H. A. TANAKA: PHY 489/1489 LECTURE 4: PHASE SPACE IN DECAYS LAST TIME: • We reviewed some basic quantum mechanics • hamiltonian, time evolution • angular momentum, Pauli matrices • Transition rates, Born approximation, golden rule • Decay rates • Today: • phase space • decay rate formula • (scattering rate) • sorry, no pizza SCATTERING RATES • Send in particles on a “target” and study what comes out • if particles are “hard spheres”,projectile is infinitesimal incoming Target • Probability of interaction: area of target/unit area: • area of target particle = “cross section” σ • Rate ∝ rate of incoming particles: • Flux � = particles/unit area/time ~ ni v 3 MORE THAN ONE TARGET • More than one “layer” of target particles • More than one target per unit area. incoming Targets l • Rate ∝ targets in the column swept by the incoming beam • Rate = NT/Unit Area x σ x � = n l σ � • n = number density of target particles, l = length of target • Rate/volume = n σ � 4 COLLIDING BEAMS: nA, vA nB, vB • In this case, the relative velocity between the two particles determines the flux: • � = na(va+vb) • Rate =na (va+vb) nb σ D IFFERENTIAL CROSS SECTION • In hard sphere scattering, something “happening” is binary: • If the balls hit each other, then something happened • otherwise, nothing happened • Generalize: considering “differential cross section.” • Probability that particle ends up in a particular part of phase space • e.g.. a particular momentum/angle range. d3σ polar angle σ dΩ = sin θdθ d⇥ = d cos θ d⇥ ⇒ dΩ dp “solid angle” azimuthal angle • Notation lends itself to “integrating” over a phase space variable: say we don’t care about the momentum but only the angle: dσ d3σ = p2dp dΩ dΩ dp 6 TOTAL CROSS SECTION • “total cross section” • integrate over all phase space d3⇥ ⇥ = p2dp d⇤ d cos θ TOT dΩ dp • cross section for a particle to end up anywhere • For “infinite range” interactions, the total cross section can be infinite; i.e. “something” always happens • Reflects the fact that no matter how far you are away, there is still some electric field that will deflect your particle. 7 GOLDEN RULE • Fermi’s golden rule states that the probability of a transition in quantum mechanics is given by the product of: • The absolute value of the matrix element (a k a amplitude) squared • The available density of states. P 2 ρ ⇥ |M| × 2 2 P P 2 P (E) 2 ρ(E) dE ∝ |M| ⇥ × |M| ∝ |M | • Typically a decay of a particle into states with lighter product masses has more “phase space” and more likely to occur. • Let’s see how to calculate the phase space • we’ll learn how to calculate amplitudes later. PRODUCT OF PHASE SPACE Initial State Particle 3 Particle 1 Particle 2 • What is net phase space for the particle 1,2,3 to end up in particular places? • 0 if energy and momentum are not conserved • 0 if particles are not on “mass shell” • Otherwise, the product of the individual phase spaces: µ µ µ ρ = ρ1(p1 ) ρ2(p2 ) ρ3(p3 ) each component of the four- × × momentum is independent integral extends over 4 4 4 d p1 d p2 d p3 µ µ µ region satisfying ⇥tot = 4 4 4 ⇥1(p1 )⇥2(p2 )⇥3(p3 ) allowed (2π) (2π) (2π) kinematic constraints PHASE SPACE IN DECAYS Product over all Symmetry factor outgoing particles Energy must be positive S N d4p Γ = 2 (2⇡)4δ4(pµ pµ) 2⇡(p2 m2 ) ⇥(p0 ) f 2m ⇥ |M| ⇥ 1 − f ⇥ f − f f (2⇡)4 Z Xf jY=f distributed Lorentz Invariant Energy and Matrix element factor evenly in phase momentum must Outgoing (function of kinematics, space be conserved particles must polarizations, etc.) be on mass shell • Complicating looking, but represents a basic statement: • apart from matrix element, phase space is distributed evenly among all particles subject to mass requirements, E/p conservation • “dynamics” like parity violation, etc. incorporated into matrix element. THE SYMMETRY FACTOR: • Consider the integration of phase space for two particles in the final state where the particles are of the same species. 