Topics in Contemporary Physics
Basic concepts 5
Luis Roberto Flores Castillo
Chinese University of Hong Kong
Hong Kong SAR
January 28, 2015
PART 1
• Brief history
• Basic concepts
• Colliders & detectors
5σ
• From Collisions to
papers
S
ATLA
(*)
15
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s=7
2000
1800
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s=8
ò
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ò Ldt =
-1
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1600
1400
1200
1000
800
600
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400
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2011
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s=7
s=8
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• The Higgs discovery
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150
140
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160
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120
110
• BSM
• MVA Techniques
• The future
L. R. Flores Castillo
CUHK
January 28, 2015
2
… last time: Basic concepts 4
• Review of relativistic kinematics
– Lorentz transformations
– Consequences
– A few examples
– Four-vector notation
• Lorentz transformation in matrix form
• The metric, the scalar product, invariants
• Energy and momentum
– Relativistic collisions
• Examples
– Sticky collision
– Explosive collisions
–…
L. R. Flores Castillo
CUHK
January 28, 2015
3
Reminder: interactions
QED:
QCD:
SM Particle Content
Cabibbo-Kobayashi-Maskawa matrix
Weak:
NO Flavor-Changing-Neutral-Currents
W/Z:
L. R. Flores Castillo
W/Z/γ:
CUHK
January 28, 2015
4
Reminder: Relativistic kinematics
Maxwell
equations
æ
v ö
t ' = g çt - 2 x ÷
è c ø
x ' = g (x - vt)
y' = y
z' = z
Lorentz
transformations
c for all observers
x 0 ' = g (x 0 - b x1 )
Four-vector
x m ' = Lnm xn
x1 ' = g (x1 - b x 0 )
x2 ' = x2
x 0 = ct,
x = x,
1
x3 ' = x3
x 2 = y,
x =z
3
bº
time-position:
proper velocity:
energy-momentum:
v
c
E = g mc2 = mc 2 + 12 mv2 + 83 m vc4 +...
4
contravariant
I º (x 0 )2 - (x1 )2 - (x 2 )2 - (x3 )2
I = xm x
is Lorentz-invariant
Energy-momentum
a × b º am bm
Useful:
p = g mv
pm pm = (mhm )(mh m ) = m2 c2
E2
pm p = 2 - p2
c
m
L. R. Flores Castillo
Scalar product:
m
covariant
CUHK
E = m c +p c
2
xμ = (ct, x, y, z)
ημ=dxμ/dτ = γ(c, vx, vy, vz)
pμ = mημ = (E/c, px, py, pz)
2 4
2 2
E = g mc
2
For v=c,
January 28, 2015
p / E = v / c2
v = pc 2 / E
E = hv
5
Collisions
• Classical
– Mass is conserved: mA+mB=mC+mD
– Momentum is conserved: pA+pB=pC+pD
– Kinetic energy may or may not be conserved:
• Sticky (kinetic energy decreases):
TA+TB > TC+TD
• Explosive (kinetic energy increases):
TA+TB < TC+TD
• Elastic (kinetic energy decreases):
TA+TB = TC+TD
• Relativistic
– Energy is conserved: EA+EB=EC+ED
m
m
m
m
p
+
p
=
p
+
p
A
B
C
D
– Momentum is conserved: pA+pB=pC+pD
– Kinetic energy may or may not be conserved
• Sticky (T decreases):
rest energy and mass increase
• Explosive (T increases): rest energy and mass decrease
• Elastic (T is conserved): rest energy and mass are conserved
L. R. Flores Castillo
CUHK
January 28, 2015
6
Example 1
Two lumps of clay, each of mass m, collide head-on at 3c/5; they
stick together. What is the mass M of the final composite lump?
Conservation of momentum: trivial. p1 + p2 = pM = 0
Conservation of Energy:
2mc 2
5
Mc = 2Em =
= (2mc 2 )
1- (3 / 5)2 4
2
5
M= m
2
Notice that M > m1 + m2
L. R. Flores Castillo
CUHK
January 28, 2015
7
Example 2
A particle of mass M, initially at rest, decays into two
pieces, each of mass m. What is the speed of each piece
as it flies off?
Conservation of momentum: equal and opposite speeds
Conservation of energy:
Mc = 2g mc =
2
2
2mc 2
1- v 2 / c 2
v = c 1- (2m / M )
L. R. Flores Castillo
CUHK
2
Only makes sense if M>2m
For M=2m, v=0
For M>>2m, vc
β depends only on m/M
January 28, 2015
8
Example 3
A pion at rest decays into a
muon plus a neutrino. What is
the speed of the muon?
(mμ=105.66 MeV/c2; mπ=139.57 MeV/c2; mν=0)
Ep = Em + En
pp = pm + pn
pp = 0 ® pm = -pn
mp c 2 = pm c + c mm2c 2 + pm2
pm =
mp2 - mm2
2mp
c
Subst. in E2=m2c4+p2c2:
E 2 = m2 c 4 + p2 c 2 :
Ep = mp c 2
En = pn c = p m c
Em = c mm2 c 2 + p m2
L. R. Flores Castillo
CUHK
Em =
mp2 + mm2
2mp
c2
Subst. in v = p c2 / E :
mp2 - mm2
vm = 2
c = 0.27c
2
mp + mm
January 28, 2015
9
Example 3. Second solution.
A pion at rest decays into a
muon plus a neutrino. What is
the speed of the muon?
(mμ=105.66 MeV/c2; mπ=139.57 MeV/c2; mν=0)
pp = pm + pn
Scalar product w/itself:
In all frames:
pn = pp - pm
pn2 = pp2 + pm2 - 2 pp × pm
pn2 = 0; pp2 = mp2c2 ; pm2 = mm2 c2
With the pion at rest: pp × pm =
Similarly, from
:
2 2
Em =
Ep Em Ep
= 2 Em = mp Em
c c c
mp2 + mm2
2mp
c2
pm = pp - pn
mm c = mp2c 2 - 2mp En
Similarly, from En = pn c = pm c and v = p c2 / E:
L. R. Flores Castillo
0 = mp2c 2 + mm2 c2 - 2mp Em
CUHK
mp2 - mm2
vm = 2
c = 0.27c
2
mp + mm
January 28, 2015
10
Example 4
The Bevatron at Berkeley aimed at producing antiprotons via
p+ p ® p+ p+ p+ p
What is the minimum energy of the incident proton that allows it?
Lab frame:
CM frame:
L. R. Flores Castillo
CUHK
January 28, 2015
11
Example 4
The Bevatron at Berkeley aimed at producing antiprotons via
p+ p ® p+ p+ p+ p
What is the minimum energy of the incident proton that allows it?
Minimum = all particles at
rest in the CM frame.
m
æ E + mc 2
ö
=ç
, p , 0, 0 ÷
è c
ø
Before the collision:
pTOT
After the collision:
m
pTOT
' = ( 4mc, 0, 0, 0)
m
m
pTOT
pm TOT = pTOT
' pm TOT '
æ E + mc 2 ö
2
2
ç
÷ - p = (4mc)
è c ø
2
Eliminating p w/
E2=m2c4+p2c2:
E = 7mc
2
i.e., the incident proton should have a kinetic energy above 6TeV
(and, indeed, the antiproton was discovered with E ~ 6000MeV)
L. R. Flores Castillo
CUHK
January 28, 2015
12
Example 5
Two identical particles, each with mass m and kinetic energy T,
collide head-on. What is their relative kinetic energy, T’?
(the kinetic energy of one on the rest system of the other)
Fixed target
Colliding beams
m
pTOT
( p ) = ( p ')
m
2
TOT
æ E '+ mc 2 ö
pTOT ' = ç
, p'÷
è c
ø
æ 2E ö
= ç , 0÷
è c ø
m
TOT
m
2
æ 2E ö æ E '+ mc 2 ö
2
ç ÷ =ç
÷ - p'
è c ø è c ø
2
2
:
(
Eliminating p w/ E2=m2c4+p2c2: 2E 2 = mc 2 E '+ mc 2
In terms of
L. R. Flores Castillo
T=E-mc2
and
CUHK
T’=E’-mc2:
)
æ
T ö
T ' = 4T ç1+
÷
è 2mc 2 ø
January 28, 2015
Classical result: only T’=4T
Reduces to it when T<<mc2
In B’s frame, A has twice the
speed, hence 4 times the E.
13
Symmetries
14
What can we infer from symmetry?
• No cosines in its Fourier expansion
• Only odd powers on its Taylor series
• …
L. R. Flores Castillo
CUHK
January 28, 2015
15
Symmetry in nature
• Symmetry is a powerful tool
• Symmetries need not show up in objects or their motion
– Stars do not move in circles
– As much as Kepler wished for it, neither planets in their orbits
• Six planets, five platonic solids …
• Rather, symmetry may lay in the set of all possible motions
• Manifest in the equations of motion
L. R. Flores Castillo
CUHK
January 28, 2015
16
Symmetry and conservation laws
• 1917: Emmy Noether’s theorem:
Every symmetry yields a conservation law
Conversely, every conservation law reflects an underlying
symmetry
L. R. Flores Castillo
CUHK
January 28, 2015
17
What is a symmetry?
• It is an operation on a system that leaves it invariant.
i.e., it transforms it into a configuration indistinguishable
from the original one.
f(x)  -f(-x)
L. R. Flores Castillo
CUHK
January 28, 2015
18
What is a symmetry?
• It is an operation on a system that leaves it invariant.
i.e., it transforms it into a configuration indistinguishable
from the original one.
•
•
•
•
•
Clockwise 120° rotation (R+)
Counterclockwise (R-)
Flipping it about vertical axis a (Ra)
Around axes b or c (Rb, Rc)
Doing nothing (I)
• We can also combine them
– R+2 = RL. R. Flores Castillo
CUHK
January 28, 2015
19
Symmetry groups
On a given system, the set of all symmetry operations has
the following properties:
• Closure: If Ri and Rj are in the set, then their product RiRj
is also in the set (product: apply Rj, then Ri)
• Identity: there is an element I such that RiI = IRi = Ri for
all elements Ri.
• Inverse: For every element R there is an inverse, Ri-1,
such that RiRi-1 = Ri-1Ri = I
• Associativity: Ri(RjRk) = (RiRj)Rk
These are the properties of a mathematical group.
Products need not commute (in general RiRj ≠ RjRi);
if they do, the group is called an Abelian group
L. R. Flores Castillo
CUHK
January 28, 2015
20
Symmetry groups
• Most of the groups of interest in physics can be formulated
as groups of matrices.
• e.g., the Lorentz group: set of 4x4 Λ matrices
• In particle physics, the most common groups are of the
type called U(n) ~ the collection of the unitary n×n matrices
– A unitary matrix is one for which U-1=U*T
• Unitary matrices with determinant 1: SU(n) (S for ‘special’)
• Restriction to real unitary matrices: O(n) (‘orthogonal’)
• Real, orthogonal, n×n matrices of determinant 1: SO(n)
L. R. Flores Castillo
CUHK
January 28, 2015
21
Symmetry groups
• Every group G can be represented by a group of matrices:
For every group element a there is a corresponding matrix Ma,
and the correspondence respects group multiplication (if ab=c,
then MaMb=Mc)
L. R. Flores Castillo
CUHK
January 28, 2015
22
Angular Momentum
• Classically, an object may have two kinds of angular
momentum:
– orbital (rmv)
– spin (Iω)
Not different in essence for macroscopic objects.
• Quantum-mechanically
– the “spin” interpretation is no longer valid
– The three components of L = r × mv cannot be measured
simultaneously
– At most we can measure
• the magnitude of L ( L2 = L L )
• one component (usually labeled “z”)
L. R. Flores Castillo
CUHK
January 28, 2015
23
Angular Momentum
… in QM:
• Orbital (L)
Allowed values:
– magnitude (L2): l(l+1)ħ2 where l is a nonnegative integer (0, 1, 2,…)
where ml is an integer between –l and l:
– one component: ml ħ
ml = -l, -l+1, …, -1, 0, +1, …, l-1, l
• Spin (S)
Allowed values:
– magnitude
– one component: msħ
(S2):
s(s+1)ħ2
æ 1 3 5 ö
with s half-integer or integer çè 0, 2 ,1, 2 , 2, 2 ,...÷ø
where ml is an integer between –l and l:
ms = -s, -s+1, …, -1, 0, +1, …, s-1, s
2l+1 or 2s+1 possibilities for the measured component.
L. R. Flores Castillo
CUHK
January 28, 2015
24
Angular momentum
Example: l=2
• L2 = 2(2+1)ħ2; L=√6 ħ = 2.45ħ
• Lz can be 2ħ, ħ, 0, -ħ, -2ħ
• Lz cannot be oriented purely in z
Similarly, with spin ½:
• S2 = (1/2 *3/2 ) ħ2; S = 0.866 ħ
• Sz can be –ħ/2, +ħ/2
L. R. Flores Castillo
CUHK
January 28, 2015
25
Angular momentum
An important difference:
• A particle can have any value of orbital angular
momentum l
• But, the value of s is fixed for each type of particle
A few examples:
– Every pion has s=0
– Every kaon has s=0
– Every electron, proton, neutron and quark has s=1/2
– The ρ, the ψ, the γ and the gluon: s=1
– Δ, Ω- : s=3/2
– etc.
It is part of our definition of each particle type.
We call s the “spin” of a particle (notice: it is NOT √S2)
L. R. Flores Castillo
CUHK
January 28, 2015
26
Angular momentum
Bosons (integer spin)
Fermions (half-integer spin)
Spin 0
Spin 1
Spin 1/2
Spin 3/2
-
Mediators
Quarks, Leptons
-
Elementary
Pseudoscalar
mesons
Vector mesons
Baryon octet
Baryon decuplet
Composite
• “Fermion” and “Boson” refer to the rules for constructing
composite wavefunctions for identical particles
– Boson wavefunctions should be symmetric
– Fermion ones should be antisymmetric
• Consequences:
– Pauli exclusion principle
– Statistical properties
L. R. Flores Castillo
CUHK
The connection between
spin and statistics is a
deep result from QFT.
January 28, 2015
27
Addition of Angular Momenta
• Angular momentum states
represented by ‘kets’:
l ml ,
s ms
• Example: an electron in a hydrogen atom occupying:
– orbital state | 3 -1 > :
– spin state |½ ½ > :
l=3,
s=½,
ml=-1
ms=½
(although specifying s would be unnecessary)
• We may need the total angular momentum L+S
How to add the angular momenta J1 and J2? J = J1 + J2
• Again, we can only work with one component and the
magnitude
L. R. Flores Castillo
CUHK
January 28, 2015
28
Addition of Angular Momenta
What do we get if we combine states |j1 m1> and |j2 m2> ?
• The z components are simply added: m = m1 + m2
• But the magnitudes depend on J1, J2 relative orientations
– If they are parallel, the magnitudes add
– If antiparallel, they subtract
In general, in between:
j = |j1-j2|, |j1-j2|+1, …, (j1+j2)-1, (j1+j2)
Examples:
– A particle of spin 1 in an orbital state l=4:
j=5 (J2=30ħ2), j=4 (J2=20ħ2), j=3 (J2=12ħ2)
– a quark and an antiquark bound in a state with zero orbital angular
momentum: j=½+½=1, j=½-½=0.
“Vector” mesons
(ρ, K*, φ, ω)
L. R. Flores Castillo
CUHK
“Pseudoscalar” mesons
(π, K’s, η, η’)
January 28, 2015
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Addition of Angular Momenta
Adding three angular momenta: first two, then the third
• For the two quarks, adding orbital
angular momentum l>0:
l+1, l, l-1.
• Orbital quantum number has to be an integer, hence:
– All mesons carry integer spin (and are bosons)
– All baryons (3 quarks) must have half-integer spin (fermions)
L. R. Flores Castillo
CUHK
January 28, 2015
30
Addition of Angular Momenta
Adding three angular momenta: first two, then the third
Adding three quarks in a state with zero orbital angular
momentum:
From two quarks
(each with spin ½)
Adding the third one
(also spin ½)
1 + ½ = 3 /2
½+½=1
Decuplet
1–½= ½
Octets
½-½=0
L. R. Flores Castillo
0+½= ½
CUHK
January 28, 2015
31
Addition of Angular Momenta
Besides total angular momentum, sometimes we need the
specific states:
j1 m1 j2 m2 =
j1+ j2
å
Cmj jm1 1j2m2 j m , with m = m1 + m2
j= j1 - j2
Clebsch-Gordan
coefficients
(Particle Physics
Booklet, internet,
books, etc.)
L. R. Flores Castillo
CUHK
January 28, 2015
32
Addition of Angular Momenta
(a square root sign over
each number is implied)
m=5/2
m=3/2
m=1/2
m= -1/2
m= -3/2
m= -5/2
j1 m1 j2 m2 =
j1+ j2
å
Cmj jm1 1j2m2 j m , with m = m1 + m2
j= j1 - j2
L. R. Flores Castillo
CUHK
January 28, 2015
33
Addition of Angular Momenta
(a square root sign over
each number is implied)
m=5/2
m=3/2
Example:
m=1/2
e in a H atom in
orbital state |2 -1>,
m= -1/2
spin state |½ ½>.
If we measure J2,
what values might we get, and
what is the probability of each?
m= -3/2
m= -5/2
Possible values: l+s = 2+½ = 5/2, l-s = 2 – ½ = 3/2
z component: m = -1 + ½ = –½
L. R. Flores Castillo
CUHK
January 28, 2015
34
Addition of Angular Momenta
(a square root sign over
each number is implied)
m=5/2
m=3/2
Example:
m=1/2
e in a H atom in
orbital state |2 -1>,
m= -1/2
spin state |½ ½>.
If we measure J2,
what values might we get, and
what is the probability of each?
m= -3/2
m= -5/2
Possible values: l+s = 2+½ = 5/2, l-s = 2 – ½ = 3/2
z component: m = -1 + ½ = –½
L. R. Flores Castillo
CUHK
January 28, 2015
35
Addition of Angular Momenta
(a square root sign over
each number is implied)
m=5/2
m=3/2
Example:
m=1/2
e in a H atom in
orbital state |2 -1>,
m= -1/2
spin state |½ ½>.
If we measure J2,
what values might we get, and
what is the probability of each?
2 -1
L. R. Flores Castillo
CUHK
1
2
1
2
=
2 5
5 2
m= -3/2
m= -5/2
- 12 -
3 3
5 2
January 28, 2015
- 12
36
Addition of Angular Momenta
Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find
the explicit Clebsch-Gordan decomposition.
1
2
1
2
1
2
1
2
1
2
1
2
- 12
- 12
- 12
=11
= 12 1 0 +
1
2
0 0
1
2
=
1
2
0 0
1
2
1
2
1
2
1
2
1
2
1
2
10 -
- 12 = 1 -1
Equivalently (solving for states with j=0,1):
11 =
1
2
1
2
10 =
1
2
éë 12
1 -1 =
1
2
0 0 =
L. R. Flores Castillo
1
2
1
2
1
2
1
2
- 12
éë 12
Symmetric under 12
1
2
1
2
1
2
CUHK
- 12 +
1
2
- 12
1
2
1
2
ùû
Spin 1 states
“triplet”
- 12
1
2
- 12 -
1
2
- 12
1
2
1
2
ùû
Antisymmetric
Spin 0 state
“singlet”
January 28, 2015
37
Addition of Angular Momenta
Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find
the explicit Clebsch-Gordan decomposition.
1
2
1
2
1
2
1
2
1
2
1
2
- 12
- 12
- 12
=11
= 12 1 0 +
1
2
0 0
1
2
=
1
2
0 0
1
2
1
2
1
2
1
2
1
2
1
2
10 -
- 12 = 1 -1
Equivalently (solving for states with j=0,1):
11 =
1
2
1
2
10 =
1
2
éë 12
1 -1 =
1
2
0 0 =
1
2
L. R. Flores Castillo
1
2
1
2
1
2
- 12
éë 12
1
2
1
2
1
2
CUHK
- 12 +
1
2
- 12
1
2
1
2
ùû
1
2
- 12
1
2
1
2
ùû
- 12
1
2
- 12 -
N.B.: we could have read
these coefficients from
the Clebsch-Gordan table
(it works both ways)
January 28, 2015
38
Spin ½
• Proton, neutron, electrons, all quarks, all leptons
• For s=1, ms=½ (“spin up”, ) or ms= –½ (“spin down”, )
• Spinors:
1
2
L. R. Flores Castillo
1
2
=
CUHK
January 28, 2015
39