part 3: scattering theory - jyväskylän yliopistousers.jyu.fi/~tulappi/fysh300sl11/l3.pdf1...

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Introduction Free particles Cross section 2 → 2 Decay width

FYSH300, fall 2011

Tuomas [email protected]

Office: FL249. No fixed reception hours.

kl 2011

Part 3: Scattering theory

2


Basic idea

Particle physics experiment:Know incoming particle(s) — measure outgoing particle properties:

Momenta (|p| and direction) Charges (electric, other conserved), masses, spins/polarizations

=⇒ identify particle type Repeat experiment and measure number/frequency of events

=⇒ collect statistics , form spectra (jakauma) dN/ d3p, correlations Compare to theory to test/learn about fundamental interactions.

3


Terminology IWhat goes in

Beam Consists of projectile (ammus) particles a (e.g.e±, µ±, ν, p, π, . . . ).

∼ monoenergetic & collimated (|∆p|/|p| ≪ 1). Sufficient (1011 . . . 1013 particles/s) but not too high

intensity : want many scatterings, but no interactionsbetween beam particles

May be polarized or not Consists of bunches (LHC: 7.4cm long, 7.5m = 25ns

apart)

Target Macroscopic sample at rest in lab (kohtio)

Large number of particle b. distance between scattering centers db ≫ λa = h/pa:

independent scatterings Thin, so that each a scatters most once.

4


Terminology IIWhere it collides

Collider experiment: two beams collide ; LHC, Tevatron, LEP,+ Achieve higher energy/$ (

√s = E∗

a + E∗b ,

$ ∼ Ebeam ∼ √s)

- Hard to align beams: less events(Not always symmetric: e.g. HERA 30GeV e− + 920GeV p )

Fixed target experiment: beam+target+ Can make target big enough so every beam particle

scatters =⇒ better statistics- Lower energy/$ (

√s ≈

p

2mbETRFa , $ ∼ Ebeam ∼ s )

An experimentalist cannot boost the lab to v = 0.9999 but a theorist can, sowe can do the following derivations in the TRF (= fixed target lab frame).

5


Terminology IIIWhat comes out

Exclusive experiment measure and identify all final state particles.Examples

π− + p → π+ + π− + n e+ + e− → µ+ + µ−

Inclusive experiment: only care about/measure certain particles(rest denoted as “X ”). Completely random examples:

p + p → X π− + p → π+ + X p + p → µ+ + µ− + X

Elastic scattering: a + b → a + b (initial and final particles same).

Inelastic scattering: final state particles 6= initial state particles.Examples:

e+ + e− → e+ + e−, elastic e+ + e− → µ+ + µ−, inelastic e+ + e− → µ+ + µ− + γ, inelastic p + p → X , with X 6= p + p inelastic (total inelastic cross

section)

6


Luminosity and cross section

Number of events in scattering a + b → X depends on Experimental setup: Intensity of incoming projectile particles a + number

of target particles b encountered Properties of fundamental interaction

Number of events per unit time

dNa+b→Xev

dt= σa+b→X

Lz |

ΦaNb, where

Φa is the flux of particle a (particles/(time*area))

Nb is the number of target particles b “under” the beam

L is the luminosity

σ is the cross section : this encodes everything about theparticle interaction.

Dimensions:»

dNdt

–

=1s

= m2 1sm2

=⇒ Cross section is an area.

(Remember we chose to work in TRF, but you can always boost.)

7


Some interpretationCross section

σ can be thought of as size of particle; but “size” depends on interaction . . .σ of billiard ball scattering is π(2R)2, (scatter if centers less than 2R apart.)

— but σ of neutrino-billiard ball scattering is much much less.

Loosely cross section = area × probability to scatter

Probability is dimensionless — Billiard balls have P = 1 for distance < 2R.

Cross section is Lorentz-invariantIt does not change in boost along the beam axis direction(unlike luminosity, because of time dilation!)

σ usually is differential, e.g.dNa+b→c+X

ev

dt d3pc(pc) =

dσa+b→c+X

d3pc(pc)

Lz |

ΦaNb

Number of scattering events where final state has particle c in momentumspace cube between pc = (px , py , pz) and (px + ∆px , py + ∆py , pz + ∆pz)— divided by time interval and ∆px∆py∆pz .

