Standard Model/Quantun Field Theory Solution Set 7
Due: 4 December 2019
Suggested reading: QFT Notes Ch 25; Textbook: Sec. 83
22. The Higgs particle. As we have seen in class, the way particles are given mass in the standard model is through coupling to the complex scalar Higgs field doubletφwhich has a vacuum expectation value which we describe by writing
φ= v+ (h+ia)/ √ 2 φ2 ! (1) where we take the vacuum valuev to be a real number, andh and a are real fields. The complex fieldφ2 and the real fieldaprovide the zero helicity fields for the massive vector bosons W andZ. The remaining real field h describes a new massive spin zero particle, the Higgs particle. Its mass
Mh and its quartic self coupling λare independent parameters in the standard model.
a) Identify the cubic terms in the standard model Lagrangian that contain one factor ofhand two other fields: two vector bosons, quark-antiquark, and lepton-antilepton. Hint: since h enters the Lagrangian withvin the combinationv+h/√2, you can find these terms by substituting
v→v+h/√2 in the mass (quadratic) terms of the Lagrangian and expanding to first order inh. Use the fact that the particle masses are all proportional to v.
Solution: The vector boson mass terms are
MZ2 2 Z 2+M2 WW †· W = g 2 2v2 2 W †· W + g 2 2v2 4 cos2θ W Z2 → g 2 2(v+h/ √ 2)2 2 W †·W +g22(v+h/ √ 2)2 4 cos2θ W Z2 → M 2 Z 2 Z 2+M2 WW †· W +h " g2 2v √ 2 2 W †· W + g 2 2v √ 2 4 cos2θ W Z2 # = M 2 Z 2 Z 2+M2 WW †· W +h g2MWW†·W + g2MW 2 cos2θ W Z2 (2) where we eliminatedv=MW √
2/g2 in the last line. The fermion mass terms are linear in v:
mff f¯ =Gfvf f¯ →mff f¯ +h Gf √ 2 ¯ f f =mff f¯ +h g2mf 2MW ¯ f f (3)
In summary the cubic terms in the Lagrangian are:
−h g2MWW†·W + g2MZ 2 cosθW Z·Z+X f g2mf 2MW ¯ f f (4)
b) Extract the Feynman rules for the cubic vertices corresponding to each of these terms.
Solution: We simply read off:
−ig2MW, −i g2MZ cosθW , −ig2mf 2MW (5) for the three terms respectively.
c) Calculate the decay rate into each of these two body final states, (assuming Mh is large enough for the decay to proceed in each case). Now that the Higgs particle mass is known to beMh ≈125GeV, the decay into a pair of vector bosons is purely an academic exercise!
Solution: Γh→W W = g22Mh3 64πMW2 " 1−4M 2 W Mh2 + 12 MW4 Mh4 # s 1−4M 2 W Mh2 (6) Γh→ZZ = g22Mh3 128πM2 Zcos2θW " 1−4M 2 Z M2 h + 12M 4 Z M4 h # s 1−4M 2 Z M2 h (7) Γh→ff¯ = g22Mhm2f 32πMW2 1−4 m2f Mh2 !3/2 (8) When the fermion is a quark, the corresponding rate is a factor ofNc= 3 larger because they come in three colors.
23. Calculating the effective potential. Consider the field theory of a single real scalar fieldφ
with Lagrangian density
L=−1 2(∂φ)
2− λ 4!(φ
2−a2)2, (9)
which is invariant underφ→ −φ. Define the actionS(φ)≡R
d4xL. Recall that the effective action Γ(ϕ)≡W(J)−R
d4xϕJ, whereeiW(J) is the functional integral overφof exp
iS(φ) +iR
d4xφJ , and ϕ(x) ≡ hφiJ = δW/δJ. For the effective potential it will be enough to introduce a constant sourceJ, soW(J) =V T w(J), andϕ=∂w/∂J, and Γ(ϕ) =−V T Vef f(ϕ).
a) In class we have observed that W(J) is the sum of all connected vacuum graphs (0-point functions) for the actionS(φ) +R
d4xφJ. By changing variables in the path integral foreiΓ(ϕ), show that Γ(ϕ) is the sum of all connected vacuum graphs for the actionS(φ+ϕ) +R
d4xφJ, whereJ is chosen so thathφiJ = 0.
