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Elementary Particle and Nuclear PhysicsSummary
Jacob Shapiro
September 4, 2013
Abstract
A very rough translation of Prof. Klaus Stefan Kirch’s lecture notesfrom German to English in an attempt to prepare for his exam.
Contents1 Introduction 3
1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Clebsch-Gordan Coefficients . . . . . . . . . . . . . . . . . . 91.6 Particles in the Standard Model . . . . . . . . . . . . . . . . 111.7 Elementary Particles . . . . . . . . . . . . . . . . . . . . . . 131.8 Radioactive Decay . . . . . . . . . . . . . . . . . . . . . . . 151.9 Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
I Particle Physics 17
2 Basics 172.1 Cross-Section and Luminocity . . . . . . . . . . . . . . . . . 172.2 Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . 182.3 Feynman Diagrams . . . . . . . . . . . . . . . . . . . . . . . 182.4 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3 Scattering 213.1 Bethe-Bloch Formula . . . . . . . . . . . . . . . . . . . . . . 213.2 Elastic Scattering . . . . . . . . . . . . . . . . . . . . . . . . 233.3 Inelastic Scattering . . . . . . . . . . . . . . . . . . . . . . . 273.4 Summary and Remarks . . . . . . . . . . . . . . . . . . . . 30
1
4 Collider Physics and Particle Detectors 304.1 Particle Accelerators . . . . . . . . . . . . . . . . . . . . . . 304.2 Important Terms . . . . . . . . . . . . . . . . . . . . . . . . 324.3 Detectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.4 Up to Date Examples–LHC at CERN . . . . . . . . . . . . 36
5 Baryons 375.1 Isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.2 Classification and Nomenclature . . . . . . . . . . . . . . . 385.3 Conservation and Electric Charge . . . . . . . . . . . . . . . 395.4 Baryon Multiplets . . . . . . . . . . . . . . . . . . . . . . . 395.5 Magnetic Moment . . . . . . . . . . . . . . . . . . . . . . . 425.6 Mass of the Baryons . . . . . . . . . . . . . . . . . . . . . . 435.7 π-Nucleon-Systems and Isospin Considerations . . . . . . . 43
6 Symmetries and Symmetry Breaking 456.1 Conserved Quantities . . . . . . . . . . . . . . . . . . . . . . 456.2 Parity Violation . . . . . . . . . . . . . . . . . . . . . . . . . 476.3 Charge Conjugation . . . . . . . . . . . . . . . . . . . . . . 516.4 Helicity of the Neutrino–the Goldhaber Experiment . . . . 516.5 The CPT Theorem . . . . . . . . . . . . . . . . . . . . . . . 536.6 Kaons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.7 Parity Invariant 2-state system . . . . . . . . . . . . . . . . 546.8 Indirect CP-Breaking with Neutral Kaons . . . . . . . . . . 56
7 Particle Physics at PSI 567.1 Proton Accelerator . . . . . . . . . . . . . . . . . . . . . . . 567.2 Muonic-Hydrogen Lambshift . . . . . . . . . . . . . . . . . 577.3 CP-Violation of Ultra Cold Neutrons . . . . . . . . . . . . . 57
8 Properties of the Weak Interaction 588.1 Universality of the Charged Weak Interaction . . . . . . . . 588.2 Quark Mixing . . . . . . . . . . . . . . . . . . . . . . . . . . 618.3 ElectroWeak Unification . . . . . . . . . . . . . . . . . . . . 628.4 Neutrino Oscillations . . . . . . . . . . . . . . . . . . . . . . 62
II Nuclear Physics 63
9 Nucleon Model 639.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 639.2 Semi-Empirical Mass Formula (Liquid Drop Model) . . . . 659.3 Fermi Gas Model . . . . . . . . . . . . . . . . . . . . . . . . 669.4 Nuclear Shell Model . . . . . . . . . . . . . . . . . . . . . . 68
10 Deformed Nucelei and Collective Nuclear Exciations 7210.1 Quadrupole Moments . . . . . . . . . . . . . . . . . . . . . 7210.2 Rotations Model . . . . . . . . . . . . . . . . . . . . . . . . 7210.3 Multipol Transitions . . . . . . . . . . . . . . . . . . . . . . 7210.4 Giant Resonance . . . . . . . . . . . . . . . . . . . . . . . . 72
2
11 Nuclear Fission, Fusion and Nucleosynthesis 7311.1 Nuclear Fission . . . . . . . . . . . . . . . . . . . . . . . . . 7311.2 Nuclear Fusion . . . . . . . . . . . . . . . . . . . . . . . . . 7311.3 Nucleosynthesis . . . . . . . . . . . . . . . . . . . . . . . . . 73
12 Star Evolution, Cosmology and Dark Matter 7412.1 Star Evolution . . . . . . . . . . . . . . . . . . . . . . . . . 7412.2 Basics of Cosmology . . . . . . . . . . . . . . . . . . . . . . 7712.3 Time Evolution of Density in the Universe . . . . . . . . . . 7712.4 Dark Matter . . . . . . . . . . . . . . . . . . . . . . . . . . 77
III From Test Protocols 80
13 Things to Memorize 8013.1 Memorize Leisurely . . . . . . . . . . . . . . . . . . . . . . . 80
14 Things To Do and Open Questions 8114.1 Experimental Setups . . . . . . . . . . . . . . . . . . . . . . 8614.2 Particle Physics . . . . . . . . . . . . . . . . . . . . . . . . . 8614.3 Nuclear Physics . . . . . . . . . . . . . . . . . . . . . . . . . 8714.4 Early Universe . . . . . . . . . . . . . . . . . . . . . . . . . 87
1 Introduction
1.1 MotivationI was forced to study this course as a prerequisite for my degree.
1.2 General1.2.1 Units and Orders of Magnitudes
• 1eV = 1.602 · 10−19J
• 1fm = 10−15m
• 1u = 112M12C ≈ 1.66 · 10−27kg ≈ 931.5MeV/c2
1.2.1.1 Natural Units
• ~ = 6.582 · 10−16eV · s, but if we set ~ != 1 ⇐⇒ 6.582 · 10−16eV ·
s!= 1 from which we could get the relationship between s and eV :
s = 1.519 · 1015 (eV )−1 .
• c = 2.99 · 108ms . Setting c != 1, we get a relationship between m and s,
which in turn gives us a relationship betweenm and eV : m = 5.08 · 106 (eV )−1 .
3
• 1eV = 1.6 · 10−9Joules = 1.6 · 10−19 kg·m2
s2 , from which we infer that
kg = 5.58 · 1035eV .
• 1barn = 10−24cm2 =⇒ (GeV )−2
= 0.389mb .
• 1e(V 2)= 1.68 · 10−2Tesla
1.2.2 Constants and Relations
• ~ = 6.582 · 10−22MeV s = 197MeV fmc−1 =⇒ ~c ≈ 200MeV fm
• 1fm = 5.08 · (GeV )−1.
• α =e2
4π~c≈ 1
137
• de Broglie wave length: λ =h
p=
2π~cpc
.
• Compton wave length: λ =2π~mc
.
1.2.3 Table of Nuclides
• N =number of neutrons
• Z =number of protons
• Black = stable nuclei
• Colored: unstable nuclei
4
– orange = β+ decay
– blue = β− decay
– yellow = α-decay
• above about N = Z = 20, more neutrons are necessary to balance therepulsive Coulomb force between the protons.
1.2.3.1 Isotopes of Z protons The collection of all nuclei with Z protonsand different number of neutrons.
1.2.3.2 Isotones of N neutrons The collection of all nuclei with N neu-trons and different number of protons.
1.2.3.3 Isobars of mass A The collection of all nuclei with the same sumA = N + Z, but different numbers of protons and neutrons.
1.2.3.4 β−-DecayLet A (Z, N) denote a nucleus of Z protons and N neutrons. The β−-decay
is the following process:A (Z,N) → A (Z + 1, N − 1) + e− + νe
1.2.3.4.1 Example–Neutron Decay A (0, 1) → A (1, 0) + e− + νe isthe process of decay of one free neutron.
1.2.3.5 β+-decay A (Z,N) → A (Z − 1, N + 1) + e+ + νe
1.2.3.6 Double β-decay Using the semi-empirical mass formula we canemploy the inequality M (A, Z) ≥M (Z + 2, A) + 2me and thus get a relationon when a double beta decay is possible:
Z ≤ 65.77·A1/3−11−130.46A−2/3
5
1.2.3.7 α-decay A (Z, N) → A (Z − 2, N − 2)+4He (leptons are not men-tioned in this reaction)
1.3 Special Relativity
• Etotal = Ekinetic +m = γm =m√1− β2
• p = Ev = γmv
• β =
√1−
(m
Etotal
)2• xµ =
[tx
]
• pµ =
[Ep
]• aµb
µ = a0b0 − a · b
• pµpµ = E2 − p2 = m2 is the invariant mass. It stays the same regardless
of reference frame.
• 4-momentum is conserved in reactions.
6
• Λβx :=
γ −γβ 0 0
−γβ γ 0 00 0 1 00 0 0 1
=⇒ x′ = Λx
• For a charged particle in a magnetic field, R = pqB = m
qB
√Tm +
(Tm
)21.3.0.8 Mandelstam Variables They are used for scattering processes oftwo particles to two particles.
s := (p1 + p2)2= (p3 + p4)
2
t := (p1 − p3)2= (p2 − p4)
2
u := (p1 − p4)2= (p2 − p3)
2
• Easy to prove that s+ t+ u = m21 +m2
2 +m23 +m2
4.
• For highly relativistic, t ≈ −4E2E4 sin2 (θ/2).
1.3.0.9 Motion of charged particle in Magnetic Field p = qBR =⇒
R = pqB =
√(m+Ekin)
2−m2
qB
1.4 StatisticsTerms:
7
• FWHM: Width between two half maximum points.
• Expectation Value: E [X] :=´Ωxf (x) dx
• Standard Deviation: STD :=
√E[(X − E [X])
2]
• The error in n experiments, where each experiment has error σ is σ√n.
• Covariance: Cov (X, Y ) := E [(X − E [X]) (Y − E [Y ])]
1.4.1 Gaussian Distribution
f (x) := 1√2π
1σ e
− 12
x2
σ2
Standard deviation = σ, expectation value = 0, fwhm = 2√
2 log (2)σ.
1.4.2 Binomial Distribution
f (k; n, p) :=
(nk
)pk (1− p)
n−k
Probability to get k “successes” in n trials, where p is the probability to geta single isolated success.
1.4.3 Poisson Distribution
Discrete distribution. If the (theoretical) average rate of events per time periodT is given by λ, the probability that k actual events take place (where k ∈ N ∪0) during time period T (a random variable X) is given by PPoissonλ [X = k] =
e−λ λk
k! .
• E [X] = λ
• V ar [X] = λ
• In the limit of λ→ ∞, this distribution converges to the normal distribu-tion.
1.4.4 Error Propagation
If u = f (x, y) is some variable that depends on two independent variables, then
σu ≈√[∂xf (x, y)]
2σ2x + [∂yf (x, y)]
2σ2y.
1.4.5 Principle of Least Squares
TODO
8
1.4.6 χ2-distribution and χ2-test statistics
χ2 =∑Ni=1
[yi−f(xi)]2
σ2i
Minimize χ2 by the parameters of f to get them.Can also get the variance on these parameters as the covariance matrix.E(χ2min
)= N − P where N is the number of measurements and P is the
number of parameters to be estimated. The probability that χ2 exceeds a certainvalue can be calculated.
1.4.7 Confidence Intervals
Confidence Intervals (CIs) are statistics tools used to express our estimate ofa certain unknown parameter (which has some definite fixed value). The CIwith confidence level CL for the estimate of the parameter w according to ourobserved outcome X can be defined as a random interval [u(X), v(X)] for whichthe probability to include the real value w is equal CL.
1.4.8 Method of Maximum Likelihood
In statistics, maximum-likelihood estimation (MLE) is a method of estimatingthe parameters of a statistical model. When applied to a data set and givena statistical model, maximum-likelihood estimation provides estimates for themodel’s parameters.
We would like to estimate a parameter α from n measurements. The proba-bility distribution function is f (x, α). Define L (α) :=
∏ni=1 f (xi, α), differen-
tiate ∂L∂α = 0, find αmax.
1.4.9 Method of Least squares
χ2 :=∑Ni=1
[yi−f(xi; a)
σi
]2dχ2
da = 0
1.5 Clebsch-Gordan Coefficients1.5.1 Theorem
Let j1 > j2. Then Dj1 ⊗Dj2 = Dj1+j2 ⊕Dj1+j2−1 ⊕ · · · ⊕ D|j1−j2| .Thus
| j1, j2, j,m 〉 =∑
m1,m2,m1+m2=m
〈 j1, j2,m1,m2 | j1, j2, j,m 〉 | j1, j2,m1,m2 〉
where 〈 j1, j2,m1,m2 | j1, j2, j,m 〉 are the Clebsch-Gordan coefficients.As an example, adding two spin- 12 angular momenta:
9
We can read off this table for example that | j = 0,m = 0 〉 = 1√2|m1 =
12 ,m2 = − 1
2 〉 − 1√2|m1 = − 1
2 ,m2 = 12 〉.
This table can also be constructed with the usage of the lower and raisingoperators.
10
1.6 Particles in the Standard Model
Lifetimes are mean lifetimes in seconds (P (t) = e−t/τγ):
11
12
1.7 Elementary ParticlesAn elementary particle is such that cannot be decomposed into constituentparticles.
1.7.1 Fermionic Elementary Particles
There are two types of particles: quarks and leptons. They are arranged intothree families according to mass.
Muons have been discovered in cosmic rays.Each of these particles has an anti-particle, marked with a bar above its
symbol, or opposite electric charge.
13
1.7.2 Bosonic Elementary Particles
The Gauge-Bosons are the mediators of interactions.
• The W-boson was first detected at CERN with the UA1 experiment bythe decays W+ e+ + e and W e + e. The mass of the W boson canbe determined from the distribution of transverse momentum pt,e of theelectron, or out of the mt-distribution.
