Physics 214 UCSD/225a UCSB Lecture 12 • Halzen & Martin ... › 2007 › Fall ›...

Physics 214 UCSD/225a UCSB

Lecture 12

• Halzen & Martin Chapter 6

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(e2)

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#µuB )

pA " pC( )2

"(u D#

µuA )(u C#µuB )

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 ininitial and final state.

• In many experimental situations, in particularin 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

Spin Averaging in non-relativistic limit

• Incoming e- :

• Outgoing e- :

!

" 0 =I 0

0 #I

$

% &

'

( )

" j=1,2,3=

0 * j=1,2,3

#* j=1,2,3 0

$

% &

'

( )

!

u f "µui =2m if (µ = 0)# si = sf

0 otherwise

$

% &

' &

!

us= +1/ 2( ) = 2m

1

0

0

0

"

#

$ $ $ $

%

&

' ' ' '

u s=(1/ 2( ) = 2m 0 1 0 0( )

Reminder:

Invariant variables s,t,uExample: e- e- -> e- e-

• s = ( pA + pB )2

• = 4 (k2 + m2)

• t = ( pA - pC )2

• = -2 k2 (1 - cosθ)

• u = ( pA - pD )2

• = -2 k2 (1 + cosθ)

k = |ki|= |kf| m = me θ = scattering angle, all in com frame.

A

B

C

D

M for Different spin combos

!

M = "e2(u

c# µ

uA)(u

D#µu

B)

t"(u

D# µ

uA)(u

C#µu

B)

u

$

% &

'

( )

(u c# µ

uA)(u

D#µu

B)[ ]*+,*+

= 4m2

(u D# µ

uA)(u

C#µu

B)[ ]+*,*+

= 4m2

(u c# µ

uA)(u

D#µu

B)[ ]+*,*+

= 0

etc.

!

M2

=1

44m

2e2( )2

21

t"1

u

#

$ %

&

' ( 2

+1

t2

+1

u2

)

* +

,

- .

Let’s do one relativistically

• Pick the easiest one: e- mu- -> e- mu-– Easiest because it has only one diagram !!!

!

M = "e2

u c ( # k )$ µuA (k)[ ] u D ( # p )$µuB (p)[ ]

t

M2

=e4

t2"k"[ ]

µ" p"[ ]

µ"k"[ ]

%" p"[ ]

%( )&

=e4

t2"k"[ ]

µ%" p"[ ]

µ%

This is a sensible way to proceed because it groupselectron and muon parts together into one tensor each.We then worry only about spin averaging those tensors.After all, only same flavor spinors appear at the same vertex,and all reductions of spin averages happen at a vertex.

!

M = "e2

u c ( # k )$ µuA (k)[ ] u D ( # p )$µuB (p)[ ]

t

M2

=e4

t2

Lelectron

µ%L

muonµ%

Lelectron

µ%=1

2u c ( # k )$ µ

uA (k)[ ] u c ( # k )$% uA (k)[ ]*

e"spins

&

Lmuon

µ%=1

2u D ( # p )$ µ

uB (p)[ ] u D ( # p )$% uB (p)[ ]*

muon"spins

&

We will now look at the electron tensor only, and try to simplify!

!

u D ( " p )#$ uB (p)[ ]*

= ( ) 4x4( )

%

&

' ' ' '

(

)

* * * *

+

,

- - - -

.

/

0 0 0 0

$

= 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 c"# u

A[ ]T *

= uC

T *" 0"# uA[ ]

T *

= uA

T *"#T *" 0uC[ ] = u

A(k)"# u

C( $ k )[ ]

Where in the last step , we used the commutation properties.

At this point we get the electron tensor:

!

Lelectron

µ"=1

2u

c( # k )$ µ

uA(k)[ ] u

A(k)$" u

C( # k )[ ]

s, # s

%

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

!

Lelectron

µ"=1

2u c ( # k )$ µ

uA (k)[ ] u A (k)$"uC ( # k )[ ]

s, # s

%

=1

2u i

# s ( # k C )$ ij

µu j

s(kA )[ ]

ijlm

%s, # s

% u ls(kA )$ lm

"um

# s ( # k C )[ ]

=1

2$ ij

µ$ lm

"

ijlm

% u i# s ( # k C )um

# s ( # k C )[ ]

# s

% u ls(kA )u j

s(kA )[ ]

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.

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( )%"[ ]

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% µ

k&%&%" + m

2% 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.

!

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

%)(k&$

&)(b'$

')[ ] =

= 4 ( " k ( a)(k ( b) ) ( " k ( k)(a ( b) + ( " k ( b)(k ( a)[ ]

= 4 " k µk* ) ( " k ( k)gµ* + k

µ " 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.

!

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’Cp’D

electron muon

muonelectron

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

k’C - pB

s = (k+p)2

t = (k’ - k)2 t = (k+p)2

s = (k’ - k)2 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

= 2e48k

41+ cos2"( )16k

4=>

!

d"

d#cm

$% 2

4s1+ cos2&( )

"(e+e'( µ+µ'

) $4)% 2

3s

Relativisticlimit

Result worthy of discussion

1. σ ∝ 1/s must be so on dimensional grounds

2. σ ∝ α2 two vertices!

3. At higher energies, Z-propagator alsocontributes:

More discussion

4. Calculation of e+e- -> q qbar is identical aslong 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 discussion

5. 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$11(") =

1

21+ cos"( ) #

$u

s

d$111(") =

1

21$ cos"( ) #

$t

s

d1$1

1(") =

1

21$ cos"( ) #

$t

s

Cross products cancel inSpin average:

!

M2

" 1+ cos2#( )

dJ1-1

initial final Jz

Conclusion on relativistic limit

• Dependence on scattering angle is givenentirely by angular momentum conservation !!!

• This is a generic feature for any vector oraxialvector current.

• We will thus see the exact same thing also forV-A coupling of Electroweak interactions.