Higgs decay to two gluons

Transcript

1 Higgs ecay to two gluons Eleftherios Moschanreou (a) The ecay rate An unstable particle has a probability to ecay to a certain number of final states uring a given time interval. The process of ecaying is escribe by the ecay rate as Γ Number of ecays per unit time N umber of unecaye particles present One of the channels, the scalar Higgs ecays through is h g (two gluons) In the Higgs rest frame, the ecay rate in the neighborhoo of the final momenta p an k is: Γ m A 3 p (π) 3 E p 3 k M(m (π) 3 h p, k) (π) 4 δ (4) (m h p k) E k where M(m h p, k) is the the ecay amplitue, a quantity that escribes the interactions that take place uring the ecay. (b) From the Feynman iagram to the ecay amplitue. The ecay involves a fermion (quark) loop. For each quark there are two possible Feynman iagrams that contribute to the ecay amplitue. For the first iagram. We recognize three vertices an three fermion propagators: Each fermion propagator contributes: i( p + m) p m + iɛ

2 The Higgs-fermion-antifermion (quark-antiquark) vertex contribution to the amplitue is: i m q υ δab υ is the vacuum expectation value for the Higgs an the Kronecker elta ensures color conservation. The gluon-quark-anti quark vertex contribution is: igγ µ t α The iagram contailns a loop momentum l. We nee to intergrate over this loop momentum. So finally the matrix element can be written as: M i m q υ ( ) 4 l (π) T r[ i( l k + m q ) 4 (l k) m q + iɛ ig sγ ν i( l + m q ) l m q + iɛ ig sγ µ i( l + p + m q ) (l + p) m q + iɛ T r[t b t a ] ɛ λ µ (p) where ɛ λ ν (k), ɛ λ µ (p) the polarization vectors of the outcoming gluons. (c) Calculating the ecay amplitue: newline The numerator N of the loop momentum integral contains terms. Using trace technology we can eventually simplify it: T r[n] (i) 5 g s T r[( l k + m q )γ ν ( l + m q )γ µ ( l + p + m q )] i g s T r[( l k + m q ) (γ ν l + γ ν m q )(γ µ l + γµ p + γ µ m q )] i g s T r{( l k + m q ) [γ ν l(γ µ l + γµ p + γ µ m q ) + γ ν m q (γ µ l + γµ p + γ µ m q )]} i g s T r{( l k + m q ) [γ ν lγ µ l + γ ν lγµ p + γ ν lγ µ m q +γ ν m q γ µ l + γ ν m q γ µ p + γ ν m q γ µ m q ]} i g s T r{ l [γ ν lγ µ l + γ ν lγµ p + γ ν lγ µ m q + γ ν m q γ µ l + γ ν m q γ µ p + γ ν m q γ µ m q ] k[γ ν lγ µ l + γ ν lγµ p + γ ν lγ µ m q + γ ν m q γ µ l + γ ν m q γ µ p + γ ν m q γ µ m q ] +m q [γ ν lγ µ l + γ ν lγµ p + γ ν lγ µ m q + γ ν m q γ µ l + γ ν m q γ µ p + γ ν m q γ µ m q ]} To simplify things we rop all the terms with o number of gamma matrices using the property T r[o number of γ s]

