Measurable Quantities, Conservation Laws and Hadrons Flashcards Preview

PHYS3543 Theoretical Elementary Particle Physics > Measurable Quantities, Conservation Laws and Hadrons > Flashcards

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What quantities can be measured in particle physics?

-decay rate of unstable particles
-cross sections of collisions


Fermi's Golden Rule

-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


Decay Rates

-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


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

-golden rule becomes:
dΓi = |Mfi|²/2m d³p1_/(2π)³2E1...d³pn_/(2π)³2En (2π)^4 δ^4 (po - Σpk)


Cross Section

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


Cross Section

-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


Cross Section
Golden Rule

-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


How to find Mfi?

-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...


Feynman Diagrams

-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


How to know which vertices are allowed?

-conservation laws


Conservation Laws
Noether's Theorem

-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


Conservation Laws
Symmetries and Conserved Quantities

-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


Conservation Laws
Feynman Diagrams

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


How do we find conserved quantities?

-by looking for forbidden processes


Baryon Number

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


Allowed Interactions of the Standard Model
Fermions and Gauge Bosons

-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


Allowed Interactions of the Standard Model
Gauge Bosons

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


Allowed Interactions of the Standard Model
Higgs Interactions

-any fermion to any fermion via Higgs



-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



-isospin is determined by multiplicity:
-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


Third Component

-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



-certain particles in particle collision experiments are always produced in pairs at high rate BUT long lifetimes / low decay rate
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


Quantifying Strangeness

-if a particle decays to an 'ordinary' particle via one step:
-if by two steps:


Strangeness-Isospin Diagrams

-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


The Fundamental Representation

-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:


Hadron Wavefunctions

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


Hadron Wavefunctions

-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_⟩


Hadron Wavefunctions
πo, η, η'

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


Hadron Wavefunctions

-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_


Hadron Wavefunctions

-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_