Fundamental Concepts in high energy/Pa!icle Physics
Lecture I : Introduction
Géraldine SERVANT CERN-Th
1 Plan
Lecture 1: intro to QFT: Relativity, kinematics, symmetries
Lecture 2: Towards Gauge theories
Lecture 3: Towards the Standard Model
Lecture 4: Towards beyond the Standard Model
2 why high energies
The Large Hadron Collider ➜ The LHC: the most gigantic microscope ever built Going to higher energies ➾ allows to study finer details
L = 400 km L = 800 km
Particle Physics: High energy physics study of short distances resolution limited by the de Broglie high resolution p>>m: wavelength λ =h/p necessitates large p relativistic regime
3 What is a particle?
A smal, quantum and fa -moving object
Quantum Mechanics Special Relativity
duality space-time wave-particle
Heisenberg inequalities energy = mass
energy non-conservation on time intervals t energy fluctuations E t E ⇥ particle number non constant
4 Creation of matter from energy Chemistry : rearrangement of matter the different constituents of matter reorganize themselves
CH4 +2O2 −→ CO2 +2H2O Particle physics : transformation energy ↔ matter W + e− µ+ u W − γ µ− d e+ E = mc2 d Equivalence between mass and energy (Einstein’s idea) plays a very fundamental role in particle physics 5 Natural units in high energy physics
The fundamental units have dimension of length (L), mass (M) and time (T). All other units are derived from these.
The two universal constants in SI units
-h= 1.055×10-34 J s = 1.055×10-34 kg m2/s and c=3 108 m/s
In particle physics we work with units ~ = c =1
Thus, velocity of particle is measured in units of the speed of light, very natural in particle physics where 0 ≤ v <1 for massive particles and v =1 for massless particles
In the c=1 unit: [velocity]=pure number [energy]=[momentum]=[mass]
notation: dimension of quantity P is [P] 6 Natural units in high energy physics
~ has dimension of [Energy]×[time]. ~ /mc~length (from uncertainty principle p x ~ / 2 ) or de Broglie’s formula λ =h/p ~ =1 ---> [length]=[mass]-1 Thus all physical quantities can be expressed as powers of mass or of length. e.g. energy density, E/L3~M4 e2 ↵ = pure number 4⇡~c
We specify one more unit taken as that of the energy, the GeV.
mass unit: M c2/c2 =1 GeV
length unit: ~ c/M c2 =1 GeV-1=0.1975 fm time unit: ~ c/M c3 =1 GeV-1=6.59 10-25s.
7 Natural units in high energy physics
1 eV = 1.6 10-19 J
--> -h c = 1.055×10-34 J s × 3 108 m/s= 1.978×10-7 eV m
Using 1 fm = 10-15 m and 1 MeV = 106 eV: -h c = 197.8 MeV fm
So in natural units: 1fm 1/(200 MeV) ⇡ -1 -25 -1 also, c=1 --> 1 fm ~ 3×10-24 s-1 --> GeV ~ 6 × 10 s
-27 -h= 1.055×10-34 kg m2/s ----> GeV ~ 1.8 × 10 kg
8 The elementary blocks of matter
Matter is made of molecules
Molecules are built out of atoms
Atoms are made of nuclei and electrons
Nuclei are assemblies of protons and neutrons
Protons and neutrons are quarks bound together...
The volume of an atom corresponds to 1024 times the volume of an electron
Classically, matter contains a lot of void Quantum mechanically, this void is populated by pairs of virtual particles 1 TeV = 1012 eV
1 electron volt The energy of an electron accelerated by an electric potential 19 ( eV) = difference of 1 volt. One eV is thus equal to ... 1.6 10− J
1 kg of sugar = 4000 kCalories= 17 millions of Joules ≈ 1014 TeV but 1 kg sugar ≈ 1027 protons --> 0.1 eV / protons
To accelerate each proton contained in 1 kg of matter at 14 TeV, we would need the energy of of 1014 kg of sugar* i.e. 1% of the world energy production *world annual production of sugar=150 millions of tons≈1011 kg How impressive is tis? 24 energies involved at CERN: 1 TeV = 1000 billions of eV= 10 − kg compared with the kinetic energy of a mosquito 10−3 J ∼ 1016 eV ∼ 104TeV
...however, in terms of energy density... this corresponds to the mass of the Earth concentrated in a 1 mm3 cube ! Classical versus quantum Collision
Compton wave length λ =h/mc
strawberry : m~30 g ~1025 GeV/c2 ➲ λ~10-40 m
λ << R classical: e- : m~9.1x10-31 kg ~0.5 MeV/c2 ➲ λ~4x10-13 m λ >> R p : m~1.6x10-27 kg ~1 GeV/c2 ➲ λ~10-16 m quantum :
11 Why relativity
Particle physics is all about creating and annihilating particles. This can only occur if we can convert mass to energy and vice-versa, which requires relativistic kinematics
A bit of historyContemplating the unusual invariance of Maxwell’s equations under Lorentz transformation, Einstein stated that Lorentz invariance must be the invariance of our space and time.