3 So T r[n] i gs T r{ lγ ν lγ µ m q + lγ ν m q γ µ l + lγ ν m q γ µ p kγ ν lγ µ m q kγ ν m q γ µ l kγ ν m q γ µ p +m q [γ ν lγ µ l + γ ν lγµ p + γ ν m q γ µ m q ]} Grouping some terms: T r[n] i gs m q T r[ lγ ν { l, γ µ } + { l, γ ν }γ µ p kγ ν { l, γ µ } kγ ν γ µ p +γ ν lγ µ l] + i g s m 3 q T r[γ ν γ µ ] Furthermore using an T r[γ ν γ µ ] 4g µν {γ µ, γ ν } g µν For the first term in the square bracket the calculations yiel: T r[ lγ ν { l, γ µ }] T r[ lγ ν {l α γ α, γ µ }] T r[ lγ ν l α {γ α, γ µ }] T r[ lγ ν l α g αµ ] T r[ lγ ν l µ ] T r[l ρ γ ρ γ ν l µ ] l µ l ρ T r[γ ρ γ ν ] 8l µ l ρ g ρν 8l µ l ν So the trace can now be written: T r[n] i gs m q [8l µ l ν + 8l ν p µ 8k ν l µ + T r[ k α γ α γ ν γ µ p β γ β + γ ν l α γ α γ µ l β γ β ]] +i gs m 3 q 4g µν i gs m q [8l µ l ν + 8l ν p µ 8k ν l µ + T r[ k α p β γ α γ ν γ µ γ β + l α l β γ ν γ α γ µ γ β ]] +i gs m 3 q 4g µν Also an other ientity that shoul be use: T r[γ µ γ ν γ ρ γ σ ] 4(g µν g ρσ g µρ g νσ + g µσ g νρ ) Giving: T r[n] i gs m q [8l µ l ν + 8l ν p µ 8k ν l µ 4k α p β (g αν g µβ g αµ g νβ + g αβ g µν ) +4l α l β (g να g µβ g νµ g αβ + g νβ g µα )] + i gs m 3 q 4g µν i gs m q [8l µ l ν + 8l ν p µ 8k ν l µ 4k ν p µ + 4k µ p ν 4 kp g µν +4l ν l µ 4l g νµ + 4l µ l ν ] + i gs m 3 q 4g µν i gs m q [6l µ l ν + 8l ν p µ 8k ν l µ 4k ν p µ + 4k µ p ν 4 kp g µν 4l g νµ ] +i gs m 3 q 4g µν i gs 4m q [4(l µ l ν l g νµ ) + (l ν p µ k ν l µ ) k ν p µ + k µ p ν + (m kp) g µν ] 3

4 The next task is to treat the enominator: ((l k) m q + iɛ)(l m q + iɛ)((l + p) m q + iɛ) The integration is over l. We see that we have a prouct l l l in terms of powers of l. So we can rewrite the enominator in the form of a quaratic polynomial raise in the thir power. This proceure is calle Feynman parametrization. We start from observing that the integral x x x [xa + ( x)b]. For three terms in the enominator the appro- is easily calculate an is equal with priate formula is: ABC x y z AB x x xy δ(x + y ) [xa + yb]! xyz δ(x + y + z ) [xa + yb + zc] 3 With A ((l k) m q + iɛ), B (l m q + iɛ), C ((l + p) m q + iɛ) : ABC x x,y,z x x,y,z x x,y,z δ(x + y + z ) xyz [xa + yb + zc] 3 δ(x + y + z ) xyz [x((l k) m q + iɛ) + y(l m q + iɛ) + z((l + p) m q + iɛ)] 3 δ(x + y + z ) xyz [(x + y + z)l + xk xlk + zlp + zp (x + y + z)m q + (x + y + z)iɛ)] 3 The elta function imposes the constraint x + y + z so ABC x z x z xz [l + xk + l(zp xk) + zp m q + iɛ)] 3 We complete the square by aing an subtracting the term (zp xk) ABC x x z x x z x x z xz [l + xk + l(zp xk) + (zp xk) (zp xk) + zp m q + iɛ)] 3 xz [(l + zp xk) + xk (zp xk) + zp m q + iɛ)] 3 xz [(l + zp xk) + xk (z p + x k xzpk) + zp m q + iɛ)] 3 4

5 k an p are zero since they express the masses of the gluons which are massless. So the enominator becomes: ABC x x z xz [(l + (zp xk)) + xzkp m q + iɛ)] 3 p + k is the four momentum of the Higgs scalar an p an k are the four momenta of the gluons. Also the square of the four momentum expresses the mass square. So we have (p + k) m h p + k + pk m h pk m h Finally by setting l + (zp xk) l an xzm h m q + iɛ the enominator becomes ABC x x xz [l + )] 3 Going back to the numerator: {}}{ N i gs 4m q [ 4(l µ l ν l g νµ ) + (l ν p µ k ν l µ ) k ν p µ + k µ p ν + (m q kp) g µν ] We will call the term uner the overbrace N(L), enoting the terms of the numerator that contain the loop momentum l. These terms will be shifte to the the new momentum l l + zp xk l l zp + xk N(L) 4(l µ l ν l g νµ ) + (l ν p µ k ν l µ ) 4[(l zp + xk) µ (l zp + xk) ν (l zp + xk) g µν ] +((l zp + xk) ν p µ k ν (l zp + xk) µ ) 4[(l µ zp µ + xk µ )(l ν zp ν + xk ν ) (l + l(xk zp) + x k + z p + xzkp)g µν ] Droping k an p : +[(l ν zp ν + xk ν )p µ k ν (l µ zp µ + xk µ )] 4[(l µ l ν + l µ (xk ν zp ν ) + l ν (xk µ zp µ ) + (xk ν zp ν )(xk µ zp µ ) (l + l(xk zp) xzkp)g µν )] + +(l ν p µ zp ν p µ + xk ν p µ k ν l µ + zk ν p µ xk ν k µ ) To simplify things even more we observe that we have some linear terms with respect to l. For example the term l µ (xk ν zp ν ) insie the integral will be x x x z xz 4 l lµ (xk ν zp ν ) [l + )] 3 5