8


Some cross sections

Unit of cross section1barn = 10−28m2

E.g. proton-proton totalinelastic σ at√

s = 200GeV is∼ 40mb = (2fm )2.

Named by Enrico Fermi:( The area corresponding to a

barn is the approximate area of

a typical atomic nucleus which

has a size of 10−12cm. For

most sub-nuclear processes,

this is a very large

cross-section, so Fermi

suggested ”it was as big as a

barn”. )

9


More cross sections I

10

-8

10-7

10

-6

10-5

10

-4

10

-3

10-2

1 10 102

σ[m

b]

ω

ρ

φ

ρ′

J/ψ

ψ(2S)Υ

Z

σ(e+e− → hadrons) vs.√

s/GeV :

10


More cross sections II

10

10 2

10-1

1 10 102

103

104

105

106

107

108

⇓

Plab GeV/c

Cro

ss s

ectio

n (m

b)

10

10 2

10-1

1 10 102

103

104

105

106

107

108

⇓

Plab GeV/c

Cro

ss s

ectio

n (m

b)

√s GeV

1.9 2 10 102 103 104

p p

p−p

total

elastic

total

elastic

pp and pp total and elastic cross sections

11


Collider/experiment luminosity

The luminosity is the “figure of merit” of a particle accelerator/collider.Bigger luminosity is better, because you can see rarer events.LHC design luminosity ∼ 1034cm−2s−1 is very big for a collider.Often quoted is

Integrated luminosityZ

dtL

Has units 1/m2: inverse of cross section.

Interpretation:

When LHC has delivered 1fb−1 of integrated lumi, a process with crosssection σ = 1fb should have happened one time (± statistical fluctuations).(LHC design luminosity would give ∼ 300fb−1 in full year of nonstop running.)

12


ATLAS measured integrated luminosity at CERN/LHC

Day in 2011

27/02 29/03 29/04 30/05 30/06 31/07

]-1

Tot

al In

tegr

ated

Lum

inos

ity [f

b

00.20.40.60.8

11.2

1.4

1.61.8

22.2

Day in 2011

27/02 29/03 29/04 30/05 30/06 31/07

]-1

Tot

al In

tegr

ated

Lum

inos

ity [f

b

00.20.40.60.8

11.2

1.4

1.61.8

22.2 = 7 TeVs ATLAS Online Luminosity

LHC Delivered

ATLAS Recorded

-1Total Delivered: 1.75 fb-1Total Recorded: 1.66 fb

13


Quantum mechanical scattering theory

We now want to relate the cross section to a quantum mechanical matrixelement that we can compute from the Hamiltonian (in practice the Lagrangian) ofa given particle physics theory.

For this we need the state vectors |i〉 and |f 〉 corresponding to the initialand final particle content.These are many particle states created from the vacuum |0〉 by particlecreation operators that should be familiar if you have taken QMII.We will, however, not go into the second quantization formalism here; youneed it to derive the Feynman rules, but we will in this course not do that.

Relativistic plane waves

Contrary to many QM courses we want to have a completely relativisticnormalization (So some things may differ from nonrelativistic treatment QM I/II, conventions in

atomic physics, quantum optics etc.)

We will just assume that the in and out-states are described by a completeset of orthonormal Fock (=multiparticle) states |n〉.We now need to state our normalization of the wave functions (plane wavestates) that correspond to these multiparticle states.

14


S-matrix

Consider a scattering process a + b → 1 + 2 + · · · + nInitial state |i〉 is system of two free particles (a + b) at t → −∞.Final state |f 〉 is system of n free particles at t → ∞

All information on scattering in formally encoded into S-operator.Matrix elements of S are transition amplitudes

Sfi ≡ 〈f |S|i〉,in terms of which the transition probability is

Pfi = |Sfi |2 = 〈i |S†|f 〉〈f |S|i〉.Probability is conserved, sum over all final states is 1 (I=identity operator)

1 =X

f

Pfi = 〈i |S†

=I,complete setz |

X

f

|f 〉〈f | S|i〉 = 〈i |S†S|i〉 = 1,

i.e. S is unitary .Generally S-matrix contains also the trivial part |i〉 = |f 〉, subtracting this weget the “T -matrix” (T like transition) :

