Particle Physics WS 2012/13 ( )

Transcription

1 Particle Physics WS 2012/13 ( ) Stephanie Hansmann-Menzemer Physikalisches Institut, INF 226, 3.101

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6 Where are we? W fi = 2π 4 LI matrix element M i (2Ei) fi 2 ρ f (E i ) LI phase space i> f> M fi = M a c+x M(b+x d) q 2 mx 2 e.g. e + μ e + μ Matrix element describing two interaction vertices and one intermediate particle can be computed as the product of the matrix elements of the both vertices and 1 a factor for the intermediate particle (propagator) q 2 mx 2 Next step: compute matrix element of three prongue vertex (e.g. e e + γ) Up to now, everything was independent of type of interaction, to compute vertex, we need to study a specific interaction 6

7 Where do we want to go? Program of next two lectures Derive potential of electromagnetic IA Derive general rules for computation of Matrix element (here will come back to the free spin=1/2 particle wave functions) e μ + Calculate cross-section for e + e + μ + μ + Helicity vs. Chirality Calculate cross-section for related process exploiting Mandelstamm variables e + μ e + μ e + e μ e and e + e + e + e + e e e e + + μ μ e + e e + e + 7

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10 Matrix Element for e - τ - Scattering e e M fi = M a c+x M(b+x d) q 2 mx 2 V = qγ 0 γ μ A μ p 1 p 2 p 4 p 3 M = u e (p 2 ) q e γ 0 γ μ u e (p 1 ) λ ε μ λ (ε v λ ) q 2 u τ (p 4 )q τ γ 0 γ ν u τ (p 3 ) τ τ <Ψ V Ψ > e <Ψ V Ψ > τ Ψ = u(p)ei(px Et) Ψ = u(p) e i(px Et) = -q e q τ u e (p 2 )γ μ u e (p 1 ) g μν q 2 u τ (p 4 )γ ν u τ (p 3 ) u e (p 2 )γ 0 u τ (p 2 )γ 0 10

11 Reminder Probability Density and Currents x ψ (iγ μ μ m )ψ = 0 i μ ψγ μ + m ψ= 0 x ψ iψγ μ μ m ψψ + i( μ ψ)γ μ ψ+ m ψψ= 0 i μ (ψγ μ ψ)= 0 Continuity equation: μ j μ = 0 with j μ = (ρ, j) j μ = ψγ μ ψ ρ = ψγ 0 ψ = ψ ψ = particle current 4 4 i=1 ψ i ψi = i=1 ψ i 2 > 0 Historically the positive density ρ was the original motivation for Dirac s work Charge current for an electron: j μ = -eψγ μ ψ 11

12 Feynman Rules e e M = -q e q τ u e (p 2 )γ μ u e (p 1 ) g μν q 2 u τ (p 4 )γ ν u τ (p 3 ) = - q e q τ q 2 j eμ g μν j τ v p 1 p 2 p 3 p 4 M = - q e q τ q 2 j e j τ τ τ Matrix element is four-vector scalar product thus it is Lorentz invariant! This expression hides a lot of complexity. Needed to sum over all possible time-orderings and summed over all polarization states. Fortunately don t need to redo this for each new Feynman diagram! We can write down matrix elements using a set of simple rules Feynman rules! Propagator factor for each internal line (e.g. each internal virtual particles) Dirac Spinor for each external line (e.g. each real real incoming/outgoing particle) Vertex factor for each vertex 12

13 Feynman Rules 13

14 Feynman Rules e p 1 e p 3 -im = i q e v e p 2 γ μ u e (p 1 ) igμν q 2 iqeu e p 3 γ ν v e (p 4 ) p 2 p 4 e + e + u p 1 u p 3 -im = i q u u u (p 2 ) γ μ u u (p 3 ) igμν q 2 iqdv d p 4 γ ν v d (p 2 ) p 2 p 4 d d e - p 1 p 3 -im = i q e ε(p 3 ) μ* γ μ u e (p 1 ) i(γ ν q ν +m) q 2 m 2 iqeε p 4 ξ γ ξ v e (p 3 ) e + p 2 p 4 14

15 Calculation of cross section e + e - μ + μ - Lecture 3, diff. cross-section in CMS: e μ p 1 p 3 dσ = 1 p f M dω 64 π 2 s p fi 2 i p 1 = p 2 = pi p 3 = p 4 = pf s = (E 1 + E 2 ) 2 p 2 e + μ + p 4 lowest order diagram M ~ e 2 ~α em (NU: e= 4πα; α ~ 1 ) 137 & many second order diagrams e μ + p 1 p 3 e μ + p 1 p 3 e + p 2 M = Γ 1 + Γ 2 + μ p 4 Summing amplitudes not amplitudes square, therefore interference effects possible. Higher order in elm. IA however suppressed (α~ ), thus consider only 1st order here e + p 2 μ p 4 15

