Particle Physics lecture 5 -Weak Interactions (Continued) · PDF fileWeak Interactions...

Weak Interactions

Particle Physicslecture 5

-Weak Interactions (Continued)

Yazid Delenda

Departement des Sciences de la matiereFaculte des Sciences - UHLB

http://delenda.wordpress.com/teaching/particlephysics/

Batna, 26 October 2014

1/25 Particle Physics - lecture 5

Weak Interactions Muon and pion decay

Muon decay

The muon decay:

µ−(p)

νµ(k)

e−(p′)

νe(k′)

µ−(p)→ e−(p′) + νe(k′) + νµ(k),

is the model reaction for weak decays. The particle four-momentaare defined in the above equation.

Muon decay

The muon decay:

µ−(p)

νµ(k)

e−(p′)

νe(k′)

µ−(p)→ e−(p′) + νe(k′) + νµ(k),

is the model reaction for weak decays. The particle four-momentaare defined in the above equation.

Muon decay

According to Feynman rules, the Feynman diagram must be drawn

using only particle lines; and so the outgoing νe is 2equivalent to

an incoming νe. The invariant amplitude for muon decay is:

M = GF√2

[usνµ(k)γα(1− γ5)uSµ(p)

e (p′)γα(1− γ5)us′

νe(k′)

where s, s′, S, S′ are the spins of the corresponding particles.The matrix element squared is:

|M|2 =G2F

] [uSµ(p)γβ(1− γ5)usνµ(k)

′e (p′)γα(1− γ5)us

′νe(k

′)] [us

′νe(k

′)γβ(1− γ5)uS′

e (p′)].

Muon decay

M = GF√2

e (p′)γα(1− γ5)us′

νe(k′)

|M|2 =G2F

′e (p′)γα(1− γ5)us

′νe(k

′)] [us

′νe(k

′)γβ(1− γ5)uS′

e (p′)].

Muon decay

M = GF√2

e (p′)γα(1− γ5)us′

νe(k′)

|M|2 =G2F

′e (p′)γα(1− γ5)us

′νe(k

′)] [us

′νe(k

′)γβ(1− γ5)uS′

e (p′)].

Muon decay

M = GF√2

e (p′)γα(1− γ5)us′

νe(k′)

|M|2 =G2F

′e (p′)γα(1− γ5)us

′νe(k

′)] [us

′νe(k

′)γβ(1− γ5)uS′

e (p′)].

Muon decay

Performing the summation over final-state spins and averagingover initial-state spins to obtain:

|M|2 =1

s,s′,S,S′

we use the fact that:∑

us(p)us(p) =6p+m,

where m is the mass of the corresponding particle, we arrive at:

|M|2 =1

2Tr [γα(1− γ5)(6p+mµ)γβ(1− γ5) 6k]×

Tr[γα(1− γ5) 6k′γβ(1− γ5)(6p′ +me)

Muon decay

|M|2 =1

s,s′,S,S′

us(p)us(p) =6p+m,

|M|2 =1

2Tr [γα(1− γ5)(6p+mµ)γβ(1− γ5) 6k]×

Tr[γα(1− γ5) 6k′γβ(1− γ5)(6p′ +me)

Muon decay

|M|2 =1

s,s′,S,S′

us(p)us(p) =6p+m,

|M|2 =1

2Tr [γα(1− γ5)(6p+mµ)γβ(1− γ5) 6k]×

Tr[γα(1− γ5) 6k′γβ(1− γ5)(6p′ +me)

Muon decay

|M|2 =1

s,s′,S,S′

us(p)us(p) =6p+m,

|M|2 =1

2Tr [γα(1− γ5)(6p+mµ)γβ(1− γ5) 6k]×

Tr[γα(1− γ5) 6k′γβ(1− γ5)(6p′ +me)

Muon decay

|M|2 =1

s,s′,S,S′

us(p)us(p) =6p+m,

|M|2 =1

2Tr [γα(1− γ5)(6p+mµ)γβ(1− γ5) 6k]×

Tr[γα(1− γ5) 6k′γβ(1− γ5)(6p′ +me)

Muon decay

Exercise: Prove the following relations:

Tr [γα(1− γ5) 6pγβ(1− γ5) 6k]