S = 1 + i T , i.e. for components Sfi = δfi + iTfi ,

15


Klein-Gordon equation and plane waves

Remember from nonrelativistic QM: E → i∂0, p → −i∇.Nonrelativistic dispersion relation is

E =1

2mp2

=⇒ Schrodinger i∂0ψ(t , x) = − 12m

∇2ψ(t , x)

Relativistic dispersion relation

E2 = p2+m2=⇒ Klein-Gordon eq (i∂0)

2ϕ(t , x) =h

(−i∇)2 + m2i

ϕ(t , x)

In covariant notation (recall ¤ = ∂µ∂µ = ∂20 − ∇2) relativistic plane wave

(¤ + m2)ϕ(x) = 0, solution Ne−ip·x = Ne−iEp t+ip·x

(E2, not E , because √ of ∇ awkward. Physically: antiparticles, E < 0 solutions.)

Note: Ep =p

p2 + m2, the energy corresponding to momentum p.

16


Normalization and box density of states

Problem: want to work with plane waves (exact ) momentum eigenstates.Heisenberg uncertainty relation =⇒ completely delocalized in space andunnormalizable .

Way around: assume we are in a large box of size V = L3, periodicboundary cond’s.

Normalization choice for plane waves

N =1√V

17


Box density of states

Periodic boundary allows us to count discrete states:

ϕ(x , y , z) = ϕ(x + L, y , z) = ϕ(x , y + L, z) = ϕ(x , y , z + L)

ei(p1x+p2y+p3z) = ei(p1(x+L)+p2y+p3z) = ei(p1x+p2(y+L)+p3z) = ei(p1x+p2y+p3(z+L))

1 = eip1L = eip2L = eip3L

only true if (p1, p2, p3) = 2πL (n1, n2, n3); ni ∈ Z.

NowX

states

=X

n1,n2,n3

≈V→∞

Z

d3n =

Z

V(2π)3

d3p

(P

n f (n) ≈R

dnf (n), when f (n) varies slowly.R n+1/2

n−1/2 dx = 1.

“Slowly varying” requires V → ∞ because 2πL (n1, n2, n3) close to 2π

L (n1 + 1, n2, n3).)

Recall normalization choice for plane waves, now it is

N =1√V

=⇒ 〈p|q〉 =

Z

d3x(Ne−ip·x)∗(Ne−iq·x) = δp,q ,

(Discrete Kronecker delta for discrete momenta =⇒ Now well defined matrices S and T . )

18


Conserved particle number current, nonrelativistic

In nonrelativistic theory:Particle density ρNR = ψ∗(t , x)ψ(t , x), current jNR = − i

2m (ψ∗∇ψ − ψ∇ψ∗).Probability conservation = continuity equation: ∂tρNR + ∇ · jNR = 0.For a plane wave ψ(t , x) = Ne−iEp t+ip·x

these are ρNR = |N |2; jNR = |N |2p/m.

This does not directly work in relativistic case

A relativistic current is jµ = (ρ, j)It should also have a continuity equation ∂µjµ = 0.But it should be a proper 4-vector;

thus for a plane wave we should have jµ ∼ pµ

In particular: Density of free particles in a box should be proportional tothe energy in a Lorentz-covariant formulation.(Thought experiment: Start from fixed box V , 1 particle in rest, energy m, density 1/V .Boost to another frame, γ = 1/

p

1 − v2; box size contracted to V/γ,

particle energy now mγ, density γ/V .)

19


Conserved current, relativistic case

Definition

jµ ≡ i(ϕ∗∂µϕ − ϕ∂µϕ∗), for plane wave ϕ = Ne−ip·x=⇒ jµ = 2pµ|N |2

In particular, for our choice N = 1√V

the particle density is ρ ≡ j0 = 2EV

The density now has funny dimensions (of energy density, not number density) , butthis is just a question of normalization of the (unnormalizable) plane wave states.We could have chosen E/m in stead of E (some books do this) , but then wewould need a separate treatment for massless particles.

Density of states: now “ 2E” particles in V

The number of states per particle isZ

V2Ep(2π)3

d3p = VZ

d4pδ(p2 − m2)

(2π)4θ(p0)

This is the Lorentz-invariant phase space measure .