6 Since the interval of integration is symmetric an the integran is an o function, the integral vanishes. So we can rop all linear (l) terms from the numerator, which now becomes. N(L) 4[(l µ l ν + (xk ν zp ν )(xk µ zp µ )] [l xzkp)g µν ] +( zp ν p µ + xk ν p µ + zk ν p µ xk ν k µ ) 4[(l µ l ν + x k ν k µ zxk ν p µ xzp ν k µ + z p ν p µ ] (l xzkp)g µν ) + ( zp ν p µ + xk ν p µ + zk ν p µ xk ν k µ ) 4l µ l ν l g µν + (4z z)p ν p µ + (4x x)k ν k µ 4xzp ν k µ + (x + z 4zx)k ν p µ + xz m hg µν Now we have everything that we nee to calculate the integral 4 l (π) T r[ i( l k + m q ) 4 (l k) m q + iɛ ig sγ ν i( l + m q ) l m q + iɛ ig sγ µ i( l + p + m q ) (l + p) m q + iɛ ] Which now becomes x xz With pk m h. x xz 4 l (π) 4 ig s4m q { [4lµ l ν l g µν + (4z z)p ν p µ + (4x x)k ν k µ 4xzp ν k µ ] + [l + )] 3 + (x + z 4zx)kν p µ + xz m h gµν k ν p µ + k µ p ν + (m kp) g µν [l + )] 3 } {}}{ 4 l (π) 4 ig s 4m q { [ 4l µ l ν l g µν +(4z z)p ν p µ + (4x x)k ν k µ ] [l + )] 3 [( 4xz)p ν k µ + (x + z 4zx)k ν p µ + (m q + m h + xz m h )gµν ] } [l + )] 3 Lets eal with the terms uner the overbrace, containing l an l µ l ν first. 4 l I(l) (π) ( 8l µ l ν 4 [l + )] l g µν 3 [l + )] ) 3 First the integral 4 l l µ l ν will vanish if µ ν this can be seen through the example (π) 4 [l +)] 3 4 l l l (π) 4 [l + )] 3 l 4 l l[ l l 3 (π) 4 [l + )] 3 ]l The integral in the brackets has an o integran in a symmetric integration interval so it vanishes. So only the terms where µ ν really matter. A tensorial quantity that has the µ ν elements zero is g µν. So assume that l µ l ν is proportional to g µν : l µ l ν λg µν g µν l µ l ν λg µν g µν 6

7 But g µν g µν is equal with, the imension of the space. So λ l. Now the integral I(l) can be written: 4 4 l ( I(l) )gµν l (π) 4 [l + )] 3 To procee we employ the technique calle imensional regularization. First we perform a Wick rotation l il. By oing this we get ri of the minus sign of the Minkowski metric an now 4 l Ω 3 3 l in polare coorinates can be written We start from the -imensional integral (in polar coorinates, D Eukliean) : l i l (π) [l )] i 3 Ω l (π) l l [l )] 3 Note that the factor i came from the replacement l il. Also notice that the sign of has change. The area of a -imensional sphere is Ω π / Γ( /) The secon integral is l + l [l )] 3 l (l (l ) ) [l ] 3 l l ( l ) (l ) l + ) ( l )(l l )(l ( + ) l y ( y + ) y l (l (l ) ) [l ] (l ) + ) ( l )(l l ( l ) ( l + ) l l )(l ( + ) l y y y (y + ) ( y) ( ) ( y) ( ) ( y ) x (x) (x ) x x (x ) + 7