Tr[γα(1− γ5) 6k′γβ(1− γ5) 6p′

]= 256(p.k′)(k.p′),

Tr [γα(1− γ5)γβ(1− γ5) 6k] =

Tr[γα(1− γ5)γβ(1− γ5) 6p′

Hence we may write:

|M|2 = 64G2F (p.k′)(k.p′)

The muon decay rate can now be obtained using:

dΓ = 12E|M|2dQ,

Muon decay

Tr [γα(1− γ5) 6pγβ(1− γ5) 6k]

Tr[γα(1− γ5) 6k′γβ(1− γ5) 6p′

]= 256(p.k′)(k.p′),

Tr [γα(1− γ5)γβ(1− γ5) 6k] =

Tr[γα(1− γ5)γβ(1− γ5) 6p′

Hence we may write:

|M|2 = 64G2F (p.k′)(k.p′)

dΓ = 12E|M|2dQ,

Muon decay

Tr [γα(1− γ5) 6pγβ(1− γ5) 6k]

Tr[γα(1− γ5) 6k′γβ(1− γ5) 6p′

]= 256(p.k′)(k.p′),

Tr [γα(1− γ5)γβ(1− γ5) 6k] =

Tr[γα(1− γ5)γβ(1− γ5) 6p′

Hence we may write:

|M|2 = 64G2F (p.k′)(k.p′)

dΓ = 12E|M|2dQ,

Muon decay

Tr [γα(1− γ5) 6pγβ(1− γ5) 6k]

Tr[γα(1− γ5) 6k′γβ(1− γ5) 6p′

]= 256(p.k′)(k.p′),

Tr [γα(1− γ5)γβ(1− γ5) 6k] =

Tr[γα(1− γ5)γβ(1− γ5) 6p′

Hence we may write:

|M|2 = 64G2F (p.k′)(k.p′)

dΓ = 12E|M|2dQ,

Muon decay

where E is the energy of the incoming muon,and the

Lorentz-invariant phase space of the decay products is

dQ = d3~p ′

(2π)32E′d3~k

(2π)32ωd3~k′

(2π)32ω′

(2π)4δ(4)(p− p′ − k − k′),over which we shall integrate.Using the easily-derived relation:

∫d3~k

∫d4kΘ(ω)δ(k2).

we can turn the 3-momentum integration over k into 4-momentumintegration.We thus have:

dΓ =32G2

(2π)5E(p.k′)(k.p′)

d3~p ′

2E′d3~k′

2ω′d4kΘ(ω)δ(k2)δ(4)(p−p′−k−k′).

Muon decay

dQ = d3~p ′

(2π)32E′d3~k

(2π)32ωd3~k′

(2π)32ω′

∫d3~k

∫d4kΘ(ω)δ(k2).

dΓ =32G2

(2π)5E(p.k′)(k.p′)

d3~p ′

2E′d3~k′

2ω′d4kΘ(ω)δ(k2)δ(4)(p−p′−k−k′).

Muon decay

dQ = d3~p ′

(2π)32E′d3~k

(2π)32ωd3~k′

(2π)32ω′

∫d3~k

∫d4kΘ(ω)δ(k2).

dΓ =32G2

(2π)5E(p.k′)(k.p′)

d3~p ′

2E′d3~k′

2ω′d4kΘ(ω)δ(k2)δ(4)(p−p′−k−k′).

Muon decay

dQ = d3~p ′

(2π)32E′d3~k

(2π)32ωd3~k′

(2π)32ω′

∫d3~k

∫d4kΘ(ω)δ(k2).

dΓ =32G2

(2π)5E(p.k′)(k.p′)

d3~p ′

2E′d3~k′

2ω′d4kΘ(ω)δ(k2)δ(4)(p−p′−k−k′).

Muon decay

dQ = d3~p ′

(2π)32E′d3~k

(2π)32ωd3~k′

(2π)32ω′

∫d3~k

∫d4kΘ(ω)δ(k2).

dΓ =32G2

(2π)5E(p.k′)(k.p′)

d3~p ′

2E′d3~k′

2ω′d4kΘ(ω)δ(k2)δ(4)(p−p′−k−k′).

