Production of Light SUSY Hadrons at LHC Ding-Yu Shao shaodingyu@mail.nankai.edu.cn School of Physics, Nankai University Co-workers: Prof. Xue-Qian Li and Prof. Mao-Zhi Yang School of Physics, Nankai University First Previous Next Last 1
Motivation The energy and luminosity of LHC may be sufficient to produce the SUSY particles. Creation of sleptons provides the cleanest and unambiguous signals, however, sleptons can only be produced via electromagnetic or weak interactions, so that their production rates are much suppressed. Instead, production of squark can be realized via quark-antiquark annihilation and gluon fusion sub-processes governed by QCD. This offers us a better opportunity of observing SUSY particles. School of Physics, Nankai University First Previous Next Last 2
Background Supersymetric Hadrons top squark: two mass eigenstates: t 1 and t 2 ; t 1 is lighter. We assume m t 1 120 150GeV width Γ t 1 Λ QCD. Finally, top-squark can form superhadrons ( t q), where (q = u,d,s,c, b) School of Physics, Nankai University First Previous Next Last 3
How to Study Superhadrons double heavy mesons heavy superhadrons ( t Q) Q : (c, b) heavy light mesons light superhadrons ( t q) q : (u, d,s) School of Physics, Nankai University First Previous Next Last 4
Heavy Superhadrons Fragmentation approach was adopt to study heavy superhadrons in Eur. Phys. J. C50,969 (2007). The cross section of T( t Q)-production at hadronic colliders can be factoriaed into three factors as below: dσ h H 1 H 2 TX = dx 1 dx 2 dzf i/h1 (x 1,µ f )f j/h2 (x 2,µ f ) ijk dσ ij kx (x 1,x 2,z; µ f )D k T (z,µ f) School of Physics, Nankai University First Previous Next Last 5
Since the fragmentation function D k T (z, µ f) of T-production is universal, it can be derived by pqcd through other processes. Suppose a fictious Z boson which is so heavy that it can decay to a pair of t 1 and t 1, the corresponding Feynman diagram is School of Physics, Nankai University First Previous Next Last 6
Light Superhadrons Since the SM quark in ( t q) is light, we can not use pqcd to derive the fragmentation function D k T (z, µ f). How to solve this problem? School of Physics, Nankai University First Previous Next Last 7
Production of Superhadrons at Hadronic Colliders q q t 1 T q Quark-antiquark annihilation produces superhadron T Gluon-gluon fussion produces superhadron T School of Physics, Nankai University First Previous Next Last 8
Setting the process quark-antiquark annihilation as an example, σ tot = σ hard θ(q 2 Λ 2 ) + σ soft θ(λ 2 Q 2 ) σ hard : fragmentation approach; σ soft : superflavor symmetry; To avoid double-counting, we set a cut-off parameter Λ School of Physics, Nankai University First Previous Next Last 9
superflavor symmetry Superflavor symmetry establishes a symmetry between a heavy meson which contains a heavy quark and a light anti-quark and a heavy baryon which contains a heavy scalar in color triplet and a light anti-quark. There is only one universal Isgur-Wise type parameter ξ in the model at the leading order. h q χ q superflavor symmetry h :heavy quark; χ :heavy scalar particle; q :light anti-quark School of Physics, Nankai University First Previous Next Last 10
Concretely, the current hγ µ h and iχ µ χ h q χ q The corresponding matrix elements can be written as < H(v ) hγ µ h H(v) > = ξ(v v)m h (v µ + v µ ) the same < X(v ) iχ µ χ X(v) > = ξ(v v) 1 2 (vµ + v µ )ū u School of Physics, Nankai University First Previous Next Last 11
The total cross section σ tot (p + p t q + X) = σ hard θ(q 2 Λ 2 ) + σ soft (ξ)θ(λ 2 Q 2 ) we use the data for inclusive b q meson production p + p b q + X measured at TEVATRON as an input to determine ξ. Concretely, we have σ tot (p + p b q + X) = σ hard θ(q 2 Λ 2 ) + σ soft (ξ)θ(λ 2 Q 2 ) and then we can fix the value ξ. School of Physics, Nankai University First Previous Next Last 12
High Energy Process The cross section dσ h H 1 H 2 TX = ijk dx 1 dx 2 dzf i/h1 (x 1,µ f )f j/h2 (x 2,µ f ) dσ ij kx (x 1,x 2,z;µ f )D k T(z,µ f ) School of Physics, Nankai University First Previous Next Last 13
The DGLAP evolution eq. of fragmentation function D t 1 T (z,q2 ): dd i T(z,Q 2 ) dτ = j α s (Q 2 ) 2π 1 z dy y P i j(z/y)d j T (y,q2 ) The solution is D t 1 T (z,q2 ) = 1 z dy y D t 1 T (z y,q2 0) ( ln(y))8 Γ( 8 3 κ) 3 κ 1 where D t 1 T (z y,q2 0) is the boundary condition. Here Q 0 = m t School of Physics, Nankai University First Previous Next Last 14
