Topics in Contemporary Physics Basic concepts 6 Luis Roberto Flores Castillo Chinese University of Hong Kong Hong Kong SAR January 28, 2015.

Topics in Contemporary Physics

Basic concepts 6

Luis Roberto Flores CastilloChinese University of Hong Kong

Hong Kong SARJanuary 28, 2015

L. R. Flores Castillo CUHK January 28, 2015

PART 1 • Brief history

• Basic concepts

• Colliders & detectors

• From Collisions to papers

• The Higgs discovery

• BSM

• MVA Techniques

• The future

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5σ


… 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

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Reminder: interactions

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QED:

QCD:

Weak:

W/Z: W/Z/γ:

Cabibbo-Kobayashi-Maskawa matrix

NO Flavor-Changing-Neutral-Currents

SM Particle Content


Reminder: Relativistic kinematics

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Maxwell equations

c for all observers Lorentz transformations

is Lorentz-invariant covariant

contravariant Scalar product:

Four-vector

time-position: xμ = (ct, x, y, z)

proper velocity: ημ=dxμ/dτ = γ(c, vx, vy, vz)

energy-momentum: pμ = mημ = (E/c, px, py, pz)

Energy-momentum Useful:

For v=c, E = hv


Symmetry, conservation laws, groups

• 1917: Emmy Noether’s theorem:

Every symmetry yields a conservation law

Conversely, every conservation law reflects an underlying symmetry

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• 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


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– “Spin” interpretation no longer valid– All 3 components cannot be measured simultaneously; and most we can measure: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:

• Ket notation:

• “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

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Orbital angular momentum (l) Spin angular momentum (s)

Allowed values integer integer or half integer

For each particle type any (integer) value fixed


Angular Momentum

… in QM:

• Orbital (L)Allowed values:

– magnitude (L2): l(l+1)ħ2 where l is a nonnegative integer (0, 1, 2,…)

– one component: ml ħ

• Spin (S) Allowed values:

– magnitude (S2): s(s+1)ħ2 with s half-integer or integer

– one component: msħ

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where ml is an integer between –l and l:

ml = -l, -l+1, …, -1, 0, +1, …, l-1, l

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.


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

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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

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• “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

The connection between spin and statistics is a deep result from QFT.


Addition of Angular Momenta

• Angular momentum states represented by ‘kets’:

• Example: an electron in a hydrogen atom occupying:– orbital state | 3 -1 > : l=3, ml=-1

– spin state |½ ½ > : s=½, ms=½ (although specifying s would be unnecessary)

• We may need the total angular momentum L+SHow to add the angular momenta J1 and J2? J = J1 + J2

• Again, we can only work with one component and the magnitude

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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.

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“Vector” mesons

(ρ, K*, φ, ω)

“Pseudoscalar” mesons

(π, K’s, η, η’)



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)

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Adding three angular momenta: first two, then the third

Adding three quarks in a state with zero orbital angular momentum:

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From two quarks (each with spin ½)

½ + ½ = 1

½ - ½ = 0

Adding the third one (also spin ½)

1 + ½ = 3/2

1 – ½ = ½

0 + ½ = ½

Decuplet

Octets



Besides total angular momentum, sometimes we need the specific states:

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Clebsch-Gordan coefficients

(Particle Physics Booklet, internet, books, etc.)



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m=5/2

m=3/2

m=1/2

m= -1/2

m= -3/2

m= -5/2

(a square root sign over each number is implied)



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m=5/2

m=3/2

m=1/2

m= -1/2

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

Example:

e in a H atom inorbital state |2 -1>, spin state |½ ½>. If we measure J2, what values might we get, and what is the probability of each?




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m=5/2

m=3/2

m=1/2

m= -1/2

m= -3/2

m= -5/2

Example:


Possible values: l+s = 2+½ = 5/2, l-s = 2 – ½ = 3/2

z component: m = -1 + ½ = –½




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m=5/2

m=3/2

m=1/2

m= -1/2

m= -3/2

m= -5/2

Example:





Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition.

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Equivalently (solving for states with j=0,1):

Spin 1 states

Spin 0 state

“triplet”

“singlet”

Symmetric under 12

Antisymmetric




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Equivalently (solving for states with j=0,1):

N.B.: we could have read these coefficients from the Clebsch-Gordan table (it works both ways)


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 column vectors:

• “a particle of spin ½ can only exist in one of these states”

Wrong! its general state is

Where α and β are complex numbers, and

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Spin ½

• 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?

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Spin ½

In general, in QM:

• Construct the matrix Â representing the observable A

• The allowed values of A are the eigenvalues of Â– Eigenvalues: ei, eigenvectors:

• Write the state of the system as a linear combination of the eigenvectors of Â

• The probability that a measurement of A would yield the value ei is |ci|2

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Spin ½

• Eigenvalues of Sx are ±ħ/2, corresponding to normalized eigenvectors:

• any spinor can be written as a linear combination of these eigenvectors:

by choosing

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• Construct matrix Â 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

• In terms of the Pauli spin matrices:

the spin operators can be written as

• Effect of rotations on spinors:

where

θ is a vector pointing along the axis of rotation, and its magnitude is the angle of rotation.

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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.

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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:

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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.

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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.

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Flavor symmetries

• Horizontal rows: very similar masses but different charges• We assign an isospin I to each multiplet • And a particular I3 to each member of the multiplet:

• Pion: I=1:

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Flavor symmetries

• For the Λ, I=0:

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Flavor symmetries

• For the Δ’s, I = 3/2:

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Isospin of a multiplet: multiplicity = 2l+1

I3=I for the maximum Q


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:

(all other flavors carry isospin 0)

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Flavor symmetries

It has dynamical implications; for example:

• Two nucleons (hence I=1) can combine into – a symmetric isotriplet: – an antisymmetric isosinglet

– 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.

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Flavor symmetries

• Nucleon-nucleon scattering:

• From the one in the middle, only the I=1 combination contributes. As a result, the scattering amplitudes are in the ratio:

• and the cross sections (~ square of s.a.’s):

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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.

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Backup

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Eigenvectors and eigenvalues

A nonzero vector χ is called an eigenvector of a matrix M if

M χ = λ χ

for some number λ (the eigenvalue).

(notice that any multiple of χ is also an eigenvector of M, with the same eigenvalue)

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Topics in Contemporary Physics Basic concepts 6 Luis Roberto Flores Castillo Chinese University of Hong Kong Hong Kong SAR January 28, 2015.

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