Mechanism in the Standard Model
To construct the electroweak sector of the standard model we will begin from the following Lagrangian that is invariant under transformations of the SU(2)×U(1)-group: L = ((DµΦ)1)†((DµΦ)1) − ((DµΦ)2)†((DµΦ)2) − 1 4λ(Φ †Φ −1 2v 2)2 −1 4(F 1)µν(F1) µν− 1 4(F 2)µν(F2) µν− 1 4(F 3)µν(F3) µν− 1 4B µνB µν = −((DµΦ)1)†((DµΦ)1) − ((DµΦ)2)†((DµΦ)2) − 1 4λ(φ † 1φ1+ φ†2φ2− 1 2v 2)2 −1 4(F 1)µν(F1) µν− 1 4(F 2)µν(F2) µν− 1 4(F 3)µν(F3) µν− 1 4B µνB µν (44) where: Φ = φ1 φ2
The F :s and the B contain the gauge fields and are defined as: (F1)µν = ∂µ(A1)ν− ∂ν(A1)µ+ g 2((A2)µ(A3)ν− (A2)ν(A3)µ) (F2)µν = ∂µ(A2)ν− ∂ν(A2)µ+ g 2((A3)µ(A1)ν− (A3)ν(A1)µ) (F3)µν = ∂µ(A3)ν− ∂ν(A3)µ+ g 2((A1)µ(A2)ν− (A1)ν(A2)µ) Bµν = ∂µBν− ∂νBµ (45)
where the A:s and the B on the right side are the gauge fields, while g2 is a
coupling constant. The covariant derivative, (DµΦ)i, in Eqn. (44) is given by:
(DµΦ)i= ∂µΦi− i(g2(Aa)µTa+ g1BµY )ijΦj (46)
where g1is another coupling constant, Taare the generators of SU(2) discussed
in the previous section 4, but with a slightly different condition, namely:
T r(TaTb) = 1 2δ
Y is the generator for U(1). In this case Y = −12I. Y is called the hypercharge, which is a conserved quantity in particle physics.[1]
The SU(2)×U(1)-group has been chosen specifically to yield the right con-served quantities in the electroweak theory.
Studying the Lagrangian in Eqn. (44), it is seen that the potential is:
V (φ1, φ†1, φ2, φ†2) = 1 4λ(φ † 1φ1+ φ†2φ2− 1 2v 2)2 (48)
Due to the positive semi-definiteness of this potential the minimum value is zero. Expanding around the real minima in the φ1-direction as follows:
φ
1(x) = √12(v + H(x))
φ2(x) = 0
(49)
where H is real valued and the unitary gauge has been chosen to simplify further calculations. Then inserting this into Eqn. (44) yields:
L = −(∂µ 1 √ 2H − i(g2(A a) µσa+ g1BµY )11 1 √ 2(v + H)) †(∂µ√1 2H − i(g2(Aa)µσa+ g1BµY )11 1 √ 2(v + H)) − (−i(g2(A a) µσa+ g1BµY )21 1 √ 2(v + H))†(−i(g2(Aa)µσa+ g1BµY )21 1 √ 2(v + H)) − 1 4λ( 1 2(v + H) 2 −1 2v 2)2 −1 4(F 1)µν(F1) µν− 1 4(F 2)µν(F2) µν− 1 4(F 3)µν(F3) µν− 1 4B µνB µν = −1 2(∂µH − i( g2 2(A 3) µ− 1 2g1Bµ)(v + H)) †(∂µH − i(g2 2(A 3)µ −1 2g1B µ)(v + H)) −1 2((−i g2 2(A 1) µ+ g2 2(A 2) µ)(v + H))†((−i g2 2(A 1)µ+g2 2(A 2)µ)(v + H)) −1 4λ( 1 2H 2+ vH)2 −1 4(F 1)µν(F1) µν −1 4(F 2)µν(F2) µν− 1 4(F 3)µν(F3) µν− 1 4B µνB µν (50)
where Eqn. (46) has been used to expand D. Next, introduce the following gauge fields:
Wµ±:=√12((A1)µ∓ (A2)µ)
Zµ:= cos(θW)(A3)µ− sin(θW)Bµ
Aµ:= sin(θW)(A3)µ+ cos(θW)Bµ
(51)
where θW is called the weak mixing angle and is defined by:
θW := tan−1
g1
g2
(52)