– Jacobian Peak: A peak in a probability distribution which can be un-derstood as due to the variation of a Jacobi determinant. For exam-ple, in the decayW → e+ν, dσ
dpt= dσ
d[cos(θ)]dσdpt
= dσd[cos(θ)]
2ptMW
1√(MW
2
)2−p2t
This will give a peak at pt =MW /2.
• There are 8 gluons, because SU(3) is represented by an adjoint represen-tation of 8 dimensions.
1.7.3 Hadrons–Composite Particles
Hadrons are composed of several quarks (or anti-quarks).
• Hadrons, along with the valence quarks (q v) that contribute to theirquantum numbers, contain virtual quark–antiquark (qq) pairs known assea quarks (q s). Sea quarks form when a gluon of the hadron’s color fieldsplits; this process also works in reverse in that the annihilation of twosea quarks produces a gluon. The result is a constant flux of gluon splitsand creations colloquially known as "the sea".[77] Sea quarks are muchless stable than their valence counterparts, and they typically annihilateeach other within the interior of the hadron. Despite this, sea quarks canhadronize into baryonic or mesonic particles under certain circumstances.
1.7.3.1 Baryons Baryons are Fermionic Hadrons composed of 3 quarks.The nucleons, hyperons, etc are baryons.
• proton = uud
• neutron = udd
1.7.3.2 Mesons Mesons are Bosonic Hadrons compsoed of one quark andone-antiquark. The pions, kaons, etc are mesons.
• The pions:
– π+ = ud
– π− = ud
– π0 = 1√2
(uu− dd
)
14
• The kaons:
– K+ = us
– K− = us
– K0 = ds
1.8 Radioactive Decay
dN
dt= −λN
1.9 Experiments1.9.1 Annihilation
e− + e+ → γ + γ
2211Na→22
10 Ne∗ + e+ + νe
2211Na undergoes β+ decay, which releases e+. The material surround the
source is abundant with e− and so we get plenty of e− − e+ pairs which canannhilate. Two photomultipliers detect the incident photons and are connectedto a coincidence box which only makes a sound when the two photomultipliersare triggers simultaneously (hence, stemming from the same event). To conservemomentum, the photons will come out back to back, because, approximately,the momentum before the annhilation is zero.
1.9.2 Cherenkov Counter
A device to detect particles.
1.9.2.1 Cherenkov Effect When a charged particle passes through dialec-tric material at a speed greater than the phase velocity of light in that mediumcn , electromagnetic radiation is emitted. The velocity of the particle is related
15
to the angle of the emitted EM radiation (between the path of the charged par-
ticle) by β =1
n cos (θ), where n is the refractive index of the incident material
and β is the velocity (in units of c) of the particle. If the particle were movingslowlier than c
n then there’d be destructive interference and we’d observe no ra-diation. However, when the speed of the particle is over c
n there is constructiveinterference.
So this can be exploited to detect certain particles, for example, which havethe same energy, but different masses (and hence different β), because only ifnβ > 1 will we see Cherenkov radiation).
1.9.3 Electron Turbine
r =√
2mUeB2
1.9.4 Electron Diffraction
Bragg diffraction shows that electrons are also waves.nλ = 2d sin (θ)
1.9.5 Cloud Chamber / Wilson Chamber
On top of cold platter there’s alcohol steam and under the platter, there’s a wiregrid hole, which “sucks up” the charges of the gas. Then accelerated particlescan be sent through this appartus, or, as shown in the lecture, inject radioactivematerial.
Particle detector for ionizing radiation: charged particles ionize gas atoms,which creates a mist. Apply magnetic field to detect charge sign.
“In its most basic form, a cloud chamber is a sealed environment containing asupersaturated vapor of water or alcohol. When a charged particle (for example,an alpha or beta particle) interacts with the mixture, it ionizes it. The resultingions act as condensation nuclei, around which a mist will form (because themixture is on the point of condensation). The high energies of alpha and betaparticles mean that a trail is left, due to many ions being produced along thepath of the charged particle. These tracks have distinctive shapes (for example,an alpha particle’s track is broad and shows more evidence of deflection by colli-sions, while an electron’s is thinner and straight). When any uniform magneticfield is applied across the cloud chamber, positively and negatively charged par-ticles will curve in opposite directions, according to the Lorentz force law withtwo particles of opposite charge.”
1.9.6 Gamma Spectroscopy
Observe the energy levels of something.
16
1.9.7 Radioactive Decay and Poisson Statistics
Part I
Particle Physics2 Basics
2.1 Cross-Section and LuminocityRead Griffiths chapter 6.
2.1.1 Cross-Section
We would like to describe a reaction of a beam of particles incident on sometarget.
• na is the particle per unit volume density of the incident beam.
• va is the velocity of each particle in the beam.
• φa := nava is thus the flux of incident particles.
• nt is the particle per unit volume density of the target.
• At is the surface area of the target.
• d is the thickness of the target.
• Nt = ntAtd is the number of particles in the target.
• σ is defined to be the projective surface of a single target particle. Thetotal, or geometric cross-section.
• The number of reactions per unit time is thus N = φaNtσ
• Thus σ can also be thought of as NφaNt
which is a measure of the probabilityof a reaction if an incident particle comes near a target particle.
2.1.2 Luminocity
Luminosity: the number of particles passing down the line per unit time perunit area.
• L := φaNt
• N = σL
17
2.1.3 Differential Cross-Section
• σ :=´dσdΩdΩ
2.1.3.1 Doubled Differential Cross-Section Also for energy values:
• d2σ(E, E′, θ
)dΩdE′
2.2 Fermi’s Golden Rule“A transition rate is given by the product of the phase space and he (absolute)square of the amplitude.”
The square of the transition matrix element is proportional to the probabilitythat the particles of state i goes over to the state f :
Mfi := 〈ψf |Hint |ψi 〉where Hint is the Hamiltonian of the perturbation.Thus the probability of transition between the states is given by:
Γi→f =2π
~|Mfi|2 ρ (Ef )
where ρ (Ef ) is the energy state density.The density of states is proportional to p2 dpdE usually.
2.3 Feynman Diagrams• Horizontal axis is time.
• Vertical axis is space.
• This sometimes gets reversed.
• Fermions are straight lines.
• Particles going agains time are anti-particles.
• Photons are wavy-lines.
• Bosons of the Weak Interaction are dashed lines.
• Gluons are lines with loops.
1. Charge is conserved
2. Lepton number is conserved.
3. Baryon number is conserved.
4. Causality.
• Particles that appear only inside the diagram are virtual particles.
18
• Charge is conserved along solid (Fermionic) lines.
• Virtual particles don’t obey the energy-momentum relation E2 = m2+p2
• Each vertex’ probability is proportional to a factor of√α, where α is the
coupling-constant of the interaction.
• In electron-positron annihaltion, photons have a virtual mass.
Pinguin Diagram
2.4 Interactions2.4.1 Model of Strong Interactions
The strong force between protons and neutrons, a residual force, is to the strongforce, what the van der Waals force is to EM.
Nucleons exchange pions and thus form bound states.
19
Since the mass of the pion is ≈ 140MeV , we get a scale for the strength ofthe nuclear force.
One can fit a potential that obeys this scale, the Yukawa potential:
UY (r) = αs~ce−
rλ
r
2.0 ´107 4.0 ´107 6.0 ´107 8.0 ´107 1.0 ´108 1.2 ´108 1.4 ´108
2. ´10-8
4. ´10-8
6. ´10-8
8. ´10-8
where λ = hmπc
is the Compton wave length of the exchanged particle andαs ≈ 1/5 is the coupling constant of the strong interaction.
However, this is only an approximation: other mesons than pions can be
20
exchanged by the nuclear force, and the nuclear force is not a central force,there is also spin-spin and spin-orbit couplings.
The weak interaction is 10−12 orders of magnitude weaker than the stronginteraction.
EM is 1137 weaker than the strong interaction
Graviation is 10−39 orders of magnitude smaller than the strong interaction.
2.4.2 Weak Interactions
The weak interaction is responsible for decay and transition of particles. It hasa very short range.
2.4.3 Gravitation
Graviation has no effect on the phenomena we’re dealing with.
3 ScatteringScattering is used to probe the inner structure of matter.
“The backscatter peak is caused by such primary gamma rays that have firstinteracted by Compton scattering in the surrounding material.” It is usuallythe case that most of the photons emerge with angle π and energy 1
2me.
• Compoton scattering spectrum: E′ = E1+ E
me[1−cos(θ)]
or λ− λ′ =1− cos (θ)
me.
Incident electron is at rest.
3.0.3.1 Three Body Collision A→ B + C +D
• EB =m2
A+m2B−(PA−PB)2
2mA
• Maximal value for EB is EB =m2
A+m2B−(mC+mD)2
2mA.
• Minimal value for EB is EB = mB .
3.1 Bethe-Bloch Formula“charged particles moving through matter interact with the electrons of atomsin the material. The interaction excites or ionizes the atoms. This leads to anenergy loss of the traveling particle. The Bethe formula describes the energyloss per distance travelled of swift charged particles (protons, alpha particles,atomic ions, but not electrons–their mass is too small for this approximation)traversing matter (or alternatively the stopping power of the material). “
For a particle with speed v, charge z, and energy E, traveling a distance xinto a target of electron number density n and mean excitation potential I, therelativistic version of the formula reads:
21
−dEdx
=4π
mec2· nz
2
β2·(
e2
4πε0
)2
·[ln
(2mec
2β2
I · (1− β2)
)− β2
]where c is the speed of light and ε0 the vacuum permittivity, = v/c, e and
me the electron charge and rest mass respectively.
• Observe that in the Bethe-Bloch formula the mass of the incident particledoes not appear.
• Memorize that −dEdx ∝ z2
β2 or
−dEdx
= C1Z21
β21
[ln
(C2
β21
1− β21
)− β2
1
]
• If two charged particles of mass m1 and m2 go and charge Z1e and Z2e gothrough the same material, their energy losses are related by the equation:dE2
dx =Z2
2
Z21
dE1(p·m1/m2)dx
• For low energies, the range is proportional to EZ2 .
3.1.1 Radiation Length and Mean Free Path
“In physics, the radiation length is a characteristic of a material, related to theenergy loss of high energy, electromagnetic-interacting particles with it.”
A property of the material that describes how much energy is lost for a highenergy electron travelling in a medium:
−⟨dE
dx
⟩=
E
X0
Thus the energy of the particle is given by:
E (x) = E0e− x
X0
Thus the mean free path is the average length an electron goes through in amedium until it collides ionizes an atom.
There’s also the pair-building mean free path, which is the length a photongoes before doing pair-production: Xp =
97X0.
Usually the mean free path is given together multiplied particle density al-ready. Thus the actual mean free path, for lead for example, if it is given as6.4g/cm2 and the density of lead is 11.3g/cm3, is X0 ≈ 0.566cm.
There is a critical energy under which it no longer is possible to make pair-production or to do Bremsstrahlung and the energy of a particle is simply de-posited as ionization. The number of particles at the end of such a cycle isroughly E0/EC .
In an EM calorimeter, ∆E ≈ 0.05√E.
22
3.2 Elastic Scattering3.2.1 Rutherford Scattering
In 1909 Rutherford sent α particles at a thin gold foil.(dσ
dΩ
)Rutherford
=
(Z1Z2α~c4Ekinetic
)2
sin−4
(θ
2
).
Can be derived quantum mechanically from Fermi’s golden rule by assumingplane-waves and delta-function charge distribution.
3.2.2 Mott-Scattering (no recoil, with e-spin, no nucleon-spin)
“Mott scattering, also referred to as spin-coupling inelastic Coulomb scattering,is the separation of the two spin states of an electron beam by scattering thebeam off the Coulomb field of heavy atoms. It is mostly used to measure thespin polarization of an electron beam.”
Mott scattering adds the effect of the spin of the electron to the Rutherfordscattering : (
dσ
dΩ
)∗
Mott=
(dσ
dΩ
)Rutherford
[1− β2 sin2
(θ
2
)]The asteric means recoil of the target has not been taken into account.
3.2.2.1 Helicity h := s·p|s||p| = ±1 When h = 1 the particle is right handed.
When h = −1 the particle is left handed. For highly relativistic particles,helicity is a conserved quantity.
3.2.3 Form Factor and Charge Distribution in the Nucleon
The form factor is the Fourier transform (into momentum-space) of the chargedistribution of the target.
• ρ (r) = ρ0exp(− r−a
b
)where a = 1.07A1/3 and b ≈ 0.54fm.
• Thus Ze =´dV ρ and F
(q2)= 1
Ze
´dV ρ (r) e−iq·r
It turns out, that in experiments, we don’t measure(dσdΩ
)∗Mott, but something
else. It is sensible to assume that what we measure is:(dσ
dΩ
)measured
=
(dσ
dΩ
)∗
Mott|F (q)|2
we could assume fit the measurements for example to a charge distributionof the form:
23
f (r) =f0
1 + er−ca
a kind of Fermi distribution.From that we can deduce the relationship between the nucleus radius and
the number of nucleons in it:
R ≈ 1.21fm ·A 13
3.2.4 Mott Scattering (with recoil, with e-spin, no nucleon-spin)
If we take into account the recoil of the target:(dσ
dΩ
)Mott
=
(dσ
dΩ
)∗
Mott
E
E′
where E is the initial energy of the incident particle and E′ is the final energyof the incident particle.
3.2.5 Scattering of Point Spin-Particles (with Recoil, with both spins)
If we take the spin-spin interactions into account we get:(dσ
dΩ
)Point Spin-1/2
=
(dσ
dΩ
)Mott
[1 + 2τ tan2
(θ
2
)]where τ :=
(p−p′)2−(
E−E′)24M2c2 and M is the mass of the target. If we denote
Q2 as minus the transferred 4-momentum we get τ = Q2
4M2c2 .
3.2.5.1 Nuclear Magneton
• µN = e~2Mp
≈ 3.1525 · 10−14MeV/T
• µp = +2.79µN for protons.
• µn = −1.91µN for neutrons.
3.2.6 Rosenbluth Formula (Griffiths Chapter 8.2)
But the nucleon is not a point particle, so we must correct for that with two formfactors: one for the electric interaction and another for the magnetic interaction.
