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The Standard Model
Thomas J. LeCompte
High Energy Physics Division
Argonne National Laboratory
The Direction of These
Lectures
 I’m an experimenter
 I will start with simple models, and
build up to the Standard Model
– Models will be progressively
more predictive, but…
– This isn’t strictly chronological
 I will touch on QCD only lightly
 Themes
– What data needs a new model to explain it?
– Where might the Standard Model be wrong?
• And how could we tell?
I’ll (over)emphasize tests that might be performed at the LHC.
2
Part I: Quantum Electrodynamics
A Theory with Just One Parameter
3
Local Gauge Invariance – Part I
 In quantum mechanics, the probability density is the square of the
wavefunction: P(x) = |Y|2
– If I change Y to –Y, anything I can observe remains unchanged
 P(x) = |Y|2 can be perhaps better written as P(x) = YY*
– If I change Y to Yeif anything I can observe still remains unchanged.
– The above example was a special case (f = p)
 If I can’t actually observe f, how do I know that it’s the same everywhere?
– I should allow f to be a function, f(x,t).
– This looks harmless, but is actually an extremely powerful constraint
on the kinds of theories one can write down.
4
Local Gauge Invariance – Part II
 The trouble comes about because the Schrödinger equation
(and its descendants) involves derivatives, and a derivative
of a product has extra terms.
d
dv
du
uv  u  v
dx
dx
dx
 At the end of the day, I can’t have any leftover f’s – they all have to
cancel. (They are, by construction, supposed to be unobservable)
 If I want to write down the Hamiltonian that describes two electrically
charged particles, I need to add one new piece to get rid of the f’s: a
massless photon.
5
Local Gauge Invariance & QED
 Add a phase to the electron
wavefunction
 The derivative brings a phase out
front, but…
 There has to be an extra term in
the covariant derivative
representing the photon field
 And the photon field has to
transform properly.
6
Why Massless?
 A massive spin-1 particle has three spin states
(m = 1,0,-1)
 A massless spin-1 particle has only two.
– Hand-wavy argument: Massless particles
move at the speed of light; you can’t boost to
a frame where the spin points in another
direction.
 To cancel all the f’s, I need just the two m = ± 1
states (“degrees of freedom”)
– Adding the third state overdoes it and
messes up the cancellations
– The photon that I add must be massless
m = ±1 “transverse”
m = 0 “longitudinal”
Aside: this has to be just
about the most confusing
convention adopted since
we decided that the current
flows opposite to the
direction of electron flow.
We’re stuck with it now.
7
U(1), Lie Algebras and All That
 The key abstraction is the same equations have the same solutions
– “You’ve seen one U(1) theory, you’ve seen them all”
 U(1) stands for “unitary group of degree one”
– We used to call those things “numbers”
– The U(1)-ness comes in at the qAm term
• Here the charge is just a number
 Instead of a number, we can make charge a matrix
– e.g. QCD color: {red, blue, green}
– Then the SU(3)-ness manifests as matrix multiplication: QaAam
 SU(2) is the same group that governs the algebra of angular momentum
– Why we have “spin” and “isospin”
– Again, the same equations have the same solutions
8
Fermions – The Red-Headed Stepchildren of SM Talks
 e  m   
   
 e  m   
 u  c  t 
   
 d  s  b 
   
 12 massive spin-½ particles
– 3 charged leptons
– 3 neutral leptons
– 6 colored quarks
 Masses vary by 11-13 orders of
magnitude
 There are three families – nobody
knows why
– There is no evidence whatever that
the families are as ordered in the
left. (e.g. the tau could be the
partner of the u & d quarks)
9
Why Are Atoms Neutral?
 Why are atoms neutral?
– Or, why is q(p) = -q(e)
– Or, why do quarks and leptons have charges
related by small integer ratios?
 This is not an accident
– One needs to avoid “anomalies” – quantum
mechanical inconsistencies (e.g infinities) in the
calculations
– The troublesome terms have simple algebraic
relations between the charges (e.g. Sq) as
coefficients
• Make the coefficients zero, and all is well.
e




e


u u u 
B
G
 R

d d d 
B
G
 R
1




0


 3 2 / 3 


 3  1 / 3


10
QED
 The Good:
– Incredible precision: ge = 2.00231930436 (and every digit is significant)
 The Bad:
– Doesn’t explain nuclear b decay
– Inconsistent at very high energy
• So-called “Landau Pole”
• Very high energy means E > m(visible universe)
The next step: adding a theory of weak interactions that doesn’t spoil
what has already been accomplished…
11
Part II: Fermi’s Four Fermion Theory
A Theory with Two Parameters
12
Four Fermion Theory
 The idea is just like the name says
– Four fermions couple directly
– The coupling strength is GF
GF  1.16637 105 GeV -2 

