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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. Useful Sections in Martin & Shaw:. Chapter 9, Chapter 10. Regeneration. - PowerPoint PPT Presentation
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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!)