particle physics phenomenology 2. phase space and matrix...

Particle Physics Phenomenology2. Phase space and matrix elements

Torbjorn Sjostrand

Department of Astronomy and Theoretical PhysicsLund University

Solvegatan 14A, SE-223 62 Lund, Sweden

NBI, Copenhagen, 3 October 2011

Four-vectors

four−vector : p = (E ;p) = (E ; px , py , pz)

vector sum : p1 + p2 = (E1 + E2;p1 + p2)

vector product : p1p2 = E1E2 − p1p2

= E1E2 − px1px2 − py1py2 − pz1pz2

= E1E2 − |p1| |p2| cos θ12

square : p2 = E 2 − p2 = E 2 − p2x − p2

y − p2z = m2

transverse mom. : p⊥ =√

p2x + p2

y

transverse mass : m⊥ =√

m2 + p2x + p2

y =√

m2 + p2⊥

E 2 = m2 + p2 = m2 + p2⊥ + p2

z = m2⊥ + p2

z

Warning: No standard to distinguish p = (E ; px , py , pz) and

p = |p| =√

p2x + p2

y + p2z , but usually clear from context.

When we remember, we will try to use p = |p|, since p = p.

Torbjorn Sjostrand PPP 2: Phase sapce and matrix elements slide 2/48

Decay widths and cross sections

Decay width at rest, 1 → n:

dΓ =|M|2

2MdΦn

Integrated it gives exponential decay rate

dPdt

= Γe−Γt and 〈τ〉 = 1/Γ

Collision process cross section, 2 → n:

dσ =|M|2

4√

(p1p2)2 −m21m

22

dΦn

Integrated it gives collision rate

N = σ

∫L(t) dt with L ≈ f

n1n2

A

in a theorist’s approximation of the luminosity L for a collider.Torbjorn Sjostrand PPP 2: Phase sapce and matrix elements slide 3/48

Phase space

n-body phase space:

dΦn = (2π)4δ(4)(P −n∑

i=1

pi )n∏

i=1

d3pi

(2π)32Ei

Lorentz covariant:

d4pi δ(p2i −m2

i ) θ(Ei ) = d4pi δ(E2i − (p2

i + m2i )) θ(Ei )

=d3pi

2Ei

with Ei =√

p2i + m2

i and using

δ(f (x)) =∑

xj ,f (xj )=0

1

|f ′(xj)|δ(x − xj)

Application: Lorentz invariant production cross sections E dσ/d3pTorbjorn Sjostrand PPP 2: Phase sapce and matrix elements slide 4/48

Spherical symmetry

Spherical coordinates:

d3p

E=

dpx dpy dpz

E=

p2 dp dΩ

E=

p EdE dΩ

E= p dE dΩ

where Ω is the unit sphere,

dΩ = d(cos θ) dφ = sin θ dθ dϕ

px = p sin θ cos ϕ

py = p sin θ sin ϕ

pz = p cos θ

and E 2 = p2 + m2 ⇒ E dE = p dp.

Convenient for use e.g. in resonance decays,but not for standard QCD physics in pp collisions.Instead:


Cylindrical symmetry and rapidity

Cylindrical coordinates:

d3p

E=

dpx dpy dpz

E=

d2p⊥ dpz

E= d2p⊥ dy

with rapidity y given by

y =1

2ln

E + pz

E − pz=

1

2ln

(E + pz)2

(E + pz)(E − pz)=

1

2ln

(E + pz)2

m2 + p2⊥

= lnE + pz

m⊥= ln

m⊥E − pz

The relation dy = dpz/E can be shown by

dy

dpz=

ddpz

(ln

E + pz

m⊥

)=

ddpz

(ln(√

m2⊥ + p2

z + pz)− lnm⊥

)

=

12

2p⊥√m2⊥+p2

z

+ 1√m2⊥ + p2

z + pz

=pz+E

E

E + pz=

1

E


Lightcone kinematics and boosts

Introduce (lightcone) p+ = E + pz and p− = E − pz .Note that p+p− = E 2 − p2

z = m2⊥.