2
1


2  246 GeV 
2
e
m
e
m
 I write it this way (rather than 292 GeV) because this number will be
important later – and I want you to remember where it came from.
 The seeds of the theory’s destruction are already planted
…but let’s talk about the theory’s successes before talking about its failures.
13
Some successes
 Nuclear b decay rates should go as ~Q5
– Observed in 1933 as “Sargent’s Rule”.
– Explained by simple power counting
• Any decay rate ~ width G ~ dimensions of energy
• The 4-fermion interaction has 2 powers of GF (and thus four
powers of energy) in the denominator
• Therefore you need five powers of Q in the numerator
 The same GF that explains nuclear b decays explains
– m decay
– p decay
–  interactions & cross-sections
14
The Pion Decay: Evidence of handedness (chirality)
p
(Anti-)
Lepton
Neutrino
 The pion is spin-0, so the daughters must have the same helicity
– The decay rate depends on whether the interaction can couple to
both left and right handed electrons, or only one
Only one

Both
m m  m
2 2G
p
G  fp

p 
mp3
2
F
2
l
2
 
2 2
l
B(p  m )
 7800
B(p  e )


2
2

m

m
2
G
l
G  fp2
 p
p 
mp
2
F
 
2

B(p  m )
 0.182
B(p  e )
Measurement: 8100
15
Which Handedness?
 If the weak interaction is left-handed (V-A)
– The neutrino is left-handed
– The anti-lepton has left-handed helicity
• But the weak interaction needs to couple
to the right-handed chiral projection
• This costs us a ml/mp in the amplitude
(squared in the rate)
 Suppose the weak interaction were right-handed (V+A)
– Now the neutrino is right-handed
– The anti-lepton has right-handed helicity
• But the weak interaction needs to couple to the left-handed chiral projection
• This still costs us a ml/mp in the amplitude (again, squared)
The pion branching fractions tell us that there is a handedness to the weak
interaction. But it doesn’t tell us which hand it is.
16
Measuring the Neutrino Helicity
 K-capture on the right nucleus has
the helicity of the daughter nucleus
match the neutrino’s.
 Next, the daughter emits a photon
– Carries the spin of the parent
– Transfers the neutrino spin to the photon
M. Goldhaber, 1958
152
63
*
Eu  e- 152
Sm

62
0 
152
62
1
1
 1 
2
2
Sm* 152
62 Sm  
1  0   1
The essential idea is that the neutrino’s helicity (hard to measure) is transferred
to the photon (easy – well, easier – to measure).
Outcome: the weak interaction is left handed.
17
Strangeness and the Cabibbo Angle
 Kaon decays have all the same
qualitative properties as other weak
decays
– Except they proceed ~30x slower
 N. Cabibbo explained this by
theorizing:
– 1. The weak interaction only allows
transitions within a doublet
– 2. The weak interaction
eigenstates are rotated with
respect to the mass eigenstates
 The s-quark mass eigenstate contains
only ~3% dW, so these decays are
suppressed by this factor of 30.
B( K  m )
 41000
B( K  e )
For the same reason – the
charged lepton has the “wrong”
helicity. Experiment agrees.
 dW   cos C
   
 sW    sin C
sin C  d 
 
cos C  s 
 C  13o
Weak interactions are universal – the
same force that governs nuclear b
decay governs heavy flavor decays.
18
Kobiyashi & Maskawa: Extending this to Three Families
 dW   cos C
   
 sW    sin C
Vud Vus Vub 


Vcd Vcs Vcb  
V

 td Vts Vtb 
 cos 1

 sin 1 cos  2
 sin  sin 
1
2

sin C  d 
 
cos C  s 
 dW  Vud Vus Vub  d 
  
 
 sW   Vcd Vcs Vcb  s 
 b  V
 b 
V
V
ts
tb  
 W   td
Can be expressed in terms of three angles and
one phase – the 9 terms are not independent
 sin 1 cos 3
cos 1 cos  2 cos 3  sin  2 sin 3ei
cos 1 sin  2 cos  3  cos  2 sin 3ei
 .974 .227 .004 