16 Calculation of cross section e + e - μ + μ - e μ + e + p 1 p 2 μ p 4 p 3 -im = i q e v e p 2 γ μ u e (p 1 ) igμν iq q 2 μ u μ p 3 γ ν v μ (p 4 ) M = - e 2 [v s e p 2 γ ν u e (p 1 )] [ u μ p 3 γ ν v μ (p 4 )] In general electron and positron will not be polarized, i.e. there will be equal numbers of positive and negative helicity states there are four possible combinations of helicities in the initial state! (right handed helicity helicity=+1; left handed helicity helicity=-1) Similarly there are four possible helicity combinations in the final state In total there are 16 combinations e.g. RL RR, RL RL, Need to sum over all 16 possible helicity combinations and then average over number of Initial helicity states: < M 2 > = 1 4 ( M RL RR 2 + M RL LR 2 + ) elicity 16 spin averaged matrix element particle helicity antiparticle helicity

17 Calculation of cross section e + e - μ + μ - Helicity eigenstates: u h=1 = N c e iφ s p E+m c p E+m eiφ s u h=-1 = N s e iφ c p E+m s p E+m eiφ c v h=1 = N p E+m s p E+m eiφ c s e iφ c v h=-1 = N p E+m c p E+m eiφ s c e iφ s N = E + m, c=cosθ/2, s=sinθ/2 z p = p sin θ cos φ sin θ sin φ cos θ φ θ y x In limit of E>>m: u h=1 = E c e iφ s c e iφ s u h=-1 = E s e iφ c s ei φ c v h=1 = E s ei φ c s e iφ c v h=-1 = E c e iφ s c e17 iφ s

18 x p 1 z p 4 Calculation of cross section e + e - μ + μ - p 3 e e + θ μ + u h=1 = E π-θ c e iφ s c e iφ s μ p 2 φ=0 u h=-1 = E p 1 = (E, 0, 0, E) p 2 = (E, 0, 0, -E) p 3 = (E, Esin θ, 0, E cos θ) p 4 = (E, -Esin θ, 0, E cos θ) s e iφ c s ei φ c v h=1 = E s ei φ c s e iφ c [θ in CMS, E>>m μ ] v h=-1 = E c e iφ s c e iφ s Initial state electron (φ=0, θ=0): u h=1 = E u h=-1 = E Final state muon (φ=0,θ): c u h=1 = E s c s u h=-1 = E s c s c Initial state positron (φ=0, θ=π): v h=1 = E v h=-1 = E Final state anti-muon (φ=0,θ-π): v h=1 = E c s c s v h=-1 = E s c s c 18

19 muon current: j μ = u p 3 γ ν v(p 4 ) Muon Current e μ + p 1 p 4 For arbitrary spinors it is straight forward to show: e + p 2 μ p 3 Ψ γ 0 γ 0 Φ Ψ 1 Φ1 + Ψ 2 Φ2 + Ψ 3 Φ3 + Ψ 4 Φ4 Ψ γ ν Φ= Ψ γ 0 γ 1 Φ Ψ γ 0 γ 2 Φ = Ψ 1 Φ4 + Ψ 2 Φ3 + Ψ 3 Φ2 + Ψ 4 Φ1 i(ψ 1 Φ4 Ψ 2 Φ3 + Ψ 3 Φ2 Ψ 4 Φ1 ) [eq.1] Ψ γ 0 γ 3 Φ Consider μ - R, μ + L using Ψ=u R, φ=v L Ψ 1 Φ3 Ψ 2 Φ4 + Ψ 3 Φ1 Ψ 4 Φ2 u h=1 = E j μ RL= u p 3 R γ ν v p 4 L = 2E(0,sinθ/2 2 -cosθ/2 2,i,2Esinθ/2 cosθ/2) = 2E(0,-cosθ, i,sinθ) c s c s v h=-1 = E s c s c 19

20 Muon Current analogously compute: j μ RL= u p 3 R γ ν v p 4 L = 2E(0,-cosθ,i,sinθ) j μ RR= u p 3 R γ ν v p 4 R = (0,0,0,0) j μ LL= u p 3 L γ ν v p 4 L = (0, 0, 0, 0) j μ LR= u p 3 L γ ν v p 4 R = 2E(0, -cosθ,-i, sinθ) In the limit of E>>m, only two non vanishing helicity combinations for final state! Due to angular momentum conservation, need to look at 4 combinations of initial and final state only: Next step: needs to compute, j e RL j e LR 20

21 Electron Current The electron current can either be directly derived from eq[1] or from the expression of the muon current: j μ RL = [u p 3 R γ ν v p 4 L ] = [u(p 3 ) R γ 0 γ ν v p 4 L ] = v(p 4 ) L γ ν γ 0 u p 3 R = v(p 4 ) L γ 0 γ ν u p 3 R = v p 4 L γ ν u p 3 R p 1 e μ + p 4 γ μ = -γ μ p 1 p 3 μ e e + θ p 2 e + p 2 μ p 3 π-θ p 4 φ=0 j e RL = v p 2 L γ ν u p 1 R = Θ =0 2E(0,-cosθ,-i,sinθ) = 2E(0,-1,-i,0) j e LR = v p 2 R γ ν u p 1 L = 2E(0,-cosθ,i,sinθ) = 2E (0, -1,i,0) 21