Muon decay

dQ = d3~p ′

(2π)32E′d3~k

(2π)32ωd3~k′

(2π)32ω′

∫d3~k

∫d4kΘ(ω)δ(k2).

dΓ =32G2

(2π)5E(p.k′)(k.p′)

d3~p ′

2E′d3~k′

2ω′d4kΘ(ω)δ(k2)δ(4)(p−p′−k−k′).

Muon decay

Performing the integration over d4k using the Dirac-δ(4) functionforces k = p− p′ − k′ in the integrand (which meansω = E − E′ − ω′), with k0 = ω, p0 = E, k0 = ω and k′0 = ω′.Thus:

dΓ =G2F

π5E(p.k′)((p−p′−k′).p′)d

3~p ′

2E′d3~k′

2ω′Θ(E−E′−ω′)δ((p−p′−k′)2).

Now since mµ > 200me, we can safely neglect the mass of theelectron.One can show that in the muon rest frame, wherep = (mµ, 0, 0, 0):

(p.k′)((p− p′ − k′).p′) = (p.k′)(p− k′).p′ == (p.k′)(p− k′)2/2 = mµω

′(m2µ − 2mµω

′)/2,

where we treat all particles massless except the muon.In the thirdequality we used the relation for the massless neutrino:

Muon decay

dΓ =G2F

π5E(p.k′)((p−p′−k′).p′)d

3~p ′

2E′d3~k′

2ω′Θ(E−E′−ω′)δ((p−p′−k′)2).

′(m2µ − 2mµω

′)/2,

Muon decay

dΓ =G2F

π5E(p.k′)((p−p′−k′).p′)d

3~p ′

2E′d3~k′

2ω′Θ(E−E′−ω′)δ((p−p′−k′)2).

′(m2µ − 2mµω

′)/2,

Muon decay

dΓ =G2F

π5E(p.k′)((p−p′−k′).p′)d

3~p ′

2E′d3~k′

2ω′Θ(E−E′−ω′)δ((p−p′−k′)2).

′(m2µ − 2mµω

′)/2,

Muon decay

dΓ =G2F

π5E(p.k′)((p−p′−k′).p′)d

3~p ′

2E′d3~k′

2ω′Θ(E−E′−ω′)δ((p−p′−k′)2).

′(m2µ − 2mµω

′)/2,

Muon decay

dΓ =G2F

π5E(p.k′)((p−p′−k′).p′)d

3~p ′

2E′d3~k′

2ω′Θ(E−E′−ω′)δ((p−p′−k′)2).

′(m2µ − 2mµω

′)/2,

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

and we used p′2 = 0 (massless electron).Thus:

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

The decay rate in the muon rest frame is (E = mµ):

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

×Θ(mµ−E′−ω′)δ(m2µ− 2mµE

′− 2mµω′+ 2E′ω′(1− cos θ)),

where θ is the angle between ~p ′ and ~k′. We have the freedom tofix the polar angle of say ~p ′ and multiply by a factor 2 to accountfor the averaging

∫ π0 sin θ′ = 2,and so we can express the phase

space d3~p ′d3~k′ by:

4πE′2dE′2πω′2dω′d cos θ.

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

k = p−k′−p′ ⇒ k2 = 0 = (p−k′−p′)2 = (p−k′)2−2(p−k′).p′,

(p− k′).p′ = (p− k′)2/2 = m2µ/2− 2mµω

′/2.

dΓ =G2F

2π5mµ

d3~p ′

2E′d3~k′

2ω′mµω

′(m2µ − 2mµω

′)×

′− 2mµω′+ 2E′ω′(1− cos θ)),

Muon decay

where we integrated over the azimuthal angles (giving 2πfactors).Thus we have:

dΓ =G2F

π3E′dE′ω′dω′d cos θω′(m2

µ − 2mµω′)Θ(mµ − E′ − ω′)×

2E′ω′δ

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

where we used the relation:

δ(m2µ − 2mµE

′ − 2mµω′ + 2E′ω′(1− cos θ)) =

2E′ω′δ

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

in order to perform the integration over the opening angle θbetween the emitted e− and νe. Hence:

Muon decay

dΓ =G2F

µ − 2mµω′)Θ(mµ − E′ − ω′)×

2E′ω′δ

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

δ(m2µ − 2mµE

′ − 2mµω′ + 2E′ω′(1− cos θ)) =

2E′ω′δ

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

Muon decay

dΓ =G2F

µ − 2mµω′)Θ(mµ − E′ − ω′)×

2E′ω′δ

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

δ(m2µ − 2mµE

′ − 2mµω′ + 2E′ω′(1− cos θ)) =

2E′ω′δ

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

Muon decay

dΓ =G2F

2π3dE′dω′d cos θω′(m2

µ − 2mµω′)Θ(mµ − E′ − ω′)×

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

The integration over θ is straightforward (gives 1)forcingcos θ = 1 + (m2

µ − 2mµE′ − 2mµω

′)/2E′ω′,but since−1 < cos θ < 1 then:

−1 < 1 + (m2µ − 2mµE

′ − 2mµω′)/2E′ω′ < 1.

Thus we obtain the following constraints on E′ and ω′ togetherwith the step function above:

0 ≤ E′

mµ− 1

2≤ 2

E′ + ω′ ≤ mµ,

Muon decay

dΓ =G2F

µ − 2mµω′)Θ(mµ − E′ − ω′)×

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

−1 < 1 + (m2µ − 2mµE

′ − 2mµω′)/2E′ω′ < 1.

0 ≤ E′

mµ− 1

2≤ 2

E′ + ω′ ≤ mµ,

Muon decay

dΓ =G2F

µ − 2mµω′)Θ(mµ − E′ − ω′)×

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

−1 < 1 + (m2µ − 2mµE

′ − 2mµω′)/2E′ω′ < 1.

0 ≤ E′

mµ− 1

2≤ 2

E′ + ω′ ≤ mµ,

Muon decay

dΓ =G2F

µ − 2mµω′)Θ(mµ − E′ − ω′)×

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

−1 < 1 + (m2µ − 2mµE

′ − 2mµω′)/2E′ω′ < 1.

0 ≤ E′

mµ− 1

2≤ 2

E′ + ω′ ≤ mµ,

Muon decay

dΓ =G2F

µ − 2mµω′)Θ(mµ − E′ − ω′)×

(m2µ − 2mµE

′ − 2mµω′

2E′ω′+ (1− cos θ)

−1 < 1 + (m2µ − 2mµE

′ − 2mµω′)/2E′ω′ < 1.

0 ≤ E′

mµ− 1

2≤ 2

E′ + ω′ ≤ mµ,

Muon decay

where we respectively subtracted 1, then multiplied by −E′ω′/m2µ

in the first inequality.We can rewrite these constraints as:

mµ− 2

mµ≤ 1

2mµ ≤ E′ + ω′ ≤ mµ.

From the first inequality we may write:

(1− 2

)≤ 1

(1− 2

Now we either have

1− 2ω′

mµ≤ 0 thus E′ ≥ 1

which means ω′ ≥ mµ/2 and E′ ≥ mµ/2, which impliesω′+E′ ≥ mµ,which contradicts the second inequality in the above;

Muon decay

mµ− 2

mµ≤ 1

2mµ ≤ E′ + ω′ ≤ mµ.

(1− 2

)≤ 1

(1− 2

Now we either have

1− 2ω′

Muon decay

mµ− 2

mµ≤ 1

2mµ ≤ E′ + ω′ ≤ mµ.

(1− 2

)≤ 1

(1− 2

Now we either have

1− 2ω′

Muon decay

mµ− 2

mµ≤ 1

2mµ ≤ E′ + ω′ ≤ mµ.

(1− 2

)≤ 1

(1− 2

Now we either have

1− 2ω′

Muon decay

mµ− 2

mµ≤ 1

2mµ ≤ E′ + ω′ ≤ mµ.

(1− 2

)≤ 1

(1− 2

Now we either have

1− 2ω′

Muon decay

mµ− 2

mµ≤ 1

2mµ ≤ E′ + ω′ ≤ mµ.