(dσ
dΩ
)nucleon
=
(dσ
dΩ
)Mott
[A(Q2)+B
(Q2) Q2
2M2tan2
(θ
2
)]
24
• limQ2→0
[GE
(Q2)]2 is the electric charge.
– limQ2→0
[GE
(Q2)]2
= 1 for the proton.
– limQ2→0
[GE
(Q2)]2
= 0 for the neutron.
• limQ2→0
[GM
(Q2)]2 is the magnetic moment.
– limQ2→0
[GM
(Q2)]2
= 2.79 for the proton
– limQ2→0
[GM
(Q2)]2
= −1.91 for he neutron.
3.2.7 Experimental Determination of the Form Factors
The cross-section, for a fixed momentum transfer is measured, in various an-gles, and is fitted against the Rosenbluth-Formula: It is divided by the calcu-lated Mott cross section and then plotted as a function tan2
(θ2
): the slope and
intersect can give the magnetic and electric form factors then:
25
By making the experiment with a fixed value of Q2, one can try to fit amodel of the ansatz for G
(Q2), for instance, G
(Q2)∝ 1(
1+Q2a
~2
)2 . Thus by
making a Fourier transform we can get the charge distribution of a nucleon:ρ (r) = ρ0e
−ar where a ≈ 4.3 1fm .
26
3.3 Inelastic Scattering
Griffiths: “Resonance - a special energy at which the particles invovled ’like’to interact, forming a short-lived semibound state before breaking apart. Such’bumps’ in the graph of σ vs. Ekin are in fact the principal means by whichshort-lived particles are discovered.”
The name of a resonance denotes it invariant mass.If P is the 4-momentum of a target proton and q is the 4-momentum of an
exchanged particle, then the invariant mass, W , is defined as W 2 := (P + q)2
– the total mass available for the creation of new particles.
Define ν :=Pq
Mwhere M is the mass of the proton. Then ν is Lorentz
invariant, and (because Q2 := −q2):
W 2 = (P + q)2= P 2 + 2Pq + q2 =M2 + 2Mν −Q2
It is possible to show that ν is equal to the energy transfer from the electronto the proton, in the proton’s rest frame ν = E − E′ .
27
3.3.0.1 The Bjorken Scale Variable The Bjorken Scale Variable x is de-
fined as x :=Q2
2Mνwhere x ∈ [0, 1].
• For elastic scattering W =M =⇒ Q2 = 2Mν =⇒ x = 1
• For inelastic scattering E′ < E.
The Bjorken scale variable is a measure of the elasticity of the scattering:
• For elastic scattering x = 1.
• For inelastic scattering x < 1.
W 2 := (P ′)2= (P + q)
2= P 2+2Pq+q2 =M2+2Mν−Q2 =M2+2Mν (1− x)
Thus if x = 1 =⇒W =M , elastic.If x < 1 =⇒W > M , inelastic.
• The variable x can be interpreted as the fraction of the proton momentumcarried by the parton.
3.3.1 Width and Lifetime of Resonance
g
-100 -50 0 50 100
1
2
3
4
The cross section of a resonance with an energy ER can be approximatedvia the Breit-Wigner Distribution:
dσ
dΩ∝ Γ2
(E − ER)2+(Γ2
)2where Γ is the width of the distribution.From the Heisenberg uncertainty principle we can estimate Γ · τ ≈ ~ , where
τ is the lifetime of the resonance.
• Γ ∝ 1τ ∝ m, so the bigger the mass, the bigger the resonance width.
28
3.3.2 Modified Rosenbluth Formula and the Callan-Gross Relation
We would like to have a relation for the cross-section and form factors in nonelastic scattering.
We make the Ansatz:(dσ
dΩ
)=
(dσ
dΩ
)Mott
[W2
(Q2, ν
)+ 2W1
(Q2, ν
)tan2
(θ
2
)]where W1 and W2 are the structure functions as before.We define dimen-
sionless functions:
F1
(x, Q2
):=MW1
(Q2, ν
)F2
(x, Q2
):= νW2
(Q2, ν
)where x := Q2
2Mν is the Bjorken scale variable.If we put these definitions in the above anstaz we get the Modified Rosenbluth
Formula:
(dσ
dΩ
)=
(dσ
dΩ
)Mott
[1
νF2
(x, Q2
)+
2
MF1
(x, Q2
)tan2
(θ
2
)]If we compare this formula to the scattering cross section of a spin-1/2 point
particle we get:
2τ
1=
2ν
M
F1
F2
Which, if we define the τ := Q2
4m2 we get:
F2 (x) = 2xF1 (x)
Which is the Callan-Gross Relation.If we were to experimentally find that this relation indeed holds, we could
conclude that the constituents of the proton are point particles of spin-1/2.
3.3.3 Scale Refraction and Scale Invariance
? Not sure about this section...When F2 does not depend on Q2 there is “scale breaking”.The higher the Q2, the better the resolution and more likely it is to discern
constituent particles of the proton.
29
3.4 Summary and Remarks• Nucleons have point particle constituents. The inner structure of nucle-
ons can be prescribed with the structure factors, which depend on theexchange of energy and momentum.
• If the constituents have spin-1/2 the Callan-Gross relation is observed.
• The relation between scattering neutrinos and electrons on neutrons is:
F ν,N2 =18
5F e,N2
The factor 185 comes from the average of square of charge of proton vs.
neutron, and then the average of that over proton and neutron. TODO
• From the fact that´ 10F ν,N2 (x) dx ≈ 0.5, which is the part of the nucleon
momentum which is carried by the quarks, we conclude that there are moreconstituents in the nucleon which don’t interact electromagnetically–thegluons. TODO
• There are 3 constituent quarks, each of mass approximately 300MeV .This stands in contradiction ot the pions which have two quarks and havea mass of less than 600MeV , thus, there are also Valence Quarks whichcontribute to the observed properties such as charge and spin. TODO
• Scale Invariance / Scale violation: F2 depends on Q2!
4 Collider Physics and Particle Detectors
4.1 Particle Accelerators4.1.1 Van-De-Graff Accelerator
An electrostatic accelerator, can accelerate up to energies of 30− 40MeV .A motor loads up charges (from earth) on a belt by spinning rubber near felt,
this rubber belt then trasmit the charges to a dome by the same mechanism.
4.1.1.1 Center of Mass Energy The center of mass energy is defined asthe Lorentz-Invariant scalar product of the total four-momentum of the system:
s := (p1 + p2)2= m2
1 +m22 + 2E1E2 − 2p1 · p2
√s :=
√PµPµ
where Pµ is the total 4-momentum of the system.
30
If the particles are in a frame of reference in which there’s no relative3-momentum, then the momentum cannot reduce the total center of mass en-ergy. Thus these type of experiments always provide larger center of massenergy.
4.1.1.2 Cyclotron “charged particles accelerate outwards from the centeralong a spiral path. The particles are held to a spiral trajectory by a staticmagnetic field and accelerated by a rapidly varying (radio frequency) electricfield.”
The cyclotron frequency is: f = qB2πm . This frequency is independent of the
radius of motion or the velocity. Thus in each revolution of the particle, thesame frequency can be used to alternate the electric field so that the particlealways gains kinetic energy and never loses it. The magnetic field makes theparticle move in a spiral trajectory.
The upper limit on the energy of a cyclotron is 15MeV for protons and10MeV for electrons.
4.1.1.2.1 Synchocyclotron “A synchrocyclotron is a cyclotron in whichthe frequency of the driving RF electric field is varied to compensate for rela-tivistic effects as the particles’ velocity begins to approach the speed of light.“
f =f0γ
= f0√1− β2
4.1.1.2.2 Isochronous Cyclotron “An alternative to the synchrocy-clotron is the isochronous cyclotron, which has a magnetic field that increaseswith radius, rather than with time.”
4.1.2 Synchrotron
“the guiding magnetic field (bending the particles into a closed path) is time-dependent, being synchronized to a particle beam of increasing kinetic energy.”
4.1.2.1 Examples
1. LEP (Large Electron-Positron Collider) at CERN. Up to 206GeV .
2. LHC (Large Hadron Collider) Proton-Proton collisions of up to 7TeV . By2014, the energies should reach 14TeV .
3. Tevatron at Fermilab in Chicago. Proton-anti-proton collisions.
4. DESY (Deutsches Elektronen-Syncrotron) in Hamburg. Also called HERA(Hadron-Elektron-Ring Anlage).
31
4.1.3 Usages outside of Particle Physics Experiments
• Synchrotron acceleration is a very good source of high energy UV andX-ray photons. These can be used to study atoms, molecules, and crystals.
• Age studies of C12 and C14: Accelerator Mass Spectroscopy (also at theETH).
• In medicine: radiation therapy.
4.2 Important Terms• For a collider experiment,
√s =
√4Ebeam1Ebeam2
• For a fixed target experiment,√s =
√2Ebeammp
• The specific luminosity L, is a measure of the frequency of a collision atthe interaction point of an experiment. For a collider experiment, wheretwo wave-packets with density n1 and n2 respectively, with frequency fof collision, the specific luminosity is L = f n1n2
4πσxσywhere σx and σy are
the cross sections in the x and y directions, corresponding to the collisionplane.
4.2.1 Example–Peak Luminosity LHC
In a real collider particles get lost and are inject, the luminosity varies withtime. The highest reachable luminosity is called the “peak luminosity”. Atthe LHC it is around 2 · 107 1
bs . We can estimate the effective luminsotiy asLeffective ≈ 1
2Lpeak.The number of events is given by N = σL.
4.2.2 Example
The cross section for proton-proton collision at the LHC is around 100mb.The cross section for Higgs to four leptons is around 1fb. That would mean(with the above mentioned luminosity) around ten events per year.
4.3 Detectors4.3.1 Structure
4.3.1.1 Tracking Chamber Works like a cloud chamber. “We alreadyknow about the cloud chamber in the section about the experiments in the firstchapter. The tracking chamber works exactly like that. Electrically charged par-ticles ionize the gas in the chamber and thus bestowed upon ionization "clouds".Usually, however, semiconductor detectors are used today. Incident generatecharged particles in a diode electron-hole pairs, which then generate the voltagewhich is applied a measurable current impulse. Additionally, a magnetic field isapplied, so that the particles are deflected on circular paths.”
32
4.3.1.1.1 The Muon For a muon with momentum 100GeV in a mag-netic field of 1T we get a radius of 33m. The curvature is thus very weak. Thesagitta of the muon is s = l2
8r = 380µm.
4.3.1.2 Electromagnetic Calrometer–ECAL
• An example for the radiation length is X0 ∼ 1cm.
• After 25X0, around 95% of the energy of the electrons and photons withan initial energy of 100GeV is absorbed.
• The number of produced particles per shower is about five to six timesthe radiation length.
• A shower has a transverse spread of only 3cm to 6cm.
4.3.1.3 Hadronic Calorimeter–HCAL The Hadron Interaction Length,Λ, is a used term instead of radiation length.
• The Hadron Interaction Length is in real experiments is around Λ = 10−20cm
• After 8Λ around 95% of the energy of Hadrons with initial energy of100GeV is absorbed.
• The maximum number of produced particles per shower is around 2− 3Λ.
• A shower has a transverse spread of 30− 60cm.
4.3.1.4 Muon Chamber The muon chamber detects charged particles aswell. Up to that point muons have not been absorbed. The Muon chamberallows to differentiate between muons and electrons and to determine the mo-mentum of the muons.
4.3.2 Measurements
4.3.2.1 Neutrinos Neutrinos cannot be detected directly so well, becausethey interact very seldom. But through the analysis of “missing energy” and“missing momentum” many parameters of the neutrinos can be determined. Forthat everything else about the reaction has to be measured, in order to invokethe conservation of energy and momentum. Other events can also account forthe missing energy or momentum:
• Imperfect detector geometry means measurable particles don’t get mea-sured.
• Failure of detector equipment.
• General measurement inaccuracies.
33
4.3.2.1.1 The Large Electron Collider at CERN Electron-PositronCollisions. There were searches for Z-boson with ECMS = 100GeV andW -bosonphysics with ECMS = 206GeV . Through these experiments the cross sectionand energy spectrum of W -W bosons and certain decay branches were deter-mined. In addition Supersymmetry was researched and the search for the HiggsBoson was conducted.
4.3.2.2 Measuring the Neutron Mass The decay of tritium 3H 3 He+e+ e is suitable to determine the neutrino mass.
c) The decay of tritium 3H!3 He+e+ e is suitable to determine the neutrino mass (Fig. 1).Give a list of possible experimental diculties to conduct such a measurement?
Abbildung 1: An early plot of the energy spectrum in tritium -decay, by Langer and Mo↵att(1952), showing the linearity of the Kurie plot and the turnover expected for various neutrinomasses. At low energies the Fermi correction factor F(Z, p) is included to account for theinteraction between the electron and the coulomb field of the nucleus and retain the linearityat low energies.
Hint : The nucleus recoil can be neglected O(103MeV) and thus E0 = E+Ee+Ep E+Ee. Theelectron phase factor when the wave-function is normalized in unit volume and integratedover space angles is dNe =
4(2~)3 p2
e dpe; an analogous factor should be used for the neutrino.
5. Double beta decay
In a double decay a nucleus with atomic mass numbers (Z,A) transforms to:
(Z,A)! (Z + 2,A) + 2e + 2e (2)
These decays are due to second order weak interactions and have a rate proportional to G4F
where GF is the Fermi coupling constant having a typical life time of the order of 1020y. Thepossibility of having a neutrinoless double decay is under current experimental research.
a) What conditions must nuclei fulfill so that it is possible to maesure reaction 2? Nameone of nuclei, with which the double beta decay is possible. Take into account the dropmodel of nuclei (see series 9, eq.4 Fission):
M(A,Z) = NMn + Z(Mp +me) avA + asA2/3 + acZ2
A1/3 + aa(N Z)2
4A+
A1/2 ,
Hint: Consider the shape of mass parabula of isobaric nuclei for even-even and odd-oddnuclei.
b) Why is the observation (or not) of a neutrinoless double decay is important? Whatdoes it imply for the nature of the neutrinos?
