Physics 214 UCSD/225a UCSB Lecture 12 • Halzen & Martin ... › 2009 › Fall ›...

Physics 214 UCSD/225a UCSB

Lecture 12• Halzen & Martin Chapter 6

– Spin averaged e-mu scattering• Introduce traces, and allude to trace theorems.

– Use crossing to derive spin averaged e+e- -> mu+mu-

– Detailed discussion of the result.

Spinless vs Spin 1/2

! t,x( ) = Ne"ipµ x

µ

! t,x( ) = u(p)e"ipµ x

µ

Jµ = !ie "#($µ" ) ! ($µ"

#)"( )

Tfi = !i J fiµAµd

4x + O(e2)%

Jµ

= !e"# µ"

Tfi = !i J fiµAµd

4x$ + O(e

2)

We basically make a substitution:

pf + pi( )µ! u f "µui

And all else in calculating |M|2 remains the same.

Example: e- e- scattering

M = !e2(pA + pC )

µ(pB + pD )µ

pA ! pC( )2

+(pA + pD )

µ(pB + pC )µ

pA ! pD( )2

"

#

$ $

%

&

' '

For Spinless (i.e. bosons) we showed:

M = !e2(u c"

µuA )(u D"µ uB )

pA ! pC( )2

!(u D"

µuA )(u C"µ uB )

pA ! pD( )2

#

$

% %

&

'

( (

For Spin 1/2 we thus get:

Minus sign comes from fermion exchange !!!

Spin Averaging• The M from previous page includes spinors in

initial and final state.• In many experimental situations, in particular

in hadron collissions, you neither fix initial norfinal state spins.

• We thus need to form a spin averagedamplitude squared before we can comparewith experiment:

M2

=1

2sA +1( ) 2sB +1( )M

2

spin

! =1

4M

2

spin

!A

B

C

D

Last time we did thenonrelativistic case.

This time we do the completederivation.

Note: This is quite possibly the mostpainful derivation we do this quarter.

“Easiest”: e- mu- -> e- mu-• Easiest because it has only one diagram !!!

M = !e2u ( "k )# µ

u(k)$% &' u ( "p )# µu(p)$% &'t

e- e-

µ- µ-

k k’

p’p

k for electron momenta.p for muon momenta.prime for outgoing momenta.

t = q2 = (k’-k)2 = (p’-p)2 = scalar product of 4-momenta

Spin averaged |M|2 (1)

M2=

1

2sA +1( ) 2sB +1( )M

2

spin

! =1

4M

2

spin

!

M2=e4

4t2

u ( "k )# µu(k)( )gµ$ u ( "p )# $

u(p)( ) u ( "k )# %u(k)( )g%& u ( "p )# &

u(p)( )'( )*+

spin

!

This is a Scalar product of 4-vectors, multiplied by its complex conjugate, to get a positive definite number.

Unfortunately, the scalar product mixes e and mu currents.

To form the spin average, we separate e and mu,to execute the spin average on e and mu independently

Spin averaged |M|2 (2)M

2=e4

4t2

u ( !k )" µu(k)( )gµ# u ( !p )" #

u(p)( ) u ( !k )" $u(k)( )g$% u ( !p )" %

u(p)( )&' ()*

spin

+

M2

=e4

t2Lelectronµ!

Lmuon"#

gµ"g!#

Where we defined:

Lelectronµ! =

1

2u ( "k )# µ

u(k)( ) u ( "k )# !u(k)( )

*

spin

$

We can now focus on doing the sum over spins for this tensor!

Summary of where we are

M = !e2u ( "k )# µ

u(k)$% &' u ( "p )# µu(p)$% &'t

M2=e4

t2Lelectron

µ(Lmuon

µ(

Lelectronµ(

=1

2u ( "k )# µ

u(k)$% &' u ( "k )# (u(k)$% &'

*

e! spins)

Lmuonµ(

=1

2u ( "p )# µ

u(p)$% &' u ( "p )# (u(p)$% &'

*

muon! spins)

All that’s left to do is the sum over spins.

u ( !p )" #u(p)$% &'

*

= ( ) 4x4( )

(

)

****

+

,

----

$

%

.

.

.

.

&

'

////

#

= 1x1( )#

Note the structure of this expression:

The complex conjugate of this object is thus identical to the hermitian conjugate of this !!!

We can use the latter in order to rearrange terms,while ignoring the lorentz index for the moment.

u ( !k )" #u(k)$% &'

T *

= uT *( !k )" 0" #

u(k)$% &'T *

= uT *(k)" #T *" 0

u( !k )$% &' = u (k)" #u( !k )$% &'

Where in the last step , we used the commutation properties.To be explicit:

At this point we get the electron tensor:

Lelectron

µ!=1

2u ( "k )# µ

u(k)$% &' u (k)# !u( "k )$% &'

s, "s(

Next, we are going to make all summations explicit, by writing out the gamma-matrices and spinor-vectors as components.

! 0! "#$ %&T *

= ! "T *! 0= '! "! 0

= ! 0! "

! 0#$ %&T *

= ! 0

! "#$ %&T *

= '! "

for ν=1,2,3

Lelectronµ!=1

2u ( "k )# µ

u(k)$% &' u (k)# !u( "k )$% &'

s, "s(

=1

2ui

"s( "k )# ij

µuj

s(k)$% &'

ijlm

(s, "s( ul

s(k)# lm

!um

"s( "k )$% &'

=1

2# ij

µ# lm

!

ijlm

( ui"s( "k )um

"s( "k )$% &'

"s( ul

s(k)uj

s(k)$% &'

s

(

Here we now apply the completeness relations:

us(p)u

s(p)

s

! = pµ"µ + m = 4x4( )

See H&M exercise 5.9 for more detail on completeness relation.

Lelectronµ!

=1

2" ij

µ" lm

!

ijlm

# ui$s( $k )um

$s( $k )%& '(

$s# ul

s(k)uj

s(k)%& '(

s

#

us(p)u

s(p)

s

! = pµ"µ + m = 4x4( )

Now apply to this the relationship:

And you get as a result:

Lelectronµ!=1

2"k#$

#+ m%& '(mi $ ij

µk#$

#+ m%& '( jl $ lm

!

ijlm

)

Mathematical aside:

Let A,B,C,D be 4 matrices.

AijB jlClmDmi ! Tr A " B "C "D( )ijlm

#

This weird sum is thus nothing more than the trace of the product of matrices !!!

I won’t prove this here, but please feel free to convince yourself.

Lelectron

µ! =1

2Tr "k#$

# + m( )$ µk%$

% + m( )$ !&' ()

Lelectronµ!=1

2"k#$

#+ m%& '(mi $ ij

µk#$

#+ m%& '( jl $ lm

!

ijlm

)

Lelectron

µ!=1

2Tr " k #$

# + m( )$ µk%$

% + m( )$![ ]

Lmuon

µ!=1

2Tr " p #$

# + m( )$ µp%$

% + m( )$![ ]

kA

pB

k’C

p’D

electron

muon muon

electron

M2

=e4

t2Lelectron

µ!Lmuon

µ!

“All” that’s left to do is apply trace theorems.(There’s a whole bunch of them in H&M p.123)

Lelectron

µ!=1

2Tr " k #$

# + m( )$ µk%$

% + m( )$![ ]

=1

2Tr " k #$

#$ µk%$

%$! + " k #$#$ µ$!

m + m$ µk%$

%$! + m2$ µ$![ ]

Trace of product of any 3 gamma matrices is zero!

Lelectron

µ!=1

2Tr " k #$

#$ µk%$

%$![ ] +m2

2Tr $ µ$![ ]

Next, define unit vectors a,b for µ and ν coordinate:

a!"!# " µ

b$"$# "%

This will allow us to use trace theorems to evaluate the remaining traces.

Aside• What does this mean?

a!"!# " µ

b$"$# "%

a!"!

Is a scalar product of 4-vectors.As al is a unit vector, it projects out a component of γl . The components of γl are themselves 4x4 matrices.

Tr abcd[ ] = 4 a !b( ) c !d( ) " a ! c( ) b !d( ) + a !d( ) b ! c( )#$ %&

We introduce this to be able to use this trace theorem:

Tr ( ! k "#")(a$#

$)(k%#

%)(b&#

&)[ ] =

= 4 ( ! k ' a)(k ' b) ( ( ! k ' k)(a ' b) + ( ! k ' b)(k ' a)[ ]

= 4 ! k µk) ( ( ! k ' k)g

µ) + kµ ! k

)[ ]

Tr (a$#$)(b&#

&)[ ] = 4a ' b = 4g

µ)

2 Trace Theorems to use:

(1) (2) (3) (4)

(1x2) (3x4) - (1x3) (2x4) + (1x4) (2x3)

Now put it all together …

M2

=e4

t2

Lelectron

µ!L

muonµ!

=8e

4

t2

" k " p ( ) kp( ) + " k p( ) k " p ( ) # m2p " p # M

2k " k + 2M

2m2[ ]

Electron - Muon scattering

Lelectron

µ! = 2 " k µk! + (m2

# " k $ k)gµ! + k

µ" k ![ ]

Lµ!

muon = 2 " p µ p!

+ (M 2# " p $ p)gµ! + pµ

" p ![ ]

This is the “exact” form. Next look at relativistic approx.

gµ!gµ!= 4Aside:

M2

=8e

4

t2

! k ! p ( ) kp( ) + ! k p( ) k ! p ( )[ ]

Relativistic approx. for e-mu scattering:

Let’s use the invariant variables:s = (k+p)2 ~ 2kp ~ 2k’p’u = (k - p’)2 ~ -2kp’ ~ -2k’p

M2

= 2e4s2

+ u2( )

t2

Next look at ee -> mumu, and get it via crossing.

kA

pB

k’C

p’D

electron

muon muon

electronkA pB

k’C p’D

electron muon

muonelectron

emu -> emu => ee -> mumu via crossing

k’C - pB

s = (k+p)2

t = (k’ - k)2 “s” = (k’ - k)2 = t“t” = (k+p)2 = s s t

e-mu- -> e-mu- => e+e- -> mu+mu-

M2

= 2e4s2

+ u2( )

t2

M2

= 2e4t2

+ u2( )

s2

d!

d" cm

=M

2

64#2s

pf

pi

Recall exercise 4.2 from H&M:

We will now use this for relativistic e+e- -> mu+mu- scattering.

pf ~ pi =>

t = -2 k2 (1 - cosθ)u = -2 k2 (1 + cosθ)s ~ 4k2

α = e2 /4π

d!

d"cm

#M

2

64$2s

M2

= e41+ cos

2!( )=>

d!

d"cm

#$2

4s1+ cos2%( )

!(e+e&' µ +µ&

) #4($

2

3s

Relativisticlimit

(because we can neglect masses)

(see next slide)

Aside on Algebrat = -2 k2 (1 - cosθ)u = -2 k2 (1 + cosθ)s ~ 4k2

t2+ u

2( )s2

=4k

41! 2cos" + cos

2" +1+ 2cos" + cos2"#$ %&

16k4

t2+ u

2( )s2

=1+ cos

2"( )2

M2

= 2e4t2

+ u2( )

s2

M2

= e41+ cos

2!( )

Result worthy of discussion1. σ ∝ 1/s must be so on dimensional grounds2. σ ∝ α2 two vertices!3. At higher energies, Z-propagator also

contributes:

More discussion4. Calculation of e+e- -> q qbar is identical as

long as sqrt(s) >> Mass of quark.

!(e+e"# qq ) = 3 $ eq

2!(e

+e"# µ +µ"

)q" flavor

%

# of flavorsCharge of quark

Sum over quark flavors

Measurement of this cross section was very important !!!

Measurement of R

R =!(e

+e"# qq )

!(e+e"# µ +µ"

)= 3 $ eq

2

q" flavor

%

Below charm threshold: R = 3 [ (2/3)2 + (1/3)2 + (1/3)2 ] =2

Between charm and bottom: R = 2 + 3(4/9) = 10/3

Above bottom: R = 10/3 + 3(1/9) = 11/3

Measurement of R was crucial for:a. Confirm that quarks have 3 colorsb. Search for additional quarksc. Search for additional leptons

Experimental Result

Ever more discussion5. dσ/dΩ ∝ (1 + cos2θ)

5.1 θ is defined as the angle between e+ andmu+ in com. cos2θ means that the outgoingmuons have no memory of the direction ofincoming particle vs antiparticle.Probably as expected as the e+e- annihilatebefore the mu+mu- is created.

5.2 Recall, phase space is flat in cosθ. cos2θdependence thus implies that the initial stateaxis matters to the outgoing particles. Why?

Helicity Conservation in relativistic limit

• You showed as homework that uL and uR arehelicity eigenstates in the relativistic limit, andthus:

• We’ll now show that the cross terms are zero,and helicity is thus conserved at each vertex.

• We then show how angular momentumconservation leads to the cross section wecalculated.

u ! µu = u

L+ u

R( )! µu

L+ u

R( )

u ! µu = u

L+ u

R( )! µu

L+ u

R( )

u L

= uL

T *! 0 = uT * 1

21" ! 5( )! 0 = u

1

21+ ! 5( )

uR

=1

21+ ! 5( )u

u L! µ

uR

= u 1

41+ ! 5( )! µ

1+ ! 5( )u = u ! µ 1

41" ! 5( ) 1+ ! 5( )u = 0

Let’ do one cross product explicitly:

! 5! µ= "! µ! 5

! 5 = ! 5T *

! 5! 5 =1

Here we have used:

Helicity conservation holds for allvector and axialvector currents as E>>m.

• eL- eR

+ -> muL- muR

+ Jz +1 -> +1• eL

- eR+ -> muR

- muL+ Jz +1 -> -1

• eR- eL

+ -> muL- muR

+ Jz -1 -> +1• eR

- eL+ -> muR

- muL+ Jz -1 -> -1

• Next look at the rotation matrices:

d11

1(!) =

1

21+ cos!( ) "

#u

s

d#1#1

1(!) =

1

21+ cos!( ) "

#u

s

d#11

1(!) =

1

21# cos!( ) "

#t

s

d1#1

1(!) =

1

21# cos!( ) "

#t

s

Cross products cancel inSpin average:

M2

! 1+ cos2"( )

dJ1-1

Initial Jz final Jz

Conclusion on relativistic limit• Dependence on scattering angle is given

entirely by angular momentum conservation !!!• This is a generic feature for any vector or

axialvector current.• We will thus see the exact same thing also for

V-A coupling of Electroweak interactions.