Measurable Quantities, Conservation Laws and Hadrons

Question 1

What quantities can be measured in particle physics?

Answer 1

• decay rate of unstable particles

- cross sections of collisions

Question 2

Fermi’s Golden Rule

Answer 2

-transition rate for an allowed process is:
Γif = 2π |Mfi|² * (phase space)
-where Γif is the transition rate form initial state i to final state f
-Mfi is the transition amplitude, the probability amplitude going from initial to final states
-phase space is the number of allowed final states

Question 3

Decay Rates

Answer 3

-if particles can decay through multiple channels, then:
Γ = ΣΓi
-where Γ is the total decay rate
-and Γi is the rate of decay to a particular final state

Question 4

Particle, mass m, decays to n final states through one particular channel with momenta p_k

Answer 4

-golden rule becomes:

dΓi = |Mfi|²/2m d³p1/(2π)³2E1…d³pn/(2π)³2En (2π)^4 δ^4 (po - Σpk)

Question 5

Cross Section

Definition

Answer 5

• a measure of how often two colliding particles produce a particular final state
• this is an intrinsic property

Question 6

Cross Section

Experiment

Answer 6

• a typical experiment will accelerate bunches of particles towards each other (most will miss) some collide and their collision energy can produce new particles
• the rate at which this happens depends on experimental parameters e.g. target density, beam flux etc.
• but these parameters can be factored out for the intrinsic cross section

Question 7

Cross Section

Golden Rule

Answer 7

-total cross section is denoted, σ
dσ = |Mfi|²/[4(√(k1[-k2)²-m1²m2²]]
d³p1/[(2π)³2E1] … d³pn/[(2π)³2En (2π)^4 δ^4 [k1+k1-Σpf]
-so total σ is the integral over this over p1…pn

Question 8

How to find Mfi?

Answer 8

-Mfi is the amplitude of the probability of going from i to f:
Mfi = ⟨f|Ef|i⟩
-to find Mfi, add up all of the possible Feyman diagrams for the interaction
-every aspect of a Feynman diagram represents an algebraic quantity
-this could be an infinite number of interactions…

Question 9

Feynman Diagrams

Vertices

Answer 9

-each vertex carries a factor of e for photon ‘interactions’ where:
e = EM coupling
-so when adding up the possible interactions to find Mfi, the more complex Feynman diagrams have more vertices so are proportional to higher powers of e
-since e«1, these terms can be neglected and we can take the simplest form as a good approximation

Question 10

How to know which vertices are allowed?

Answer 10

-conservation laws

Question 11

Conservation Laws

Noether’s Theorem

Answer 11

• there are lots of conserved quantities in particle physics
• many of them derive from continuous symmetries via ‘Noether’s Theorem:
• -if you have a continuous symmetry in a physical system, there is a corresponding conserved quantity

Question 12

Conservation Laws

Symmetries and Conserved Quantities

Answer 12

• spatial translation -> momentum
• time translation -> energy
• rotational -> angular momentum
• boost -> motion of centre of mass
• phase shift -> charge (not just EM charge, any kind of quantum no. can be though of as a charge)
• strong phase invariance -> colour

Question 13

Conservation Laws

Feynman Diagrams

Answer 13

-any vertex in a Feynman diagram must conserve momentum, energy etc. and also electric charge etc.

Question 14

How do we find conserved quantities?

Answer 14

-by looking for forbidden processes

Question 15

Baryon Number

Answer 15

1 for baryons
-1 for anti baryons
0 for mesons
\+1/3 quarks
-1/3 anti quarks

Question 16

Allowed Interactions of the Standard Model

Fermions and Gauge Bosons

Answer 16

• charger fermion to charged fermion via photon
• quark to quark via gluon
• any fermion to weak partner via W boson
• any fermion to any fermion via Zo

Question 17

Allowed Interactions of the Standard Model

Gauge Bosons

Answer 17

• W+, W- and photon
• three gluons
• four gluons
• Zo, W+, W-
• there also four point weak interactions (many combinations)

Question 18

Allowed Interactions of the Standard Model

Higgs Interactions

Answer 18

-any fermion to any fermion via Higgs

Question 19

Isospin

Description

Answer 19

• experimentally observed hadrons come in groups of similar mass e.g. (p,n), (Σ-,Σo,Σ+)
• know that a spin 1/2 particle has two states and a spin 3/2 has four states etc.
• introduce isospin that behaves in the same way
• i.e. that p,n and Σ-,Σo,Σ+ are different states of one particle which is not correct but the maths leads to useful properties

Question 20

Isospin

Definition

Answer 20

-isospin is determined by multiplicity:
2I+1
-so for protons and neutrons multiplicity is 2:
2 = 2I+1 => I=1/2
-we say n,p are a multiplet with I=1/2
-for Σ-,Σo,Σ+ :
3 = 21+1 => I=1
-Σ-,Σo,Σ+ are a multiplet with I=1