√N(pe)p2e
(Ee) ∝[1−
(mν
E0−Ee
)2]1/4(E0 − Ee)
34
Experimental difficulties:
• The beta spectrum is steeply falling at the end point. As a consequence thecounting rate is low, and the statistical error is big while the backgroundbecomes significant.
• Due to the finite detector resoltion the Kurie plot is distorted (smeared)and has a high energetic tail, which extends beyond E0.
• Electrons change their energy via internal scattering inside the sourcewhich increases the intrinsic spread of the measured energies of the elec-trons. This problem can be minimized using small sources but leads to adecreased counting rate.
• After the decay the daughter ions or molecules can be in an excited statewith different binding energies. As a consequence we have a superpositionof beta decay spectra.
35
• Sargent rule: in weak interactions the total transition rate is proportionalto Q5 where Q is energy available in the reaction.
4.4 Up to Date Examples–LHC at CERN4.4.1 The Higgs Boson
“The whole begins with the development of QED (quantum electrodynamics),which the electromagnetic interaction of massless particles (photons) completelydescribes. On the other hand you discover the short-range weak interaction,which is produced by massive spin-1 (vector) bosons (W and Z bosons). How-ever, a consistent theoretical formulation of the weak interaction is possible onlyfor massless interaction particles (Yang-Mills 1954). Peter Higgs solves this byintroducing a symmetry breaking (Higgs mechanism) and a new spin-0 particle,the Higgs boson, predicted that gives the vector bosons of the weak interactionmass. Later succeeds then Glashow, Weinberg and Salam to unite the electro-magnetic and weak interaction by mixtures of vector bosons and the photonconstruct (more on that later). These predictions are experimentally 1983/84confirmed at CERN find this bosons. What is missing to complete the standardmodel is thus the Higgs boson. Some Possible actions in which a Higgs parti-cle could arise, samples are given in Figure 4.6. Conversely, the theory predictsalso decay modes and Relations, for a given energy / mass of the Higgs advance.The mass of the Higgs is unknown, so you have to search the whole energy spec-trum. The theoretical branching ratios are shown in Figure 4.7. Two importantobservables in the Standard Model are the QED fine structure constant ondthe Weinberg angle (W), which is the mixing angle between photon and Z0describes (we will introduce this detail later). According to the standard model,the experimentally measurable size sin2 (W) of the mass of the Higgs boson andthe QED Feinstruktukkonsante (and the mass of the top quark) depends. Wecan thus testify about the Higgs boson make (because if the model should becorrect).”
• Why is the decay channel H ZZ llll often referred to as the “goldenchannel”?
1. Almost no background
2. No missing energy, the invariant mass is the mass of the Higgs Boson.
3. Easy to trigger.
36
4.4.2 Measurements at the LHC
5 BaryonsLight baryons, made up of u, d or s quarks.
5.1 IsospinIn the study of nucleons, it was found that protons and neutrons operate asidentical particles under the strong interaction. In light of this, Heisenbergconsidered both particles as two states of the same particles, differing only byelectric charge. The isospin is only defined for particles containing u and dquarks.
5.1.1 Definition of Isospin “z-component”
Iz = I3 =1
2
[(nu − nu)−
(nd − nd
)]where by ni is the number of quarks of type i.
5.1.2 Example
• Proton: I = 12 , Iz = 1
2 .
• Neutron: I = 12 , Iz = − 1
2 .
The z-component of the isospin is additive for a whole system: Iz,total =∑i Iz,i.
37
5.1.3 Example
5.1.3.1 Mirror Nuclei “Mirror nuclei are nuclei where the number of pro-tons of element one (Z1) equals the number of neutrons of element two (N2),the number of protons of element two (Z2) equal the number of neutrons inelement one (N1) and the mass number is the same.”
5.1.3.2 148 C6 and 14
6 O8 are Mirror Nuclei Mirror nuclei should have sim-ilar energy levels, because the strong interaction is invariant under isospin, andthe only differences should stem from EM forces.
5.1.4 Remark
Experimentally it has been found that isospin is conserved by strong interac-tions. Thus it is a good “quantum number”.
5.2 Classification and Nomenclature5.2.1 Strangeness
A strange quark has strangeness of −1. Thus the strangeness is the number ofanti-quarks in the Baryon!
S := − (ns − ns)
• It has been experimentally established that strangeness is aconservedquantum number in strong interactions, but not in weak interactions.
• Baryons with strangeness are called hyperons (but no charm or bottom).
5.2.2 Nucleons and Deltas
Nucleons are Baryons with isospin I = 12 whereas ∆-particles have isospin I = 3
2 .
5.2.3 Hyperon
5.2.4 The ∆-particle
38
5.3 Conservation and Electric Charge5.3.1 Baryon Number
The Baryon number is a quantum number, which, (at least as far as could bedetermined up until today), is always conserved. It is defined as
B :=nq − nq
3
whereby nq is total number of (any) quarks and nq is the total number of (any)anti-quarks.
• For example, pp → π+π− is not allowed because baryon number is notconserved. But pp→ π+π− is allowed, because baryon number before andafter is zero.
5.3.2 Strangeness
The strageness is always conserved in the strong interaction. That means intransitions with exchange of gluons S is constant.
5.3.3 Electric Charge of the Baryons
The electric charge of the Baryons is given by:
Q = I3 +1
2(B + S)
5.4 Baryon MultipletsOut of the sss delta particle, we see that there must another quantum number(due to Pauli’s exclusion principle): the color.
5.4.1 Baryon Decuplet–The Ten Fold Way
Baryons with total spin S = 32 , orbital angular momentum of L = 0 and thus
total angular momentum of J = 32 form a Baryon decuplet.
L = 0, so the space-wave-function is symmetric.In order to reach S = 3
2 , all spins have to be aligned at the same direction,and so the spin wave function is also symmetric.
If we have three quarks of the same flavor, then the color wave function mustbe anti-symmetric. This can be achieved in the following way:
φ (color) =1√6
∑α∈r, g, b
∑β∈r, g, b
∑γ∈r, g, b
εαβγ |qαqβqγ〉
where εαβγ is the anti-symmetric tensor (Levi-Civita tensor).
39
5.4.1.1 Example–Symmetrising a Wave Function Symmetrizing the∆+ = |uud〉 wave function:
ζ (flavor)χ (spin) =1√3(|uupuupdup〉+ |uupdupuup〉+ |dupuupuup〉)
5.4.1.2 Remark For three different quark flavors, there are 10 possible com-binations:
5.4.2 Baryon Octet–The Eight Fold Way
Baryons with total spin S = 12 , orbital angular momentum L = 0 and thus total
angular momentum J = 12 form a Baryon octet.
L = 0, so the space-wave-function is symmetric.
40
In order for the total wave function to be anti-symmetric, we thus make,as before, the color wave function anti-symmetric and so the flavor-spin wavefunction should be symmetrized.
The spin is S = 12 , so the spin wave function is up,up,down, which is mixed,
so the flavor wave function has tobe mixed as well.
5.4.2.1 Example–Anti-symmetrizing the Proton We consider the twospin-up u’s as one spin-1 system and the one spin-down d as a spin-1/2 system(via the Clebsch-Gordan coefficients):
χproton
(J =
1
2, mJ =
1
2
)=
√2
3χuu (1, 1)χd
(1
2, −1
2
)−√
1
3χuu (1, 0)χd
(1
2,1
2
)where
Thus
Now the state is symmetric for exchange of 1 and 2. To completely sym-metrize it, we use the flavor symmetry:
Similarly the wave function for the neutron could be achieved.
41
5.4.2.2 The Octet
5.5 Magnetic MomentFor a spin-1/2 point particle with l = 0 the magnetic moment is given by:
µ =q~2M
Thus for quarks:
Using the above wave function for the proton, we can calculate the expecta-tion value for the magnetic moment:
42
Thus for protons:
µp =1
3(4µu − µd)
and analogously for neutrons:
µn =1
3(4µd − µu)
Thus assuming that mu ≈ md we get:
µnµd
= −2
3
The experimental result is:
µnµp
= −0.685
Thus the eight-fold way is a cool theory.
5.6 Mass of the BaryonsThe mass of the Baryons is estimated partly by the mass of their constituentsand party by the spin-spin interactions of their constituents. Thus the mass ofa Baryon is:
M = m1 +m2 +m2 +A
(s1 · s2m1m2
+s1 · s3m1m3
+s3 · s2m3m2
)whereby A is a constant and si is the spin of the ith quark.
Through experimentally measuring the mass of a Baryon we could “fit” themasses of the various quarks: mu ≈ md ≈ 363MeV , ms ≈ 538MeV .
(Derivation in POVH chapter 13.4 or 15.3)
5.7 π-Nucleon-Systems and Isospin Considerations5.7.1 Pions
The pions are mesons composed of:
43
All pions have spin 0 and isospin 1. Thus the three pions form an isospintriplet:
The mass of the pions is approximately 130− 140MeV .
5.7.2 π-N-Systems
The possible reactions for pion exchange between nucleons:
5.7.2.1 Isospin Mixing as Angular Momentum Addition -
Thus the wave function of the proton and neutron are:
44
5.7.3 Pion Exchange
We can also calculate the same result as above in the following way:The exchange of pions by two nucleons can be considered as a scattering
process. We now proceed to estimate the transition matrix element of thisscattering process.
Then requiring that a2 = b2 = ab + c2 and also that c2 6= 0 (for chargeconservation)
So then for example, the neutron is ,which gives us the same result as above.
6 Symmetries and Symmetry Breaking
6.1 Conserved Quantities6.1.1 Noether’s Theorem
“any differentiable symmetry of the action of a physical system has a corre-sponding conservation law. “
6.1.2 Conserved Quantities in Quantum Mechanics
An observable O is a conserved quantity if ddt
⟨O⟩= 0.
45
Schroedinger’s equation states that i~dψdt = Hψ =⇒ −i~dψ∗
dt =(Hψ
)∗=
ψ∗HThus:
6.1.3 Examples for Symmetries and their Corresponding ConservedQuantities
• Space translation symmetry: momentum conservation
• Rotation symmetry: angular momentum conservation
• Time symmetry: energy conservation
• Local gauge symmetry: electric charge conservation
• Inversion symmetry: Parity conservation
The last three symmetries are denoted by the operators T , C, P:
• Time reversal: T : t 7→ −t
• Charge conjugation: C : particle 7→ anti-particle
• Parity: P : r 7→ −r
– “The parity of an anti-fermion is opposite to the parity of a fermion,whereas the parity of an anti-boson is the same as the parity of aboson.”
46
6.2 Parity ViolationBy mirroring, momentum changes sign, however, the spin does not (as it is across product) and so the helicity changes sign under parity mirroring.
• Maximal parity violation:
– V-A: W couples only left handed Fermions and right handed anti-Fermions.
– V+A: W couples only right handed Fermions and left handed anti-Fermions.
• No partiy violation:
– W couples to both left handed and right handed particles.
• From the SPS experiment of proton-anti-proton collisions, examining theangle distribution between the resultant positron and the anti-protonproves that the W has spin 1 and it couples with V ± A that is, max-imal parity violation.
6.2.1 The Wu-Experiment
47
The Coblat nuclei were cooled down, to insure that the spins are reallyaligned with the magetic field (that is, without thermal distribution)–that iswhat energetically economical for them.
Thus:
Griffiths chapter 4.4: “Radioactive cobalt-60 nuclei were carefully aligned,so that their spins pointed in what was chosen to be the z-direction. Cobalt60 undergoes beta decay: 60Co →60 Ni+ e+ νe. Wu recorded the direction ofthe emitted electrons. What she found was that most of them came out in the’southerly’ direction, opposite to nuclear spin.”
This proves that there is no parity symmetry in this process, for if therewere, then there should be an approximately equal amount of electrons goingoff in the same direction as the nuclear spin.
Wikipedia: “It is now the number of in negative z-direction of emitted elec-trons is measured. One must distinguish here the following two orientations ofthe nuclei: Forward : The nuclear spins are aligned in the positive z-direction.In the negative z-direction detected electrons were emitted thus counter to thedirection of spin (i.e. with negative helicity). This can be illustrated as follows(here the double arrow indicates a Spin-1/2 orientation, the single arrows fordirection of motion):
Reverse : The nuclear spins have intrinsic angular momentum, under mirrorsymmetry the angular momentum vector, p, is still in the same direction relativeto the direction of motion, r: . To set up a "mirror" of the experimental set-up,it is therefore sufficient to rotate the nuclear spins with the magnetic field.There are then detected electrons, the spin vector in the same direction of the
48
momentum vector, with positive helicity.[citation needed] The nuclear spins arenow aligned in the negative z-direction, with the electrons being emitted in thepositive z-direction in the direction of the spin (i.e. with positive helicity) asfollows:
As shown in the diagrams above, in both cases the spin and angular momen-tum vectors are conserved. If the parity operation was conserved, both scenariosabove would be equally likely to happen: It would be the same number of elec-trons emitted in the direction of the nuclear spins in the opposite direction.Wu experimentally observed, however, that almost all the electrons are emittedagainst the direction of spin of the nuclei, which corresponds to a maximumparity violation.”
• Difficulties:
– Detector has to be placed near the Cobalt source since the e− havelow energy, and thus a short range.
– High polarization requires:
∗ low temperaturs∗ high magnetic field.
• The polarization of the Cobalt source was determined by the anisotropyof the gamma rays. A maximum polarization of 0.6 was achieved.
• If the weak interaction is parity invariant, the expected result is for thereto be equal probabilities for an electron to be emitted in the same directionas the nuclear spin as to be emitted opposite the direction of the nuclearspin.
• Momentum gets flipped with partiy transform, but angular momentumdoesn’t, so if we make a partiy transformation, the electrons should beemitted with the direction of the spin, but the spin should still pointto the same direction, since its direction wouldn’t change under partiytransformation.
6.2.2 The Parity of the Pion
The parity of the pion was measured in 1954 by stopping negative pions in adeuteron target, they form a pionic atom and the pion decays into the atomicS-state before the reaction d nn takes place.