8 Comparing with efinition of the beta function : the last integral can be written: x (x) α (x ) β B(α, β) Γ(α)Γ(β) Γ(α + β) l + l [l )] 3 So the integral we wish to calculate can be written: 4 l ( )l ( 4 (π) [l )] 3 ) ( 4 ) π / Γ( /) Γ( )Γ( + ) Γ(3) Γ( ) Γ( ) Γ(3) Ω l (π) (π) l l [l )] 3 ( 4 ) 4 Γ( ) π / Γ(3) Γ( ) Γ( ) Γ(3) This is potentially ivergent. We nee to examine the behavior in the neighborhoo of 4 which the imension of our Minkowski space. To o so we introuce ε 4. So This gives Γ( ) Γ(ε) ( 4 ) 4 π / Γ( ) Γ(3) Using now the expansion of the Γ function: (4 ) (4π) ( 4π ) Γ( ) Γ(3) ε (4π) ( Γ(ε) ) ε 4π Γ(3) Γ(ε) ε γ + O(ε) the above becomes ε (4π) ( Γ(ε) ) ε 4π Γ(3) ε (4π) (4π γ + O(ε) )ε ε 8

9 where γ the Euler - Mascheroni constant an ( 4π )ε. So 4 l ( )l (π) [l )] 3 ε (4π) [ ε γ + O(ε)] εγ + εo(ε) (4π) Taking the limit ε we see that for 4 the integral oes not iverge an its value i is. 6π The part of the integral in M that oes not contain any l (we enote it as I ): I x xz i g s 4m q 4 l (π) [ [(4z z)p ν p µ + (4x x)k ν k µ + ( 4xz)p ν k µ ] 4 [l + )] 3 [(x + z 4zx)k ν p µ + (m q + m h + xz m h )gµν ] ] [l + )] 3 xz i gs 4m q [(4z z)p ν p µ + (4x x)k ν k µ x +( 4xz)p ν k µ + (x + z 4zx)k ν p µ + (m q m h + xz m h)g µν ] 4 l (π) 4 [l + )] 3 9

10 Lets repeat the calculation for the integral: 4 l (l il) (π) 4 [l + )] 3 Ω 4 l 4 i l (π) 4 [l )] 3 Ω 4 l 4 i l (π) 4 [l )] 3 i π l 3 l 6π 4 [l ] 3 i 6π i 6π [ i 6π [ l l [l ] 3 6π l [l ] + x x + x x 3 ] l [l ] 3 ] (l + )l [l ] 3 i 6π [( x ) + ( x )] 6π [ + ] i 6π This quantity will be multiplie with the polarization vectors of the gluons ɛ λ ν (k), ɛ λ µ (p). But having in min the tranversality of the polarization: p µ ɛ λ µ (p) an k ν ɛ λ ν (k), many terms vanish. The terms that will survive are [( 4xz)p ν k µ + (m q m h + xz m h )gµν ]. So the initial expression of the 6π ecay amplitue M can now be written:

11 M i m υ ( ) i g s 4m q x x ɛ λ µ (p) T r[t b t a ] M i m υ ( ) i g s 4m q ɛ λ µ (p) T r[t b t a ] M m υ g s m q x x [( xz[ igµν 4xz)pνkµ + (m 6π + i q m h + xz m h )gµν ] ] 6π ɛ λ µ (p) T r[t b t a ] M m υ g s m q M m υ g s m q M m q υ g s x x x x M g s υ T r[tb t a ] x x g µν + [( 4xz)p ν k µ + (m q xz[ m h + xz m h )gµν ] ] 6π g µν ( m q + xzm h ) + [( 4xz)pν k µ + (m q xz[ m h + xz m h )gµν ] ] 4π g µν ( xz[ m h + xzm h ) + ( 4xz)pν k µ 4π ( m q + xzm h ) ] ɛ λ µ (p) T r[t b t a ] ( 4xz)[ xz[ m h gµν p ν k µ ] ] ɛ λ 4π m q( xz m µ (p) T r[t b t a ] h ) m q m h gµν p ν k µ 4π m q m h gµν p ν k µ 4π x xz[ 4xz ] xz m h x m q }{{} I( m h m q ) ɛ λ µ (p) ɛ λ µ (p) T r[t b t a ] From the GWS theory of weak interactions we have that υ m W g with g e sin θ W υ m W sin θ W. Also the trace T r[t b t a ]: e. So T r[t b t a ] δαb Plugging υ, g an T r[t b t a ]: M e g s m W sin θ W δαb m h gµν p ν k µ 4π I( m h m q ) ɛ λ µ (p)