-> completely changed our view of space and time, so intertwined that it is now called spacetime, leading to exotic phenomena such as
- time dilation
-prediction of antimatter when special relativity is married with quantum mechanics
12 Relativistic transformations
The two postulates - Speed of light is the same in all reference frames of Special relativity - Laws of physics are unchanged under a galilean transformation, i.e. in all reference frames moving at constant velocity with respect to each other
x x’ Look for coordinate transformations that S’ satisfy thesect requirements0 ct0 ctct = = z 0 !z0 ! !! zz !! Unique choice: S z’ Lorentz transformations v v = = z c c ct0 ct 1 = = 1 z0 ! ! z ! = v2 y y’ 1 2 2 1 v c q c2 q
13
1
1
1 ct0 ct = z0 z ! ! ! ct0 ct = v Implicationsz 0of Lorentz transformation z = ! ! ! c v 1 ct0 ct = = = v2 z z c 1 2 0 ! ! ! c q 1 consider time interval ⌧ = = t 0 t 0 in S’, the rest frame Time dilation 2 1 v2 1 2 of a particle located at z 0 = z 0 c =0 . 1q 2 then in frame S where the particle is moving: t t = ⌧ 2 1 -->The observed lifetime of a particle is ⌧ so it can travel over a distance c ⌧⇥ 6 -->muons which have a lifetime ⌧ 2 10 s produced by reaction of ⇠ ⇥ cosmic rays with atmosphere at 15-20 km altitude can reach the surface
length an object at rest in S’ has length L = z0 z0 0 2 1 contraction It measures in S z z = L / 2 1 0 --> densities increase ⇢0 = n/( x0 y0 z0) ⇢ = n/( x y z)= ⇢0
is invariant x y z t 14
1
1 1 ct0 ct = z0 z ! ! ! v = c 1 = 4-vectors 1 v2 c2 q Time and space get mixed-up under Lorentz ct x0 1 0 transformations. They are considered as µ 0 x 1 0 x 1 x x = = 2 = y x ~x ! different components of a single object, a B C B C B z C B x3 C four-component spacetime vector: B C B C @ A @ A
µ 2 02 2 By construction: x x = x = x ~x is invariant µ d⌧ = dt2 d~x2 is invariant
Lorentz invariant p action built with the S = m d⌧ = dt proper time d ⌧ . L Z Z { = m 1 x˙ 2 L @ p~ = L p= m ~ p~ = E ~ @x˙ E = p~ . x˙ = m L 15
1 Energy-momentum four-vector
2 2 2 so we find: m = E p~ µ E This suggests to define the four-vector p =( ,p ,p ,p ) c x y z
m =0 =1 = ⌧ = ! ! 1! 1
a massless particle cannot decay
16 Conservation of energy-momentum
Consider collision between A and B Define center of mass (CM) frame as where p~ A + p~ B =0
Energy available in center of mass frame 2 2 ptot = E is an invariant:sqrt(s)=E* = EA + EB ⇤ 1) Collision on fixed target
B is at rest in lab frame, EB =mB and EA is energy of incident particle 2 2 2 E = mA + mB +2mBEA ⇤ 2) Colliding beams A and B travel in opposite directions 2 2 2 E = mA + mB + 2(EAEB + pA pB ) 4EAEB ⇤ | || | ⇡ if mA, mB << EA, EB
So for fixed target machine E 2mBEA ⇤ ⇠ while for colliding beam accelerators E 2E p⇤ ⇠ To obtain 2 TeV in the CM with a fixed proton target accelerator the energy of a proton beam would need to be 2000 TeV! 17 ct0 ct = z0 ! ! z ! Conservation of energy-momentum v = ct0 ct c Consider the interaction e + p -> =e + p due to exchange of electromagnetic field z0 ! ! z ! 1 = v 1 v2 = c2 p1 p3 a photon is massless, however, for a shortc amountq of time, an “exchanged” photon (virtual photon) q 1 ct x0 =⇤ can have a mass (Heisenberg inequalities)2 1 0 µ 0v x 1 0 x 1 x x 1= c2 = 2 = p p y x ~x p! p q B C B C 2 4 p1 + p2 = p3 + p4 B C B 3 C B z C B x C ct x0@ A @ A q = p p = p 1 p 0 1 µ 30 x 1 40 x 1 2 x x = = 2 = y x ~x !m q is the transfer energy-momentumB C Bfour-vectorC p = B C B 3 C 2 ~ B z C B x C 0 ! 2 2 @ A @ A q = m : invariant, can be computed in any frame ⇤ m E4 E = m + T in frame where proton is at rest: p2 = ~ p4 = 4 0 ! p~4 ! kinetic energy 2 2 2 2 2 T p /2m q =(p4 p2) = p + p 2p4.p2 ⇡ 4 2 2 2 2 q = m + m 2E4 m = 2mT < 0 virtual photon 18
1
1 Range of an interaction ~c R = m p1 p3 | ⇤| q reminder: ~c 200 MeV fm ⇠ p p2 p4p To probe the proton, we need 1 R R 1fm ⌧ proton ⇠ m 200 MeV | ⇤|
19 Next step : marry quantum mechanics and relativity
(Schrodinger’s equation cannot account for creation/annihilation of particles) non relativistic 2 Schrödinger Equation (1926): i + V ⇥ =0 t 2m p2 ⇥ E = + V classical ↔ quantum E i & p i 2m correspondance t x
relativistic 1 2 m2c2 Klein-Gordon Equation (1927): + ⇥ =0 c2 t2 2 E2 ⇥ = p2 + m2c2 c2 2 2 1/2 negative energies E = (p + m ) and ± does not admit a positive probability density and does not describe fermions 20 Antimatter and Dirac equation
µ mc Dirac Equation (1928): i ⇥µ =0 + p2c2 + m2c4 matter ⇥ E = µ, ⌫ =2⌘µ⌫ ⌅p2c2 + m2c4 antimatter { } ⇤ ⇥ ⌅ i(p.x Et)/ plane wave solution (x, t)=u(p)e ~ a particle of energy -E travelling backward in time -> antiparticle
positron (e+) discovered by C. Anderson in 1932
conservation of fermion number: +1 for particles and -1 for antiparticles. fermions can only be created or destroyed in pairs
21 The necessity to introduce fields for a multiparticle description
Relativistic processes cannot be explained in terms of a single particle. Even if there is not enough energy for creating several particles, they can still exist for a short amount of time because of uncertainty principle W + e− µ+ u W − γ µ− d e+ E = mc2 d
We need a theory that can account for processes in which the number and type of particles changes like in most nuclear and particle reactions
quantization of a single relativistic particles does not work, we need quantization of fields -> Quantum Field Theory (QFT)
22 Quantum Field Theory
We want to describe A-> C1 + C2 or A+B-> C1 + C2 + ...
1) Associate a field to a particle
4 2) Write action S = d x ( ,@ ) L i µ i Z 3) invariant under Poincaré (Lorentz+translations) Ltranformations and internal symmetries
The symmetries of the lagrangian specify the interactions
4) Quantization of the fields
23 Symmetries and conservation laws: the backbone of particle physics
Noether’s theorem (from classical field theory) : A continuous symmetry of the system <-> a conserved quantity
I- Continuous global space-time symmetries: translation invariance in space <-> momentum conservation translation invariance in time <-> energy conservation rotational invariance <-> angular momentum conservation
Fields are classified according to their transformation properties under Lorentz group: µ µ µ ⌫ x x0 =⇤ x (x) 0(x0) ! ⌫ ! 0(x)= (x) scalar µ µ ⌫ vector V ⇤⌫ V ! i (x) exp( ! J µ⌫ ) (x) spinor ! 2 µ⌫ The true meaning of spin arises in the context of a fully Lorentz-invariant
theory (while it is introduced adhoc in non-relativistic quantum mechanics) 24 A field transforms under the Lorentz transformations in a particular way.
Picking a particular representation of the Lorentz transformation specifies the spin.
After quantizing the field, you find that the field operator can create or annihilate a particle of definite spin
The spin is part of the field
25 II- Global (continuous) internal symmetries:
acting only on fields
conservation of baryon number and lepton number
W + e− µ+ u W − e, Lμ, B = conservedγ quantities NB: Q, L − µ d e+ E = mc2 d
26 Quantum numbers and Conservation laws
When the positron was discovered, it raised a naive question: + why can’t a proton decay into a positron and a photon p e ? !
This process would conserve momentum, energy, angular momentum, electric charge and even parity
This can be understood if we impose conservation of baryon number
Similarly, when the muon was discovered, it raised the question: why doesn’t a muon decay as µ e ? !
This led to propose another quantum number: lepton family number
27 The Standard Model: matter
the elementary blocks: leptons quarks
tau − each of the 6 muon τ quarks ντ − exists in three electron µ colors νµ e− νe ⤴ mas mas x20 ⤴ x1000 ⤴ ⤴ x1000 x200 composite states (white objects) no composite states p = (u, u, d) made of leptons baryons proton neutron n = (u, d, d) + antpartcles mesons 129A Midterm (due Oct 31) 129A Midterm (due Oct 31) 129A Midterm (due Oct1. On129A 31) February Midterm 23, 1987, a (due supernova Oct (named 31) SN1987A) exploded in Large 1. On February 23, 1987, a supernova (named SN1987A) explodedMagellanic in Large Cloud, which is about 50 kpc away from the Earth. Kamiokande 1. On February 23, 1987, a supernova (named SN1987A) exploded in Large Magellanic Cloud, which is about 50 kpc away from1. On the February Earth. 23,KamiokandeCollaboration 1987, a supernova and IMB Collaboration (named SN1987A) detected a exploded burst of neutrinos in Large and re- Collaboration and IMBMagellanic Collaboration Cloud, detected which is about a burstMagellanic 50 kpc of away neutrinos Cloud, fromported which the and them Earth. re- is in about K. Kamiokande Hirata 50 kpcet al away., Phys. from Rev. the Lett. Earth.58,1490(1987);R.M. Kamiokande Collaboration and IMB Collaboration detected a burst of neutrinos and re- ported them in K. Hirata et al., Phys. Rev. Lett.Collaboration58,1490(1987);R.M. andBionta IMBet al Collaboration, Phys. Rev. detected Lett. 58,1494(1987).Answerthefollowing a burst of neutrinos and re- ported them in K. Hirata et al.,portedPhys.129A them Rev. Midterm in Lett.questions. K. Hirata58 (due,1490(1987);R.M. Octet al 31)., Phys. Rev. Lett. 58,1490(1987);R.M. Bionta et al, Phys. Rev.Bionta Lett.et al, Phys.58,1494(1987).Answerthefollowing Rev. Lett. 58,1494(1987).Answerthefollowing Bionta et al, Phys. Rev. Lett. 58,1494(1987).Answerthefollowing questions. questions. 1. On February 23, 1987, a supernova129A(a) Calculate Midterm (named SN1987A) the (due energy exploded Oct of neutrinos 31) in Large using the table of events in Hirata Magellanic129Aquestions. Cloud, Midterm which is about (due 50paper, kpc Oct away from from31) the the observed Earth. Kamiokande electron (actually believed to positron) energy Collaboration1. On February and IMB Collaboration23, 1987, a detected supernova a burst (named of neutrinos SN1987A) and re- exploded in Large (a) Calculate1. theOn energy February ofported neutrinos 23, them 1987, in K. a using supernova Hirata theet al., table (namedPhys.and the of Rev. SN1987A) events angle Lett. from58 in,1490(1987);R.M. exploded Hirata the SN1987A in Large direction for each of them. Assume (a) Calculate the energy of neutrinos using the tableMagellanic of events Cloud, which in Hirata is about 50 kpc away from the Earth. Kamiokande+ MagellanicBionta Cloud,et(a) which al, CalculatePhys. is about Rev. 50 Lett. the kpcthat energy58 away,1494(1987).Answerthefollowing all from events of neutrinos the are Earth. caused Kamiokande using by the the reaction tableν ¯ ofep eventse n,andneglectthe in Hirata paper, from the observed electronCollaboration (actually and believed IMB Collaboration to positron) detected energy a burst of neutrinos and→ re- paper, from the observed electronCollaboration (actuallyquestions. believed and IMBpaper, Collaboration to positron) from the detectederror energy observed bars. a burst electron of neutrinos (actually and re- believed to positron) energy and the angle from the SN1987Aported them direction in K. for Hirata eachet of al., them.Phys. AssumeRev. Lett. 58,1490(1987);R.M. and the angle from the SN1987Aported direction them(a) in for Calculate K. each Hirataand the ofet energy the al them.., angle ofPhys.(b) neutrinos Obtain Assume from Rev.+ using the Lett. the the SN1987A upper table58,1490(1987);R.M. of bound events direction on in Hirata the neutrino for each mass. of them. Since Assume the energy of that all events are caused byBionta the reactionet al, Phys.ν ¯ep Rev.e n Lett.,andneglectthe58,1494(1987).Answerthefollowing Bionta et al, Phys.paper, from Rev.+ the observedLett. 58 electron,1494(1987).Answerthefollowing→ (actually believed to positron) energy + that all events are causederror by bars. the reactionν ¯ep questions.ethatn,andneglectthe all eventsneutrinos are caused varied by from the one reaction event toν ¯ep another,e n a,andneglectthe massive neutrino would questions. and→ the angle from the SN1987Ahave direction different for velocities each of them. and Assume hence di→fferent arrival times. 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From the have different velocities and hence differentobserved arrivalerror spread times. bars. in the arrival From times, the onee can place an upper bound on The following→ processes have not been seen. the neutrino mass.error (The bars. lastthe neutrino threeobserved events mass. (The spread arrived last three in more events the than arrival arrived 9 more seconds times, than 9 one seconds can place an upper bound on observed spread in the arrival times, one can place(b) Obtain an upper the2. The upper bound following bound Explain on which on processes the conservation neutrino have law forbids mass. not been each Since of seen. them the This energy is of understood as conse- later from the(b) first Obtain event, thelater and upper fromneutrinosthe there bound the neutrino first is on varied event, aquences the dispute mass. and from neutrino there of one if certain (The theyis event mass. a dispute last conservation came to Since another, three if they from the events came a energylaws. massive from Explainarrived of neutrino what more would conservation than 9 seconds law forbids the neutrino mass. (The last three events arrivedSN1987A. more Discard than them and 9 useseconds only the first 9 events.)+ 0 + SN1987A. Discardneutrinos them and varied use fromhavelater only one di the fromfferent event firsteach the velocities to 9 process. firstanother, events.) and event, (a) a hencep massive ande diπff neutrino thereerent,(b) arrivalµ is− would ae dispute times.−e−e ,(c) From ifn they thepν cameeν¯e,(d) fromτ − µγ, later from the first event, and there is2. The a dispute following processes if they have came not been from seen. This→ is0 understood+ as conse-→ → → have different velocitiesobservedSN1987A. and spread(e) hence Discardn in the dipµfferent themarrival−ν¯µ,(f) arrival and times,K use times. oneµ only cane−,(g) From place theµ first an the− upperπ 9− events.)νµ bound. on 2. The following processesobserved havequences not spread of been certain inthe the seen. neutrino conservation arrival This mass. times, laws. is→ (The understood Explain one can last what place three conservation as an events→ conse- upper arrived law bound forbids more on→ than 9 seconds SN1987A. Discard them and use only the first 9 events.)+ 0 + each process. (a) p e 3.π ,(b)Theµ− isospine−e−e is,(c) a goodn p quantumνeν¯e,(d)τ − numberµγ, in strong interaction. There are quences of certain conservationthe neutrino2. laws.The mass.later following Explain (The→ from0 last what the+ processes three first conservation→ events event, have arrived and not law there→ more been forbids is than a seen. dispute 9 seconds→ This if they is came understood from as conse- (e) n pµ−ν¯ ,(f)K µ e−,(g)µ− π−ν . + 0 → µSN1987A.→+ Discardspin 1 them mesons→ and (µρ use, ρ only0, ρ+ the)whichformisospin1multipletandanotherspin first 9 events.) 2. The following processeseach have process. not been (a) p seen.latere π ,(b) from This theµquences− is first understoode− event, ofe−e certain,(c) and asn there conservation conse-p isνe aν¯e dispute,(d)− laws.τ − if they Explainµγ came, what from conservation law forbids 3. The isospin is a good quantum number in strong interaction. There are →0SN1987A.+ Discard→ them and1meson use only→ theω+which first0 9 has events.) isospin→ 0. They+ are understood as the bound states (e) n pµ−ν¯µ,(f)K µ e−2.,(g)eachTheµ following process.− 0π−+ν processesµ (a). p havee π not,(b) beenµ seen.e Thise e is understood,(c)n aspν conse-ν¯ ,(d)τ µγ, quences of certain conservation→ laws. Explain→ whatspin 1 mesons conservation (ρ−→, ρ , ρ )whichformisospin1multipletandanotherspin(d lawu,¯ (u forbidsu¯ dd¯)/√2,ud¯−), and (−uu¯−+ dd¯)/√2, respectively.e e − + 0 + 1mesonquencesω which of has certain isospin conservation 0. They→0 are understood laws.+ Explain as→ the what bound conservation states → law forbids → 2. The following processes(e) n havepµ− notν¯µ,(f) beenK seen.− Thisµ e− is,(g) understoodµ− asπ− conse-νµ. each process. (a) p 3. eTheπ isospin,(b)µ− is a goode−e− quantume ,(c)(du,¯ (uu¯n numberdd¯)/p√ν2,ueν¯ ind¯),e,(d) and strong (uu¯τ+− interaction.d+d¯)/0√µ2,γ respectively., There+ are →0 + →quences of certaineach− conservation→ process.→ (a) laws.p(a) Explaine Even→π ,(b)→ though whatµ− conservation bothe−eρ−eand,(c)→ω lawsharen forbidsp theνeν¯e same,(d)τ constituents,− µγ, they decay very 0 + →0 + → → → (e) n pµ−ν¯µ,(f)K spinµ 1 mesonse−,(g) (ρµ−−, ρ , ρπ−)whichformisospin1multipletandanotherspinνµ. (e) n + pµ0 −ν¯µ,(f)K µ +e−,(g)µ− π−νµ. → → each→ process.(a)3. (a) EvenThep thoughe isospin→π both,(b)ρ isandµ− aω share goodedi−→eff the−erently.e quantum same,(c) constituents,n Lookp number→ν upeν¯e they,(d) the decay Bookletτ in− strong veryµγ and, interaction. find the most There dominant are decay 1mesonω which has isospin 0.differently. They→0 are Look+ understood up the→ Booklet0 as and+ the find bound the→ most states dominant decay→ (e) n pµ−ν¯µ3.,(f)spinTheK isospin 1 mesonsµ e is−,(g) a ( goodρµ−−,modesρ quantum,πρ−ν)whichformisospin1multipletandanotherspin forµ. them. number in strong interaction. There are 3. The isospin is a good quantum¯ number¯ in→ strongmodes interaction. for¯ → them. There→ are (du,¯ (uu¯ dd)/√2,ud), and (uu¯ + dd)/√2, respectively.0 + 0 + 0 0 0 0 + − 1mesonspin 1 mesons0ω which (+ρ−(b), ρ has, Explainρ )whichformisospin1multipletandanotherspin isospin why0 0.ρ 0 They0 π π− areis allowed understood but ρ as theπ π boundis not. states spin 1 mesons (ρ , ρ , ρ )whichformisospin1multipletandanotherspin3. The isospin(b) is Explain a good why quantumρ π π number− is allowed in but strongρ interaction.π π→is not. There are → − 1meson0 + ω →which¯ has isospin¯ 0. They→ are¯ understood as the bound states (a) Even thoughspin both 1ρ mesonsand(c)ω Construct(shareρ(−d,u,¯ρ ,( theρu wavefunctionsu¯)whichformisospin1multipletandanotherspin samedd)/ constituents,√(c) for2,u two-pion Constructd), and states they wavefunctions (u bothu¯ decay+ ford totald) very/√ isospin for2, respectively. two-pionI =1 states both for total isospin I =1 (du,¯ (uu¯ dd¯)/√2,ud¯), and (uu¯ + dd¯)/√2, respectively. 29 1mesonω which has isospin 0. They1meson are understoodω whichand has 0, and isospin as determine− the 0. bound the They allowed are states values understood for the relative as the orbital bound angular states ¯ ¯ differently.¯ Look up the Booklet and− find theand most 0, and dominant determine decay the allowed values for the relative orbital angular (du,¯ (uu¯ dd)/√2,ud), and (uu¯ + dd)(/d√u,¯ 2,(uu¯ respectively.dd¯)/momentum√2(a),u(a)d¯), Even Even andL. though ( thoughuu¯ + d bothd¯) both/momentum√ρ2,and respectively.ρ ωandshareωL.share the same the constituents, same constituents, they decay very they decay very modes for them. − − didifffferently.erently. Look Look up the up Booklet the Booklet and find and the find most the dominant most decay dominant decay 0 + 0 0 0 (b) Explain why ρ(a) Evenπ π though− is allowed both ρ and butωρshareπ theπ sameis not. constituents, they decay very (a) Even though both ρ and ω share the→ same constituents,modesmodes forthey for them.→ them. decay very differently. Look up the Booklet0 and+ find the most dominant0 0 decay0 differently. Look up(c) Construct the Booklet wavefunctions and find for the two-pion(b) most Explain states dominant why ρ both0 π for decayπ− total+is allowed isospin butI ρ=1 π π 0is not.0 0 modes for them.(b) Explain why →ρ π π− is allowed→ but ρ π π is not. and 0, and determine the allowed(c)0 Construct values+ for wavefunctions the relative→ 0 for orbital two-pion0 0 angular states both for→ total isospin I =1 modes for them. (b) Explain why ρ π π− is allowed but ρ π π is not. L (c)→and Construct 0, and determine wavefunctions the allowed→ for values two-pion for the states relative both orbital for angular total isospin I =1 0 +momentum . 0 0 0 (b) Explain why ρ π π− is allowed(c) but Constructρ wavefunctionsπ π momentumandis not. 0, for and two-pionL. determine states the both allowed for total values isospin forI =1 the relative orbital angular → and 0, and→ determinemomentum the allowedL. values for the relative orbital angular (c) Construct wavefunctions for two-pionmomentum states bothL. for total isospin I =1 and 0, and determine the allowed values for the relative orbital angular momentum L. 129A Midterm (due Oct 31) 129A Midterm (due Oct 31) 129A Midterm (due Oct1. On129A 31) February Midterm 23, 1987, a (due supernova Oct (named 31) SN1987A) exploded in Large 1. On February 23, 1987, a supernova (named SN1987A) explodedMagellanic in Large Cloud, which is about 50 kpc away from the Earth. Kamiokande 1. On February 23, 1987, a supernova (named SN1987A) exploded in Large Magellanic Cloud, which is about 50 kpc away from1. On the February Earth. 23,KamiokandeCollaboration 1987, a supernova and IMB Collaboration (named SN1987A) detected a exploded burst of neutrinos in Large and re- Collaboration and IMBMagellanic Collaboration Cloud, detected which is about a burstMagellanic 50 kpc of away neutrinos Cloud, fromported which the and them Earth. re- is in about K. Kamiokande Hirata 50 kpcet al away., Phys. from Rev. the Lett. Earth.58,1490(1987);R.M. Kamiokande Collaboration and IMB Collaboration detected a burst of neutrinos and re- ported them in K. Hirata et al., Phys. Rev. Lett.Collaboration58,1490(1987);R.M. andBionta IMBet al Collaboration, Phys. Rev. detected Lett. 58,1494(1987).Answerthefollowing a burst of neutrinos and re- ported them in K. Hirata et al.,portedPhys.129A them Rev. Midterm in Lett.questions. K. Hirata58 (due,1490(1987);R.M. Octet al 31)., Phys. Rev. Lett. 58,1490(1987);R.M. Bionta et al, Phys. Rev.Bionta Lett.et al, Phys.58,1494(1987).Answerthefollowing Rev. Lett. 58,1494(1987).Answerthefollowing Bionta et al, Phys. Rev. Lett. 58,1494(1987).Answerthefollowing questions. questions. 1. On February 23, 1987, a supernova129A(a) Calculate Midterm (named SN1987A) the (due energy exploded Oct of neutrinos 31) in Large using the table of events in Hirata Magellanic129Aquestions. Cloud, Midterm which is about (due 50paper, kpc Oct away from from31) the the observed Earth. Kamiokande electron (actually believed to positron) energy Collaboration1. On February and IMB Collaboration23, 1987, a detected supernova a burst (named of neutrinos SN1987A) and re- exploded in Large (a) Calculate1. theOn energy February ofported neutrinos 23, them 1987, in K. a using supernova Hirata theet al., table (namedPhys.and the of Rev. SN1987A) events angle Lett. from58 in,1490(1987);R.M. exploded Hirata the SN1987A in Large direction for each of them. Assume (a) Calculate the energy of neutrinos using the tableMagellanic of events Cloud, which in Hirata is about 50 kpc away from the Earth. Kamiokande+ MagellanicBionta Cloud,et(a) which al, CalculatePhys. is about Rev. 50 Lett. the kpcthat energy58 away,1494(1987).Answerthefollowing all from events of neutrinos the are Earth. caused Kamiokande using by the the reaction tableν ¯ ofep eventse n,andneglectthe in Hirata paper, from the observed electronCollaboration (actually and believed IMB Collaboration to positron) detected energy a burst of neutrinos and→ re- paper, from the observed electronCollaboration (actuallyquestions. believed and IMBpaper, Collaboration to positron) from the detectederror energy observed bars. a burst electron of neutrinos (actually and re- believed to positron) energy and the angle from the SN1987Aported them direction in K. for Hirata eachet of al., them.Phys. AssumeRev. Lett. 58,1490(1987);R.M. and the angle from the SN1987Aported direction them(a) in for Calculate K. each Hirataand the ofet energy the al them.., angle ofPhys.(b) neutrinos Obtain Assume from Rev.+ using the Lett. the the SN1987A upper table58,1490(1987);R.M. of bound events direction on in Hirata the neutrino for each mass. of them. Since Assume the energy of that all events are caused byBionta the reactionet al, Phys.ν ¯ep Rev.e n Lett.,andneglectthe58,1494(1987).Answerthefollowing Bionta et al, Phys.paper, from Rev.+ the observedLett. 58 electron,1494(1987).Answerthefollowing→ (actually believed to positron) energy + that all events are causederror by bars. the reactionν ¯ep questions.ethatn,andneglectthe all eventsneutrinos are caused varied by from the one reaction event toν ¯ep another,e n a,andneglectthe massive neutrino would questions. and→ the angle from the SN1987Ahave direction different for velocities each of them. and Assume hence di→fferent arrival times. From the error bars. that all eventserror are bars. caused by the reactionν ¯ p e+n,andneglectthe (b) Obtain the upper bound on(a) the Calculate neutrino the mass. energy Since of neutrinos thee → energy using the of table of events in Hirata (a) Calculateerror the energybars. of neutrinosobserved using the spread table in of the events arrival in Hirata times, one can place an upper bound on (b) Obtain the upper boundneutrinos on the varied neutrino from one mass. event(b)paper, Obtain to Since another, from the the the a upper energy observed massive bound electron of neutrino on (actually the would neutrino believed to mass. positron) Since energy the energy of paper,(b) from Obtain the observed the upper electron bound on (actuallythe the neutrino neutrino believed mass. mass. to Since (The positron) the last energy energy three of events arrived more than 9 seconds andneutrinos the angle varied from the from SN1987A one event direction to another, for each of a them. massive Assume neutrino would neutrinos varied fromhave one di eventfferent to velocities another, andneutrinos a massive hence varied diff from neutrinoerent one event arrivallater would to another, from times. the a massive From first event, neutrino the and would+ there is a dispute if they came from and the angle fromthat the all SN1987A events are direction caused by for the each reaction of them.ν ¯ep Assumee n,andneglectthe observed spread inthat the all arrival eventshave di times, arefferenthave caused velocities one diff by canerent the and place reactionSN1987A. hence velocities an diff upperν ¯erentp Discard and arrivale bound+n hence them,andneglectthe times. on and di Fromff→ useerent the only arrival the first times. 9 events.) From the have different velocities and hence differentobserved arrivalerror spread times. bars. in the arrival From times, the onee can place an upper bound on The following→ processes have not been seen. the neutrino mass.error (The bars. lastthe neutrino threeobserved events mass. (The spread arrived last three in more events the than arrival arrived 9 more seconds times, than 9 one seconds can place an upper bound on observed spread in the arrival times, one can place(b) Obtain an upper the2. The upper bound following bound Explain on which on processes the conservation neutrino have law forbids mass. not been each Since of seen. them the This energy is of understood as conse- later from the(b) first Obtain event, thelater and upper fromneutrinosthe there bound the neutrino first is on varied event, aquences the dispute mass. and from neutrino there of one if certain (The theyis event mass. a dispute last conservation came to Since another, three if they from the events came a energylaws. massive from Explainarrived of neutrino what more would conservation than 9 seconds law forbids the neutrino mass. (The last three events arrivedSN1987A. more Discard than them and 9 useseconds only the first 9 events.)+ 0 + SN1987A. Discardneutrinos them and varied use fromhavelater only one di the fromfferent event firsteach the velocities to 9 process. firstanother, events.) and event, (a) a hencep massive ande diπff neutrino thereerent,(b) arrivalµ is− would ae dispute times.−e−e ,(c) From ifn they thepν cameeν¯e,(d) fromτ − µγ, later from the first event, and there is2. The a dispute following processes if they have came not been from seen. This→ is0 understood+ as conse-→ → → have different velocitiesobservedSN1987A. and spread(e) hence Discardn in the dipµfferent themarrival−ν¯µ,(f) arrival and times,K useenergy times. oneµ only cane−,(g) From place theµ first an the− upperπ 9− events.)νµ bound. on 2. The following processesobserved havequences not spread of been certain inthe the seen. neutrino conservation arrival This mass. times, laws. is→ (The understood Explain one can last what place three conservation as an events→ conse- upper arrived law bound forbids more on→ than 9 seconds SN1987A. Discard them and use only the first 9 events.)+ 0 + each process. (a) p e 3.π ,(b)Theµ− isospine−e−e is,(c) a goodn p quantummuonνeν¯e,(d) numberτ − or number electronµγ, number in strong interaction. There are quences of certain conservationthe neutrino2. laws.The mass.later following Explain (The→ from0 last what the+ processes three first conservation→ events event, have arrived and not law there→ more been forbids is than a seen. dispute 9 seconds→ This if they is came understood from as conse- (e) n pµ−ν¯ ,(f)K µ e−,(g)µ− π−ν . + 0 → µSN1987A.→+ Discardspin 1 them mesons→ and (µρ use, ρ only0, ρ+ the)whichformisospin1multipletandanotherspin first 9 events.) 2. The following processeseach have process. not been (a) p seen.latere π ,(b) from This theµquences− is first understoode− event, ofe−e certain,(c) and asn there conservation conse-p isνe aν¯e dispute,(d)− laws.τ − ifelectric they Explainµγ charge came, what from conservation law forbids 3. The isospin is a good quantum number in strong interaction. There are →0SN1987A.+ Discard→ them and1meson use only→ theω+which first0 9 has events.) isospin→ 0. They+ are understood as the bound states (e) n pµ−ν¯µ,(f)K µ e−2.,(g)eachTheµ following process.− 0π−+ν processesµ (a). p havee π not,(b) beenµ seen.baryone Thisnumbere e is or understoodelectron,(c)n number aspν conse-ν¯ ,(d)τ µγ, quences of certain conservation→ laws. Explain→ whatspin 1 mesons conservation (ρ−→, ρ , ρ )whichformisospin1multipletandanotherspin(d lawu,¯ (u forbidsu¯ dd¯)/√2,ud¯−), and (−uu¯−+ dd¯)/√2, respectively.e e − + 0 + 1mesonquencesω which of has certain isospin conservation 0. They→0 are understood laws.+ Explain as→ the what bound conservation states → law forbids → 2. The following processes(e) n havepµ− notν¯µ,(f) beenK seen.− Thisµ e− is,(g) understoodtauµ number− or asπ muon− conse-νµ number. each process. (a) p 3. eTheπ isospin,(b)µ− is a goode−e− quantume ,(c)(du,¯ (uu¯n numberdd¯)/p√ν2,ueν¯ ind¯),e,(d) and strong (uu¯τ+− interaction.d+d¯)/0√µ2,γ respectively., There+ are →0 + →quences of certaineach− conservation→ process.→ (a) laws.p(a) Explaine Even→π ,(b)→ though whatµ− conservation bothe−eρ−eand,(c)→ω lawsharen forbidsp theνeν¯e same,(d)τ constituents,− µγ, they decay very 0 + →0 + → → → (e) n pµ−ν¯µ,(f)K spinµ 1 mesonse−,(g) (ρµ−−, ρ , ρπ−)whichformisospin1multipletandanotherspinνµ. (e) n + pµ0 −ν¯µ,(f)K µ +e−,(g)µ− muonπ− numberνµ. or electron number → → each→ process.(a)3. (a) EvenThep thoughe isospin→π both,(b)ρ isandµ− aω share goodedi−→eff the−erently.e quantum same,(c) constituents,n Lookp number→ν upeν¯e they,(d) the decay Bookletτ in− strong veryµγ and, interaction. find the most There dominant are decay 1mesonω which has isospin 0.differently. They→0 are Look+ understood up the→ Booklet0 as and+ the find bound the→ most states dominant decay→ (e) n pµ−ν¯µ3.,(f)spinTheK isospin 1 mesonsµ e is−,(g) a ( goodρµ−−,modesρ quantum,πρ−ν)whichformisospin1multipletandanotherspin forµ. them. numberenergy in strong interaction. There are 3. The isospin is a good quantum¯ number¯ in→ strongmodes interaction. for¯ → them. There→ are (du,¯ (uu¯ dd)/√2,ud), and (uu¯ + dd)/√2, respectively.0 + 0 + 0 0 0 0 + − 1mesonspin 1 mesons0ω which (+ρ−(b), ρ has, Explainρ )whichformisospin1multipletandanotherspin isospin why0 0.ρ 0 They0 π π− areis allowed understood but ρ as theπ π boundis not. states spin 1 mesons (ρ , ρ , ρ )whichformisospin1multipletandanotherspin3. The isospin(b) is Explain a good why quantumρ π π number− is allowed in but strongρ interaction.π π→is not. There are → − 1meson0 + ω →which¯ has isospin¯ 0. They→ are¯ understood as the bound states (a) Even thoughspin both 1ρ mesonsand(c)ω Construct(shareρ(−d,u,¯ρ ,( theρu wavefunctionsu¯)whichformisospin1multipletandanotherspin samedd)/ constituents,√(c) for2,u two-pion Constructd), and states they wavefunctions (u bothu¯ decay+ ford totald) very/√ isospin for2, respectively. two-pionI =1 states both for total isospin I =1 (du,¯ (uu¯ dd¯)/√2,ud¯), and (uu¯ + dd¯)/√2, respectively. 30 1mesonω which has isospin 0. They1meson are understoodω whichand has 0, and isospin as determine− the 0. bound the They allowed are states values understood for the relative as the orbital bound angular states ¯ ¯ differently.¯ Look up the Booklet and− find theand most 0, and dominant determine decay the allowed values for the relative orbital angular (du,¯ (uu¯ dd)/√2,ud), and (uu¯ + dd)(/d√u,¯ 2,(uu¯ respectively.dd¯)/momentum√2(a),u(a)d¯), Even Even andL. though ( thoughuu¯ + d bothd¯) both/momentum√ρ2,and respectively.ρ ωandshareωL.share the same the constituents, same constituents, they decay very they decay very modes for them. − − didifffferently.erently. Look Look up the up Booklet the Booklet and find and the find most the dominant most decay dominant decay 0 + 0 0 0 (b) Explain why ρ(a) Evenπ π though− is allowed both ρ and butωρshareπ theπ sameis not. constituents, they decay very (a) Even though both ρ and ω share the→ same constituents,modesmodes forthey for them.→ them. decay very differently. Look up the Booklet0 and+ find the most dominant0 0 decay0 differently. Look up(c) Construct the Booklet wavefunctions and find for the two-pion(b) most Explain states dominant why ρ both0 π for decayπ− total+is allowed isospin butI ρ=1 π π 0is not.0 0 modes for them.(b) Explain why →ρ π π− is allowed→ but ρ π π is not. and 0, and determine the allowed(c)0 Construct values+ for wavefunctions the relative→ 0 for orbital two-pion0 0 angular states both for→ total isospin I =1 modes for them. (b) Explain why ρ π π− is allowed but ρ π π is not. L (c)→and Construct 0, and determine wavefunctions the allowed→ for values two-pion for the states relative both orbital for angular total isospin I =1 0 +momentum . 0 0 0 (b) Explain why ρ π π− is allowed(c) but Constructρ wavefunctionsπ π momentumandis not. 0, for and two-pionL. determine states the both allowed for total values isospin forI =1 the relative orbital angular → and 0, and→ determinemomentum the allowedL. values for the relative orbital angular (c) Construct wavefunctions for two-pionmomentum states bothL. for total isospin I =1 and 0, and determine the allowed values for the relative orbital angular momentum L.