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Physics & Astronomy International Journal

Review Article Volume 7 Issue 2

On the synopsis of the Higgs boson

Manyika Kabuswa Davy, Likolo Anabiwa George, Katongo Judith

Mulungushi University, School of Natural and Applied Sciences, Department of Physics, Zambia

Correspondence: Manyika Kabuswa Davy, Mulungushi University, School of Natural and Applied Sciences, Department of Physics, Zambia

Received: April 17, 2023 | Published: May 12, 2023

Citation: Davy MK, George LA, Judith K. On the synopsis of the Higgs boson. Phys Astron Int J. 2023;7(2):113-116. DOI: 10.15406/paij.2023.07.00294

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Abstract

In this paper, we analyze the Standard Model (SM) Higgs boson by computing both the partial width and the amplitude of a number of decay channels that the Higgs Boson can undergo. In our computations, we treat the Higgs boson as a free parameter despite its estimated mass of around 125GeV discovered at the Large Hadron Collider (LHC).1

Keywords: Higgs boson, decay channels, Standard Model, Large Hadron Collider.

Introduction

The Higgs boson can undergo several decays via a number of decay channels.2 The main decay channels that we are going to concentrate on in this paper are h0f.ˉfh0f.¯f , h0W+Wh0W+W , h0Z0Z0h0Z0Z0 , h0ggh0gg , and h0γγh0γγ .3 From the five decay channels listed, the first three occur at tree level while the other two at one-loop.

In this paper, we will calculate the decay widths by treating our decay system as a body of a final state system. Beginning with rotational symmetry and considering our momentum to be conserved, we parametrize our variables in the centre-of-mass (CM) frame as p1=(E,0,0,p)p1=(E,0,0,p)   and p2=(E,0,0,p)p2=(E,0,0,p)  , with E=12mhE=1/2mh . Here we note that the amplitude will not depend on the angular parameters.4 Thus the integral of the phase space reads

dΠ2|M|2=14πpmh|M|2dΠ2|M|2=14πpmh|M|2   (1)

From Equation 1, we determine the decay width as5

Γ=12mhdΠ2|M|2=18πpm2h|M|2Γ=12mhdΠ2|M|2=18πpm2h|M|2   (2)

Furthermore, the momenta of our final state particles becomes k1=(E,0,0,k)k1=(E,0,0,k) and k2=(E,0,0,k)k2=(E,0,0,k)  with E2=k2+m2iE2=k2+m2i  and 2E=mf2E=mf .6 Thus, the cross section becomes

σ=12βsd3p(2π)312Ep|M|2(2π)2δ4(pk1k2)σ=12βsd3p(2π)312Ep|M|2(2π)2δ4(pk1k2)
=14mfβs|M|2(2π)δ(2k1mf)   

=πβm2f|M|2δ(sm2f)   (3)

with β=1(4mi/mf)2 being the initial particle’s velocity magnitude.

The h0f.ˉf decay channel

This is the simplest channel to compute and its corresponding amplitude reads as

iM(h0f.ˉf)=imfvˉu(p1)v(p2)   (4)

It is easy to compute the square amplitude as

|M(h0f.ˉf)|2=m2fv2tr[(p1+mf)(p2mf)]
=2m2fv2(m2f4m2f)   (5)

Using the momenta for final states, the decay width becomes

Γ(h0f.ˉf)=18πpm2h|M|2=mhm2f8v2(14m2fm2h)3/2   (6)

The h0W+W,Z0Z0 decay channels

In this section, we describe the amplitude for h0W+W  as

iM(h0W+W)=igμνg2v2μ(p1)v(p2)   (7)

and the square amplitude reads

|M|2=g4v24(gμvp1μp1vm2W)(gμvp2μp2vm2W)
=παsin2θwm4hm2W(14m2Wm2h+12m4Wm2h)   (8)

Finally, the decay width becomes

Γ(h0W+W)=18πp1m2h|M|2   

=αm3h16πm2Wsin2θw(14τ1W+12τ2W)(14τ1W)1/2   (9)

where τW(mh/mW)2 . In order to compute for h0Z0Z0 , we just swap the masses of the vector boson and add an extra factor of half and arrive at

Γ(h0Z0Z0)=αm3h32πm2Zsin2θw(14τ1Z+12τ2Z)(14τ1Z)1/2   (10)

with  τZ(mh/mZ)2 .

The h0gg decay channel

The amplitude takes the form

iM(h0gg)=imqv(igs)2μ(p1)v(p2)tr(tatb)
×ddp(2π)d{(1)tr[γμiqmqγviq+p2mqiqp1mq]  

+(1)tr[γviqmqγμiq+p1mqiqp2mq]}   (11)

From Equation 11, we can simplify the trace as

tr[γviqmqγμiq+p2mqiqp1mq]=itr[(q+mq)(q+p2mq)(qp1mq)](q2m2q)[(q+p2)2m2q][(qp1)2m2q]
=2i10dx1x0dyNμν(q2Δ)3   (12)

where

qμ=qμxp1μ+yp2μ   (13)

Δ=m2qx(1x)p21y(1y)p222xyp1p2=m2qxym2h   (14)

Nμν=4mq(pv1pv2pμ1pv2+2pv2qμ2pμ1qv+4qμqv+(m2qp1p2q2)ημν)   (15)

After rigorous computations, the final decay width becomes

Γ(h0gg)=(αmh8sin2θw)m2hm2wα2s9π2|If(τq)|2   (16)

Generally, for Nq quarks, the decay width takes the form

Γ(h0gg)=(αmh8sin2θw)m2hm2Wα2s9π2|qIf(τq)|2   (17)