12 This is our first achievement, the calculation of the transition amplitue for the first iagram: M λλ αb q e gs 8π m W sin θ W δαb ( m h g µν p ν k µ ) I( m h m q ) ɛ λ µ (p) We nee now to calculate the transition amplitue for the secon iagram. Lets compare an observe the two iagrams: We see that if we change in the first iagram p k, µ ν an λ λ we get the secon iagram. But these three transformations leave invariant the final (boxe) expression of M λλ αb q. So we conclue that the two iagrams contribute equally to the total amplitue M which is the boxe expression. Thus finally M λλ αb q e gs {}}{ 4π m W sin θ W δαb I( m h ) ( m h g µν p ν k µ ) ɛ λ µ (p) m q Now to fin the ecay rate we nee to sum this expression over all possible gluon polarizations λ an λ, over all colors α an b an over all possible intermeiate loop quarks an then square it. To simplify things lets eal first with the tensorial structure uner the overbrace (lets call it T λλ ).

13 T λλ λ,λ ( m h g µν p ν k µ ) ɛ λ µ (p) λ,λ ( m h g µ ν p ν k µ ) ( m h g µ ν p ν k µ ) ɛ λ ν (k) ɛ λ µ (p) ɛ λ (k) ɛ λ µ (p) λ,λ ν [ m4 h 4 gµ ν g µ ν m h gµ ν p ν k µ m h pν k µ g µ ν + p ν k µ p ν k µ ] ɛ λ ν (k) ɛ λ µ (p) ɛ λ (k) ɛ λ µ (p) λ,λ m4 h 4 gµ ν g µ ν λ ν m h gµ ν p ν k µ λ m h pν k µ g µ ν λ +p ν k µ p ν k µ λ ɛ λ µ (p) ɛ λ µ (p) λ ɛ λ ɛ λ µ (p) ɛ λ µ (p) ν (k) ɛ λ (k) λ ɛ λ µ (p) ɛ λ µ (p) ɛ λ µ (p) ɛ λ µ (p) λ λ ɛ λ ɛ λ ν ν (k) ɛ λ (k) ɛ λ ν ν (k) ɛ λ (k) ν ν (k) ɛ λ (k) When we have a sum over the polarization vectors of massless gauge bosons, we can use the replacement originate from War ientity: ɛ λ µ(p) ɛ λ ν (p) g µν λ So the above can be written: T λλ αb m4 h 4 gµ ν g µ ν g µ µ g ν ν m h gµ ν p ν k µ g µ µ g ν ν m h pν k µ g µ ν g µ µ g ν ν + p ν k µ p ν k µ g µ µ g ν ν m4 h 4 4 m h pν k µ g µ ν m h pν k µ g µ ν + p k m 4 h m hpk m 4 h m h m4 h So the total amplitue M λλ αb q (unerline expression) can be square: M e gs 4π m W sin θ W δαb I( m h ) T λλ m q e gs 4 6π 4 m W sin θ W 4 a,b m h δ αb a,b 3 q q ν λ,λ I( m h ) m4 h m q

14 The are 8 gluon colors so the inices a, b {, 8}. Thus: δ αb 8 a,b So finally M e gs 4 m 4 h I( m h ) 6π 4 m W sin θ W m q q Last thing is to go back to the ecay rate formula: Γ 3 p 3 k M(m m h (π) 3 E p (π) 3 h p, k) (π) 4 δ (4) (m h p k) E k In the rest frame of the scalar Higgs the energies of the two (massless) gluons are the same E p E k p k. Also the elta function can be analyze as: δ (4) (m h p k) δ(m h E p E k )δ (3) ( p h p + k) δ(m h p )δ (3) ( p + k) δ[ ( p m h )]δ(3) ( p + k) Using the ientity of the elta function δ[a(x x )] a δ(x x ). We have δ (4) (m h p k) δ( p m h )δ(3) ( p + k) Then Γ becomes: 3 p 3 k Γ m h (π) 3 (π) 3 4 p M(m h p, k) (π) 4 δ( p m h )δ(3) ( p + k) 64m h π M 3 p p δ( p m h ) 64m h π M Ω p p δ( p m h ) p 64m h π M 4π 4

15 The final result for the ecay rate is: Γ M 6m h π e gs 4 m 4 h I( m h ) 6m h π 6π 4 m W sin θ W m q If we introuce the fine structure constant α e 4π α s g s 4π : Γ 4 α α s m 3 h π m W sin θ W q q an its analogue in strong interactions I( m h ) m q Also the form factor implies that the heavier quarks contribute significantly to the ecay rate. So we expect only the top quark to play an important role. I ( m h m q ) x x z xz[ 4xz ] xz m h m q 5