The overall symmetry factor for spin and angular momentum is (−1)L+S+1.
The parity of a two-particle state is (−1)L times the intrinsic parities.
6.2.3 Pion Decay (Griffiths page 132)
π± are the lightest hadrons, except for π0, so they decay into leptons.
49
If the muon or electron had no mass, then this decay wouldn’t have beenpossible, because then it wouldn’t have been possible to switch to a frame ofreference in which the particles are actually left handed. The portion of themuons that are left handed is (1− β).
6.2.3.1 Pion Production In collision between protons and neutrons thefollowing are possible:
Observe: the formation of π+ and π0 is more probable than π−.In π+ and π− the parity is also violated:
Neutrinos are left handed and anti-neutrinos are right handed. Thus there
50
is only one type of reaction possible:
Measurements of the helicity of the muon allows us to determine the helicityof the anti-neutrino, by conservation laws.
6.3 Charge ConjugationOnly neutral particles can be eigenstates of the charge conjugation operator, forexample, the neutral pion, or the photon.
for n photons.
6.3.1 Decay of the Pion
This is because the initial state has C eigenvalue 1 whereas that of threephotons has eigenvalue −1, and the transition is C-invariant.
6.3.2 Neutrino Symmetries
By combining C and P together, we get the CP operator:
as can be seen, the neutrino is CP-invariant.
6.4 Helicity of the Neutrino–the Goldhaber ExperimentThe Goldhaber experiment allowed the determination of the helicity of the neu-trino.
51
6.4.1 The Steps of the Experiment
6.4.1.1 Electron Capture Through a Weak ProcessThen the excited Samarium emits a photon.
6.4.1.2 Angular Momentum
6.4.1.3 Spin Orientation The Samarium and Neutrino could either bothhave helicity 1 or both have helicity −1.
6.4.1.4 Photon Polarization 152Sm∗ →152 Sm+ γWe are interested in the case where the photon has the same momentum
direction as the Samarium, in which case they have the same helicity.
6.4.1.5 Polarization Filter Only photons with certain helicity pass througha magentized iron.
6.4.1.6 Experimental Construction
6.4.1.7 Results Only L.H. photons were measured. But the helicity of thephotons is the same as the helicity of the neutrinos. Thus Hν = −1.00± 0.3.
52
6.4.2 Measuring the Weak Interaction
H = HEM +Hstrong +Hweak
The transition probability is proportional to H2. Thus even though the weakint. is so weak, we can study it via the term HweakHstrong.
6.4.3 Eigen Parity
The eigen value of the parity operator for Bosons is the same as for anti-Bosons.For Fermions it is minus the anti-fermions.
6.5 The CPT TheoremAccording to QFT ∃ CPT symmetry in nature.
6.6 KaonsK0 = ds
K− = usK+ = us
6.6.1 Production of Neutral Kaons
6.6.2 Interaction with Material
Possible reactions with various energies, in which baryon number and strangenessmust be conserved:
53
??? TODO
6.7 Parity Invariant 2-state system6.7.1 Double Potential Well
Solution to the double potential well as a finite perturbation to the infinite twowells.
We get two states, symmetric and anti-symmetric, both of which are eigen-states of the parity operator.
6.7.2 K0 −K0-System
(Griffiths 4.4.3.1)We take the K0 −K0 to be two states which are CP eigenstates.
Because (CP)2= 1, we can choose η = η′ = −1.
The Kaon can decay into two or three pions. Thus there was the idea thatthe kaons and anti-kaons can go back and forth from one another.
54
Two pions have parity 1 whereas three pion have parity −1. Both haveC = 1. Thus K1 decays into two pions whereas K2 decays into three pions.
These two eigenstates have different lifetimes, because the 2π is much faster,because the energy released is greater. So if we start with a beam of K0 =1√2(K1 +K2) the K1 component will decay away quickly and down the line
there will be a beam of pure K2.Observe: K1 and K2 are not anti-particles of one another, rather, each is its
own anti-particle.Kaons are produced in the strong interaction, as eigenstates of strangeness,
but they decay in the weak interaction, as eigenstates of CP.
• CP(π0)= −1, CP
(π0π0
)= 1
• CP (π±π±) = 1
• CP(π0π0π0
)= −1
• CP(π±π±π0
)= −1
55
• Thus K1 decays weakly into two pions and K2 decays weakly into threepions, the phase space available for two pions is larger, so the decay of theK1 is faster, hence it has a shorter lifetime.
• τK1≈ 8.96 · 10−11s
• τK2 ≈ 5.18 · 10−8s
6.7.3 Regeneration of Kaons
If you wait long enough, you get onlyK2 in your beam. But you could regenerateK1. If you shoot the K2 beam at a material that has different cross sectionbetween K0 and K0.
6.8 Indirect CP-Breaking with Neutral KaonsExperimental data shows that for a pure K2 there are still some 2π decays: theK2 is not a pure eigenstate of CP.
With B-mesons direct CP violation was observed.
7 Particle Physics at PSI
7.1 Proton AcceleratorThe proton accelerator at the PSI can accelerate particles to energy of up to590MeV . The current of the resulting particle-current is Ip = 2.2mA. Theresulting power is P = 1.3MW . Protons are shot at a target, so that pions areproduced, which can then used for further experiments or the following reactioncan be researched:
p+N → x+ y + π
Provided the produced particles are charged, they can be deflected by a magneticfield. At the PSI the following questions are examined:
• What makes leptons so fundamental? Something about being point par-ticles: a muon cannot be distinguished from an electron, other than beingheavier.
• The process µ → e + γ is not allowed in the standard model. But theprocess µ → e + ν + ν exists. Thus there must be a conserved quantitywhich is forbidden in the first process and allowed in the second.
56
• Up to now the following relation for both processes has been observed:R(µ→e+γ)R(µ→e+ν+ν) < 10−11.
• Investigation of the muon decay allows us to determine the lifetime of themuon:
1
τµ=G2Fm
5µ
192π3F (p)
(1 +
3
5
m2µ
M2W
)
whereby GF is the Fermi constant and MW is the effective coupling con-stant for the weak interaction.
• With these known constants the lifetime of the muon was found to beτµ = 2.2µs.
7.2 Muonic-Hydrogen LambshiftIn 2010 the radius of the proton was newly determined at the PSI. The resultdeviates strongly from the value measured up to that point.
7.2.1 Lamb Shift
Vacuum fluctuations lift the degeneracy of the levels at the n, l energy level inthe hydrogen atom. The shift is proportional to the mass of the electron. Thusreplacing the electron in the atom with the muon, we get a much larger shift.
7.2.2 The Proton Radius
Because the proton is not a point particle, the Coulomb potential 1r does not
describe it well for r → 0. So we use another potential. If we include thehyperfine structure, we can shoot a laser to raise the muonic hydrogen from 2Sto 2P with the right energy and through the Lamb shift ascertain the protonradius.
7.3 CP-Violation of Ultra Cold NeutronsBased on the inspection of the electric dipole moment of neutrons it is possible toverify the CP-violation. Despite the fact that the neutron is electrically neutral,it can still have an electric dipole moment, independently of the distributionof charge. If the neutron has an electric dipole moment, then the parity andtime-reversal are violated.
7.3.1 Matter-Anti-Matter Assymetry
Sacharow postulated 3 criteria to explain the assymetry between matter andanti-matter:
• Thermodynamic out-of-equilibrium during the big bang
57
• Violation of the baryon number conservation
• Violation of the C and CP symmetry.
7.3.2 Ultra Cold Neutrons (UCN)
“To demonstrate the CP violation, so the electrical dipole moment has to bedetermined. These neutrons are slowed down to a few meters per second. Thisis done at the moment at PSI (see slides). Wherein the dipole moment isthen measured by placing the UCN in a uniform electric and magnetic field.The magnetic field which oscillates time-reversal is generated. By reversing thepolarity of the electric field, a space inversion is created. By measuring theso-called Larmor frequency, the dipole moment can be determined. So far, thedipole moment of about 2.9 ·10−26e cm could be determined accurately. At PSI,the UCN to help this upper bound further and correct site down.”
8 Properties of the Weak Interaction(chapter 9 in Griffiths)
Because the cross section for scattering experiments of the weak interaction isvery small, the properties of the weak interaction are measured through decays.
8.1 Universality of the Charged Weak Interaction8.1.1 β+ decay
(du)u→ (du) d+W+ → (du) d+ e+ + νe with lifetime of around 103s.
58
8.1.2 µ+-decay
µ+ → +νµ +W+ → νµ + e+ + νe with lifetime of around 2.2µs.
59
8.1.3 τ+-decay
τ+ → +ντ +W+ → ντ + e+ + νe or τ+ → +ντ +W+ → ντ + µ+ + νµ withlifetime of around 3 · 10−13s. The branching ratios for both is around 0.18.
8.1.4 The Universality
The matrix element of the weak interaction is all the same for the above inter-actions. The only difference is in the phase space volume.
The relation between lifetime and transition probablity is τ ∝ 1Γi→f
, thus,τ ∝ 1
p5maxwhere pmax is the maximal momentum of the positron.
Γi→f = 2π~ |Mfi|2 ρ (Ef )
the matrix element is the same for all three decays. Thus the only differenceis in the phase space room, which is proprotional to pmax. If we plot log (τ) ∝pmax we will see the slope is indeed −5 which confirms that fact.
Mfi ∝ g · 1Q2+M2
W· g, where g is the weak charge. From that we find that
g ∼ e.
8.1.4.1 The Fermi Constant The Fermi constant GF is defined as GF :=√2(πα2g2
e2(~c)3M2
W c4
)It is the effective coupling constant if the decay was a 4-particle
decay (without the W) (Griffiths 9.2).
60
8.1.5 Summary
The universality of the weak charge is the same for quarks and leptons (verifiedexperimentally)!
8.1.6 Example–The Decay of the Tau
The decay of the tau is about 3 times more likely into a pion and somethingthan it is for a lepton (either electron or muon), which can be explained by thefact that quarks come in three colors so they are three times as likely.
8.2 Quark Mixing(Griffiths 9.5)
There are six types of quarks, by rising mass:
They are eigenstates of the strong force. In the strong force only transitionswithin a family are allowed. But the weak force does not respect that.
8.2.1 Quark Mixing Matrix
8.2.2 Cabbibo-Kobayashi-Masukawa Matrix Theory and Experiment
We can estimate the ratios between different entries in the CKM matrix byconsidering the branching ratios of different reaction:∣∣∣Vub
Vcb
∣∣∣ ≈√Γ(reaction involving b→u
)Γ(reaction involving b→c
)This is because at each vertex, the coupling constant will be gW · |Vbc|, for
an exchange between b→ c for example.The CKM matrix is mostly diagonal, but not quite.
8.2.3 GIM Mechanism
In quantum field theory, the GIM mechanism (or Glashow–Iliopoulos–Maianimechanism) is the mechanism by which flavour-changing neutral currents (FC-NCs) are suppressed. It also explains why weak interactions that change strangenessby 2 (S = 2 transitions) are suppressed while those that change strangeness by1 (S = 1 transitions) are allowed. The mechanism was put forth by SheldonLee Glashow, John Iliopoulos and Luciano Maiani in their famous paper "WeakInteractions with Lepton–Hadron Symmetry" published in Physical Review Din 1970.[1] At the time the GIM mechanism was proposed, only three quarks(up, down, and strange) were thought to exist. Glashow and James Bjorken pre-dicted a fourth quark in 1964,[2] but there was little evidence for its existence.
61
The GIM mechanism however, required the existence of a fourth quark, and theprediction of the charm quark is usually credited to Glashow, Iliopoulos, andMaiani.
8.2.4 CP-Violation Revisited
8.3 ElectroWeak Unification8.3.1 Nuetral Weak Interaction: The Z0 Boson
Detected in the decay of Z0 → e+ + e−.
8.3.2 Weak Isospin and the Weinberg Angle
Weak iso-spin is zero for right-handed fermions. This involves weak-quark eigen-states (rotated by the CKM matrix). It’s + 1
2 for the “not-negatively” chargedfermions and − 1
2 for the negatively charged fermions. It’s −1 for W− and +1for W+. As a triplet, we expect there to be also a W 0 with weak iso-spin |1, 0 >and B0 with weak iso-spin |0, 0 >.
|γ >= cos (θW ) |B0 > +sin (θW ) |W 0 >|Z0 >= − sin (θW ) |B0 > +cos (θW ) |W 0 >where θW ≈ 30.Then e = g · sin (θW ).
8.4 Neutrino OscillationsToday we know of three different species of neutrinos, depending on whetherthey are produced in association with an electron (e), a muon () or a tau ()lepton. For simplicity, let us consider only the first two species. The |e and the |states are created by weak interaction (weak eigenstates). But it is now knownthat free neutrinos do not stay in the state in which they were originally createdbut they oscillate between the different flavours. The weak eigenstates (|e and|) are thus not identical with the eigenstates of the free neutrino hamiltonian(energy eigenstates). The energy eigenstates are denoted |1 and |2, with thecorresponding eigenvalues E1 and E2. The relation between the weak eigenstates(|e and |) states and the energy eigenstates (|1 and |2) is given by:
6. Z! qqg
7. Z! qq
8. WW ! µ!qq
b) LEP performed e+e"-collisions with a center of mass energy#
s up to 209 GeV in 2000.An integrated luminosity of 536 pb"1 for collisions with 206 GeV $
#s $ 209 GeV was
recorded. Considering Fig. 8 and 9, why are there no event displays of events includingHiggs bosons on the cafeteria tables at CERN, and which of the events from a) showsa probable background process? (Assume a mass of the Higgs boson of about 125 GeVand a main production process e+e" ! HZ. Data and plots from [1].)