21


T -matrix elementThe last, and most important, thing we need

Number of scatterings, T -matrix

The number of scatterings is given by the T -matrix element between finaland initial states. We separate out some factors:

Tfi = −iNaNbN1 . . .Nn(2π)4δ(4)

pa + pb −n

X

i=1

pi

!

Mfi ,

Why the factors?

energy, momentum always conserved in amplitude : useful to extract δ(4)

to explicitly treat (δ(4))2 in |Tfi |2 plane wave normalizations: conventions for N vary; for Mfi not.

(Justification: Tfi is matrix element between our box-normalized plane waves

Tfi ∼ 〈f |T |i〉 ∼ (Nae−ipa·xNbe−ipb·x )∗T (N1e−ip1·x . . .Nne−ipn·x )

The “invariant amplitude” Mfi , o.t.o.h. is defined using unnormalized plane waves

Mfi ∼ (e−ipa·x e−ipb·x )∗T (e−ip1·x . . . e−ipn·x )

Why this complication: needed the box to understand delta functions, count particles. . . )

22


Putting things together

What is then (δ(k))2? Integral representation of δ:Z

dx eikx = (2π)δ(k), thus δ(k = 0) =1

2π

Z

dx =L

2πand (2π)4δ(4)(0) = ∆tV

Now we have

σ =1

ΦaNb

Ns

∆t=

V

2q

λ(s, m2a, m2

b)

Z nY

m=1

»

V2Ep.(2π)3

d3pm

–

|−iNaNbN1 . . .Nn|2h

(2π)4δ(4)(pa + pb − Pni=1 pi)

i2|Mfi |2

∆t

=V

2q

λ(s, m2a, m2

b)

Z nY

m=1

»

V2Ep.(2π)3

d3pm

–

V−n−2(2π)4δ(4)(pa + pb − Pni=1 pi) [∆tV ]|Mfi |2

∆t

Now volume cancels — time interval cancels

23


Final master formula for cross section

The final formulae

Total σ(ab → 1 . . . n) =1

2q

λ(s, m2a, m2

b)

"

Z nY

m=1

d3pm

(2π)32Em

×(2π)4δ(4)“

pa + pb −Xn

i=1pi

”

#

|Mab→1...n|2 ,

Differentialdσab→1...n

d3p1 . . . d3pn=

1(2(2π)3E1) · · · (2(2π)3En)

1

2q

λ(s, m2a, m2

b)

×(2π)4δ(4)“

pa + pb −Xn

i=1pi

”

|Mab→1...n|2

Everything is Lorentz-invariant (Derived in TRF, but now works in any frame.)

The invariant amplitude M(ab → 1 . . . n) from Feynman rules. Part in [·] is Lorentz-invariant n-particle phase space .

24


Special case: 2 → 2 scattering

First work in general frame, start from the general formula

σab→cd =1

2q

λ(s, m2a, m2

b)

Z

d3pc

2Ec(2π)3

d3pd

2Ed(2π)3(2π)4δ(4)(pa+pb−pc−pd )|M|2.

Now δ(4)(p) = δ(E)δ(3)(p) =⇒ integrate d3pd . Other variable d3pc = |pc |2 d|pc | dΩc

σab→cd =(2π)−2

8q

λ(s, m2a, m2

b)

Z

dΩc

Z |pc |2 d|pc |EcEd

δ(Ea + Eb − Ec − Ed)|M|2

Get rid of δ(Ea + Eb − Ec − Ed), usingR

d|pc |.Remember, now Ec =

p

|p2c | + m2

c , Ed =q

|pa + pb − pc |2 + m2d .

=⇒ δ-function has |pc |-dependence in both Ec and Ed .Complicated in general frame, specialize to CMS.

25


Special case: 2 → 2 scattering in CMS

In CMS |pc | = |pd | ≡ p∗f , E∗

a + E∗b =

√s

Get rid of δ(Ec + Ed −√s), using

R

d|pc |.

pa

pb

pc

pd

θ∗

θ∗

Remember δ(f (x)) =δ(x − x0)

|f ′(x0)|, where f (x0) = 0

Nowd

dpE(p) =

ddp

p

p2 + m2 =p

p

p2 + m2=

pE(p)

So δ(E∗c (p∗

f ) + E∗d (p∗

f ) −√

s) =δ(p∗

f − . . . )p∗fE∗

c+

p∗fE∗

d

=E∗

c E∗d

p∗f

√s

δ(p∗f − . . . )

and σab→cd =(2π)−2

8q

λ(s, m2a, m2

b)

Z

dΩc

Z

(p∗f )2 dp∗

f

E∗c E∗

d

E∗c E∗

d

p∗f

√s

δ(p∗f − . . . )|M|2

=(2π)−2

8q

λ(s, m2a, m2

b)

p∗f√s

Z

dΩ∗c |M|2.

p∗f =

q

λ(s,m2c ,m2

d )

(2√

s)(Remember:

q

λ(s, m2a, m2

b) came from p∗

i .)