The cross section for ggh0

By using the results in section 4 as well comparing Equations 2 and 3, we deduce that

σ(ggh0)=π28mhδ(ˆsm2h)Γ(h0gg)   (18)

By using Equation 17, we obtain

σ(ggh0)=αα2s576sin2θwm2hm2W|qIf(τq)|2δ(ˆsm2h)   (19)

Finally, the gluon-gluon fusion cross section at proton level becomes

σGGF(p(P1)p(P2)h0)   

=10dx110dx2fg(x1)fg(x2)σ(g(x1P1)g(x2P2)h0)   

=dM2Y|(x1,x2)(M2,Y)|fg(x1)fg(x2)σ(g(x1P1)g(x2P2)h0)   

=dM2Y1M2x1fg(x1)x2fg(x2)σ(g(x1P1)g(x2P2)h0)   (20)

The h02γ decay channel

In this decay channel, the contribution is as a result of both the W boson and the fermion loop. It is easier to compute for the later contribution and our beginning point is to pick the result in Equation 17. By including the internal fermion electric charge to our result we arrive at

iM(h02γ)=(αmh8sin2θw)m2hm2wα2s18π2|fQ2fNc(f)If(τf)|2   (21)

For the W boson contribution, we have to compute for all the thirteen loop diagrams.

Thus, we have

iM(a)=12igρσg2v2(ie2)(2ημνηρσημρηνσημσηνρ)μ(p1)v(p2)   

×ddq(2π)dDW(q)DW(kq)   

=2i(4π)d/2e2m2Wv(p1)(p2)(d1)Γ(2d/2)
×10dx[m2Wx(1x)m2h]2d/2   (22)

iM(b)=12(2iλv)(2ie2)(p1)(p2)ddq(2π)dDs(q)Ds(kq)   

=i(4π)d/2e2m2Wv(p1)(p2)Γ(2d/2)   

×10dx[m2Wx(1x)m2h]2d/2   (23)

iM(c)=iM(d)=ig2sinθw2ig2vsinθw212(p1)(p2)ddq(2π)dDs(q)DW(p2q)
=i(4π)d/2e2m2Wv(p1)(p2)Γ(2d/2)1(m2W)2d/2   (24)

iM(e)=ig2v2(ie)2ηρσμ(p1)v(p2)ddq(2π)dDW(q)DW(qp1)DW(qp2)   

×[ηρλ(2qp1)μ+ημρ(2p1q)ληλμ(p1+q)ρ]   

×[ησλ(2qp2)vηvλ(qp2)σησv(2p2+q)λ]   

=i(4π)d/2e2m2Wv(p1)(p2)[dxdy(πxy+4xy)m2hm2Wxym2h
+6(d1)Γ(2d/2)10dxdy(m2Wxym2h)2d/2]   (25)

iM(f)=(2iλν)(ie)2μ(p1)v(p2)ddq(2π)d(2qp1)μ(2qp2)μ
×Ds(q)Ds(qp1)Ds(q+p2)   

=i(4π)d/2e2m2hv(p1)(p2)Γ(2d/2)2dxdy(m2Wxym2h)2d/2   (26)

iM(g)=(im2Wv)(ie)2μ(p1)v(p2)ddq(2π)d(1)(qp1)μqv   

×Ds(q)Ds(qp1)Ds(q+p2)   

=i(4π)d/2e2m2Wv(p1)(p2)Γ(2d/2)dxdy(m2Wxym2h)2d/2   (27)

iM(h)=iM(i)=ig2igμλg2vsinθw2(ie)2μ(p1)v(p2)ddq(2π)d(qp1k)σ   

×[ησλ(2q+p2)vηvλ(qp2)σηvσ(2p2+q)λ]   

×DW(q)Ds(qp1)DW(q+p2)   

=i(4π)d/2e2m2Wv(p1)(p2)[dxdy(1x)(1+y)m2hm2Wxym2h
12(πd1)Γ(22/d)dxdy(m2Wxym2h)22/d]   (28)

iM(j)=ig2v2(ig2vsinθw2)2(p1)(p2)   

×ddq(2π)dDs(q)DW(qp1)DW(q+p2)   

=i(4π)d/2e2m2Wv(p1)(p2)dxdy2m2Wm2Wxym2h   (29)

iM(k)=iM(l)=ig2ig2vsinθw2(ie)2μ(p1)v(p2)   

×ddq(2π)d(p1+2p2+q)μ(2q+p2)vDs(q)DW(qp1)DW(q+p2)   

=i(4π)d/2e2m2Wv(p1)(p2)Γ(22/d)dxdy(m2Wxym2h)22/d   (30)

iM(m)=(2iλv)(ig2vsinθw2)2(p1)(p2)   

×ddq(2π)dDW(q)Ds(qp1)Ds(q+p2)   

=i(4π)d/2e2m2Wv(p1)(p2)dxdym2Wm2Wxym2h   (31)

Results

Our results for the total decay width, the cross section as well as the branching ratios can be plotted to show in the figures.

Figure 1 Branching ratios vs. Higgs mass.

Figure 2 Total width vs. Higgs mass.

Figure 3 Cross section vs. center-of-mass energy.

Conclusion

Not only does the Higgs boson give mass to quarks but also undergoes decay via a number of channels from which a few have been discussed in this paper. Experimental discovery of these decay modes have given room for unknown potential particles that may contribute mass to be discovered in future experiments. Studying the Higgs boson tells us that it is capable of interacting with other particles like quarks and give them mass.

Acknowledgments

None.

Conflicts of interest

None.

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