4. Neutrino oscillations
Today we know of three di!erent species of neutrinos, depending on whether they areproduced in association with an electron (!e), a muon (!µ) or a tau (!") lepton. For simplicity,let us consider only the first two species. The |!e% and the |!µ% states are created by weakinteraction (weak eigenstates). But it is now known that free neutrinos do not stay in thestate in which they were originally created but they oscillate between the di!erent flavours.The weak eigenstates (|!e% and |!µ%) are thus not identical with the eigenstates of the freeneutrino hamiltonian (energy eigenstates). The energy eigenstates are denoted |!1% and |!2%,with the corresponding eigenvalues E1 and E2. The relation between the weak eigenstates(|!e% and |!µ%) states and the energy eigenstates (|!1% and |!2%) is given by:
|!e% = cos (#) |!1% " sin (#) |!2% (2)
|!µ% = sin (#) |!1% + cos (#) |!2% (3)
thus defining the mixing angle #.
a) At the time t = 0 an electron neutrino is created. What is the state of this neutrino at alater time in the basis of |!1% and |!2%?
b) At a later time t this neutrino interacts weakly with a nucleus. What is the probabilitythat it creates a muon in this interaction? What is the (minimal) time T after which thisprobability is maximal? What is the relation between the maximal probability and themixing angle?
c) The masses of the neutrinos are known to be non-zero, but very small: only upper limitshave been set on their masses so far. Neutrinos are therefore ultra-relativistic (m& p),and the mass is a mere
”correction“ to the momentum. Let m1 and m2 be the masses of
the two energy eigenstates E1 and E2, of momentum p0. Express the energy di!erenceas a function of the masses and derive the length L for which the mixing probability ismaximal.
Literatur
[1] A. G. Holzner, Nucl. Phys. Proc. Suppl. 115 (2003) 369 [hep-ex/0208045].
If, for example, at time 0 there was an electron neutrino, the probability tohave a muon neutrino vs. time is:
|〈νµ|ψ (t)〉|2 = [sin (2θ)]2
[sin
(E1 − E2
~t
)]2The length at which the probability to find a muon neutrino is thus:
62
Lmax =~π2p0c
(m21 −m2
2)
Part II
Nuclear Physics9 Nucleon Model
9.1 Introduction9.1.1 Charge Density Distribution in Nucelon
The charge distribution of a nuclide looks like this:ρ (r) = ρ0
1+e(r−c)/a
whereby ρ0, a and c are experimental parameters.For big nuclides c = 1.07fm ·A1/3 and a = 0.54fm.
63
Because the nucleus is basically incompressible, it can be deduced that thestrong interaction is repulsive at very small distances.
9.1.2 Binding Energy in Nucelon
Binding energy per nucleon in the nucleus:
64
The lightest stable nucleus is the deuteron (neutron and a proton). 4He hasa very high binding energy.
9.2 Semi-Empirical Mass Formula (Liquid Drop Model)
3. Spin-orbit coupling
Nuclear potential can be written as,
V(r) = Vcentre(r) +Vs.o.(r),
where Vs.o.(r) describes the coupling of the spin and the orbital angular momentum,
Vs.o.(r) = Vls(r) (l · s) =e2
8!"0m2e c2
1
r3(l · s).
Evaluate the energy splitting caused by the spin-orbit coupling for the 2p state of hydrogenatom with the following steps:
a) Express the expectation value of the product of the electron’s orbital angular momentumand spin, < l · s >, in terms of the eigenvalues of the total angular momentum, j2, etc.and show that the energy splitting can be written as,
!Es.o. =< Vls(r) >!2 (2l + 1)
2,
b) Compute the expectation value of the first term, < V#s(r) >, using the wave function forthe hydrogen 2p state of the form,
$nlm(r,%,&) = Rnl(r)Ym=0j (%,&).
(The corresponding radial and spherical harmonic wave functions can be found in textbooks.)
c) Finally, calculate the value of !Es.o. in units of [eV]. (The algebra can be simplified usingfine structure constant, ' ! 1/137)
4. Fission
In 1935 C.F.v. Weizsacker derived a phenomenological formula to describe the masses ofthe nuclei in analogy to a droplet of water:
M(A,Z) = NMn + Z(Mp +me) " avA + asA2/3 + ac
Z2
A1/3+ aa
(N " Z)2
4A+
(
A1/2
with N = A " Z. Assume the following values for the parameters:volume term av = 15.67 MeV ,surface term as = 17.23 MeV ,Coulomb term ac = 0.714 MeV ,asymmetry term aa = 93.15 MeVpairing term ( = "11.2, 0, 11.2 MeV for (Z,N) even-even, odd-even, odd-odd nuclei.
M (A, Z) := (A− Z)mn + Zmp −∆M∆M is called the mass defect, which is also the definition of the binding
energy.The mass of the nucleus is lower than the mass of its constituents, implying
that there should a positive binding energy.The semi empirical mass formula has several terms:
1. The Volume Term
65
The nucleus is considered to be a liquid-drop, and so, the binding energyshould be proportional to the volume of the drop. As the density is moreor less homogenous, the volume is proportional to A.
aV ·A
where av ≈ 15.7MeV
2. The Surface Term
TODO
3. The Coulomb Term
TODO
4. The Assymetry Term
TODO
5. The Paring Term
TODO
• For example, using the formula, we can calculate that the reaction nthermal+23592
U −→13150 Sn +102
42 Mo + 3n releases 176.59MeV . The kinetic energy ofthermal neutrons is ∼ 0.025eV .
• We could also compute the minimal ration Z2
A for spontaneous fission:Z2
A > 48. This is assuming a deformation in the nucleus of the form:
– Ellipsoid with a := R (1 + ε), b := R(1− 1
2ε), where ε is the defor-
mation parameter. Thus the volume and surface term change:
a) Use the above formula and compute the released energy, Q, of the following inducedfission reaction:
nthermal +23592 U!131
50 Sn +10242 Mo + 3n.
Remark: Q =Minitial "Mf inal is also called heat tone. Is the reaction exo- or endo-thermic?
b) Heavy nuclei (Z # 40) can in principle disintegrate into two lighter nuclei. But thepotential barrier, which has to be tunnelled in such processes, is so high, that nuclei nor-mally do not fission spontaneously. For very heavy nuclei spontaneous fission becomespossible again if the total energy of the nucleus is reduced under certain deformations.
Derive the minimal ratio Z2/A from which one we can expect spontaneous fission tooccur. Assume an ellipsoidal deformation of a normally spherical nucleus (radius R):
a = R(1 + !)
b = R!
1 "!
2
"
where ! is called the deformation parameter. The relevant changes in the mass formulaare:
asA2/3 ! asA
2/3!
1 +2
5!2 + ...
"
and
acZ2
A1/3! ac
Z2
A1/3
!
1 "1
5!2 + ...
"
Z is the number of protons and A the number of nucleons. Derive the critical value Z2/Afor which every small deformation results in spontaneous fission.
– Then we require that the binding energy of the deformed state ishigher than the binding energy with zero deformation (ε = 0).
9.2.1 Binding energy
9.2.2 Bethe-Weizsaecker-Formula
9.3 Fermi Gas ModelAssumptions of the model:
66
• The nucleons move freely in a potential box with no collisions. The po-tential on any one nucleon is caused by all other nucleons.
• Nucleons are spin-1/2 particles, and so, they must obey the Fermi-Diracstatistics, and correspondingly the Pauli exclusion principle.
• The potential is more or less constant in the inner of the nucleus and atthe edge rapidly falling, so it is approximately by a box potential.
The protons and neutrons are filled separately.The Fermi eneegy is approximately 30MeV , thus (as in the above picture)
the total energy of the box is about 40MeV .
• dNdE = a3m3/2
√E√
2π2~3
• EF = ~2
2m
(32π
2n) 3
2 The factor of 1/2 is because we put both proton andneutrons into the same state.
9.3.0.1 Explain the Assymetry term TODO (not a very illuminatingexplanation...)
9.3.1 Average Energy of a Particle in a Fermi Gas
〈E〉 ≈ (20MeV )×[A+ 5
9(N−Z)2
A
]
67
9.4 Nuclear Shell ModelThere are magic numbers in which nucleuns are more stable (for either thenumber of protons or neutrons or both) 2, 8, 20, 28, 50, 82, 126. There are alsohalf magic numbers, 40 and 64.
Evidence for nuclear shell model:
1. Large number of nuclides with proton or neutron magic numbers.
2. Small cross section for neutron absoption.
3. First excited level is high.
4. Separation energy should be higher.
9.4.1 Separation Energy
The separation energy is the energy required to remove one proton or neutron.
9.4 Schalenmodell der Kernphysik Kapitel 9
9.4 Schalenmodell der Kernphysik
Berucksichtigt man beim Fermi-Gas-Modell noch die Spin-Bahn-Kopplung und fuhrt realistischere Po-tentiale ein, so kommt man zum Schalenmodell der Kernphysik. Bisher haben wir gesehen, dass durchdas Pauliprinzip eine Art Schalencharakter vorliegt: die Nukleonen besetzten bestimmte Energieniveausim Einklang mit dem Pauli-Prinzip; analog wie man beim Atom vorgeht. Ferner werden wir beobachten,dass bei bestimmten Nukleonenzahlen die Kernkonfiguration besonders stabil ist. Diese Zahlen lauten(2,8,20,28,50,82,126) und werden magische Zahlen genannt. Es gibt ausserdem sogenannte halbmagischeZahlen 40 und 64.Im Schalenmodell wird versucht, die Wechselwirkung der vielen Nukleonen untereinander bzw. der ein-zelnen Nukleonen mit jedem anderen in Form eines gemittelten Potentials zu schreiben. Dazu verwendetman das Hartee-Fock -Verfahren. Insbesondere wegen der beobachteten magischen Zahlen lag es nahe,Systeme mit mehr oder weniger einem Potential zu beschreiben, in dem ein zusatzliches Teilchen immittleren Potential des Kerns (abgeschlossenen Systems) lebt.
9.4.1 Separationsenergie
In Abbildung 9.7 ist grob dargestellt wie sich die Energie eines Kerns verhalt, wenn man ein Protonbzw. Neutron entfernen will. Die benotigte Energie um ein Nukleon zu entfernen nennt man Separa-tionsenergie. Die Separationsenergie der Protonen nimmt bei zunehmender Protonenzahl ab (grossere
Abbildung 9.7: Separationsenergie fur Protonen (Sp) und Neutronen (Sn) bei zunehmenderNeutronen- und Protonenzahl.
Coulombabstossung) und bei steigender Neutronenzahl zu, da Proton und Neutron gerne miteinanderwechselwirken. Analog gehen die Uberlegungen fur die Separationsenergie der Neutronen.
Betrachtet man die Separationsenergie fur einige Elemente etwas genauer, findet man, dass sich beizwischen gewissen Isotonen und Isotopen grossere Lucken zeigen. In Abbildung 9.8 ist die Lucke beiN = 82 und Z = 82 gezeigt. Dies findet man auch bei weiteren bestimmten Zahlen, eben genau die obenaufgezahlten (halb-)magische Zahlen und sie sind ein Hinweis auf mogliche Schalenabschlusse, wenn wirein Schalenmodell zur Hilfe nehmen. Dies entspricht in etwa den gefullten Schalen der Edelgasen, welchesich ebenfalls sehr gut durch ein Schalenmodell fur Atome beschreiben lassen.Weitere Hinweise auf mogliche Schalenabschlusse bei denselben Zahlen findet man auch bei Untersuchun-gen der Masse und des Quadrupolmoments der Kerne.Wir wollen nun etwas mathematischer vorgehen und ein Schalenmodell fur Kerne entwickeln. Man be-nutzt dazu die Hartree-Fock-Methode, mit welcher man die Losung eines vereinfachten Hamiltonian fureinen Kern approxmieren kann.
9.4.2 Hartree-Fock-Methode
Wir versuchen einen vereinfachten Hamiltonion fur unser Kernmodell zu finden. Man versucht ein mitt-leres Potential zu finden, in dem die Wellenfunktion ein Produkt aus Einteilchenwellenfunktionen ist.
110
The picture is explained due to the fact that protons “happily” interact withneutrons, but not with themselves.
There is a gap in the separation energy around the magic numbers.
9.4.2 Hartee-Fock Method
A many-particle perturbative method. Not sure how much is expected to knowof this–it is highly technical.
9.4.3 Schroedinger Equation
It is assumed about the potential that:
1. It is radially symmetric: V (r) = V (r)
2. The nucleons are considered to be independent of each other. Each one isaffected by its own potential.
68
Due to 1., there could be no quadrupole moment: [H, L] =[H, L2
]= 0
Hψ =
[p2r2m
+L2
2mr2+ V (r)
]ψ = Eψ
Thus:
ψ (r) = Rnl (r)Yml (θ, ϕ)
and so:[− ~2
2m
∂2
∂r2+ ~2
l (l + 1)
2mr2+ V (r)
]· r ·Rnl (r) = Enl · r ·Rnl (r)
9.4.4 The Potential: 3D Harmonic Oscillator
V (r) =1
2mω2r29.4 Schalenmodell der Kernphysik Kapitel 9
Abbildung 9.9: Gezeigt sind drei verschiedene rotationalssymmetrische Potentiale.
wobei Nx, Ny, Nz ganzzahlig sind. Wir mochten nun fur verschiedene Quantenzahlen N die entsprechen-den Orbitale betrachten. Die Orbitale bzw. Kernkonfigurationen werden wie bei den Elektronen notiert,also 1s2s2 . . . . Der Zusammenhang zwischen den Quantenzahlen n, N und l ist gegeben durch (ohneHerleitung):
N = 2(n 1) + l, (9.43)
wobei n = 1, 2, 3, . . . . Da fur jedes Energieniveau, das durch die Quantenzahl N gegeben ist, zwei Nukleo-nen Platz haben, ergibt sich die Gesamtanzahl an Nukleonen fur die Quantenzahl l als 2
Pnl=1(2l + 1).
Der Faktor zwei kommt von der Spin-Entartung. Zur Erinnerung nochmals die Beziehung zwischen derQuantenzahl l und der spektroskopischen Notation:
s p d f g h ...l : 0 1 2 3 4 5 ...
