Exercise 1: Decay amplitudes in the center of mass

Quantum Field Theory

Set 23

Exercise 1: Decay amplitudes in the center of mass

Consider the decayA→C D. The decay width is given by dΓA→CD= 1

2EA

|MA→CD|²dΦ2. Compute the above expression in the rest frame ofA.

Consider then the 3-body decayA→B C D. The only difference here is the presence of the 3-body phase space dΦ3. Let us consider the case where we only deal with scalar particles in the initial and final state or we sum and average over all possible polarization of non scalar particles. Hence the Lorentz invariant matrix element

|M_A→BCD|² can only depend on the three invariants, t, u, where this time

s= (PA−PB)²= (PC+PD)², t= (PA−PC)²= (PB+PD)², u= (PA−PD)²= (PB+PC)². Notice that under these assumptoins the system is invariant under rotations. Therefore we can freely integrate over three angular variables since |M_A→CD|² won’t depend on them. Performing all the integrations required, show that in the rest frame ofAwe have:

dΓ_A→BCD= 1

2M |M_A→BCD|²dEBdEC

4(2π)³ ,

whereM²=P_A². Finally express the differential decay amplitude in terms ofs, t: _{ds dt}^dΓ .

Exercise 2: Free scalar propagator

Consider the scalar free propagator

D(x−y) =h0|T(φ(x)φ(y))|0i , where the time ordered product is defined by

T(φ(x)φ(y))≡θ(x₀−y₀)φ(x)φ(y) +θ(y₀−x₀)φ(y)φ(x). Making use of the Klein-Gordon equation for the fieldφshow that:

x+m²

D(x−y) = ∂

∂x⁰δ(x₀−y₀)h0|[φ(x), φ(y)]|0i+ 2δ(x₀−y₀)h0|[∂₀φ(x), φ(y)]|0i.

Consider the above expression as a distribution that makes sense only when inserted in an integral. Hence

”integrate by parts” theδ⁰(x0−y0) and setx0=y0 everywhere. Finally, make use of the canonical commutation relations to show that

^x+m²

D(x−y) =−iδ⁴(x−y).

One can solve the above equation in momentum space. Show that the solution is D(x−y)_±=

Z d⁴p (2π)⁴

i

p²−m²±iεe^−ip(x−y). (1) The above uncertainty in the sign ofiεcomes from the fact that there are two ways to regularize the integral in the complex plane in a Lorentz invariant way. We need to choose the right one. In order to do this, expand the fieldφina a^† and show that

D(x−y) =θ(x0−y0) Z

dΩpe^−ip(x−y)+θ(y0−x0) Z

dΩpe^+ip(x−y).

Finally show that onlyD(x−y)₊ reproduces the above result: computeD(x−y)₊ from equation (1) integrating overp₀ separately in the two cases (x₀−y₀)>0 and (x₀−y₀)<0 and closing the contour in the complex plane appropriately.

Exercise 3: Free fermion propagator

Consider the free propagator of a fermion

S(x−y) =h0|T(ψ(x) ¯ψ(y))|0i,

where the time ordered product is defined by

T(ψ(x) ¯ψ(y))≡θ(x0−y0)ψ(x) ¯ψ(y)−θ(y0−x0) ¯ψ(y)ψ(x). Making use of the Dirac equation for the fieldψshow that:

(i6∂x−m)S(x−y) =iδ⁴(x−y).

One can solve the above equation in momentum space. Show that the solution is S(x−y) =

Z d⁴p (2π)⁴

S(p)e˜ ^−ip(x−y), with S(p) =˜ i

6p−m ≡i 6p+m p²−m²+iε.

Exercise 4: Free vector propagator

Consider the Lagrangian of a massless vector field:

L=−1

2∂µAν∂^µA^ν+1 2

1−1

α

(∂µA^µ)². Find the equation of motion for the fieldAµ. Consider the free vector propagator:

Dρσ(x−y) =h0|T(A_ρ(x)A_σ(y))|0i, where the time ordered product is defined by

T(A_ρ(x)A_σ(y))≡θ(x₀−y₀)A_ρ(x)A_σ(y) +θ(y₀−x₀)A_σ(y)A_ρ(x). Making use of the equation for the fieldAshow that:

δ_ν^µ−

1− 1

α

∂^µ∂ν

Dµσ(x−y) =iδ⁴(x−y)ηνσ. One can solve the above equation in momentum space. Show that the solution is

Dµν(x−y) =

Z d⁴p (2π)⁴

D˜µν(p)e^−ip(x−y), with D˜µν(p) = −i p²+iε

ηµν−(1−α)pµpν

p²

.

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