  .227 .973 .042 
 .008 .042 .999 


 sin 1 sin 3


i
cos 1 cos  2 sin  3  sin  2 cos 3e 
cos 1 sin  2 sin  3  cos  2 cos 3ei 
Aside: the phase here gives
rise to CP violation. Three is
the minimum number of
families for this to happen.
19
Cabibbo, Kobiyashi & Maskawa Matrix II
 .974 .227 .004 


  .227 .973 .042 
 .008 .042 .999 


Numbers don’t give me a very good intuition
for what’s going on
Here the shading reflects the magnitude
of the components: black = 1 and white = 0.
 Because the CKM matrix appears squared in any observable, it acts even
more like a diagonal matrix
– The weak interaction apparently does not like to cross family
boundaries
20
Cross-sections
 A cross-section has units of area: length2 (or energy-2)
 Dimensional analysis tells us any exclusive cross-section must
(eventually) fall as 1/s, i.e. 1/E2.
What are we to make of the following
data? (The cross-section for p-p
scattering)
The cross-section is actually growing
with energy.
Energy
21
Extending The Energy
 The cross-section grows because it’s the left side of a resonance, the
D(1232)
 The high energy behavior of elastic scattering is OK: 1/E2
 The total high energy behavior is also OK – it’s the sum of many channels
Energy
22
Lessons for the Weak Interaction
 Any weak cross-section must grow as E2 and not 1/E2.
 Any cross section has two powers of GF (1/Energy4), so it needs two
additional powers of E to get the right units.
– This doesn’t depend on any of the details of the calculation, or even
on field theory
– It’s simply dimensional analysis
 That means that the theory will get into trouble – and violate unitarity –
somewhere around 300 GeV.
1
 292 GeV
GF
23
Fermi Four-Fermion Theory Scorecard
 The Good:
– Explains b decay, weak decays and neutrino interactions in detail
• Surprise – it’s a chiral theory
– Left and right handed fermions behave differently
– Parity is violated
 The Bad:
– Theory breaks down somewhere before ~300 GeV
1
 292 GeV
GF
The next step: add new physics (the W’s and Z) to the theory to fix it
up past 300 GeV
24
Part III: Glashow-Weinberg-Salam Theory
A Theory with Three Parameters
25
A Troublesome Event
m
m
A charged current event – the outgoing
muon has a different charge than the
incoming neutrino.
A picture of an event from the CERN
Gargamelle bubble chamber.
This incomprehensible plot from 1973
shows “neutral weak currents”.
m
m
A neutral current event – the outgoing
neutrino has the same charge as the
incoming neutrino.
26
Some Experimental Issues
The trick is to distinguish this
from this.
m
m
m
It’s the presence of a long muon track
that distinguishes charged current from
neutral current events. But what if the
muon ends up with relatively little
momentum?
m
m
m
“How long
is long”
becomes a
key
question.
A second issue is “How do you know these are neutrinos? Might they be neutrons?”
That can be addressed by looking at where the interactions occur – are they uniform,
like neutrinos? Or are they near a detector edge, like neutrons?
27
What The Authors Were Probably Thinking Of…
 They weren’t trying to explain neutral
weak currents
– In fact, neutral weak currents were
predicted years before their
discovery.
 They were trying to unify QED
with a theory of weak interactions
– Their theory predicted weak
neutral currents and the W
and Z bosons
28
The Roadmap
 We want QED to come out
– So let’s start by putting it (or
something very much like it) in
 We want a left-handed weak theory to come out
– So let’s start by putting it (or something very much like it) in
– Let’s pick a small group that will let us have charged currents…
– …and at least the possibility of neutral currents
 Once we’ve done this, we will match terms to pull QED out
 What’s left will be our (new) theory of weak interactions
29
First Ingredient: Weak Hypercharge
 This is where the Lie Algebra formalism starts to help us:
– “We start with a U(1)”
• This means that we will end up with a theory just like QED
– Instead of a field Am, we call it (unimaginatively) Bm
– Instead of a charge q, we have a “hypercharge” Y.
– Replace qAm with YBm in the Lagrangian and we’re done.
The Same Equations Have The Same Solutions
30
Second Ingredient: Weak Isospin
 In this case, our gauge group is SU(2)
– This is the same algebra that governs angular momentum addition
– It’s a non-Abelian group – the fields themselves carry weak isospin
 We have three fields: w1, w2 and w3.
– They interact among themselves in this way
w2
w1
w1
w1
w3
w2
w2
31
Interaction Between these Fields and Matter
Weak Hypercharge
Weak Isospin


 g1 Y m
1 m1
2 m2
3 m3 
 2 jm B  g 2 jm w  jm w  jm w 
 Here g1 and g2 are the coupling constants for hypercharge and isospin
– Right now, they are arbitrary (the ½ is there by convention)
 The j’s are the fermion currents
 Our next step is to declare electric charge to be: Q = ½Y + T3.
– Essentially, this specifies which w is neutral and which are charged
– The absolute magnitude of charge is rolled up in the g’s
– I haven’t really lost any generality by doing this
32
A Change of Basis
jm  jm1  ij m2
We should now cast everything in terms
of electric charge – we want QED to fall out of this:


1
Wm 
w1m  iwm2
2



 g1 Y m
1 m1
2 m2
3 m3 
 2 jm B  g 2 jm w  jm w  jm w  
Then becomes
 g1 Y m
 1  m 1  m
3 m 3 
jm W 
j mW  j m w   
 2 jm B  g 2 
2
 2


g1 Y m 
 g2  m  g2

m
3 m3
 2 jm W  2  jmW  g 2 jm w  2 jm B 


Positive
Negative
Neutral
33
Charged Currents: What Exactly Have We Done?

m
We’ve replaced the 4-fermion
interaction with two vertices
where the fermion current
couples to a W field.
e

m
W

We just
calculated this
vertex factor: it’s
ig 2
 m 1   5 
2 2

m
W
e

Now we have everything we need to
calculate the same process in the old 4fermion theory and the new GWS theory. If
we do this, and match the results, we get:
2  g2

GF 
8  MW



2
34
Neutral Currents
We have no guarantee that the w3 and
B are the physical fields (in fact, they
aren’t) so we introduce a rotation
matrix to mix them.
g1 Y m 

3 m3
 g 2 jm w  2 jm B  
 Am   cos  w
   
 Z   sin 
w
 m 
sin  w  Bm 
 3 
cos  w  wm 
(in the new basis)
g1
g1


3
Y m
3
Y m
g
sin

j

cos

j
A

g
cos

j

sin

j
W m
W m
2
W m
W m Z
 2

2
2



35
Neutral Currents
We have no guarantee that the w3 and
B are the physical fields (in fact, they
aren’t) so we introduce a rotation
matrix to mix them.
g1 Y m 

3 m3
 g 2 jm w  2 jm B  
 Am   cos  w
   
 Z   sin 
w
 m 
sin  w  Bm 
 3 
cos  w  wm 
(in the new basis)
g1
g1


3
Y m
3
Y m
g
sin

j

cos

j
A

g
cos

j

sin

j
W m
W m
2
W m
W m Z
 2

2
2



g

EM
m
j
A
e m
But we already know from QED what this
has to be! So we simply match up
terms…
36
More On Neutral Currents
 Matching up terms gives us a relation
between the weak and EM couplings
– They are NOT INDEPENDENT
g 2 sin W  g1 cosW  ge
 I can plug this into the Z term, and get
gW  g 2 
gZ 
cos W
f
f
Z
By matching terms,
we effectively
calculated this
vertex factor: it’s
ig Z
 m cV  c A 5 
2
Note that the weak coupling is larger
than the EM coupling: aweak ~ 1/30 vs.
1/137 for aEM.
The weak force is weak not because
the coupling is small, but because the
W is heavy.
37
Enough Math…Time For Some Physics
Fermion
cA
cV
Neutrinos
½
½
Charged Leptons
-½
-½ + 2 sin2w
(-0.04)
Up-type quarks
½
½ - 4/3 sin2w
(0.19)
Down-type quarks
-½
-½ + 2/3 sin2w
(-0.34)
 The magnitude of the axial Z coupling to fermions is universal
– The photon is a vector, so it doesn’t touch the axial piece of the weak
interaction
 The more electric charge a fermion has, the less (vector) weak charge it has
– The photon is “stealing” the charge
38
Some Predictions
2  gw

GF 
8  MW
MZ 



2
 MW  4
MW
cos W
p 2a 2
2GF2
1
sin W
This evaluates to 77.5 GeV.
The measurement is 80.4 GeV.
Using the measured W mass, this evaluates to 91.6 GeV.
The measurement is 91.2 GeV.
Z decays
Calculated BF
Measured BF
Leptons (per flavor)
3.4%
3.4%
Invisible
20.5%
20.0%
All jets
69.1%
69.9%
Bottom quark jets
15.2%
15.5%
All these predictions
use W measured
independently - from
neutrino experiments.
39
The W Boson in Pictures
From the original
discovery (UA1/UA2)
This is a UA2 event
D0
CDF
Missing ET
(neutrino)
Electron momentum
40
The Z Boson In Pictures
UA1
OPAL
DELPHI
Z  qq
Z  ee
CDF
ALEPH
L3
Z  m m 
Z    
41
Additional Consequences of SU(2)

We originally introduced SU(2) because we wanted
the W to be charged (“charged current”). That
means it has to couple to the photon:
W+
W+
This is the physical manifestation of the 3-field
interaction in the unbroken SU(2) theory. The physical
states are projections of w1, w2 and w3.
W 
1
w1  w2 
2
w3
w1
  B cos  w  w3 sin W
w2
Z0
But these aren’t the only projections one can make.
Z   B sin  w  w3 cosW
The theory predicts the W couples to the Z as well.
W+
W+
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W+Z Events
CDF
D0
W electron
Z muon
Z muon
W neutrino
Not only does this theory predict the
existence of these events, the rate of these
events is completely determined.
More on this tomorrow.
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A Word on Symmetries
 This theory is often described as SU(2) x U(1)
 Since electromagnetism also has a U(1) theory, it’s tempting to associate
EM with the above U(1)
If you discipline yourself to always write
– This is wrong
SU(2)L x U(1)Y, you’ll avoid making this mistake.
 The unbroken symmetry is SU(2)Left x U(1)Hypercherge
– We explicitly broke both symmetries when we declared the w3
component to be electrically neutral
– We will discuss another way to break these symmetries tomorrow
 Often, a calculation that looks terrible in the broken symmetry is simple if
you consider the unbroken symmetry basis.
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Fixing the Unitarity Problem…
 The Fermi theory has a
problem with unitarity
violation above 300 GeV
– There must be new
physics below that.
– We added new physics
below that: the W and Z’s
at 80 and 90 GeV
 We’ve fixed that problem
– The cross-sections drop
like they are supposed to.
Compare
This is actually a picture of what happens in the neutral currents,
and I originally argued this for charged currents. The same thing
happens for both– this is simply a clearer plot.
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Fixing the Unitarity Problem…Almost
 Imagine you could collide
beams of W bosons:
WW → WW
 If you calculate this in GWS
theory, it too violates unitarity
– Specifically, one piece:
WLWL → WLWL
– New physics has to enter
in below about 1 TeV
 So we haven’t so much as
solved the problem as moved
it to higher energy.
Potential Troublemaker
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GWS Theory Scorecard
 The Good:
– Matches every test against data we could think of
– Predicts new phenomena, borne out by experiment
• W and Z bosons
• Neutral Weak Currents
• Diboson production at colliders
– Explains everything with just three numbers
• GF, the strength of the weak force, a, the strength of the EM force,
and w, how they mix
– Fixes the 300 GeV Unitarity problem of the Fermi Theory
 The Bad:
– Theory breaks down above ~1 TeV
– Masses put in by hand
Our next step – fixing these problems
• Breaks gauge invariance
by adding one more piece to the theory.
– Symmetry broken by hand
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