4 4 d p1 d p2 (2π)4 (2π)4 • At some point, say, there will be a configuration where p1 = K1 and p2 = K2 • Since the particles are identical, we should also have the reverse case: • p1= K2, p2 = K1 • the integral will contain both cases separately. • However, in quantum mechanics, the identicalness of particles of the same species means that these are the same state and we have double counted. • We need to add a factor of 1/2 to the phase space • Likewise, for n identical particles in the final state, we need a factor of 1/n! PHASE SPACE: 2-BODY DECAY Γ = S 2 (2⇡)4δ4(pµ pµ pµ) 2m1 ⇥ |M| ⇥ 1 − 2 − 3 R (2⇡) δ(p2 m2) ⇥(p0) (2⇡) δ(p2 m2) ⇥(p0) ⇥ 2 − 2 2 ⇥ 3 − 3 3 4 4 d p2 d p3 Let’s integrate over overall outgoing particle (2⇡)4 (2⇡)4 phase space to get the total decay rate d4p dp0dp1dp2dp3 • Start with the phase space factors: ≡ δ(p2 m2c2)=δ((p0)2 p⇧2 m2c2) − − − 1 δ(p2 m2c2)= δ(p0 p2 + m2c2)+δ(p0 + p2 + m2c2) − 2p0 − h p p i • Ignore the 2nd δ function since Θ(p0) will be 0 whenever p0 is negative 1 δ(p2 m2c2)= δ(p0 p2 + m2c2) − 2p0 − p p p⌅ 2 + m2c2 0 ⇒ ENERGY/MOMENTUM CONSERVATION 0 0 • Now integrate over p 3 and p 2 using the previous relations Γ = S 2 (2⇡)4δ4(pµ pµ pµ) 2m1 ⇥ |M| ⇥ 1 − 2 − 3 R (2⇡) δ(p2 m2) ⇥(p0) (2⇡) δ(p2 m2) ⇥(p0) ⇥ 2 − 2 2 ⇥ 3 − 3 3 4 4 d p2 d p3 (2⇡)4 (2⇡)4 3 3 d p2 d p3 1 1 3 3 (2⇡) (2⇡) 2 2 2 2 2 2 2 p⌅2 + m c 2 p⌅ + m c × 2 × 3 3 0 0 note p 2 and p 3 are now set according to E/p conservation by the δ function 4 µ µ µ S 2 δ (p1 p2 p3 ) 3 3 Γ = 2 − − d p2d p3 32⇡ m1 ⇥ |M| ⇥ p2 + m2 p2 + m2 Z 2 2 3 3 p p DECAY AT REST: • Decompose the product delta function (particle 1 at rest) δ4(pµ pµ pµ)=δ m p2 + m2 p2 + m2 δ3(p p p ) 1 − 2 − 3 1 − 2 2 − 3 3 1 − 2 − 3 ✓ q q ◆ 3 • Perform the d p3 integral 4 µ µ µ S 2 δ (p1 p2 p3 ) 3 3 Γ = 2 − − d p2d p3 32⇡ m1 ⇥ |M| ⇥ p2 + m2 p2 + m2 Z 2 2 3 3 p p 2 2 p2 + m3 q 4 µ µ µ S 2 δ (p1 p2 p3 ) 3 Γ = 2 − − d p2 32⇡ m1 ⇥ |M| ⇥ p2 + m2 p2 + m2 Z 2 2 2 3 p p INTEGRAL IN SPHERICAL COORDINATES d3p dφ d cos ✓ p 2d p 2 ) | 2| | 2| Assume no dependence of M on p 2 2 2 2 2 2 S 2 2 δ(m1c p2 + m2c p2 + m3c ) Γ = 2 d⌅ d cos ⇥ p2 d p2 − − 32⇤ m1 ⇥ | | | ||M| ⇥ p2 + m2c2 p2 + m2c2 ⇥2 2 2 ⇥ 3 2π +1 ⇥ ⇥ d⇤ 2⇥ d cos θ 2 Problem 3.10 0 → 1 → − u = p2 + m2c2 + p2 + m2c2 2 2 2 3 do a change of variables ⇥ u p ⇥ to enact the final integral | 2| du = 2 2 2 2 2 2 d p2 ⇤p2+m2c ⇤p2+m3c | | S p Γ = du 2 δ(m u)| 2| 8⇡m ⇥ |M| ⇥ 1 − u 1 Z The final integral over u sends u=m and makes p2 consistent with E conservation FINAL RESULT: TWO-BODY DECAY RATE: S p Γ = | 2| 2 2 why |p2| and 8⇡m1 |M| not |p3|? • Now need to calculate the matrix element ℳ • We’ll use Feynman diagrams and the associated calculus to calculate amplitudes for various elementary processes 16 SCATTERING • Phase space expression for scattering of two particles p4 p1 and p2 are p3 4-vectors! p1 p2 p6 p5 S 2 4 4 µ µ µ σ = 2 2 (2⇡) δ (p1 + p2 f pf ) 4p(p1 p2) (m1m2) · − ⇥ |M| ⇥ − R P 4 N 2⇡(p2 m2) ⇥(p0) d pj ⇥ j=3 j − j j (2⇡)4 • It hasQ almost the same form as the decay phase space Γ = S 2 (2⇡)4δ4(pµ pµ) 2m ⇥ |M| ⇥ 1 − f f R 4P N 2⇡(p2 m2) ⇥(p0) d pj ⇥ j=2 j − j j (2⇡)4 Q LAB-FRAME SCATTERING • consider e + p → e + p, initial proton at rest p3 p2 Since me ≪ mp, p1 assume me ~ 0 p4 S 2 4 4 µ µ µ σ = 2 2 (2⇡) δ (p1 + p2 f pf ) 4p(p1 p2) (m1m2) · − ⇥ |M| ⇥ − R P 4 N 2⇡(p2 m2) ⇥(p0) d pj ⇥ j=3 j − j j (2⇡)4 Q S 2 4 4 µ µ µ σ = 2 2 (2⇡) δ (p1 + p2 f pf ) 4p(p1 p2) (m1m2) · − ⇥ |M| ⇥ − R P 3 N 1 d pj j=3 2 2 (2⇡)3 ⇥ 2ppj +mj Q SUMMARY • Please read 4.1-4.5 for next time.