26


2 → 2 scattering in CMS, final result I

Not integrating over the angle gives differential σ

dσab→cd

dΩ∗c

=|Mab→cd |2

64π2s

s

λ(s, m2c , m2

d )

λ(s, m2a, m2

b)=

|Mab→cd |264π2s

p∗f

p∗i

.

(Differential: scattering events with final particle in angle Ω∗

c , divided by angle dΩ∗

c )

Special case (ma = mc and mb = md ) or (ma = md and mb = mc)

Leads to p∗i = p∗

f =⇒

dσab→cd

dΩ∗c

=|Mab→cd |2

64π2s

Remarks Energy-momentum conservation constraints now all implicitly in |M|

27


2 → 2 scattering, invariant dσ/ dt

Convenient to also express dΩ∗c in a Lorentz-invariant way, via t .

In general dΩc = dφc d(cos θc). If scattering is unpolarized (summed, averaged

over spins) , cross sections do not depend on φc =⇒dσ

d(cos θc )= 2π dσ

dΩc.

t = (pa − pc)2 = m2

a + m2c − 2E∗

a E∗c + 2|p∗

a ||p∗c | cos θ∗

c

=⇒ dt = 2p∗i p∗

f d(cos θ∗c )

=⇒dσ

dt=

2π

2p∗i p∗

f

dσ

dΩ∗c

pa

pb

pc

pd

θ∗

θ∗

Using the expressions from the previous slide:dσab→cd

dΩ∗

c= |Mab→cd |2

64π2s

p∗fp∗i

and p∗i =

p

λ(s, m2a, m2

a)/(2√

s), this leads to

Differential cross section in manifestly Lorentz-invaria nt form

dσab→cd

dt=

|Mab→cd |216πλ(s, m2

a, m2b)

If s ≫ m2a, m2

b: λ(s, m2a, m2

b) = s2 + m4a + m4

b − 2sm2a − 2m2

am2b − 2m2

bs ≈ s2

=⇒ this is the easiest version to remember.

28


Example: unpolarized e+e−→ µ+µ− scattering

A little flavor of things to come

e− µ−, q

e+ µ+, q

γ, Z

Assume√

s small enough so we can neglectZ0-boson =⇒ only one Feynman diagram at LO(=leading order)

Use Feynman rules for QuantumElectroDynamics= QED =⇒ get expression for|Me+e−→µ+µ− |2.

unpolarized = summed over spins of µ±;averaged over the spins of e±, denote |M|2

Assuming√

s ≫ me, mµ

|M|2 ≡ 12

12

X

spins

|M|2 = 2e4 t2 + u2

s2.

Origin of Mandelstams: M contains dot productsof pa, pb, pc and pd =⇒ written in terms of s, tand u.

29


Example: e+e−→ µ+µ− continued

The differential cross section for the above process becomes(p∗

i ≈ p∗f in

√s ≫ me, mµ limit)

dσ

dΩ∗c

=α2

4s(1 + cos2 θ∗

µ).

From this, the total cross section is

σ(e+e− → µ+µ−) =4πα2

3s.

e− µ−, q

e+ µ+, q

γ, Z Two interactions vertices ∼ e, so

M(e+e− → µ+µ−) ∝ e2, and |M|2 ∝ e4 ∝ α2;α = e2/(4π) ≈ 1/137.

Check dimensions: [s] = GeV 2

30


e+e−→ µ+µ−, comments

Angular distribution depends on spin structure

Angular distribution 1 + cos2 θ∗ characteristic of spin-1 particle (virtualphoton γ∗) decaying into two spin 1/2 particles.

For spin interpretation, there are two spin states with amplitudes:∼ (1 ± cos θ). These are different final states =⇒ they do not interfere,each amplitude is squared separately, then added.