(9.44)
Die Kernkonfigurationen fur die ersten 6 Quantenzahlen lauten:
N Kernkonfiguration (Orbitale) Paritat Anzahl degenerierter Nukleonen GesamtN = 2(n 1) + l (1)l 2(2l + 1)
0 1s + 2 21 1p - 6 82 2s, 1d + 2+10=12 203 2p, 1f - 6+14=20 404 3s, 2d, 1g + 2+10+18=30 705 3p, 2f, 1h - 6+14+22=42 1126 4s, 3d, 2d, 1i + 2+10+18+26=56 168
Wie man erkennt hat man bei 2, 8, 20, 40, 70, . . . abgeschlossene Schalen, was nur fur die tiefsten dreiZahlen mit den beobachteten magischen Zahlen 2, 8, 20, 28, 50, . . . ubereinstimmt. Wir mussen also unserPotential anpassen.Eine Moglichkeit besteht darin, das Potential der Ladungsdichteverteilung, welche wir zu Beginn die-ses Kapitels kennen gelernt haben, anzupassen. Ein bekannter Ansatz ist das sogenannte Wood-Saxon-Potential , welches auch in Abbildung 9.9 gezeigt ist, es lautet
V (r) = V01
1 + e(rR)/a. (9.45)
Man verwendet es bei grossen Kernen. Leider konnen auch damit die Schalenabschlusse (magischenZahlen) nicht zufriedenstellend erklart werden. Wir brauchen also noch eine weitere Korrektur.
9.4.5 Spin-Bahn-Kopplung
Wir haben die Energieniveaus beim Oszillator analysiert und gesehen, dass dieses Potential fur leichteKerne funktioniert. Die Schalenabschlusse fur A > 20 werden so jedoch nicht erklart. Diese Problem kann
114
V (r) =
−V0
(1− r2
R2
)r < R
0 r > R
E = ~ω(N + 3
2
), N ∈ N
N = 2 (n− 1) + l, n ∈ NFor every level l the occupation possible is 2
∑nl=1 (2l + 1).
9.4 Schalenmodell der Kernphysik Kapitel 9
Abbildung 9.9: Gezeigt sind drei verschiedene rotationalssymmetrische Potentiale.
wobei Nx, Ny, Nz ganzzahlig sind. Wir mochten nun fur verschiedene Quantenzahlen N die entsprechen-den Orbitale betrachten. Die Orbitale bzw. Kernkonfigurationen werden wie bei den Elektronen notiert,also 1s2s2 . . . . Der Zusammenhang zwischen den Quantenzahlen n, N und l ist gegeben durch (ohneHerleitung):
N = 2(n 1) + l, (9.43)
wobei n = 1, 2, 3, . . . . Da fur jedes Energieniveau, das durch die Quantenzahl N gegeben ist, zwei Nukleo-nen Platz haben, ergibt sich die Gesamtanzahl an Nukleonen fur die Quantenzahl l als 2
Pnl=1(2l + 1).
Der Faktor zwei kommt von der Spin-Entartung. Zur Erinnerung nochmals die Beziehung zwischen derQuantenzahl l und der spektroskopischen Notation:
s p d f g h ...l : 0 1 2 3 4 5 ...
(9.44)
Die Kernkonfigurationen fur die ersten 6 Quantenzahlen lauten:
N Kernkonfiguration (Orbitale) Paritat Anzahl degenerierter Nukleonen GesamtN = 2(n 1) + l (1)l 2(2l + 1)
0 1s + 2 21 1p - 6 82 2s, 1d + 2+10=12 203 2p, 1f - 6+14=20 404 3s, 2d, 1g + 2+10+18=30 705 3p, 2f, 1h - 6+14+22=42 1126 4s, 3d, 2d, 1i + 2+10+18+26=56 168
Wie man erkennt hat man bei 2, 8, 20, 40, 70, . . . abgeschlossene Schalen, was nur fur die tiefsten dreiZahlen mit den beobachteten magischen Zahlen 2, 8, 20, 28, 50, . . . ubereinstimmt. Wir mussen also unserPotential anpassen.Eine Moglichkeit besteht darin, das Potential der Ladungsdichteverteilung, welche wir zu Beginn die-ses Kapitels kennen gelernt haben, anzupassen. Ein bekannter Ansatz ist das sogenannte Wood-Saxon-Potential , welches auch in Abbildung 9.9 gezeigt ist, es lautet
V (r) = V01
1 + e(rR)/a. (9.45)
Man verwendet es bei grossen Kernen. Leider konnen auch damit die Schalenabschlusse (magischenZahlen) nicht zufriedenstellend erklart werden. Wir brauchen also noch eine weitere Korrektur.
9.4.5 Spin-Bahn-Kopplung
Wir haben die Energieniveaus beim Oszillator analysiert und gesehen, dass dieses Potential fur leichteKerne funktioniert. Die Schalenabschlusse fur A > 20 werden so jedoch nicht erklart. Diese Problem kann
114
69
This only fits for the first three magic numbers.One possibility is to modify the potential, to the Wood-Saxon Potential:
V (r) := −V01
1 + e(r−R)/a
But even this potential doesn’t produce the magic numbers.
9.4.5 Spin-Orbit Coupling
Consider the interaction of the spin-orbit of a nucleon:
V (r) = Vcenter + Vls (r)〈l · s〉~2
〈l·s〉~2 =
l/2 j = l + 1/2
− (l + 1) /2 j = l − 1/2
Thus∆Els =
2l+12 〈Vls (r)〉
70
9.4 Schalenmodell der Kernphysik Kapitel 9
durch Einfuhren der Spin-Bahn-Kopplung gelost werden.Wir fuhren analog zur elektromagnetischen Wechselwirkung bei den Atomen einen zusatzlichen Termbeim Potential ein, welcher die Spin-Bahn-Kopplung beschreibt.
V (r) = VZentr + Vls(r)hlsi~2 . (9.46)
Die Kopplung von Bahndrehimpuls l und Spin s fuhrt zu einem Gesamtdrehimpuls von j = l ± 12 . Dies
fuhrt dann zu Erwartungswerten
hlsi~2 =
j(j + 1) l(l + 1) s(s+ 1)
2=
l/2 fur j = l + 1/2
(l + 1)/2 fur j = l 1/2,(9.47)
wobei man benutzt, dass Nukleonen Spin 1/2 haben. Dies fuhrt dann zu einer Aufspaltung der Energie-niveaus mit l > 0 mit
Els =2l + 1
2· hVls(r)i. (9.48)
Die resultierenden Energieniveaus sind in Abbildung 9.10 gegeben.
Abbildung 9.10: Aufspaltung der Energieniveaus durch die Spin-Bahn-Kopplung. Die Luckenzeigen sich deutlich bei den magischen Zahlen. Aus [PRSZ06].
Wir haben gesehen, dass sich durch die l · s-Kopplung die Paritat nicht andert. Wahrend beim Har-monischen Oszillator die Paritat innerhalb einer Schale gleich fur alle Zustande gleich ist, so kann es beider l · s-Kopplung zu sogenannten Intruder kommen. Dies sind Niveaus, welche durch die Herab- oderHeraufsetzung zwischen Niveaus mit entgegengesetzter Paritat rutschen.
Beispiel 9.4.1 (Intruder). Wir betrachten die Schale, die bis zu 50 Nukleonen gefullt ist, die entspre-chenden Energieschema sind in Abbildung 9.11 dargestellt.
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9.4.5.1 Intruder In the 50 shell, the 1g+9/2 level has positive parity:9.4 Schalenmodell der Kernphysik Kapitel 9
Abbildung 9.11: Ganz links sind die Energieniveaus nur fur den Harmonischen Oszillatordargestellt, in der Mitte kommt noch der Radiale Term hinzu und ganz rechts wird dieSpin-Bahn-Kopplung berucksichtigt. Das rot eingefarbte Niveau stellt den Intruder dar.
Wie man sieht sind fur Schalen mit 50 Nukleonen noch Energieniveaus mit l = 4 dabei. Somit hat maninnerhalb einer Schale beide Paritaten. Zustande mit unterschiedlichen Paritaten konnen nicht mischen.
9.4.6 Anwendungen des Schalenmodells
Wir mochten nun das Schalenmodell anwenden, indem wir ein paar Beispiele betrachten. In Abbildung9.12 sind die tiefsten angeregten Zustande der Kerne schematisch dargestellt, welche wir untersuchenwerden.
Abbildung 9.12: Grundzustande und erste angeregte Zustande von einegen leichteren Ker-nen. Die Grundzustande wurden dabei auf dasselbe Niveau normiert. Die Energiedi↵erenzensind in Mega-Elektronvolt angegeben. Aus [PRSZ06].
Beispiel 9.4.2 (Sticksto↵ und Sauersto↵). Betrachte den Kern 157N8 in Abbildung 9.12. Fur die Neu-
tronen haben wir eine abgeschlossene Schale und der Gesamtdrehimpuls fur die Neutronen ist 0. Da wiraber nur 7 Protonen haben existiert keine abgeschlossene Schale fur diese und der 1/2-Zustand ist nurzur Halfte gefullt. Die Partiat und der Gesamtdrehimpuls des Grundzustandes des Sticksto↵s ist damit
116
States with different parity cannot mix.
71
9.4.6 Applications of the Shell Model
9.4.7 Valence Nucleons
10 Deformed Nucelei and Collective Nuclear Ex-ciations
Electrical quadrupole moment means that the charge distribution is not spher-ically symmetric:
The odd moments (dipole and octupole) vanish due to parity conservation,and so the quadrupole is the measure of non-sphericity:
Qij =´ρ(3rirj − r2δij
)d3r
10.1 Quadrupole Moments
10.2 Rotations Model10.2.1 Rotations Model
10.2.2 Classical Rotor
10.2.3 No Excitations–Symmetric Case
10.2.4 Excitations–Asymmetric Case
10.3 Multipol Transitions10.3.1 Characteristic Selection Rules
10.3.2 Transition Probability
10.4 Giant ResonanceGiant Dipole Resonance
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11 Nuclear Fission, Fusion and Nucleosynthesis
11.1 Nuclear Fission11.1.1 Uranium
11.1.2 Reaction and Fission Products
11.1.3 Fission Cross Section
11.1.4 Chain Reaction
11.1.5 Reactor Types
11.1.6 Radiation Dosage
11.2 Nuclear Fusion11.2.1 Physical Aspects
11.2.2 Fusion Cycles
11.2.3 Fusion in the Sun
11.3 NucleosynthesisThe primordial synthesis of light elements in the early Universe (D, 3He, 4Heand 7Li) is a process which is sensitive to many aspects of the physics modelthat describes it, such as the number of neutrino species in the theory. Theyield of the most stable element 4He is determined by a competition betweenthe expansion rate of the Universe, the rates of the weak interactions that inter-convert neutrons and protons, and the rates of the nuclear reactions that buildup the 4He nuclei.
1. a) At high temperatures (t < 1s, kT > 1 MeV), neutrons and protons caninterconvert via weakinteractions:n+e+ p+ e,n+e p+e,andnp+e+ e.As-longas the interconversion rate of neutrons and protons is faster than theexpansion rate of the Universe, the neutron-to-proton ratio tracks its equi-librium value, exponentially decreasing with temperature (Boltzmann fac-tor). Calculate the nn/np ratio at the end of this period (use kT = 720keV).nn
np= e−(mn−mp)/kBT =⇒ nn
np' 1
6 .
2. b) Once the interconversion rate becomes less than the expansion rate,the nn/np ratio effectively freezes-out (t ~ 1 s, kT = 720 keV), thereafterdecreasing slowly due to free neutron decay n p + e + e . Calculate thenn/np ratio after ~ 100 seconds, the time after which nucleosynthesis canstart. (neutron lifetime n = 887s)
N = N0e−t/τ =⇒ nn
np(t) =
(16np
)e−t/τ
np+(16np
)(1−e−t/τ
) = 0.146 ≈ 17
73
3. c) The nucleosynthetic chain, whereby protons and neutrons fuse to D(np), and further to 4He (nnpp), starts as soon as deuterium becomesstable against photo-dissociation (t ~ 3 min, kT ~ 100 keV). Since thereare less neutrons than protons, there will be nn/2 4He nuclei formed.The mass fraction of helium in the Universe (neglecting the other lightelements) can be written as YP = MHe , where MX = nXmX, the totalmass of X in the MHe +MH universe. Calculate this fraction YP at thebeginning of nucleosynthesis. Hint: calculate nHe using the previouslyfound nn/np ratio. nH
There are two neutrons necessary for each helium, and for each nn thereexists a proton, necessary for helium. Thus, nHe
nH=
12nn
np−nn≈ 1
2 .
Yp = nHemHe
nHemHe+nHmH≈ 0.25
4. d) A full calculation of the primordial helium abundance in the StandardModel yields YP = 0.246 + 0.013(N 3) + 0.18 n 887s + 0.01 ln(/5)887s where N is the effective number of light neutrino families, and isthe baryon-to-photon ratio in units of 1010. Given that the observationalupper bound on YP is 0.248, calculate the upper bound on N for = 3.Compare this result to the LEP result from the Z width measurement.
=⇒ Nν < 3.54
The combined LEP result for N is 2.984 ś 0.008.
11.3.1 Production of Alcohol and other Light Elements
11.3.2 Generation of Heavy Ions and SYNTHESEPFADE??
11.3.3 Primordial Element Synthesis
12 Star Evolution, Cosmology and Dark Matter
12.1 Star Evolution12.1.1 The Sun
In heavy stars the dominant mechanism is the CNO (Carbon-Nitrogen-Oxygen)cycle, in which the fusion process is ’catalyzed’ by small amounts of those threeelements.
However, in the sun (and other relatively light stars), the dominant route isthe pp-chain.
• The sun shines with about 4 · 1026W .
• The energy released per 4He nucleus is ∼ 24.75MeV .
• The mass of the sun is ∼ 2 · 1030kg, of which about half are protons.
• Thus the sun could shine for ∼ 4 · 1010 more years.
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12.1.1.1 The pp-chain
1. Two protons make a deuteron:
(a) β+ decay: p+ p→ d+ e+ + νe (weak int.)
(b) e− capture: p+ p+ e− → d+ νe (weak int.)
2. Deuteron plus proton makes 3He:
d+ p→3 He+ γ (strong int.)
3. 3He makes α-particle or 7Be (Beryllium):
(a) β+ decay: 3He+ p→ α+ e+ + νe (weak int.)
