Fermi s Golden Rule and Simple Feynman Rules

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Samuel Bradley
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1 Fermi s Golden Rule and Simple Feynman Rules ; December 5, 2013

2 Outline Golden Rule 1 Golden Rule 2

3 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

4 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)

5 Fermi s Golden rules says the transition rate is essentially the product of phase space and M Decay rate for 1 -> 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 > 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

6 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 )

7 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

8 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 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 m2 2 c2 p m2 2 c2 )d 3 p 2 m 1 p 22 + m22 c2 p m2 2 c2 This can be further simplified if we use spherical coordinates

9 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 )

10 Example: Two Particle Scattering Consider 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 E 2 )/c 2 σ = (8π) 2 Sc (E 1 + E 2 ) p 1 p m2 3 c2 p m2 4 c2 ] d p p 3 p m23 c m24 c2

11 Using d 3 p = p 2 dpdω we can integrate over p and determine the differential cross section u = p m2 3 c2 + p m2 4 c2 du dp = up p 23 + m23 c2 p 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

12 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

13 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

14 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

15 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

16 So far M = ( ig) 2 i (2π) 8 δ(k q 2 mqc 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

THE FEYNMAN RULES. < f S i >. \ f >

$THE FEYNMAN RULES. < f S i >. \ f >$ THE FEYNMAN RULES I. Introduction: Very generally, interactions can be viewed as the result of a scattering operator (S), a unitary operator acting on an initial state (\i >) leading to a multitude of