Cross section for γ∗ → 2 spin-0 particles would be ∼ sin2 θ ∼ |Y 11 (θ, φ)|2;

spin has to go into m = 1, ℓ = 1 angular momentum of the pair.

See Halzen, Martin, sec 6.6.

Squaring and averaging Sum of diagrams=amplitude with one initial and final state (incl. spin) Then square each amplitude separately Only then sum/average |M|2’s from different spin states =⇒ this is

denoted by |M|2

31


Particle decay

Initial state |i〉 = unstable particle a, decaying to n particle final state |f 〉. Only natural frame is CMS = rest frame of a. Momentum conservation p2

a = (p1 + · · · + pn)2.

In CMS: m2a = (E1 + · · · + En)

2 ≥ (m1 + . . . mn)2

=⇒ threshold ma ≥ m1 + · · · + mn.(Thus a has ma > 0, there is always a rest frame.)

Terminology In general different possible final states |f 〉: decay channels . Define the decay width Γf for channel f as

Γf ≡Nf

Na∆Tin a rest frame.

Nf is the number of decays to state fNa is the number of decaying particles a

∆T is the time interval required for Nf decays. All

expe

rimen

tally

acce

ssib

le,

like

def.

ofσ

.

Now want to tie this definition with theoretically calculable quantities, skippingdetails that are similar to the derivation with cross sections.

32


Decay width and amplitude

Recall: for cross section we had

Ns

∆t=

N(ab → 1 . . . n)

∆t=

|Tfi |2∆t

× (number of possible final states),

Now define alsoNf

∆t=

|Tfi |2∆t

× (number of possible final states).

We again have the transition amplitude Tfi and the invariant amplitude Mfi :

Tfi = −iNaN1 . . .Nn(2π)4δ(4)“

pa −Xn

j=1pj

”

Mfi(a → 1 . . . n).

|Tfi |2 has a δ(4) squared =⇒ treated similarly as before.Again the number of final states per particle is 1

2EV

(2π)3 d3p

Result: decay width

Γf =1

2Ea

Z nY

m=1

d3pm

(2π)32Em(2π)4δ(4)

“

pa −Xn

i=1p i

”

|M(a → 1 . . . n)|2.

Γa and Ea depend on frame; in CMS Ea = ma. Rest is Lorentz-invariant.

33


Branching ratios

One unstable particle; many possible finals tates |f 〉, each has width Γf .Define total decay width

Γ ≡X

f

Γf ,

Fraction of each final state is branching ratio

Bf ≡Γf

Γ; often in %.

mean lifetimeτ ≡ 1

Γ.

34


Branching ratio example

Example

For the Z0 boson we have the following branching ratios:Z0 → e+e− 3.363%

µ+µ− 3.366%τ+τ− 3.370%

In addition: Z0 can decay ”invisibly” to a pair of any of the three neutrino species:

total branching ratio 20.00%. Hadronic decay modes: remaining 69.91% of the total width

Γ(Z0) = 2.4952 ± 0.0023GeV τ ≃ 0.079fm ≈ 3 · 10−25 s.

Hence the Z0 boson is very short-lived.In particle physics “long-lived particles” have lifetime ∼ 10−16s or more=⇒ can be actually seen directly in experiment.

35


Some decay times

Out of gauge bosons γ is absolutely stable, W± and Z 0 extremelyunbstable, Γ ∼ 2GeV and τ ≈ 3 · 10−25s.

gluon is confined, stability/unstability not well defined Neutrinos, e± stable τ±, µ± decay via weak interaction: τµ ≈ 2 · 10−6s and ττ ≈ 3 · 10−13 s. Quarks u, d , c, s and b form first a hadron which then decays. Out of all

hadrons only the proton is absolutely stable. Heaviest t quark decays before it has formed a hadron, τt ≈ 0.02fm ; It

would take a time of the order of 1 fm to make a hadron.

36


1 → 2 decay

Analogous to σ(2 → 2) =⇒ we’ll be brief. Frame is CMS, omit ∗.

Γa→cd =1

2ma

Z

d3pc

(2π)32Ec

d3pd

(2π)32Ed(2π)4δ(ma − Ec − Ed)δ(3)(pc + pd)|M|2.