(1− 2

)≤ 1

(1− 2

Now we either have

1− 2ω′

Muon decayor we have:

ω′ ≤ mµ

2thus E′ ≤ 1

so we just take the last inequality.Hence the above coupledinequality is equivalent to:

0 ≤ E′ ≤ 1

2mµ − E′ ≤ ω′ ≤ 1

We are now in position to perform the energy integration over ω′

to obtain the energy spectrum of the emitted electron:

dE′=G2F

2π3mµ

∫ mµ2

mµ2−E′

ω′(mµ − 2ω′)dω′ =

8π3m4µ

1−2E′/mµ

x(1−x)dx =G2F

48π3m4µ[x2(3−2x)]11−2E′/mµ

ω′ ≤ mµ

2thus E′ ≤ 1

0 ≤ E′ ≤ 1

2mµ − E′ ≤ ω′ ≤ 1

dE′=G2F

2π3mµ

∫ mµ2

mµ2−E′

ω′(mµ − 2ω′)dω′ =

8π3m4µ

1−2E′/mµ

x(1−x)dx =G2F

48π3m4µ[x2(3−2x)]11−2E′/mµ

ω′ ≤ mµ

2thus E′ ≤ 1

0 ≤ E′ ≤ 1

2mµ − E′ ≤ ω′ ≤ 1

dE′=G2F

2π3mµ

∫ mµ2

mµ2−E′

ω′(mµ − 2ω′)dω′ =

8π3m4µ

1−2E′/mµ

x(1−x)dx =G2F

48π3m4µ[x2(3−2x)]11−2E′/mµ

ω′ ≤ mµ

2thus E′ ≤ 1

0 ≤ E′ ≤ 1

2mµ − E′ ≤ ω′ ≤ 1

dE′=G2F

2π3mµ

∫ mµ2

mµ2−E′

ω′(mµ − 2ω′)dω′ =

8π3m4µ

1−2E′/mµ

x(1−x)dx =G2F

48π3m4µ[x2(3−2x)]11−2E′/mµ

ω′ ≤ mµ

2thus E′ ≤ 1

0 ≤ E′ ≤ 1

2mµ − E′ ≤ ω′ ≤ 1

dE′=G2F

2π3mµ

∫ mµ2

mµ2−E′

ω′(mµ − 2ω′)dω′ =

8π3m4µ

1−2E′/mµ

x(1−x)dx =G2F

48π3m4µ[x2(3−2x)]11−2E′/mµ

Muon decay

Hence we finally arrive at:

dE′=

12π3m2µE′2(

3− 4E′

This prediction is in excellent agreement with the observed electronspectrum.Finally, we calculate the muon decay rate:

∫ mµ/2

0dE′

dE′=G2Fm

192π3.

Inserting the measured muon lifetime τ = 2.2× 10−6 sec,we cancalculate the Fermi coupling GF .We find:

GF = 10−5/m2N .

Muon decay

dE′=

12π3m2µE′2(

3− 4E′

∫ mµ/2

0dE′

dE′=G2Fm

192π3.

GF = 10−5/m2N .

Muon decay

dE′=

12π3m2µE′2(

3− 4E′

∫ mµ/2

0dE′

dE′=G2Fm

192π3.

GF = 10−5/m2N .

Muon decay

dE′=

12π3m2µE′2(

3− 4E′

∫ mµ/2

0dE′

dE′=G2Fm

192π3.

GF = 10−5/m2N .

Muon decay

dE′=

12π3m2µE′2(

3− 4E′

∫ mµ/2

0dE′

dE′=G2Fm

192π3.

GF = 10−5/m2N .

Muon decay

dE′=

12π3m2µE′2(

3− 4E′

∫ mµ/2

0dE′

dE′=G2Fm

192π3.

GF = 10−5/m2N .

Muon decay

Comparison of the values of GF obtained in muon decay and betadecay supports the assertion that the weak coupling constant is thesame for leptons and nucleons,and hence universal.It means thatnuclear β-decay and the decay of the muon have the same physicalorigin.Indeed, when all corrections are taken into account, Gβ andGµ are found to be equal to within a few percent:

Gµ = (1.16632± 0.00002)× 10−5GeV−2,

Gβ = (1.136± 0.003)× 10−5GeV−2.

The reason for the small difference is quark mixing.

Muon decay

Gµ = (1.16632± 0.00002)× 10−5GeV−2,

Gβ = (1.136± 0.003)× 10−5GeV−2.

Muon decay

Gµ = (1.16632± 0.00002)× 10−5GeV−2,

Gβ = (1.136± 0.003)× 10−5GeV−2.

Muon decay

Gµ = (1.16632± 0.00002)× 10−5GeV−2,

Gβ = (1.136± 0.003)× 10−5GeV−2.