Solution: By definition
eiΓ(ϕ)=eiW(J)−i
R
d4xϕJ =Z DφeiRd4x[L+J(φ−ϕ)], ϕ=hφi
J (10)
Now change path integral variablesφ=ϕ+ ˆφ:
eiΓ(ϕ)= Z Dφeˆ i R d4x[L( ˆφ+ϕ)+Jφˆ, hφˆi J = 0 (11)
But this is what was to be proved: The logarithm of this path integral is i× the sum of all connected vacuum diagrams, and the condition hφˆiJ = 0 places the desired constraint on the choice ofJ.
b) In terms of Feynman graphshφ(x)iis the sum of all connected one-point diagrams (“tadpoles”) which have a propagator emerging from the pointxwith the other end hooked to an arbitrary structure with no further external legs. The condition hφiJ = 0 means that the sum of all such tadpole diagrams is zero. But it also means that all diagrams with one or more tadpole subdiagrams will sum to zero. Show that a connected vacuum diagram that is one particle reducible (i.e. can be disconnected by cutting a single line) always has at least one tadpole subdiagram, implying that all one particle reducible diagram contributions to Γ cancel out in the complete sum, so Γ can be calculated as the sum of all connected 1PIR vacuum diagrams.
Solution: Vacuum diagrams have no external legs. A vacuum diagram that can be discon-nected by cutting a single internal line has a structure which ends in at least two tadpole subdiagrams each of which have been set to zero by the condition onJ. Thus the only nonzero diagrams must be 1PIR.
c) Assuming constant J, ϕobtain the Feynman rules for the action S(φ+ϕ) +R
d4xφJ. List the propagator, as well as the values of the 1, 3, and 4 point vertices. Note that some of these couplings and the mass occurring in the propagator depend explicitly on ϕ. Also note that
J explicitly appears only in the 1-point vertex. Since this vertex only appears in one particle reducible diagrams the calculation of Γ will have no explicit dependence on J. This means that the complicated relation between ϕand J implied by hφiJ = 0 is not needed!
Solution: To get the vertices of this theory, we expand the potential term about φ= 0:
−λ 4!((φ+ϕ) 2−a2)2+J φ = −λ 4!(ϕ 2−a2)2+J− λ 3!ϕ(ϕ 2−a2)φ−λ 2ϕ 2−λ 6a 2φ2 2 −λϕ 3!φ 3− λ 4!φ 4 (12)
defining m2(ϕ) =λϕ2/2−λa2/6 andg(ϕ) =λϕthe Lagrangisn becomes
L = −1 2(∂φ) 2− λ 4!(ϕ 2−a2)2+φJ+ λ 3!ϕ(a 2−ϕ2)−m2(ϕ) 2 φ 2−g(ϕ) 3! φ 3− λ 4!φ 4 ∆ = −i p2+m2(ϕ); V1 =i J+ λ 3!ϕ(a 2−ϕ2); V 3 =−ig(ϕ); V4=−iλ
As noted,V1 only appears in reducible diagrams, which will vanish once the constraint onJ is fulfilled.
d) The zero loop value for Γ(ϕ) is clearly just S(φ) = −V T(λ/4!)(ϕ2 −a2)2. The one loop vacuum diagram is represented by a diagram with a propagator closing on itself with no vertices, and stands for the result of doing the Gaussian integral over φ using the quadratic terms in the action: det−1/2(m(ϕ)2−∂2). For constant ϕ its contribution to iΓ is the log, −(V T /2)R
(d4p/(2π)4) ln(m(ϕ)2+p2). This diagram is usually ignored when calculating in a source free theory-its just the zero-point energy of the vacuum and has no direct significance.