Consider boost along z axis with velocity β, and γ = 1/√

1− β2.

p′x ,y = px ,y

p′z = γ(pz + β E )

E ′ = γ(E + β pz)

p′+ = γ(1 + β)p+ =

√1 + β

1− βp+ = k p+

p′− = γ(1− β)p+ =

√1− β

1 + βp− =

p−

k

y ′ =1

2ln

p′+

p′−=

1

2ln

k p+

p′−/k= y + ln k

y ′2 − y ′1 = (y2 + ln k)− (y1 + ln k) = y2 − y1


Pseudorapidity

If experimentalists cannot measure m they may assume m = 0.Instead of rapidity y they then measure pseudorapidity η:

y =1

2ln

√m2 + p2 + pz√m2 + p2 − pz

⇒ η =1

2ln|p|+ pz

|p| − pz= ln

|p|+ pz

p⊥

or

η =1

2ln

p + p cos θ

p − p cos θ=

1

2ln

1 + cos θ

1− cos θ

=1

2ln

2 cos2 θ/2

2 sin2 θ/2= ln

cos θ/2

sin θ/2= − ln tan

θ

2

which thus only depends on polar angle.η is not simple under boosts: η′2 − η′1 6= η2 − η1.You may even flip sign!Assume m = mπ for all charged ⇒ yπ; intermediate to y and η.


The pseudorapidity dip

By analogy with dy/dpz = 1/E it follows that dη/dpz = 1/p.

Thus

dη

dy=

dη/dpz

dy/dpz=

E

p> 1

with limits

dη

dy→ m⊥

p⊥for pz → 0

dη

dy→ 1 for pz → ±∞

so if dn/dy is flat for y ≈ 0then dn/dη has a dip there.

η−y = lnp + pz

p⊥−ln

E + pz

m⊥= ln

p + pz

E + pz

m⊥p⊥

→ lnm⊥p⊥

when pz m⊥


Two-body phase space

Evaluate in rest frame, i.e. P = (Ecm, 0).

dΦ2 = (2π)4δ(4)(P − p1 − p2)d3p1

(2π)32E1

d3p2

(2π)32E2

=1

16π2δ(Ecm − E1 − E2)

d3p1

E1E2

=1

16π2δ(√

m21 + p2 +

√m2

2 + p2 − Ecm)p2 dp dΩ

E1E2

=1

16π2

δ(p − p∗

| p

E1+

p

E2|p2 dp dΩ

E1E2

=1

16π2

E1E2

E1 + E2

p dΩ

E1E2

=p dΩ

16π2 Ecm


The Kallen function – 1

√m2

1 + p2 +√

m22 + p2 = Ecm

gives solution

E1 =E 2

cm + m21 −m2

2

2Ecm

E2 =E 2

cm + m22 −m2

1

2Ecm

p =1

2Ecm

√(E 2

cm −m21 −m2

2)2 − 4m2

1m22 =

1

2Ecm

√λ(E 2

cm,m21,m

22)

where Kallen λ function is

λ(a2, b2, c2) = (a2 − b2 − c2)2 − 4b2c2

= a4 + b4 + c4 − 2a2b2 − 2a2c2 − 2b2c2

= (a2 − (b + c)2)(a2 − (b − c)2)

= (a + b + c)(a− b − c)(a− b + c)(a + b − c)


The Kallen function – 2

Hides everywhere in kinematics, e.g.

dσ =|M|2

4√

(p1p2)2 −m21m

22

dΦn

has

4((p1p2)2 −m2

1m22) = (p2

1 + 2p1p2 + p22 −m2

1 −m22)

2 − 4m21m

22

= ((p1 + p2)2 −m2

1 −m22)

2 − 4m21m

22

= λ(E 2cm,m2

1,m22)

so

dσ =|M|2

2√

λ(E 2cm,m2

1,m22)

dΦn


Mandelstam variables

For process 1 + 2 → 3 + 4

s = (p1 + p2)2 = (p3 + p4)

2

t = (p1 − p3)2 = (p2 − p4)

2

u = (p1 − p4)2 = (p2 − p3)

2

In rest frame, massless limit: m1 = m2 = m3 = m4 = 0,

p1,2 =Ecm

2(1; 0, 0,±)

p3,4 =Ecm

2(1;± sin θ, 0,± cos θ)

s = E 2cm

t = −2p1p3 = − s

2(1− cos θ)

u = −2p2p4 = − s

2(1 + cos θ) s + t + u = 0


Mandelstam variables with masses

β34 =

√λ(s,m2

3,m24)

s

p3,4 =

√s

2

(1± m2

3 −m24

s;±β34 sin θ, 0,±β34 cos θ

)t = m2

1 + m23 −

s

2

(1 +

m21 −m2

2

s

)(1 +

m23 −m2

4

s

)+

s

2β12 β34 cos θ

dσ =|M|2

2√

λ(s,m21,m

22)

p34√s

d cos θ dϕ

16π2=|M|2

2sβ12

β34

2

d cos θ

8π

assuming no polarization ⇒ no ϕ dependence

dσ

dt=

dσ

dcos θ

dcos θ

dt=

|M|2

16πs2β212


Mandelstam variables with final-state masses

Usually m1,2 ≈ 0, while often m3,4 non-negligible

t, u = −1

2

[s −m2

3 −m24 ∓ sβ34 cos θ

]dσ

dt=

|M|2

16πs2

s + t + u = m23 + m2

4

tu =1

4

[(s −m2

3 −m24)

