Lecture 12: Electroweak • Kaon Regeneration & Oscillation

• The Mass of the W

• The Massless Photon & Broken Symmetry

• The Higgs

• Mixing and the Weinberg Angle

• The Mass of the Z

• Z Decay

Chapter 9, Chapter 10

Useful Sections in Martin & Shaw:

KL + K

S

So what are kaons??? that depends... who wants to know?!

Ko, Ko states of definite strangeness

K1

o, K2

o states of definite CP

KS

o, KL

o states of definite lifetime

KL

strong interaction withmatter picks out Ko & Ko

which then re-mix

KL + K

S

So what are kaons??? that depends... who wants to know?!

Ko, Ko states of definite strangeness

K1

o, K2

o states of definite CP

KS

o, KL

o states of definite lifetime

KL

''Regeneration"

strong interaction withmatter picks out Ko & Ko

which then re-mix

Strangeness Oscillation:

Amplitudes for decaying states KS

o and KL

o as a function of time are

AS(t) = A

S(0) exp(im

St) exp(

St/2)

S ℏ/

S

AL(t) = A

L(0) exp(im

Lt) exp(

Lt/2)

L ℏ/

L

K1

o = 1/2 ( Ko + Ko )

AK(t) = 1/2 ( A

S(t) + A

L(t) )

AK(t) = 1/2 ( A

S(t) A

L(t) )

K2

o = 1/2 ( Ko Ko )

Ko = 1/2 ( K1

o + K2

o ) Ko = 1/2 ( K1

o K2

o )

or

≃ 1/2 ( KS

o + KL

o ) ≃ 1/2 ( KS

o KL

o )

Thus, if we start with a pure Ko beam at t=0, the intensity at time t will be

I(Ko) = 1/2 [AS(t) + A

L(t)][A

S*(t) + A

L*(t)]

= 1/4 {exp(St) + exp(

Lt) + 2 exp[(

S+

L)t/2] cosmt }

(setting AS(0) = A

L(0) = 1)

and similarly,

= 1/4 {exp(St) + exp(

Lt) 2 exp[(

S+

L)t/2] cosmt }

I(Ko) = 1/2 [AS(t) A

L(t)][A

S*(t) A

L*(t)]

where m mLm

S

= 3.49x1012 MeV (m/m ≃ 7x1015)

Ko

Ko

A B C D EF G H I J L M N OP Q R S TU V X Y

Recall that the ''matrix element" for scattering from a Yukawa potential is

f V

o = g2/(q2+M2)

In the Fermi theory of decay, this is what essentially becomes GF

or, more precisely,G

F/2 = g2/(q2+M2) = 4

W/(q2+M2)

GF

2 and the relatively small value of GF characterizes

the fact that the weak interaction is so weak

We can get this small value either by making W

small or by making M large

So what if we construct things so W

= ??? UNIFICATION !!

Assuming M ≫ q2 , M = 4 2 / GF

= 1/137G

F = 105 GeV2 M ~ 100 GeV

CERN, 1983

MW

= 80 GeV !!

uud

uud

hadrons

hadrons

W-

e-, -, -

e, ,

Electron Cooling

Stochastic Cooling

p

p

A Brief Theoretical Interlude

(electroweak theory... at pace!!)

But how can this be the ''same" force when the W’s are charged and the photon certainly isn’t !?

Is there a way we can ''bind up" the W’s along with a neutralexchange particle to form a ''triplet" state (i.e. like the pions) ??

Well, like with the pions, we seem to have a sort of ''Weak" Isospinsince the weak force appears to see the following left-handed doublets

e

e( )L

( )L

( )L

ud( )

L

cs( )

L

tb( )

L

as essentially two different spin states: IW

(3) = 1/2 (like p-n symmetry)

Thus, in the process

e e

W+The W+ must carry away +1 units of I

W(3)

so let’s symbolically denote W+

If IW

= 1 for the W’s then, similar to the o, there is also a neutral state:

Wo 1/2 ( ) (which completes the triplet)

and, similarly, W

There is, however, another orthogonal state: 1/2 ( )

If we ascribe this to the photon, then perhaps we might expectto see weak ''neutral currents" associated with the exchange ofa Wo with a similar mass to the W

Hold on... any simple symmetry is obviously very badly broken the photon is massless and the W’s are certainly not!The photon is also blind to weak isospin and also couples to right-handed leptons & quarks as well

Assume the symmetry was initially perfect and all states were massless

Then postulate that there exists some overall (non-zero) ''field" whichcouples to particles and gives them additional virtual loop diagrams :

(kind of like an ''aether" which produces a sort of ''drag")

Higgs Mechanism

so we’d have a nice''single package" which describes EM and weak forces!

but in the limit of zero momentum transfer (rest mass), so represent as

Further suppose that this field is blind to weak isospin and, thus, allows for it’s violation.