22 Matrix Element Calculation Can now calculate M =- e2 s j ej μ for the four possible helicity combinations. e.g. for M(RL RL) particle helicity antiparticle helicity j e RL = v p2 L γ ν u p 1 R = 2E (0, -1,-i,0) j μ RL= u p 3 R γ ν v p 4 L = 2E(0, -cosθ,i, sinθ) M(RL RL) = e2 s 4E2 (0, 1, i, 0)(0, cos θ, i, sin θ) = -e 2 (cosθ+1) s = 2E = -4πα(1+cosθ) M(RL RL) 2 = e cosθ 2 22

23 Matrix Element Calculation M(RL RL) 2 M(RL LR) 2 M(LR RL) 2 M(LR LR) 2 Assuming that the incoming electrons are unpolarized, all 4 possibilities are equally likely. 23

24 Differential Cross Section The cross section is obtained by averaging over the initial states and summing over all final states. (E>>m μ p f = p i ) = dσ dω = 1 4 = 1 64 π 2 s p f p i M fi 2 1 ( M(RL RL) 2 + M(LR LR) 2 + M(RL LR) 2 + M(LR RL) 2 ) 256 π 2 s 4πα π 2 s (2(1+cosθ)2 + 2(1-cosθ) 2 ) dσ = α2 dω 4s (1+cos2 θ) Example: e + e - μ + μ, s = 29 GeV pure QED (α 3 ) calculation QED + Z contribution 24

25 Total Cross Section Total cross section obtained by intregrating over θ and φ using 1 + cos 2 θ dω = 2π (1 + cos 2 θ) dθ = 16π 3 total cross section for (e + e - μ + μ ): σ = 4πα2 3s Lowest order computation provides good description of data Very impressive result: from first principle computation we have reached a 1% precise result! 25

26 Helicity and Chirality = ς 0 0 ς = ς 1 ς 1 ς 2 ς 2 ς 3 ς 3 h = Σp [H,h] = 0 p s p s h Ψ= + Ψ h Ψ= - Ψ right handed (RH) helicity right handed (RH) helicity Helicity is a conserved quantity! Gedankenexperiment: A massive particle (e.g. electron) with helicity +1 is moving with v c p,v s h=1 An observer overtakes the electron by moving with v obs > v In reference system of observer velocity and momentum will have opposite sign, while spin will be the same. Electron helicity in the frame of the observer will be LH (h=+1) Helicity is not a LI quantity! 26

27 Helicity Projector Projector on helicity eigenstates (see homework) P ± = 1 2 (1 ± Σp) unpolarized electron Ψ(e - ) = a Ψ h=+1 + b Ψ h=-1 P + (Ψ(e - )) = a Ψ h=+1 P - (Ψ(e - )) = b Ψ h=-1 P +2 (Ψ(e - )) = a Ψ h=+1 P -2 (Ψ(e - )) = b Ψ h=-1 Definition of any projector P 2 Ψ = P Ψ 27

28 Chirality chirality operator: γ 5 = i γ 0 γ 1 γ 2 γ 3 = 0 I I 0 in Dirac representation chirality projector: P ± = 1 2 (1 ± γ5 ) P ± 2 = ± γ5 2 = ± γ5 1 ± γ 5 = ± 2γ5 + γ 5 2 = ± 2γ5 + 1 = 1 2 (1 ± γ5 ) define LH chirality state: Ψ L = 1 2 (1 γ5 )Ψ define RH chirality state: Ψ R = 1 2 (1 + γ5 )Ψ [H,γ5] 0 (see homeworks) chirality is not a conserved quantity chirality is however LI! 28

29 Chirality-Eigenstates In QED, basic IA between fermion and photon is: qψ γ μ Φ φ = φ R + φ L Ψ = Ψ R + Ψ L qψ γμ Φ = q Ψ R + Ψ L γ μ Φ R + Φ L exploiting following relations: (γ 5 ) 2 =1 γ 5 =γ 5 γ 5 γ μ = -γ μ γ 5 Ψ R γμ Φ L = γ5 Ψ γ 0 γ μ γ5 Φ = 1 4 Ψ + Ψ γ 5 γ 0 γ μ [1 γ 5 ]Φ = 1 4 Ψ γ 0 γ μ Φ + Ψ γ 5 γ 0 γ μ Φ Ψ γ 5 γ 0 γ μ Φ Ψ γ 0 γ μ Φ analogous: Ψ L γμ Φ R =0 = 0 only certain chirality eigenstates contribute to IA (always true statement)! only for E>>m : γ 5 = Σp (see homework) LH chirality eigenstates LH helicity eigenstates RH chirality eigenstates RH helicity eigenstates For massive particles (not ultra relativisitc): helicity states are combination of chirality states! 29

30 Intermediate Summary 30

31 Mandelstamm Variables 31

32 Lorentz Invariant Representation of X-Section 32