2 − s2β234 cos2 θ

]=

1

4

[s2β2

34 + 4m23m

24 − s2β2

34 cos2 θ]

=1

4s2β2

34 sin2 θ + m23m

24 = sp2

⊥ + m23m

24

p2⊥ =

tu −m23m

24

s


s-, t- and u-channel processes

Classify 2 → 2 diagrams by character of propagator, e.g.

Singularities reflect channel character, e.g. pure t-channel:

dσ(qq′ → qq′)dt

=π

s2

4

9α2

s

s2 + u2

t2

peaked at t → 0 ⇒ u ≈ −s, so

dσ(qq′ → qq′)dt

≈ 8πα2s

9t2=

32πα2s

9s2(1− cos θ)2=

8πα2s

9s2 sin4 θ/2≈ 8πα2

s

9p4⊥

i.e. Rutherford scattering!Torbjorn Sjostrand PPP 2: Phase sapce and matrix elements slide 16/48

Order-of-magnitude cross sections

With masses neglected:

s−channel :dσ

dt∼ π

s2

t−channel, spin 1 :dσ

dt∼ π

t2

t−channel, spin1

2:

dσ

dt∼ π

−stu−channel : same with t → u

Add couplings at vertices:

qqg : CFαs

ggg : Ncαs

f fγ : e2f αem

f f ′W : |Vff′ |2αem

4 sin2θW

f f ′Z : (v2f + a2

f )αem

16 sin2θW cos2θWTorbjorn Sjostrand PPP 2: Phase sapce and matrix elements slide 17/48

Closeup: qg → qg

Consider q(1) g(2) → q(3) g(4):

t : pg∗ = p1 − p3 ⇒ m2g∗ = (p1 − p3)

2 = t ⇒ dσ/dt ∼ 1/t2

u : pq∗ = p1 − p4 ⇒ m2q∗ = (p1 − p4)

2 = u ⇒ dσ/dt ∼ −1/su

s : pq∗ = p1 + p2 ⇒ m2q∗ = (p1 + p2)

2 = s ⇒ dσ/dt ∼ 1/s2

Contribution of each sub-graph is gauge-dependent,only sum is well-defined:

dσ

dt=

πα2s

s2

[s2 + u2

t2+

4

9

s

(−u)+

4

9

(−u)

s

]Torbjorn Sjostrand PPP 2: Phase sapce and matrix elements slide 18/48

Scale choice

What Q2 scale to use for αs = αs(Q2)?

Should be characteristic virtuality scale of process!But e.g. for q g → q g: both s-, t- and u-channel + interference.At small t the t-channel graph dominates ⇒ Q2 ∼ |t|,at small u the u-channel graph dominates ⇒ Q2 ∼ |u|,in between all graphs comparably important ⇒ Q2 ∼ s ∼ |t| ∼ |u|.Suitable interpolation:

→ −t for t → 0

Q2 = p2⊥ =

tu

s→ −u for u → 0

→ s

4for t = u = − s

2

but could equally well be multiple of p2⊥, or more complicated

⇒ one limitation of LO calculations.


Resonances

Resonance shape given by Breit-Wigner

1 7→ ρ(s) =1

π

mΓ

(s −m2)2 + m2Γ2

7→ 1

π

sΓ(m)/m

(s −m2)2 + s2Γ2(m)/m2

where m 7→√

s in phase space and Γ(s) 7→ Γ(m)√

s/mfor gauge bosons, neglecting thresholds.Latter shape suppressed below and enhanced above peak; tilted.For s → 0 ρ(s) goes to constant or like s.PDF’s tend to be peaked at small x : convolution enhances small s.Can give secondary mass-spectrum “peak” in s → 0 region.But note that

|M|2 = |Msignal +Mbackground|2

so in many cases Breit-Wigner cannot be trusted except in theneighbourhood of the peak, where signal should dominate.