Use δ(3)(pc + pd) to doR

d3pd ; sets |pc | = |pd | ≡ pf

Write d3pc = dΩp2f dpf , want to use δ(ma − Ec − Ed) to do

R

dpf

Energy conservation

δ(ma − Ec − Ed) =EcEd

p0ma|pf =p0δ(pf − p0),

where p0 is obtained by solving ma = Ec + Ed , i.e. explicitly

p0 =

q

λ(s, m2c , m2

d)

2√

s=

q

λ(m2a, m2

c , m2d)

2ma.

Final result Γ(a → cd) =

q

λ(m2a, m2

c , m2d )

64π2m3a

Z

dΩ|Mfi(a → cd)|2.

37


Example: π− decay

Two possible final states, µ−νµ and e−νe (threshold,conservation laws).

Experiments:Γ(π− → e−νe)

Γ(π− → µ−νµ)≈ 1.23 · 10−4 very small, why?

π−

u

dW−

νℓ

ℓ

Annihilate du =⇒ charged weak current One W -boson, 2 weak vertices:

M ∼ g2W

m2W

∼ GF (“Fermi constant”)

π− ↔ du: “pion decay constant” M ∼ fπ.

Evaluation of the Feynman diagram gives:(Actually Feynman rule for Wℓνℓ vertex + information of pion wavefunction in fπ)

|M(π− → ℓ−νℓ)|2 = 4G2F f 2

πm2ℓ (pℓ · pν) = 2G2

F f 2πm2

ℓm2π

„

1 − m2ℓ

m2π

«

,

Then (exercise) the decay width is Γ(π− → ℓ−νℓ) =1

8πG2

F f 2πmπm2

ℓ

„

1 − m2ℓ

m2π

«2

,

which implies the ratioΓ(π− → e−νe)

Γ(π− → µ−νµ)=

„

me

mµ

«2 „

m2π − m2

e

m2π − m2

µ

«2

≈ 1.28·10−4.

38


Example 2: Muon decay µ−→ e−

+ νe + νµ, estimates I1 → 3 decay

µ−

νµ

e−

νe

W−

1. Conservation laws determine finalparticles:1.1 Charge conservation: e− (only negative

particle with m < mµ).1.2 µ number conservation: Lµ = 1 =⇒ νµ

1.3 e number: Le = 0: νe to compensate e−

2. Only e− detected. 2 others=⇒ continuous distribution of Ee.

As before, we can estimate the decay width as One W propagator, 2 W -fermion vertices: M ∼ g2

W /m2W ∼ GF

mµ ≫ me, mνℓ =⇒ dimensionally must have Γ ∼ G2F m5

µ

39



+ νe + νµ, calculation

The actual invariant amplitude, from S.M. Feynman rules, is

|M(µ → e−νeνµ)|2 = 64G2F (k · p′)(k ′ · p).

µ−, p

νµ, k

e−, p′

νe, k′

W−

Denote p = (mµ, 0)

p′ = (E ′, p′)

k = (ω, k)

k ′ = (ω′, k′)

Differential width is dΓ =1

2mµ

M(µ → e−νeνµ)|2 dQ,

with dQ =d3p′

(2π)32E ′d3k′

(2π)32ω′d3k

(2π)32ω(2π)4δ(4) `

p − p′ − k − k ′´ .

This can be simplified usingZ

d3k2ω

=

Z

Eats δ(4)

z|

d4k θ(ω)

1dz |

δ(k2) .

40



+ νe + νµ, result

We get dQ =1

(2π)5

d3p′

(2π)32E ′d3k′

(2π)32ω′ θ(mµ − E ′ − ω′)δ((p − p′ − k ′)2)

After intermediate steps (exercise) we get

dΓ =G2

F

2π3dE ′ dω′mµω′(mµ − 2ω′),

with restrictions12

mµ − E ′ ≤ ω′ ≤ 12

mµ

0 ≤ E ′ ≤ 12

mµ

Integrating, final result is (exercise) Γ =G2

F m5µ

192π3

Estimate ∼ G2F m5

µ true But constant of proportionality is very small!

Details: Halzen-Martin, sec. 12.5. Also in KJE’s notes, after page 90.

part 3: scattering theory - jyväskylän yliopistousers.jyu.fi/~tulappi/fysh300sl11/l3.pdf1...

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