Muon decay

Gµ = (1.16632± 0.00002)× 10−5GeV−2,

Gβ = (1.136± 0.003)× 10−5GeV−2.

Muon decay

Gµ = (1.16632± 0.00002)× 10−5GeV−2,

Gβ = (1.136± 0.003)× 10−5GeV−2.

Pion decay

Can we now also understand the lifetime of the π±-Mesons?To bespecific, we take the decay:

π−(q)→ µ−(p) + νµ(k),

which is depicted as follows:

Pion decay

π−(q)→ µ−(p) + νµ(k),

Pion decay

π−(q)→ µ−(p) + νµ(k),

νµ(k)

µ−(p)

π−(q)

Pion decay

The amplitude for the pion decay is of the form:

M = GF√2(...)αuµ(p)γα(1− γ5)uνµ(k),

where (...)α represents the weak quark current.It is tempting towrite it as uuγ

α(1− γ5)ud,but this is incorrectsince the u, dquarks are not free states (so cannot be described by Dirac freespinors),but are quarks bound into a π−–meson.We know, however, that M is Lorentz-invariant,so that (...)α mustbe a vector or axial-vector, as indicated.Also The π-meson isspin-less,so that qα is the only four-vector available to constructthe quark current (no spin structure and therefore no γmatrices).We therefore have:

(...)α = qαf(q2) ≡ qαfπ,where f is a function of q2 since it is the only Lorentz scalar thatcan be formed from q,but q2 = m2

π and f(m2π) ≡ fπ, is constant.

Pion decay

The amplitude becomes:

M =GF√

2(pα + kα)fπ[uµ(p)γα(1− γ5)uνµ(k)] =

=GF√

2fπmµ[uµ(p)(1− γ5)uνµ(k)].

Here, we have used 6kuνµ(k) = 0 and uµ(p)(6p−mµ) = 0, theDirac equations for the neutrino and muon, respectively.Theinvariant amplitude-squared is:

|M|2 =G2F f

2[uµ(p)(1− γ5)uνµ(k)][uνµ(k)(1 + γ5)uµ(p)].

Summing over final state spins (no initial state spins since the pionis spin-less)we obtain:

|M|2 =G2F f

2Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)],

Pion decay

M =GF√

=GF√

|M|2 =G2F f

2Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)],

Pion decay

M =GF√

=GF√

|M|2 =G2F f

2Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)],

Pion decay

M =GF√

=GF√

|M|2 =G2F f

2Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)],

Pion decay

where we use∑

s uνµ(k)uνµ(k) =6k and∑

S uµ(p)uµ(p) =6p+mµ,with s and S the spins of the neutrino and the muon respectively.Exercise: Show that:

Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)] = 8(p.k).

In its rest frame, the π-decay rate is

dΓ =1

2mπ|M|2 d3~p

(2π)32E

(2π)32ω(2π)4δ(4)(q − p− k).

Hence:

dΓ =G2F f

8mππ2(p.k)

ωδ(4)(q − p− k)

Pion decay

where we use∑

Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)] = 8(p.k).

dΓ =1

2mπ|M|2 d3~p

(2π)32E

(2π)32ω(2π)4δ(4)(q − p− k).

Hence:

dΓ =G2F f

8mππ2(p.k)

ωδ(4)(q − p− k)

Pion decay

where we use∑

Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)] = 8(p.k).

dΓ =1

2mπ|M|2 d3~p

(2π)32E

(2π)32ω(2π)4δ(4)(q − p− k).

Hence:

dΓ =G2F f

8mππ2(p.k)

ωδ(4)(q − p− k)

Pion decay

where we use∑

Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)] = 8(p.k).

dΓ =1

2mπ|M|2 d3~p

(2π)32E

(2π)32ω(2π)4δ(4)(q − p− k).

Hence:

dΓ =G2F f

8mππ2(p.k)

ωδ(4)(q − p− k)

Pion decay

where we use∑

Tr[(1− γ5) 6k(1 + γ5)(6p+mµ)] = 8(p.k).

dΓ =1

2mπ|M|2 d3~p

(2π)32E

(2π)32ω(2π)4δ(4)(q − p− k).

Hence:

dΓ =G2F f

8mππ2(p.k)

ωδ(4)(q − p− k)

Pion decay

In the pion rest frame (~k = −~p), hence:

p.k = Eω − ~k.~p = Eω + ~k2 = ω(E + ω).