Three-body phase space

Three-body final states has 3 · 3− 4 degrees of freedom.In massless case straightforward to show that, in CM frame,

dΦ3 = (2π)4δ(4)(P − p1 − p2 − p3)d3p1

(2π)32E1

d3p2

(2π)32E2

d3p3

(2π)32E3

=1

8(2π)5dE1 dE2 d cos θ1 dϕ1 dϕ21

with θ1, ϕ1 polar coordinates of 1 andϕ21 azimuthal angle of 2 around 1 axis (Euler angles).Phase space limits 0 ≤ E1,2 ≤ Ecm/2 andE1 + E2 = Ecm − E3 > Ecm/2.

Same simple phase space expression holds in massive case,but phase space limits much more complicated!

Higher multiplicities increasingly difficult to understand.One solution: recursion!


Factorized three-body phase space

Drop factors of 2π, and don’t write implicit integral signs.Introduce intermediate “particle” 12 = 1 + 2.

dΦ3(P; p1, p2, p3)

∼ δ(4)(P − p1 − p2 − p3)d3p1

2E1

d3p2

2E2

d3p3

2E3δ(4)(p12 − p1 − p2) d4p12

= δ(4)(P − p12 − p3) d4p12d3p3

2E3

[δ(4)(p12 − p1 − p2)

d3p1

2E1

d3p2

2E2

]= δ(4)(P − p12 − p3) d4p12 δ(p2

12 −m212) dm2

12

d3p3

2E3dΦ2(p12; p1, p2)

= dm212

[δ(4)(P − p12 − p3)

d3p12

2E12

d3p3

2E3

]dΦ2(p12; p1, p2)

= dm212 dΦ2(P; p12, p3) dΦ2(p12; p1, p2)

Note: here 4 angles + 1 mass2; last slide 3 angles + 2 energies.


Recursive phase space

Generalizes to

dΦn(P; p1, . . . , pn) = dm212...(n−1) dΦ2(P; p12...(n−1), pn)

× dΦn−1(P; p1, . . . , p(n−1))

Can be viewed as a sequentialdecay chain, with undeterminedintermediate masses.

Recall dΦ2(P; p1, p2) ∝

√λ(M2,m2

1,m22)

M2dΩ12

where dΩ12 is the unit sphere in the 1+2 rest frame.Now can write down e.g. 4-body phase space:


The M-generator

dΦ4(P; p1, p2, p3, p4) ∝

√λ(M2;m2

4,m2123)

M2m123 dm123 dΩ1234

×

√λ(m2

123;m23,m

212)

m2123

m12 dm12 dΩ123

√λ(m2

12;m21,m

22)

m212

dΩ12

Mass limits coupled, but can be decoupled: pick two randomnumbers 0 < R1,2 < 1 and order them R1 < R2. Then

∆ = M − (m1 + m2 + m3 + m4)

m12 = m1 + m2 + R1∆

m123 = m1 + m2 + m3 + R2∆

uniformly covers dm12 dm123 space with weight√λ(M2;m2

4,m2123)

M

√λ(m2

123;m23,m

212)

m123

√λ(m2

12;m21,m

22)

m12


RAMBO

For massless case a smart solution is RAMBO (RAndom Momentaand BOosts), which is 100% efficient:

RAMBO

1 Pick n massless 4-vectors pi according to

Eie−Ei dΩi

2 boost all of them by a common boost vector that brings themto their overall rest frame

3 rescale them by a common factor that brings them to thedesired mass M

Can be modified for massive cases, but then no longer 100%efficiency; gets worse the bigger

∑mi/M is.

MAMBO: workaround for high multiplicities


Efficiency troubles

Even if you can pick phase space points uniformly, |M|2 is not!A n-body process receives contributions from a large number ofFeynman graphs, plus interferences.Can lead to extremely low Monte Carlo efficiency.Intermediate resonances ⇒ narrow spikes when (pi + pj)

2 ≈ M2res.

t-channel graphs ⇒ peaked at small p⊥.

Multichannel techniques:

|M|2 =|∑

i Mi |2∑i |Mi |2

∑i

|Mi |2

so pick optimized for either |Mi |2 according to their relativeintegral, and use ratio as weight.Still major challenge in real life!


Composite beams

In reality all beamsare composite:p : q, g, q, . . .e− : e−, γ, e+, . . .γ : e±, q, q, g

Factorization

σAB =∑i ,j

∫∫dx1 dx2 f

(A)i (x1,Q

2) f(B)j (x2,Q

2) σij

x : momentum fraction, e.g. pi = x1pA; pj = x2pB

Q2: factorization scale, “typical momentum transfer scale”

Factorization only proven for a few cases, like γ∗/Z0 prodution,and strictly speaking not correct e.g. for jet production,

but good first approximation and unsurpassed physics insight .


Subprocess kinematics

If pA + pB = (Ecm; 0), A,B along ±z axis, and 1, 2 collinear withA,B then convinently put them massless:

p1 = (Ecm/2)(1; 0, 0, 1)

p2 = (Ecm/2)(1; 0, 0,−1)

such that s =(p1 + p2)2 = x1 x2 s = τ s. Velocity of subsystem is

βz =pz

E=

x1 − x2

x1 + x2

and its rapidity

y =1

2ln

E + pz

E − pz=

1

2ln

x1

x2

dx1 dx2 = dτ dy convenient for Monte Carlo.Historically xF = 2pz/Ecm = x1 − x2.Subprocess 2 → 2 kinematics for σ: s, t, u..


Matrix Elements and Their Usage

L ⇒ Feynman rules ⇒ Matrix Elements ⇒ Cross Sections+ Kinematics ⇒ Processes ⇒ . . .⇒

(Higgs simulation in CMS)


Loops and legs – 1 (Peter Skands)


Born level calculations – 1 (Frank Krauss)



Remember: to be squared for number of squared MEs.


Next-to-leading order (NLO) graphs


NLO calculations – 1

σNLO =

∫ndσLO +

∫n+1

dσReal +

∫ndσVirt

Simple one-dimensional example: x ∼ p⊥/p⊥max, 0 ≤ x ≤ 1Divergences regularized by d = 4− 2ε dimensions, ε < 0

σR+V =

∫ 1

0

dx

x1+εM(x) +

1

εM0

KLN cancellation theorem: M(0) = M0

Phase Space Slicing:Introduce arbitrary finite cutoff δ 1 (so δ |ε| )

σR+V =

∫ 1

δ

dx

x1+εM(x) +

∫ δ

0

dx

x1+εM(x) +

1

εM0

≈∫ 1

δ

dx

xM(x) +

∫ δ

0

dx

x1+εM0 +

1

εM0

=

∫ 1

δ

dx

xM(x) +

1

ε

(1− δ−ε

)M0 ≈

∫ 1

δ

dx

xM(x) + ln δ M0


NLO calculations – 2

Alternatively Subtraction:

σR+V =

∫ 1

0

dx

x1+εM(x)−

∫ 1

0

dx

x1+εM0 +

∫ 1

0

dx

x1+εM0 +

1

εM0

=

∫ 1

0

M(x)−M0

x1+εdx +

(−1

ε+

1

ε

)M0

≈∫ 1

0

M(x)−M0

xdx +O(1)M0

NLO provides a more accurate answer for an integrated cross section:

Warning!Neither approach operateswith positive definite quantities.No obvious event-generatorimplementation.No trivial connection tophysical events


Scale choices

Cross section depends on factorization scale µF

and renormalization scale µR :

σAB =∑i ,j

∫∫dx1 dx2 f

(A)i (x1, µF ) f

(B)j (x2, µF ) σij(αs(µR), µF , µR)

Historically common to put Q = µF = µR but nowadays variedindependently to gauge undertainty of cross section prediction.

Typical variationfactor 2±1 around“natural value”,but beware


Current status (N)(N)LO (Frank Krauss)


Colour flow in hard processes – 1

One Feynman graph can correspond to several possible colourflows, e.g. for qg → qg:

while other qg → qg graphs only admit one colour flow:


Colour flow in hard processes – 2

so nontrivial mix of kinematics variables (s, t)and colour flow topologies I, II:

|A(s, t)|2 = |AI(s, t) +AII(s, t)|2

= |AI(s, t)|2 + |AII(s, t)|2 + 2Re(AI(s, t)A∗II(s, t)

)with Re

(AI(s, t)A∗II(s, t)

)6= 0

⇒ indeterminate colour flow, while• showers should know it (coherence),• hadronization must know it (hadrons singlets).Normal solution:

interferencetotal

∝ 1

N2C − 1

so split I : II according to proportions in the NC →∞ limit, i.e.

|A(s, t)|2 = |AI(s, t)|2mod + |AII(s, t)|2mod

|AI(II)(s, t)|2mod = |AI(s, t) +AII(s, t)|2(

|AI(II)(s, t)|2

|AI(s, t)|2 + |AII(s, t)|2

)NC→∞


particle physics phenomenology 2. phase space and matrix...

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