The width for the fictious Z decaying to T can be factorized as: Γ( Z T + X) = 1 0 dzγ( Z t 1 + t 1,µ f )D t 1 T (z,µ f) D t 1 T (z,q2 0) = 16α2 s(λ) φ(0) 2 27πm 2 qm t 1 (1 z) 2 z 2 6(1 α 1 z) 6 School of Physics, Nankai University First Previous Next Last 15
[ ] 2α1(z 2 4)z + α1(3α 3 1 z 2z + 2)z + 3α1 2 6α 1 + 6 where Λ is a phenomenological cut-off parameter around 1GeV and φ(0) is the wave function at at origin for the supersymmetirc hadron T. Since we are dealing with a bound state containing a heavy particle and a light particle, the potential model is not available to deduce the wave function at the origin. For simplify, we use the wave function of b q meson for evaluattion. School of Physics, Nankai University First Previous Next Last 16
The subprocess differential cross sections are dσ(gg t 1 t 1 ) = 2π2 αs 2 [ 16πŝ 2 1 2A 1 ][ ] 1 2 m2 t 1 9 Aŝ (1 m2 t 1 Aŝ ) dˆt and dσ(q q t 1 t 1 ) = πα2 s[ŝ 2 4ŝm 2 t (ˆt û) 2 ] 1 9ŝ 4 dˆt where A = (ˆt m 2 t 1 )(û m 2 t 1 )/ŝ 2. ŝ, ˆt and û are Mandelstam variables. Now, we can get the corresponding cross section of high energy process. School of Physics, Nankai University First Previous Next Last 17
Low Energy Process For the low energy case, applying the superflavor symmetry, we can write down the concrete expression of the production amplitude. Similarly, both gluon-gluon fusion and quark-antiquark annihilation are involved in. In the quark-antiquark case, q q t 1 T q Quark-antiquark annihilation produces superhadron T School of Physics, Nankai University First Previous Next Last 18
im = igst 2 aa v(k 2 )γ µ 1 u(k 1 ) (k 1 + k 2 ) 2(p 1 p 3 ) µ ( ξ) 1 m t 1 v( p 2 )γ 2 u (p 1 ) In the gluon-gluon case, Gluon-gluon fussion produces superhadron T School of Physics, Nankai University First Previous Next Last 19
im A = igst 2 ab4pµ 1 pν 3ǫ µ (k 1 )ǫ ν (k 2 ) 1 (p 1 k 1 ) 2 m 2 t ( ξ) 2 )γ 1 m t 1 v( p 2 u (p 1 ) im B = igs 2 T ba4pµ 3 pν 1ǫ µ (k 1 )ǫ ν (k 2 ) 1 (p 1 k 2 ) 2 m 2 t ( ξ) 2 )γ 1 m t 1 v( p 2 u (p 1 ) im C im D = igs(t 2 ab T ba ) (k 2 k 1 ) µ g µν 2k µ 2 gµ ν + 2k1g ν µ ν (k 1 + k 2 ) 2 1 (p 1 p 3 ) µ ǫ µ (k 1 )ǫ ν (k 2 )( ξ) 2 )γ m t 1 v( p 2 u (p 1 ) = igs(t 2 ab + T ba )ǫ µ (k 1 )ǫ ν (k 2 )g µν 1 ( ξ) 2 )γ m t 1 v( p 2 u (p 1 ) School of Physics, Nankai University First Previous Next Last 20
Concretely, setting the process q + q (b q) + X as an example to instruct how to derive the amplitudes. v p 2 q 2 The untrivial part in the matrix element is < bq(b q) bγ µ b qq 0 > According to crossing symmetry, we can remove quark b and q from the final state to the initial state, < (b q) bγ µ b qq b q > As for the fermion case, performing a Fierz transformation, we can School of Physics, Nankai University First Previous Next Last 21
get the amplitude in a universal form Insert a vacuum state 0 >< 0, 1 4 < (b q) b 1 Γ 1 q 4 q 3 Γ 2 b 2 bq > 1 4 < (b q) bγ 1 q 4 0 >< 0 q 3 Γ 2 b 2 bq > then left hand side forms a bound state, and right hand side forms free states. The bound state < (b q) can be written as < (b q) = mb 2 (1 /v)γ5 School of Physics, Nankai University First Previous Next Last 22
Thus, the amplitude has the form < bq(b q) bγ µ b qq 0 >= ξ 2 mb v( q 2 )(1 /v)γ 5 γ µ u( p 2 ) Thus, we can get the cross section dσ s H 1 H 2 TX (ξ) = dx 1 dx 2 f i/h1 (x 1,µ f )f j/h2 (x 2,µ f ) ij dσ ij TX (x 1,x 2 ;µ f ) School of Physics, Nankai University First Previous Next Last 23
Finally, we obtain the cross section of T production σ tot (p + p t q + X) = σ hard θ(q 2 Λ 2 ) + σ soft (ξ)θ(λ 2 Q 2 ) Similarly, the case of (b q) production, σ tot (p + p b q + X) = σ hard θ(q 2 Λ 2 ) + σ soft (ξ)θ(λ 2 Q 2 ) School of Physics, Nankai University First Previous Next Last 24
Numerical Result We use the data of inclusive B + production at TEVTRON to evaluate the total cross section of ( tū) production. m t m t LHC ( s = 14 TeV) = 120 (GeV) ( tū) : 285.577 (pb) = 150 (GeV) ( tū) : 124.267 (pb) The results indicate that the production rate of tū is obviously larger than that for t b( c), which is about 10 2 10 3 fb. Thus, there will be of more chances to be observed at LHC. School of Physics, Nankai University First Previous Next Last 25
THE END THANKS! School of Physics, Nankai University First Previous Next Last 26