The decay rate thus is given by:

Γ =G2F f

8mππ2

∫d3~p

ωω(E + ω)δ(mπ − E − ω)δ(3)(~p+ ~k),

where we used q = (mπ,~0), i.e. ~k = −~p in the pion rest frame.Thed3~p integration is taken care of by the δ(3) function and imposing~k = −~p in the remaining integrand, and, since there is no angulardependence, we are left with only the integration over dω:

Γ =G2F f

8mππ24π

∫ω2dω(1 + ω/E)δ(mπ − E − ω),

where the 4π is the leftover from angular integrations.19/25 Particle Physics - lecture 5

Pion decay

p.k = Eω − ~k.~p = Eω + ~k2 = ω(E + ω).

Γ =G2F f

8mππ2

∫d3~p

ωω(E + ω)δ(mπ − E − ω)δ(3)(~p+ ~k),

Γ =G2F f

8mππ24π

∫ω2dω(1 + ω/E)δ(mπ − E − ω),

Pion decay

p.k = Eω − ~k.~p = Eω + ~k2 = ω(E + ω).

Γ =G2F f

8mππ2

∫d3~p

ωω(E + ω)δ(mπ − E − ω)δ(3)(~p+ ~k),

Γ =G2F f

8mππ24π

∫ω2dω(1 + ω/E)δ(mπ − E − ω),

Pion decay

p.k = Eω − ~k.~p = Eω + ~k2 = ω(E + ω).

Γ =G2F f

8mππ2

∫d3~p

ωω(E + ω)δ(mπ − E − ω)δ(3)(~p+ ~k),

Γ =G2F f

8mππ24π

∫ω2dω(1 + ω/E)δ(mπ − E − ω),

Pion decay

p.k = Eω − ~k.~p = Eω + ~k2 = ω(E + ω).

Γ =G2F f

8mππ2

∫d3~p

ωω(E + ω)δ(mπ − E − ω)δ(3)(~p+ ~k),

Γ =G2F f

8mππ24π

∫ω2dω(1 + ω/E)δ(mπ − E − ω),

Pion decay

p.k = Eω − ~k.~p = Eω + ~k2 = ω(E + ω).

Γ =G2F f

8mππ2

∫d3~p

ωω(E + ω)δ(mπ − E − ω)δ(3)(~p+ ~k),

Γ =G2F f

8mππ24π

∫ω2dω(1 + ω/E)δ(mπ − E − ω),

Pion decay

Now since ~k = −~p we have ~k2 = ω2 = ~p 2 = E2 −m2µ, thus

E =√ω2 +m2

µ.Then we write:

Γ =G2F f

2πmπ

∫ω2

1√1 +m2

µ/ω2

(√ω2 +m2

µ + ω −mπ

To perform this integration we use the fact that:

δ[f(ω)] = δ(ω − ω0)/

∣∣∣∣∂f

∣∣∣∣ω=ω0

where ω0 is the zero 1 of the function

f(ω) =√ω2 +m2

µ + ω −mπ,

1In case of many zeros we must sum over all of them.20/25 Particle Physics - lecture 5

Pion decay

E =√ω2 +m2

µ.Then we write:

Γ =G2F f

2πmπ

∫ω2

1√1 +m2

µ/ω2

(√ω2 +m2

µ + ω −mπ

δ[f(ω)] = δ(ω − ω0)/

∣∣∣∣∂f

∣∣∣∣ω=ω0

f(ω) =√ω2 +m2

µ + ω −mπ,

Pion decay

E =√ω2 +m2

µ.Then we write:

Γ =G2F f

2πmπ

∫ω2

1√1 +m2

µ/ω2

(√ω2 +m2

µ + ω −mπ

δ[f(ω)] = δ(ω − ω0)/

∣∣∣∣∂f

∣∣∣∣ω=ω0

f(ω) =√ω2 +m2

µ + ω −mπ,

Pion decay

E =√ω2 +m2

µ.Then we write:

Γ =G2F f

2πmπ

∫ω2

1√1 +m2

µ/ω2

(√ω2 +m2

µ + ω −mπ

δ[f(ω)] = δ(ω − ω0)/

∣∣∣∣∂f

∣∣∣∣ω=ω0

f(ω) =√ω2 +m2

µ + ω −mπ,

Pion decay

which can easily be found:

ω0 =m2π −m2

and we have:

∂ω=

1√1 +m2

µ/ω2

+ 1⇒∣∣∣∣∂f

∣∣∣∣ω=ω0

= 1 +1√

1 +m2µ/ω

Now we can easily perform the ω integration using the Dirac-δfunction, which forces ω = ω0 in the rest of the integrand.We notehere that the term1 +

1√1 +m2

µ/ω20

, in the integrand cancels exactly with

1∣∣∣ ∂f∂ω∣∣∣ω=ω0

Pion decay

ω0 =m2π −m2

and we have:

∂ω=

1√1 +m2

µ/ω2

+ 1⇒∣∣∣∣∂f

∣∣∣∣ω=ω0

= 1 +1√

1 +m2µ/ω

1√1 +m2

µ/ω20

1∣∣∣ ∂f∂ω∣∣∣ω=ω0

Pion decay

ω0 =m2π −m2

and we have:

∂ω=

1√1 +m2

µ/ω2

+ 1⇒∣∣∣∣∂f

∣∣∣∣ω=ω0

= 1 +1√

1 +m2µ/ω

1√1 +m2

µ/ω20

1∣∣∣ ∂f∂ω∣∣∣ω=ω0

Pion decay

ω0 =m2π −m2

and we have:

∂ω=

1√1 +m2

µ/ω2

+ 1⇒∣∣∣∣∂f

∣∣∣∣ω=ω0

= 1 +1√

1 +m2µ/ω

1√1 +m2

µ/ω20

1∣∣∣ ∂f∂ω∣∣∣ω=ω0

Pion decay

and we are left with only the ω20 term in the integrand, hence:

Γ =G2F f

2πmπω20.

Therefore, finally we obtain:

τ=G2F

8πf2πmπm

Taking the universal value of GF = 10−5m−2N obtained from β orµ-decay and assuming that fπ = mπ,(a guess which at leastguarantees the correct dimension),we indeed obtain the π− lifetime2.6× 10−8 sec.

Pion decay

Γ =G2F f

2πmπω20.

τ=G2F

8πf2πmπm

Pion decay

Γ =G2F f

2πmπω20.

τ=G2F

8πf2πmπm

Pion decay

Γ =G2F f

2πmπω20.

τ=G2F

8πf2πmπm

Pion decay

Γ =G2F f

2πmπω20.

τ=G2F

8πf2πmπm

Pion decay

If we repeat the calculation for the decay mode π− → e−νe, weobtain the same result above with the replacementmµ ↔ me.Therefore:

Γ(π− → e−νe)Γ(π− → µ−νµ)

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

where the numerical value comes from inserting the particlemasses.The charged π prefers (by a factor of 104) to decay into amuon,which has a similar mass (to the muon), rather than into themuch lighter electron.This is quite contrary to what one wouldexpect from phase-space considerations, so some dynamicalmechanism must be at work.The pion is spinless, and so, by the conservation of angularmomentum, the outgoing lepton pair (e− νe) must have J = 0.

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

)2(m2π −m2

m2π −m2

= 1.2× 10−4,

Pion decay

νeπ−

As the νe has positive helicity, the e− is also forced into a positivehelicity state.But recall that this is the “wrong” helicity state forthe electron.In the limit me = 0, the weak current only couplesnegative helicity electrons,and hence the positive helicity couplingis highly suppressed.

Pion decay

νeπ−

Pion decay

νeπ−

Pion decay

νeπ−

Pion decay

νeπ−

Pion decay

Thus, in the π-decay, the e− (or µ−) is forced by angularmomentum conservation into its “wrong” helicity state.This ismuch more likely to happen for the µ− than for the relatively lighte−,in fact, 10−4 times more likely.Experiment confirms this result,which is a direct consequence of the 1− γ5 or left-handedstructure of the weak current.ProblemPredict the ratio of the K− → e−νe and K− → µ−νµ decay rates.Given that the lifetime of the K− is, τ = 1.2× 10−8 s and theK → µν, branching ratio is 64%,estimate the decay constant fK .Comment on your assumptions and on your result.

Pion decay

Particle Physics lecture 5 -Weak Interactions (Continued) · PDF fileWeak Interactions...

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