(b) 3He+3 He→ α+ p+ p (strong int. and EM)
(c) 3He+ α→7 Be+ γ (strong int. and EM)
4. 7Be makes α-particles:
(a) e− capture: 7Be+ e− →7 Li+ νe (weak int.)
(b) 7Li+ p→ α+ α (strong or EM)
(c) 7Be+ p→8 B + γ
(d) 8B →8 Be∗ + e+ + νe
(e) 8Be∗ → α+ α
There are 5 possible reactions that result in a neutrino, and each of these neu-trinos has a characteristic spectrum:
75
12.1.1.2 The Davis Experiment Chlorine plus a neutrino turns into ar-gon, so if you count the number of argon atoms you get the number of neutrinos.It turns out (the Davis Homestake experiment) that only a third of the theoreti-cally expected neutrinos arrived at the detector. This was explained by neutrinooscillations (and the fact that the experiments are sensitive only to one type ofneutrinos–νe).
12.1.1.3 The Super-Kamiokande Experiment Uses water. This exper-iment is sensitive to all three neutrino types.
Via elastic neutrino-electron scattering (νe+e→ νe+e) the emitted electronis detected by the Cherenkov radiation it emits in water.
12.1.1.4 The Sudbury Neutrino Observatory (SNO) Uses heavy wa-ter (D2O)
76
Via elastic neutrino-electron scattering (this is sensitive to all types of neu-trinos) and two other reactions which are sensitive only to electron neutrinos:νe + d→ p+ p+ e and ν + d→ n+ p+ ν
12.1.2 Neutron Stars
12.2 Basics of Cosmology12.2.1 Birth of the Universe
12.2.2 Friedmann Equation
12.2.3 Flow Equation
12.2.4 The Hubble Law
12.3 Time Evolution of Density in the Universe12.3.1 Material
12.3.2 Radiation
12.3.3 Composition
12.4 Dark Matter12.4.1 The Problem of the Cosmological Constant
12.4.2 Contributions to the Density of the Universe
12.4.3 Evidence for Dark Matter
Opening Monologue for theTest–Neutral Kaon Decay
1. According to the standard model, there are four types of interactions:electromagnetic, gravity, weak and strong. Each type of interaction hasa force carrier–the gauge Bosons–and Fermions, which are charged forthese interactions: quarks and leptons. The most famous lepton is theelectron whereas the quarks do not seem to exist by themselves: theyusually appear in “packages” of quark anti-quark (called mesons, whichare Bosons) or three quarks (called baryons, which are Fermions). Thenucleon (proton or neutron) is a baryon. Together the mesons and baryonsare called hadrons.
2. One example of mesons are the neutral kaons: |K0 >= |ds > and |K0 >=|ds >, which have a mass of roughly 500MeV . They were first discoveredin cosmic rays but can also be produced by strong decays such as collidingtwo protons or a proton and a pion.
77
3. Charge conjugation and parity are two interesting quantum mechanicaloperators. The former swaps particles for anti-particles whereas the latterchanges the position of a particle from r to −r. It was initially believedthat physics was invariant under the operation of either one of these op-erators. Experiments, like Wu’s in the 1950s, showed us that this is notthe case for the parity operator. Then it was believed that physics shouldbe invariant under the composed operator C P , is is mostly the case butstill not precisely. Our only hope to recover symmetry is the CPT theorem(which includes the time reversal operator) of quantum field theory, whichis believed to be an exact law of nature.
4. In quantum field theory, the intrinsic parity of a Fermion must be oppositeto the intrinsic parity of an anti-Fermion. Otherwise, it is by conventionthat quarks are defined to have positive parity. Additionally parity isa multiplicative quantity of a composite system, and the parity of thespherical harmonics is (−1)
l (where l is the orbital angular momentum ofthe system)–so for a composite system we must multiply by this factor.In conclusion, for baryons the parity is positive whereas for mesons it isnegative, in the ground state (l = 0).
5. Not every particle will be an eigenstate of the charge conjugation operator.Only those particles which are their own anti-particles can be that, suchas the photon or some mesons, such as the π0. For a meson, chargeconjugation eigenvalues will be (−1)
l+s. In addition charge conjugation isalso a multiplicative number.
• The charge conjugation eigenvalue of the neutrino gives a left-handedanti-neutrino, which does not occur, so the weak interaction is notcharge conjugation invariant.
6. Let’s compute the composed C P operator eigenvalues for the K0 andK0:
• P(K0)= −K0 whereas P
(K0)= +K0
• C(K0)= K0 whereas C
(K0)= K0
• So CP(K0)= −K0 and CP
(K0)= −K0.
7. So we see K0 and K0 are not eigenstates of the CP operator, but we canconstruct linear combinations of them which will be eigenstates:
K1 := 1√2
(K0 −K0
)and K2 := 1√
2
(K0 +K0
)Then CP (K1) = K1 and CP (K2) = −K2 by direct computation.
8. Experiments have showed us that in general the neutral kaons decay intoeither two or three pions.
78
9. What are the CP values of the two and of the three pion systems?
P (π) = −1 and C(π0)= (−1)
l+s= 1
Thus CP (ππ) = C (P (ππ)) = C ((−1) (−1)ππ) = C (ππ) = (+1) (+1) =1
Similarly, CP (πππ) = −1.
10. If CP is to be a conserved quantity in the weak interaction, K1 mayonly decay to CP = 1 states(two pions) whereas K2 could only decay toCP = −1 states (three pions).
11. Experiments show that is generally the case (albeit not precisely). Observethat the decay of K1 to two pions is much faster, because the energyreleased is greater (mass of one pion is ≈ 130MeV ) and so there is morephase space available. In comparison, the mass of three pions is largerand so the decay rate of K2 will be slower. If we start with a beam of K0
particles (produced in the strong interaction, which preserves strangeness,so the products will be strangeness eigenstates) we will see after a whileonlyK2 particles left in the beam. The lifetimes are given by τK1 ≈ 10−10sand τK2 ≈ 10−8s.
12. Cronin and Fitch showed in 1964 that after a long while K2 can decayinto two pions as well. This is a very small effect, but clearly shows thatCP is not conserved in the weak interaction. Thus it would appear thatthe long-lived kaon state is actually a mixture of K1 and K2 which can bewritten as:
|KL >= 1√1+|ε2| (|K2 > +ε|K2 >) where ε is a small parameter which
determines the mixing.
This can be accomodated for by adding a small phase factor to the Cabibbo-Kobayashi-Maskawa matrix.
13. While |KL > prefers to decay to three, rather than two, pions, it can alsodecay into π++e−+νe or into π−+e++νe. Note that these two possibilitiesare CP conjugates of each other. It turns out that the branching ratiosfor these two possibilities are not the same, hence, another evidence ofCP violation by the weak interaction.
14. If we let the K2 beam interact with matter we’ll regenerate some K1 again.Note that σ
(K0)< σ
(K0).
15. By observing strangeness oscillations we can estimate the mass differencebetween the K1 and the K2 states. It turns out that ∆m u 3 · 10−6eV .The probability to find K0 if we started with K0 goes like P
(K0
)∝
exp(− 1
2 (Γ1 + Γ2) t)cos (∆mt).
79
Part III
From Test Protocols13 Things to Memorize
13.1 Memorize Leisurely1. Alpha decay, like other cluster decays, is fundamentally a quantum tun-
neling process. Unlike beta decay, alpha decay is governed by the interplaybetween the nuclear force and the electromagnetic force.
2. Age of the solar system ~ 109 years.
80
14 Things To Do and Open Questions
1. How do ν come to be? Beta decay?
2. Which ν type is most prevalent?
81
3. Are µ stable? No.
4. Ratio of ν types in the atmosphere.
5. Where do ν come from? Uranium fission? How does the reaction sequencehappen (first Uranium is excited, then ...)
6. See of quarks..?
7. Feynman diagram of e− − e+
8. Inverse β decay Feynman diagram
9. What spin does theW have? (It’s a Boson) How to measure it? Homeworkexercise.
10. Nucleons Density – Woods-Saxon form
11. Back-shifted Fermi gas model formula.
12. Electron capture?
13. Draw nuclide chart.
14. There are situations (A=76) where only double beta decay is helloed.
15. How to measure the mass of neutrinos? (HWS10E04) Why use Triton??What’s the current limit on the mass of the neutrinos?
16. How to measure neutrino oscillations? (KAMLAND)
17. What is the typical error of a particle counter?
18. How to produce pions?? cosmic rays or proton collisions.
19. How to accelerate protons? How does a linear accelerator works?
20. How does a synchrotron werk?
21. How to excite protons? (deep inelastic scattering)
22. Which delta resonances exist? 1232 (they are an isospin quadruplet)
23. How do the pions (π±, π0) decay? and with which interaction?
24. What’s the angular momentum of 3He,3H,4He?
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o~g~r JQ J""~ c-Jl'rr p, "'? ty;) 'l) '; '?)n _:=J~'"") fR.~)r(0).'";\ ~~
..J. r') 0 (D.-Jf ~~(~--l r.'I oQJ 1'::J .Q.l."F) '( 1.1-:-. ~.., '(5)o-J'7)
I 1G7/2 8///
IG <-\\\\ 1G9/2 !OJ" 2Pl/2
: 222P// 1FS/2<...;/ ' 2P3/2
IF <-""- 1F7/2 8 8
25/' ID3/2
: 12EJJ 2S
10 <"..... IDS/2
~~
IF --- IFI/2
: 6 0<: ...••1P3/2
IS IS 2 0Fig. 3.4 ~nergy levels in a single-particle shell model. The boxed integers correspondto the magIc nuclear numbers.
25. Be able to say what’s the total angular momentum of a particular nuclide(9341Nb for instance)
26. How does a given nuclide decay?
27. How to detect neutrinos? (neutrinos interact with deutrium)
28. Which conservations laws always hold? (take table from somewhere)
29. Theoretical and experimental proof of the TCP theorem.
30. CP violation with the Kaon decay.
31. Spontaneous decay: α, β, γ decays. (does it necessitate deformation?)
32. Time scales for each interaction.
33. Fermi / Gamow-Teller decays (by spin of decay)
34. Prerequisities for each scattering (Rutherford, Mott, etc..)
35. How do different form factors look like for various charge distributions..
36. Why is the form factor independent of Q2? (point-like charge distribu-tion!)
37. Dependence of Q2 on x.
38. In beta decay the coupling constant on the vertices are not the same: forthe quarks it’s g · cos (θC) whereas for the leptons it’s g.
39. Neutrinos are not massless, as can be seen by neutrino oscillations (whichcan be observed by β decay of tritium).
40. Continous energy spectrum for neutrinos in β decay.
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41. Charge density distribution of various materials (He and Pb (Fermi-distributionand Gauss Distribution)).
42. HERA experiment
43. Weinberg angle??
44. Memorize the Rosenbluth formula.
45. The potential for protons and neutrons is different in the Fermi gas model??
46. masses of different patricles.
47. weak isospin! (all interactions conserve weak isospin, all interactions gofrom one state of weak isospin to another).
48. Goldhaber experiment.
49. Bjorken scaling variable / scale “breaking??”
50. Higgs patricle.
51. In Deep Inelastic Scattering, what is the significance of the deflection angleof the electron to the nucleon?
52. How to measure the mass of the Z0?
53. Memorize the CKM matrix.
54. the spin of the W and how to measure it!
55. Nuclear _fusion_.
56. Main cycle in the sun.
57. How to measure e−? Crystals (scintillators or sandwiches, sandwicheshave more fluctuations).
58. Different resonances of e− − e+ collision?
59. Z0 production.
60. Why are there long lived and short lived vomitters? Because of availablephase space.
61. Open questions of the standard model? (dahak dark matter, neutrinos’masses)
62. Where does the mass difference between the proton and the neutron stemfrom (if one ignores the electric charge difference between the u and dquarks)? Mass difference between u and d’s, plus u and d interaction.
63. Which symmetries are exact and which are approximate in the STAN-DARD MODEL?
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64. What’s “special” about the J/ψ patricle?
65. Propagator for each Feynman-diagram vertex.
66. Can scattering and annhilation be distinguished?
67. “Fermi-point” interaction was the model before the weak interaction.
68. Mixing of B0 and W 0 with the WEINBERG angle.
69. e = g · sin (θW )???????????????
70. Why can’t free quarks be observed? (only white color, or colorless, shit,can be observed)
71. How does the electron lose energy in a material? (brehmsstrahlung, ion-ization)
72. Feynman diagram for BREMSSTRAHLUNG.
73. How does the photon lose energy in a material? (photo electric effect,compton scattering, pair-production)
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74. The reason for introducing the weak isospin.
75. Coherence length.
76. The weak decay is so slow not because the weak coupling constant is small,but rather, because the weak gauge boson are so massive, and their massesenter in the inverse form to the propagator.
14.1 Experimental Setups14.1.1 Experimental Designs
14.1.1.1 Cowan-Reines Experiment Discovered the continuous spec-trum of electrons in beta decay, and thus, neutrinos.
14.1.1.2 Cherenkov Experiment
14.1.2 Actual Experiments
14.1.2.1 DESY Typical beam energies.
• Protons 900GeV
• e− 30GeV
What does the experimental setup look like?
14.1.2.2 LHC How are ν detected?
14.2 Particle Physics14.2.1 Feynman Diagrams
Each vertex adds a factor of√α
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14.2.2 Interactions
14.2.2.1 EM Interactions
14.2.2.2 Weak Interactions Charged and nuetral currents in e − p scat-tering?????
14.2.2.2.1 β decay
14.2.2.3 Strong Interactions
14.2.3 Scattering Cross-Sections
14.2.3.1 Born Approximation Which assumptions for each model (spin/ recoil / form factors)?
14.2.3.1.1 Magnetic Form Factor
14.2.3.1.2 Electric Form Factor
14.2.3.2 Parton Model
14.2.4 CP Violation in K0
Eigenstates of the CP operator.. decay into 2 π or 3 π states..
14.2.5 Z0 resonance
One could count the number of ν families. “WG Breit-Wigner Distribution”.
14.3 Nuclear Physics
14.4 Early Universe• Number of ν per cubic centimeter??
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