But for the calculation of the effective potential the mass in the propagator depends onϕand that dependenceissignificant. Calculate the effective potentialVef f(φ) through one loop. Do the one-loop integral in Euclidean space (after Wick rotation) with the simple cutoffp2 <Λ2. Show that the cutoff dependence can be absorbed in renormalized parameters. (We know that there must also be field renomalizationϕr =ϕ/
√
Z, but with constantϕ you won’t be able to separate the part of the divergence to be absorbed in Z).
Solution: iΓ1 loop = − V T 2 Z d4p (2π)4 ln(p 2+m2(ϕ)−i) =−iV T 2 Z d4p E (2π)4ln(p 2 E +m2(ϕ)−i)(13)
after Wick rotation p0 = ip4. Note that the contour at infinity is infinite, so dropping it means we are actually defining the theory in Euclidean space. Then, remembering that the angular integral gives 2π2,
V1ef floop = 1 16π2 Z Λ 0 p3dpln(p2+m2(ϕ)) = 1 32π2 Z Λ2 0 uduln(u+m2(ϕ)) = 1 64π2 Λ4ln(Λ2+m2(ϕ))−1 2 h (Λ2+m2(ϕ))2−m4(ϕ)i +2m2(ϕ)Λ2−m4(ϕ) lnΛ 2+m2(ϕ) m2(ϕ) ∼ 1 64π2 Λ4 ln Λ2−1 2 + 2m2(ϕ)Λ2−m4(ϕ)(ln Λ2+ 1) +m4(ϕ) lnm2(ϕ)
Since the Λ dependence is a quartic polynomial inϕit can be absorbed in the parameters of the input bare lagrangian. Then we can write the renormalized effective potential through 1 l00p as Vef f(ϕr) = λr 4!(ϕ 2 r−a2r)2+ λ2r 64π2 ϕ2r 2 − a2r 6 !2 ln ϕ 2 r 2 − a2r 6 ! (14)
Note that the ln becomes complex forϕ2 < a2r/3, the inflection point of the input potential.
24. As mentioned in class chiralSU(2)×SU(2) is isomorphic toO(4). In this exercise we explore the relationship further. LetJi
L,JRi be the commuting generators of the twoSU(2)’s, rotating theL andR components of the quark fields respectively. They each satisfy the usual angular momentum commutation relations.
a) Show that Ji =JLi +JRi generates the O(3) rotations of ordinary isospin. That is, ¯qq, ¯qiγ5q are isoscalars and ¯qτq, ¯qiγ5τq are isovectors under these transformations.
Solution: Firs note that theSU(2) Lie algebra
[Jk, Jl] = [JkL, JlL] + [JkR, JlR] =iklm(JmL+JmR) =iklmJm (15)
The under Jk since qL and qR rotate the same way, ¯qq = ¯LR+ ¯RL is invariant, as is ¯qiγ5q. Similarly ¯qτq rotates as an isovector because τ does..
b) Show that Ki =JLi −JRi acts as a “boost” that transforms (¯qq,qiγ¯ 5τq) and (¯qiγ5q,q¯τq) as 4-vectors. The terminology here is to suggest an analogy with a Lorentz transformation, but of course O(4) is a compact group. The condensate for isospin conserving chiral symmetry breaking is then h¯qqi=v6= 0.
Solution: Letq→q−ik(τk/2)qL+ik(τk/2)qR=q+ik(τk/2)γ5q. Then
¯ qq → qq¯ −iaq†(τa/2)γ5γ0q+iaq†(τa/2)γ0γ5q = ¯qq+iaqτ¯ aγ5q ¯ qτbγ5q → qq¯ −iaq†(τa/2)τbγ5γ0γ5q+iaq†(τa/2)γ0τbγ52q= ¯qτbγ5q+iaq¯ {τa, τb} 2 q = qτ¯ bγ5q+ibqq¯ (16) Or ¯
qq →qq¯ +aqτ¯ aiγ5q; qiσ¯ aγ5q→qiσ¯ aγ5q−bqq¯ (17)
The four vector with an extraγ5 works in exactly the same way.
c) Work out the commutators of theO(4) generatorsJi, Ki among themselves.
Solution:
