lecture 8 - Department of Physics, The Chinese University of Hong

Report
Topics in Contemporary Physics
Basic concepts 6
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
GeV
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s=7
2000
1800
TeV,
s=8
ò
TeV,
ò Ldt =
-1
5.9 fb
1600
1400
1200
1000
800
600
10
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sample
2
126.5
and 201 fit (m H =
2011
usive
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-1
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4.8 fb
er poly
Ldt =
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fb
t = 4.8
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Ldt =
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s=7
s=8
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• The Higgs discovery
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Syst.
20
H®ZZ
Data
V
ts/5 Ge
(*)
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ZZ
round
s, tt
Backg
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round
V)
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l (m H=1
Signa
25
150
140
-100
100
160
V]
mg g [Ge
130
0
120
110
• BSM
• MVA Techniques
• The future
L. R. Flores Castillo
CUHK
January 28, 2015
2
… last time:
• Review of relativistic kinematics
– Relativistic collision examples
– Sticky collision
– Explosive collisions
– Collider vs fixed target
• Symmetries
– Symmetry groups
– SU(n)
• Angular momentum
– Orbital and spin angular momenta
– Addition of angular momenta
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
Symmetry, conservation laws, groups
• 1917: Emmy Noether’s theorem:
Every symmetry yields a conservation law
Conversely, every conservation law reflects
an underlying symmetry
• A “symmetry” is an operation on a system that leaves it invariant.
i.e., it transforms it into a configuration indistinguishable from the
original one.
The set of all symmetry operations on a given system forms a group:
• Closure: If a and b in the set, so is ab
• Identity: there is an element I s.t. aI = Ia = a for all elements a.
• Inverse: For every element a there is an inverse, a-1, such that aa-1 = a-1a = I
• Associativity: a(bc) = (ab)c
if commutative, the group is called Abelian
L. R. Flores Castillo
CUHK
January 28, 2015
6
Angular Momentum
• Classically, orbital (rmv), spin (Iω) not different in essence.
• In QM,
– “Spin” interpretation no longer valid
– All 3 components cannot be measured simultaneously; and most we can measure:
• the magnitude of L ( L2 = L L ).
Allowed values: j(j+1)ħ2
• one component (usually labeled “z”)
Allowed values: -j,…,j in integer steps
– Differences:
Orbital angular momentum (l)
Allowed values
For each particle type
• Ket notation: l ml ,
Spin angular momentum (s)
integer
integer or half integer
any (integer) value
fixed
s ms ,
j mj
• “A particle with spin 1” :
– a particle with s=1
– simple label, not the magnitude of its spin angular momentum:
– notice that the magnitude is always a bit larger than the maximum z component
L. R. Flores Castillo
CUHK
January 28, 2015
7
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
8
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
9
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
10
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
11
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
12
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
13
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
14
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
15
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
16
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
17
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
18
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
19
Addition of Angular Momenta
Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find
the explicit Clebsch-Gordan decomposition.
L. R. Flores Castillo
CUHK
January 28, 2015
20
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
21
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
22
Spin ½
• Most important case (p, n, e, all quarks, all leptons)
• Illustrative for other cases
• For s=½, 2 states:
ms=½ (“spin up”, ) or ms= –½ (“spin down”, )
• Better notation: Spinors
– two-component
æ 1 ö
æ 0 ö
1 1
1
1
÷
÷
column vectors:
2 2 =ç
2 - 2 =ç
è 0 ø
è 1 ø
• “a particle of spin ½ can only exist in one of these states”
Wrong! its general state is
æ a ö
æ 1 ö æ 0 ö
çç
÷÷ = a ç
÷+ bç
÷
è 0 ø è 1 ø
è b ø
Where α and β are complex numbers, and
L. R. Flores Castillo
CUHK
a + b =1
January 28, 2015
2
2
23
Spin ½
æ a ö
æ 1 ö æ 0 ö
çç
÷÷ = a ç
÷+ bç
÷
è 0 ø è 1 ø
è b ø
• The measurement of Sz can only return +½ħ and -½ħ
• with probabilities |α|2 and |β|2, respectively
• If we now measure Sx or Sy on a particle in this state,
– what possible results may we get?
+½ħ and -½ħ
– what is the probability of each result?
L. R. Flores Castillo
CUHK
January 28, 2015
24
Spin ½
In general, in QM:
• Construct the matrix  representing the observable A
• The allowed values of A are the eigenvalues of Â
– Eigenvalues: ei, eigenvectors:
i
• Write the state of the system as a linear combination of
the eigenvectors of Â
Y = åci i
• The probability that a measurement of A would yield the
(assuming Y is normalized)
value ei is |ci|2
L. R. Flores Castillo
CUHK
January 28, 2015
25
• Construct matrix  representing observable A
• Allowed values of A are the eigenvalues of Â
• Write state as linear combination of these eigenvectors
• The probability to measure ei is |ci|2
Spin ½
• Eigenvalues of Sx are ±ħ/2, corresponding
to normalized eigenvectors:
æ 1
2
ç
c± =
ç ±1
2
è
ö
÷
÷
ø
• any spinor can be written as a linear combination of these
eigenvectors:
æ a ö æ 12 ö æ 12 ö
÷ + bç
÷
çç
÷÷ = a ç
1
1
è b ø ç 2 ÷ ç - 2 ÷
è
by choosing
L. R. Flores Castillo
a=
CUHK
1
2
ø
è
(a + b ); b =
ø
1
2
(a - b )
January 28, 2015
26
Spin ½
• In terms of the Pauli spin matrices:
æ 0 1 ö
æ 0 -i ö
æ 1 0 ö
sx =ç
÷, s y = ç
÷, s z = ç
÷
è 1 0 ø
è i 0 ø
è 0 -1 ø
the spin operators can be written as
• Effect of rotations on spinors:
where
U (q ) = e-i(q×s )/2
æ a' ö
æ a ö
ç
÷
ç
÷
ç b ' ÷ = U (q ) ç b ÷
è
ø
è
ø
θ is a vector pointing along the axis of rotation,
and its magnitude is the angle of rotation.
e A =1+ A+ 12 A2 + 3!1 A3 +...
L. R. Flores Castillo
CUHK
January 28, 2015
27
Spin ½
• These matrices U(θ) are Unitary, of determinant 1.
i.e., they constitute the group SU(2)
• Spin-½ particles transform under rotations according to the
two-dimensional representation of the group SU(2)
• Particles of spin 1, described by vectors, transform under
the three-dimensional representation of SU(2)
• Particles of spin 3/2: described by 4-component objects,
transform under the 4d representation of SU(2)
• Why SU(2)? the group is very similar (homomorphic) to the
SO(3) (the group of rotations in three dimensions).
• Particles of different spin belong to different representations
of the rotation group.
L. R. Flores Castillo
CUHK
January 28, 2015
28
Flavor Symmetries
• Shortly after the discovery of the neutron (1932)
• Heisenberg observed that the neutron is very similar to
the proton
– mp = 938.28 MeV/c2; mn = 939.57 MeV/c2.
• Two “states” of the same particle? (the nucleon)
• Maybe the mass difference was related to the charge?
(it would be the other way around: p would be heavier)
• The nuclear forces on them are likely identical
• Implementation:
æ a ö
÷
N = çç
÷
b
è
ø
L. R. Flores Castillo
CUHK
æ 1 ö
æ 0
p = çç
÷÷ , n = çç
è 0 ø
è 1
January 28, 2015
ö
÷÷
ø
29
Flavor symmetries
• Drawing an analogy with spin, Heisenberg introduced the
isospin I (for ‘isotopic’ spin; better term: ‘isobaric’ spin)
• I is not a vector in ordinary space (no corresp. to x, y, z);
rather, in abstract ‘isospin space’.
• Components I1, I2, I3.
• Borrowing the entire machinery of angular momentum:
– Nucleon carries isospin ½
– The third component, I3, has eigenvalue +1/2, -1/2.
p=
L. R. Flores Castillo
1
2
1
2
, n=
CUHK
1
2
- 12
January 28, 2015
30
Flavor symmetries
• Can this possibly work?
Physical content: Heisenberg’s proposition that strong
interactions are invariant under rotations in isospin space.
if so, by Noether’s theorem,
isospin is conserved in strong interactions
Specifically
• Strong interactions invariant under an internal SU(2)
symmetry group
• Nucleons belong to the two-dimensional representation
(hence isospin ½).
• Originally a bold suggestion, but plenty of evidence.
L. R. Flores Castillo
CUHK
January 28, 2015
31
Flavor symmetries
• Horizontal rows: very similar masses but different
charges
• We assign an isospin I to each multiplet
• And a particular I3 to each
member0 of the multiplet:
+
p = 1 1 , p = 1 0 , p - = 1 -1
• Pion: I=1:
L. R. Flores Castillo
CUHK
January 28, 2015
32
Flavor symmetries
• For the Λ, I=0:
L. R. Flores Castillo
CUHK
L= 0 0
January 28, 2015
33
Flavor symmetries
Isospin of a multiplet:
multiplicity = 2l+1
I3=I for the maximum Q
• For the Δ’s, I = 3/2:
D++ =
L. R. Flores Castillo
3
2
3
2
,
D+ =
CUHK
3
2
1
2
,
D0 =
3
2
- 12 ,
January 28, 2015
D- =
3
2
- 23
34
Flavor symmetries
• Before 1974 (i.e., when only hadrons composed of u, d
and s were known), relation between Q and I3:
Q = I3 + ½ (A+S)
Gell-Mann–Nishijima formula
(A: baryon number; S: strangeness)
• Originally just an empirical observation; it now follows
from isospin assignment for u and d:
u=
1
2
1
2
,
d=
1
2
- 12
(all other flavors carry isospin 0)
L. R. Flores Castillo
CUHK
January 28, 2015
35
Flavor symmetries
It has dynamical implications; for example:
• Two nucleons (hence I=1) can combine into
– a symmetric isotriplet:
– an antisymmetric isosinglet
1 1 = pp
10 =
1
2
( pn + np)
10 =
1
2
( pn - np)
1 -1 = nn
– Experimentally, p & n form a single bound state (the deuteron)
– There is no bound state of two protons or two neutrons
– Therefore, the deuteron must be the isosinglet
(otherwise all three states would need to occur).
– There should be a strong attraction in the I=0 channel, and not
in the I=1 channel.
L. R. Flores Castillo
CUHK
January 28, 2015
36
Flavor symmetries
• Nucleon-nucleon scattering:
1 1
1
2
(1
0 +0 0
)
1 -1
p+ p ® d +p
+
p+n ® d +p
0
n+n ® d +p
-
1 1
1 0
1 -1
• From the one in the middle, only the I=1 combination
contributes. As a result, the scattering amplitudes are in
the ratio:
1
M a : M b : M c =1:
• and the cross sections
(~ square of s.a.’s):
L. R. Flores Castillo
CUHK
:1
2
s a : s b : s c = 2 :1: 2
January 28, 2015
37
L. R. Flores Castillo
CUHK
January 28, 2015
38
Flavor symmetries
• In the 50’s, as more particles were found, it was tempting
to extend this idea, but it became increasingly hard to
argue that they were different states of the same particle
• The Λ, Ξ’s and Σ’s could be regarded as a supermultiplet,
as if they belonged in the same representation of an
enlarged symmetry group
• SU(2) of isospin would then be a subgroup, but
• what was the larger group?
• The Eightfold way was Gell-Mann’s solution:
The symmetry group is SU(3),
the octets are 8D representations of SU(3)
the decuplet a 10D representation,
etc.
L. R. Flores Castillo
CUHK
January 28, 2015
39
Backup
40
Eigenvectors and eigenvalues
A nonzero vector χ is called an eigenvector of a matrix M if
Mχ=λχ
for some number λ (the eigenvalue).
(notice that any multiple of χ is also an eigenvector of M,
with the same eigenvalue)
L. R. Flores Castillo
CUHK
January 28, 2015
41

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