Question 21

Isospin

Third Component

Answer 21

• just as with spin, we need a convention for the third component, I3
• the particle from a multiplet with the lowest (or most negative) charge is assigned the lowest I3
• so Σ-,Σo,Σ+ all have I=1 but respectively have I3=-1,0,+1

Question 22

Strangeness

Description

Answer 22

-certain particles in particle collision experiments are always produced in pairs at high rate BUT long lifetimes / low decay rate
-but:
dΓ ∝ |M|²
dσ ∝ |M|²
-so transition amplitude is large meaning a large coupling constant so must be produced by the strong nuclear force, and decay by the weak nuclear force
-so introduce a property, strangeness, that is conserved in strong interactions and violated by weak interactions
-> i.e two particles are produced together via the strong interaction with opposite strangeness so it is conserved but then they go in different directions and are separate so when they decay they decay separately so strangeness is conserved then

Question 23

Quantifying Strangeness

Answer 23

-if a particle decays to an ‘ordinary’ particle via one step:
|S|=1
-if by two steps:
|S|=2

Question 24

Strangeness-Isospin Diagrams

Answer 24

• take all the hadrons with a certain spin
• plot on an axes x=isospin, y=strangeness
• can build any of these representations using the fundamental representation

Question 25

The Fundamental Representation

Answer 25

-starting from 0,0 on a strangeness-isospin diagram, for baryons, can get to any point in 3 moves
-these three moves correspond to constituent quarks, u, d, s
-for mesons it is a quark and an antiquark:
u_,d_,s_

Question 26

Hadron Wavefunctions

Outline

Answer 26

-can decompose into a spatial part, a spin part and a flavour part
=>
Ψ = ΨspaceΨspinΨflav

Question 27

Hadron Wavefunctions

Pion

Answer 27

-the pion is relatively stable therefore probably in the ground state
-so total angular momentum of q and q_ is 0, i.e. they are symmetric, since otherwise they would have energy to radiate away and move to an even lower energy state
=>
π+ = |ud_⟩
π- = |du_⟩

Question 28

Hadron Wavefunctions

πo, η, η’

Answer 28

-πo, η, η’ all have I3=S=0
-there are three quark-antiquark combinations that would give this:
|uu_⟩
|dd_⟩
|ss_⟩
-how do we know which is which?
-take a superposition of all three states

Question 29

Hadron Wavefunctions

πo

Answer 29

-needs to be as antisymmetric of u and d
-no strangeness since π- and π+ have zero strangeness
=>
πo = 1/√2 [|uu_⟩ - |dd_⟩]
-this means that if it was possible to pull a pion apart into its constituent quark and antiquark, half the time you would get u & u_ and the other half, d & d_

Question 30

Hadron Wavefunctions

η’

Answer 30

-want an SU(3) singlet => η’ symmetric in all 3 flavours:
η’ =1/√3 [|uu_⟩+|dd_⟩+|ss_⟩]
this means that if it was possible to pull a η’ apart into its constituent quark and antiquark, a third of the time you would get u & u_, a third of the time, d & d_ and th rest of the time s & s_

Question 31

Hadron Wavefunctions

η

Answer 31

-choose η to be orthogonal to to the others => symmetric in u and d
η = 1/√6 [|uu_⟩ + |dd_⟩ - 2|ss_⟩]
-so if we could repeatedly pull apart η mesons, in
six goes, we would find on average one uu_ pair, one dd_ pair and four ss_ pa

Question 32

Spin States

Two Quarks

Answer 32

-two quarks can combine as:
|++⟩, |+-⟩, |-+⟩, |–⟩
-where + is spin up, - is spin down
-the spins combine to form a triplet and a singlet
-the triplet is the symmetric combinations:
|++⟩
1/√2 (|+-⟩ + |-+⟩)
|–⟩
-all have spin, S=1, the z-component S3 is +1,0,-1 for each one respectively
-the singlet is the antisymmetric combinations:
1/√2 (|+-⟩ - |-+⟩)
-for this, S=0
-can’t just have |++⟩, |+-⟩, |-+⟩, |–⟩ as while they are all eigenstates of S3, |+-⟩ and |-+⟩ are not eigenstates of S^

Question 33

Three Spins

Symmetric States

Answer 33

-four states:
|+++⟩
1/√3 (|++-⟩ + |+-+⟩ + |-++⟩)
1/√3 (|+--⟩ + |-+-⟩ + |--+⟩)
|---⟩
-all of these are S=3/2
-symmetric meaning that swapping any two quarks recovers the same overall spin state

Question 34

Three Spins

Partially Symmetric States

Answer 34

-doublet:
1/√2 (|+-+⟩ + |++-⟩)
1/√2 (|–+⟩ + |-+-⟩)
-both have S=1/2
-partially symmetric meaning that they are only anti symmetric in 2 quarks, in this case, 2 and 3
-it is also possible to write down doublet states that are antisymmetric in 1 and 3
-and 1 and 2 but these are not linearly independent

Question 35

How may independent spin states are there for three quarks?

Answer 35

8

Question 36

How many independent favour states for three quarks?