while the rest can be calculated in a similar fashion): 1 √ 2((F 1)µν− i(F2)µν) = √1 2(∂ µ(A1)ν− ∂ν(A1)µ + g2((A2)µ(A3)ν− (A2)ν(A3)µ) − i(∂µ(A2)ν − ∂ν(A2)µ+ g 2((A3)µ(A1)ν− (A3)ν(A1)µ))) = (∂µ√1 2((A 1)ν−i(A2)ν)−ig 2(A3)µ 1 √ 2((A 1)ν−i(A2)ν) − ∂ν√1 2((A 1)µ− i(A2)µ) + ig2(A3)ν 1 √ 2((A 1)µ− i(A2)µ) = (∂µ(W+)ν− ig2(A3)µ(W+)ν− ∂ν(W+)µ + ig2(A3)ν(W+)µ = Dµ(W+)ν− Dν(W+)µ (53) 1 √ 2((F 1)µν + i(F2)µν) = (D†)µ(W−)ν− (D†)ν(W−)µ (F3)µν = sin(θ W)Fµν+ cos(θW)Zµν− ig2((W+)µ(W−)ν− (W+)ν(W−)µ) Bµν = cos(θ W)Fµν− sin(θW)Zµν (54) where Dµ := ∂µ− ig 2(A3)µ = ∂µ − ig2(sin(θW)Aµ + cos(θW)Zµ), Fµν :=
∂µAν − ∂µAν and Zµν = ∂µZν− ∂νZµ. Finally, substituting this into Eqn.
term for A in the Lagrangian, so its gauge boson is indeed massless just as the photon should be. There is however a mass term for Z given by:
− g 2 2v 2 8 cos2θ W ZµZµ:= − MZ2 2 Z µZ µ (56)
which corresponds to the Z0 gauge particle for weak interactions, that has a
measured mass of MZ = 91.2 GeV. There is also a mass term for W± given by:
−g 2 2v2 4 (W −) µ(W+)µ := −MW2 (W−)µ(W+)µ (57)
This corresponds to the W± gauge bosons for electroweak interactions, which have a measured mass of MW = 80.4 GeV. Lastly, there is the higgs gauge
boson connected to the higgs gauge field H with the mass term:
−λv 2 4 H 2 := −MH 2 H 2 (58)
6
Unitarity Constraints on the Higgs Mass
The previous section illustrated the necessity of the higgs particle, but it did not provide a value for its mass (λ is an unknown constant). This can be partially amended by studying constraints on the mass due to partial wave unitarity, which yields an upper bound for the mass.
A scattering amplitude in quantum field theory can be split up into partial waves in the following way:
T (s, t) = 16π
∞
X
j=0
(2j + 1)aj(s)Pj(cos θ) (59)
where the aj:s are called the partial wave amplitudes, the Pj:s are Legendre
poly-nomials, θ is the scattering angle, while s and t are two of the three Mandelstam variables:
s := −(k1+ k2)2= −(k01+ k02)2
t := −(k1− k01)2= −(k2− k20)2
u := −(k1− k20)2= −(k2− k10)2
(60)
where the k:s are four-momenta as defined in the Feynman diagrams below (fig-ures 4-10).
Since, |T |2, is a probability for the scattering to occur, it cannot be larger
than 1. Eqn. (59) then implies that the sum of the partial waves cannot exceed 1 as well. In particular |a0|≤ 1. It could be argued that, since the aj:s are
in general complex, there might be negative cross-terms allowing for a larger value of |a0|. This is not true, because the absolute value of the aj:s decrease
by increasing j. Thus, by this criteria it is possible to find an upper bound for the Higgs boson mass.
of magnitude smaller, and hence neglect-able, or they are larger, but then per-turbation theory fails and Feynman diagrams cannot be used at all. The latter case is not impossible, but is not covered by the standard model, hence it will not be considered here.[4]
In the following interactions exchanges of γ- and Z0-particles are neglected,
since their contribution to the amplitude is negligible, except in the last two processes where exchanges of Z0-particles are taken into consideration. The
Feynman diagrams and their amplitudes are as follows:
T (HH → ZZ) = k2 k1 k10 k02 H H Z Z + k1 k2 k1+ k2 H k01 k20 H H Z Z + k01 k1 k01− k1 Z k02 k2 Z H Z H + k20 k1 k20 − k1 Z k10 k2 Z H Z H = −√2GFMH2 1 + 3MH2 s−M2 H + MH2 t−M2 Z + MH2 u−M2 Z Figure 9: HH → ZZ T (HZ → HZ) = k2 k1 k10 k0 2 Z H H Z + k1 k2 k1+ k2 H k01 k20 H Z H Z + k10 k1 k01− k1 Z k02 k2 H H Z Z + k20 k1 k20 − k1 Z k01 k2 Z H H Z = −√2GFMH2 1 + 3MH2 s−M2 H + MH2 t−M2 Z + MH2 u−M2 Z Figure 10: HZ → HZ
where GF is the Fermi coupling constant. All other scatterings, not shown
Because we’re looking for an upper bound on the higgs mass it might as well be assumed that MH >> MZ, MW. It can also be assumed that s >>
M2
Z, MW2 , i.e the energy of the system is very large. With these assumptions,
and remembering that P0= 1, the zeroth partial wave amplitudes can be
cal-culated through [4, 5]: a0= 1 16πs Z 0 −s T dt, (61) together with: u = M12+ M22+ M120+ M220− s − t, (62) For example: (63) a0(ZZ → ZZ) = 1 16πs Z 0 −s T (ZZ→ZZ) dt = −GFM 2 H 8√2πs Z 0 −s 1 + s s − M2 H + t t − M2 H + u u − M2 H dt = −GFM 2 H 8√2π 1 + s s − M2 H +1 s Z 0 −s t − M2 H+ M 2 H t − M2 H +u − M 2 H+ M 2 H u − M2 H dt = −GFM 2 H 8√2π 1 + s s − M2 H + 1 + 1 +1 s Z 0 −s M2 H t − M2 H + M 2 H u − M2 H dt = −GFM 2 H 8√2π 3 + s s − M2 H +M 2 H s log M2 H s + M2 H +1 s Z 0 −s M2 H 4M2 Z− s − t − M 2 H dt ≈ −GFM 2 H 8√2π 3 + s s − M2 H +M 2 H s log MH2 s + M2 H −1 s Z 0 −s M2 H s + t + M2 H dt = −GFM 2 H 8√2π 3 + s s − M2 H +M 2 H s log M2 H s + M2 H −M 2 H s log M2 H+ s M2 H = −GFM 2 H 8√2π 3 + s s − M2 H −2M 2 H s log M2 H+ s M2 H = −GFM 2 H 8√2π 3 + s s − M2 H −2M 2 H s log 1 + s M2 H ≈ −GFM 2 H 8√2π 3 + M 2 H s − M2 H −2M 2 H s log M2 H+ s M2 H
by Eqn. (60).
In a similar manner the other partial wave amplitudes can be calculated. Eqn. (64) shows the result.
a0(W+W−→ W+W−) = − GFMH2 8π√2 2 + MH2 s−M2 H −MH2 s log 1 + s M2 H a0(ZZ → ZZ) = − GFMH2 8π√2 3 + MH2 s−M2 H −2MH2 s log 1 + s M2 H a0(HH → HH) = − GFMH2 8π√2 3 + 9MH2 s−M2 H − 18MH2 s−4M2 H log s M2 H − 3 a0(HH → W+W−) = − GFMH2 8π√2 1 + 3MH2 s−M2 H + 4MH2 q s(s−4M2 H) log s−2M 2 H− q s(s−4M2 H) 2M2 H !! a0(ZZ → W+W−) = −GFM 2 H 8π√2 1 + MH2 s−M2 H a0(HH → ZZ) = − GFMH2 8π√2 1 + 3M2 H s−M2 H + 4MH2 q s(s−4M2 H) log s−2M 2 H− q s(s−4M2 H) 2M2 H !! a0(HZ → HZ) = − GFMH2 8π√2 1 + MH2 s − 3M2 Hs (s−M2 H)2 log1 +(s−MH2)2 sM2 H − MH2s (s−M2 H)2 logs(2MH2−s) M4 H (64)
If it is further assumed that the energy of the system is so large that s >> M2 H,
then Eqn. (64) reduces to:
a0(W+W−→ W+W−) = − GFMH2 4π√2 a0(ZZ → ZZ) = − 3GFMH2 8π√2 a0(HH → HH) = − 3GFMH2 8π√2 a0(HH → W+W−) = − GFMH2 8π√2 a0(ZZ → W+W−) = − GFMH2 8π√2 a0(HH → ZZ) = − GFMH2 8π√2 a0(HZ → HZ) = − GFMH2 8π√2 (65)
which on matrix form reads:
−GFM 2 H 4π√2 1 √1 8 1 √ 8 0 1 √ 8 3 4 1 4 0 1 √ 8 1 4 3 4 0 0 0 0 12 (66) in the basis {W+W−,√1 2ZZ, 1 √
2HH, HZ}. For a0≤ 1 the largest eigenvalue of
in units of −GFMH2 4π√2 yields: (67) 0 = 1 − λ √1 8 1 √ 8 0 1 √ 8 3 4− λ 1 4 0 1 √ 8 1 4 3 4− λ 0 0 0 0 12− λ = (1 2 − λ) 1 − λ √1 8 1 √ 8 1 √ 8 3 4− λ 1 4 1 √ 8 1 4 3 4− λ = (1 2 − λ) 1 − λ 0 √1 8 1 √ 8 1 2− λ 1 4 1 √ 8 λ − 1 2 3 4− λ = (1 2 − λ) 1 − λ 0 √1 8 1 √ 8 1 2− λ 1 4 1 √ 2 0 1 − λ = (1 2 − λ) 2 1 − λ √1 8 1 √ 8 1 4 1 √ 2 1 − λ = (1 2 − λ) 2((1 − λ)2 −1 4) ⇔ λ1= λ2= 12 (1 − λ3,4)2−14= 0 ⇔ 1 − λ3,4 = ±12 ⇔ λ3,4= 12,32 Thus −GFMH2 4π√2 3
2 is the largest eigenvalue and hence by unitarity the absolute
value of this must be less than or equal to 1, which gives an upper bound for the higgs mass.
GFMH2 4π√2 3 2 ≤ 1 ⇔ M 2 H ≤ 8π√2 3GF ≈ 1(TeV)2 (68)
Which sets the upper bound on the higgs mass to 1 TeV.[5]
7
Summary
First it was shown in sections 2 and 3 that spontaneous symmetry breaking can give mass to previously massless fields through the higgs mechanism.
of the higgs scalar.
While the previous result necessitates the existence of the higgs particle it does not detail its mass. In section 6 an upper bound for the higgs mass, known as the LQT-bound, was derived by considering partial wave unitarity and the tree level Feynman diagrams contributing to the partial waves. It was shown that the mass cannot exceed 1 TeV without breaking the partial wave unitarity (or perturbation theory).
The method used in section 5 is not the only way for the gauge bosons to acquire mass, it is merely the easiest. There are other methods resulting in more types of higgs particles. The second simplest of these methods is known as the two higgs doublet model (2HDM), which utilises two doublets of complex scalar fields instead of just one as was done in section 5.
References
[1] Quantum Field Theory, M. Srednicki, Cambridge University Press, 2007 [2] SU(2) and SO(3), D. Westra, 2008 https://www.mat.univie.ac.at/
~westra/so3su2.pdf
[3] Spontaneous Symmetry Breaking, J. Zierenberg, University of Leipzig, 2008 [4] The Higgs Hunter’s Guide, J. Gunion, H. Haber, G. Kane & S. Dawson,
Perseus Publishing, 1990
[5] Weak Interactions at Very High Energies: the Role of the Higgs Bo-son Mass, B. Lee, C. Quigg & H. Thacker, Fermi National Acceler-ator LaborAcceler-atory, 1977 http://inspirehep.net/record/119348/files/ fermilab-pub-77-030-T.pdf
