Fermi s Golden Rule and Simple Feynman Rules ; December 5, 2013
Outline Golden Rule 1 Golden Rule 2
Recipe For the Golden Rule For both decays rates and cross sections we need: The invariant amplitude M Phase space available M comes from evaluating the relevant Feynman diagram Phase space depends on the masses, momentum and energy of the particles
The more phase space available to the final products the higher the probability the reaction will occur π + (139.6MeV ) e + (0.51MeV ) + ν e m = 139.1MeV (1) π + (139.6MeV ) π0 (135.0) + e + (0.51MeV ) + ν e m = 4.1MeV (2) In decay (1) there is a larger mass decrease so there are more ways to distribute momentum to the three reactants This means decay (1) has a higher probability than decay (2)
Fermi s Golden rules says the transition rate is essentially the product of phase space and M Decay rate for 1 -> 1 + 2 + 3 +... + n Γ = S M 2 (2π) 4 δ 4 (p 1 p 2 p 3... p n ) 2 m 1 n x 2πδ(pj 2 mj 2 c2 )θ(pj 0 ) d 4 p j (2π) 4 j=2 cross section for 1 + 2 -> 3 + 4 + 5 +... + n σ = S 2 4 (p 1 p 2 ) 2 (m 1 m 2 c 2 ) 2 M 2 (2π) 4 δ 4 (p 1 + p 2 p 3... p n x n j=3 2πδ(p 2 j m 2 j c2 )θ(p 0 j ) d 4 p j (2π) 4
Γ = S 2 m 1 M 2 (2π) 4 δ 4 (p 1 p 2 p 3... p n ) x n j=2 2πδ(p 2 j m 2 j c2 )θ(p 0 j ) d 4 p j (2π) 4 S is a statistical factor which corrects for double counting S = (1/s 1!)(1/s 2!)...(1/s n!) Phase space constraints Outgoing particles lie on their mass shell p 2 j = m 2 j c 2 δ(p 2 j m 2 j c 4 ) Outgoing energies are positive θ(p 0 j ) Energy and momentum are conserved δ 4 (p 1 p 2 p 3... p n )
Example: Two Particle Decay Γ = S 2 m 1 M 2 (2π) 4 δ 4 (p 1 p 2 p 3 ) x(2π) 2 δ(p 2 2 m2 2 c2 )θ(p 0 2 )δ(p2 3 m2 3 c2 )θ(p 0 3 )d 4 p 2 d 4 p 3 (2π) 8 Using the identity δ(x 2 a 2 ) = 1 [δ(x a) + δ(x + a)] 2a We can integrate over dp2 0 and dp0 3 pj 0 = p j 2 + mj 2c2
Γ = S 32 π 2 m 1 M 2 δ4 (p 1 p 2 p 3 )d 4 p 2 d 4 p 3 p 22 + m22 c2 p 3 2 + m2 3 c2 If we go to the center of mass frame p 1 = 0 and p 2 = p 3 and the integral over d 3 p 3 becomes easy S Γ = 32 π 2 M δ(m 1c p 2 2 2 + m2 2 c2 p 2 2 + m2 2 c2 )d 3 p 2 m 1 p 22 + m22 c2 p 2 2 + m2 2 c2 This can be further simplified if we use spherical coordinates
Cross Section Golden Rule σ = S 2 4 (p 1 p 2 ) 2 (m 1 m 2 c 2 ) 2 M 2 (2π) 4 δ 4 (p 1 + p 2 p 3... p n x n j=3 2πδ(p 2 j m 2 j c2 )θ(p 0 j ) d 4 p j (2π) 4 Phase space constraints Outgoing particles lie on their mass shell p 2 j = m 2 j c 2 δ(p 2 j m 2 j c 4 ) Outgoing energies are positive θ(p 0 j ) Energy and momentum are conserved δ 4 (p 1 p 2 p 3... p n )
Example: Two Particle Scattering Consider 1 + 2 3 + 4 in the CM frame p 2 = p 1 (p 1 p 2 ) 2 (m 1 m 2 c 2 ) 2 = (E 1 + E 2 ) p 1 /c Doing the same delta function manipulations as before x M δ[(e 2 1 + E 2 )/c 2 σ = (8π) 2 Sc (E 1 + E 2 ) p 1 p 3 2 + m2 3 c2 p 3 2 + m2 4 c2 ] d p p 3 p 3 23 + m23 c2 23 + m24 c2
Using d 3 p = p 2 dpdω we can integrate over p and determine the differential cross section u = p 3 2 + m2 3 c2 + p 3 2 + m2 4 c2 du dp = up p 23 + m23 c2 p 2 3 + m2 4 c2 dσ dω = 2 (8π) 2 Sc (E 1 + E 2 ) p 1 M 2 δ[(e 1 + E 2 )/c u] p u du dσ dω = ( c)2 (8π) 2 S M 2 (E 1 + E 2 ) 2 p f p i
Simple To calculate M we need to use the Feynman rules for the corresponding Feynman diagram e γ g q e + q This section will give the rules for a primitive diagram such as the following B A C Real Feynamn diagrams will have further complications mostly due to spin
1. Notation: Label all incoming and outionig four-momenta and give a direction for their momentum k 2 q k 3 k 1 k 4 Label all internal momenta (q in this case) where momentum direction is arbitary 2. Vertex factors: For each vertex write a factor -ig. Where g is a coupling constant and is considered a small parameter. 3. Propagators: For each interal line write a factor of i q 2 j m 2 j c2
4. Conservation of energy and mentum: For each vertex write a delta function (2π) 4 δ 4 (k 1 + k 2 ) Where the k s are the four momenta coming into the vertex k 2 q k 3 k 1 k 4 5. Integrate over internal momenta: For each internal line write d 4 q j (2π) 4 and integrate over the internal momenta. 6. Cancel the delta function: The result will have a factor (2π) 4 δ 4 (k 1 + k 2 k 3 k 4 ). Erase this factor, multiply by i, and the result is M
Example. Golden Rule 1. Label all momenta k 2 k 1 q k 3 k 4 2. We have to vertices ( ig) 2 3. One internal momenta i q 2 m 2 qc 2 4. Two vertices (2π) 8 δ(k 1 k 2 q)δ(q + k 4 k 3 ) 5. One internal line d 4 q (2π) 4
So far M = ( ig) 2 i (2π) 8 δ(k q 2 mqc 2 2 1 k 2 q)δ(q + k 4 k 3 ) d 4 q Simplifing and integrating M = ig 2 (2π) 4 (k 3 k 4 ) 2 m 2 qc 2 δ(k 1 k 2 k 3 + k 4 ) (2π) 4 6. Erasing the delta function and multiplying by i we arrive at M. M = g 2 (2π) 4 (k 3 k 4 ) 2 m 2 qc 2