This would allow the neutral weak isospin states to mix like with the mesons (the W are charged and cannot mix)

We will call the ''pure," unmixed states Wo and

And we will call the physical, mixed states Zo and

Think about mathematically introducing this Higgs coupling by applying some ''mass-squared" operator to the initial states (since mass always enters as the square in the propagator)

where the right-most terms represent the weak isospin - violating terms

For the W the mass would then simply be given by MW

2 = GW

2

(where G2 contains the coupling plus a few other factors)

For the latter 2 equations, we can think of M2 as an operatorwhich yields the mass-squared, M2 , for the coupled state:

M2 Wo = GW

2 Wo + GW

G

M2 = G2 + G

W G W

Assume couplings to W’s are all the same (GW

) but coupling to may be different (G)

M2 W = W + WGW

GW

M2 W = W + W+ G

W G

WG

W G

M2 = + + Wo G G

G GW

From the second of these: = Wo G

W G

(M2-G2)

Substituting into the first: M2 Wo = GW

2 Wo + Wo G

W2 G

2

(M2-G2)

M4 M2 G2 = M2 G

W2 G

W2 G

2 + GW

2 G2

M2 ( M2 G2 G

W2) = 0

M2 = 0 or M2 = GW

2 + G2

Thus, associate M2 = 0 and M

Z2 = G

W2 + G

2

Note also that MZW

We can parameterize the as a mixture of Wo and as follows:

sinW

Wo cos

W

W ''Weinberg Angle"

Thus, applying M2 : M2 = M2 ( sinW

Wo cosW

) = 0

0 = ( G2

+ G

W GW ) sin

W (G

W2 Wo G

W G

cosW

Coefficient of Wo GW

GsinW

GW

2 cosW

= 0

Coefficient of G2sin

W G

W G

cosW

= 0

MZ

2/MW

2 = (GW

2 + G2)/G

W2 = 1/cos2

W

MZ = M

W/cos

W

tan W

= G/ GW''unification condition"

Ql + 3Q

q = 0

''anomaly condition"

(leptons) (quarks)

is satisfied separately for each generation

p

Neutral Current Event (Gargamelle Bubble Chamber, CERN, 1973)

From comparing neutral and charged current rates

sin2W

= 0.226

MW

= 80 GeV

MZ = 91 GeV (observed!!)

Z e+ e

MZ = 91 GeV

(predicted)

While we’re here...

So, consider the coupling to the Z0 :

Z0

u

u

Z0

(d cosC + s sin

C)

(d cosC + s sin

C)

+

pre-ABBA weak doublet = ud cos

C + s sin

C( )u

d´ =( )

Probability ∝ product of wave functions:

S = 0 S = 1“Flavour-Changing Neutral Currents” never seen!

uu + (dd cos2C + ss sin2

C) + (sd + ds ) sin

C cos

C

Postulate 2 doublets:

ud cos

C + s sin

C( )u

d´ =( )

S = 0 S = 1

cs cos

C d sin

C( )cs´

=( )&

Z0

c

c

Z0

(s cosC d sin

C)

(s cosC d sin

C)

++

Z0

u

u

Z0

(d cosC + s sin

C)

(d cosC + s sin

C)

+

uu + cc + (dd+ss)cos2C + (ss+dd) sin2

C) + (sd + ds - sd - sd) sin

C cos

C

(Glashow, Iliopolis & Maiaini: “GIM” mechanism)

(recall = ℏ/)Blam !

W = vB / V

But recall that

Transition Rate ( )W =

dP

dN0

prob for decay toparticular final stategiven the total numberof available states

dP 1 f

dE 2 (E-E

0)2 + 2/4

=

=

0

f

q2 (E-E0)2 + 2/4

formation ''rate" of initial state

= 0 ( )dP

dE ( )dE

dN

dNdE( )1 V q2 dq d

(2)3 dE ( )=1

V q2 2 v ( ) 1

But this is non-relativistic!

From considering scattering from a Yukawa potential(which followed from the relativistic Klein-Gordon equation)we found the ''propagator" 1/(q2 + M2)

So consider the diagram:

~ exp(iE0t) = exp(iMt)

exp{i(Mi/2)}t = exp(iMt) exp(t/2)

Also note that, for a decaying state, the intermediate mass takes on an imaginary component M M i /2 since

Under a fully relativistic treatment, q is the 4-momentum transferand, if we sit in the rest frame of the intermediate state, q2 = p2 E2 = E2

Thus, the propagator goes like

1(Mi/2)2 E2

1M2 /4 iM E2

= 1M2 iM E2

(in the limit ≪ M)

≃

And the cross section will be proportional to the square of the propagator :

( )( ) 1M2 iM E2

1M2 iM E2 ~ 1

(E2 M2)2 + M2=

=

0

f

q2 (E-E0)2 + 2/4

compareso, roughly, /2 M

and we’d expectsomething like ~

M2 0

f

E2 (E2-M2)2 + M22

In fact, a full relativistic treatment yields =

M2 0

f

E2 (E2 M2)2 + M22CM CM

(e+e X) = M

Z2 ee X

E2 (E2 MZ

2)2 + MZ

22

CM CM[ ]

Thus, for the production of Z0 near resonance and the subsequent decay to some final state ''X" :

since ee can be related by time-reversal to ee

Peak of resonance MZ

Height of resonance product of branching ratios

ee X

BreeBrX=

Results:

MZ = 91.188 0.002 GeV

Z = 2.495 0.003 GeV l l = 0.0838 0.0003 GeV

hadrons= 1.741 0.006 GeV

1.741 + (3 x 0.0838) = 1.9924 ≠ 2.495 !!

So what’s left ???

''Invisible modes"

Neutrinos !!

(limit for light, ''active" neutrinos)

An End To The Generation Game ???

(not necessarily a bad thing!)