Answer 36

27 (for u, d, s)

• ten of these are fully symmetric => swapping any two of the three quarks recovers the same state
• one if fully antisymmetric => swapping any two quarks gives minus the original state of the system
• then there are two sets of 8 that are partially symmetric
• from these two sets of 8 can construct a state that is symmetric in all three quarks

Question 37

Constructing a State Symmetric in 3 Quarks

Answer 37

-considering spin and flavour
Ψ = ΨflavΨspin
-for each combination of quarks multiply partially antisymmetric states together to give an overall symmetric state:
Ψ = Ψf(12)Ψs(12) + Ψf(13)Ψs(13) + Ψf(23)Ψs(23)
-where Ψf(12) indicates a flavour state that is antisymmetric in quarks one and two etc.

Question 38

Symmetry of Ψ for Hadrons and Violation of the Pauli Exclusion Principle

Answer 38

-it fits with the observed hadrons
-symmetric favour states are only possible in physical hadrons when combined with a symmetrical spin state since:
Ψ = ΨflavΨspin
-the symmetrical spin states are S=3/2
-e.g. Ω is |sss⟩, a symmetrical flavour state so if this is in a symmetrical spin state as well |+++⟩ then all three quarks are in exactly the same state which would violate the Pauli Exclusion Principle
-to solve this we introduce a new property which the three quarks must be entirely antisymmetric in, colour

Question 39

Colour

Answer 39

-colour comes in three states; red, green and blue
-the colour state of all observed hadrons is the total antisymmetric state:
1/√6 (|rgb⟩ - |rbg⟩ + |brg⟩ - |bgr⟩ + |gbr⟩ - |grb⟩)

Question 40

Whys should all hadrons have the same colour state?

Answer 40

• colours are like charges, there is a force between them, the strong force
• the antisymmetric colour state is the only state with overall net colour neutrality
• if hadrons weren’t in this state they would have a net colour charge and would not be in ‘equilibrium’

Question 41

Evidence for Colour

Answer 41

-look at hadron production cross sections
-plot the ratio of cross section of electron + positron to quark + anti quark to cross section of electron + positron to muon + anti muon
R = Q²
-where Q is the relative charge of the product
-predicted values of R and consistently 3² too small
-implying that three times the process is happening
-i.e. when we consider:
e- + e+ -> u + u_
-we are actually looking at the process:
e- + e+ -> ur + ur_
e- + e+ -> ub + ub_
e- + e+ -> ug + ug_

Question 42

SU(3) Representation

Answer 42

-representation of hadrons consisting of three quarks, u, d, s
-SU(4) would be hadrons consisting of either u, d, s, c
etc.

Question 43

Quantum Chromodynamics

Would we expect baryons to interact with each other?

Answer 43

• each quark flavour comes in three colours, they are identical to each other apart from the new property, colour charge
• each baryon consists of three quarks, one of each colour in the fully antisymmetric colour singlet state so that they have no overall colour
• since every baryon is colour neutral we might expect that they wouldn’t interact with each other
• however residual interactions can still take place, this residual force manifests as pion exchange

Question 44

Pion Exchange

Answer 44

• quark exchange between baryons
• e.g. a neutron and proton can scatter off each other by exchange of a neutral pion and proton-neutron exchange can occur via exchange of a negative pion
• it is as if baryon number acts as a charge for pion exchange
• in the same way that a photon couples to a charged particle with non-zero electric charge, a meson is able to couple to anything with non-zero baryon number

Question 45

Potential Arising From Particle Exchange

Answer 45

• if exchange particle spin is odd:
• -if particles have like charge you get repulsion
• -if particles have opposite charge you get attraction
• if the exchange particle spin is even, this is reversed
• this means an attractive force between baryons

Question 46

Gluons

Answer 46

• gluons couple to colour
• but also carry a colour and an anticolour i.e. they have colour charge
• since gluons carry charge themselves, they can couple to each other, unlike electromagnetic forces since photons themselves are electrically neutral

Question 47

How many types of gluon are there?

Answer 47

-since a gluon is composed of a colour and an anticolour, and there are three colours red, blue and green we would expect 9 types of gluon
-but it turns out that the symmetric colour state:
1/√3 (|rr_⟩ + |gg_⟩ + |bb_⟩)
-is non-physical
-this leaves 8 independent gluons

Question 48

Confinement

Electrostatic Analogy

Answer 48

• around a charged object is a cloud of virtual photons popping in and out of existence via exchange with the charged particle
• further from the particle, these photons are spread out over a larger surface area so the force drops of by an inverse square law
• the field lines diverge

Question 49

Confinement

Gluons

Answer 49

• around a colour charged object gluons are constantly being emitted and absorbed
• however, unlike for photons the gluons will interact with each other
• instead of diverging field lines, the lines create a flux tube between two coloured particles e.g. r and r_
• since there is no divergence, force is constant with distance so any attempt to separate the two particles just increases the potential energy in the system until there is enough for pair production and the original meson becomes two mesons
• i.e. we can never observe a a free colour charged object