Neutrino LECTURE NOTES - Cornell University Physics .
1NeutrinoLECTURE NOTESLecture notes based in part on a lectures series given by PilarHernandez at TASI 2013, Neutrinos[1], and on notes written by EvgenyAkhmedov in 2000, Neutrino Physics[2].Notes Written by: JEFF ASAF DROR2014
Contents1 Preface32 Introduction2.1 Neutrino Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . .2.2 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4463 Massive ν’s3.1 Working in free space . .3.2 Interacting Case . . . . .3.2.1 Dirac mass . . .3.2.2 Majorana Masses3.3 Neutrino Masses . . . .3.3.1 Dirac Spinor . . .3.3.2 See-saw . . . . .881010111418194 Lepton Mixing4.1 Neutrino Oscillations in Vacuum . . . . . . . . . . . . . . . . . . . . . . .4.2 Neutrino Oscillations in Matter . . . . . . . . . . . . . . . . . . . . . . .212327A Useful Identities30B Solutions31.2.
Chapter 1PrefaceIn this set of notes I go over the fundamentals of neutrino physics. I assume familiaritywith the Standard Model (SM) and Quantum Field Theory. I have added in exercises inthe text and solutions can be found in the appendix. These notes will likely be expandeduntil I lose interest in neutrinos. If you have any corrections please let me know atajd268@cornell.edu.3
Chapter 2Introduction2.1Neutrino ExperimentsThe history of neutrinos is linked tightly to the history of the SM. The way which neutrinos were discovered and later on the way their masses were measured is a facinatingstory but we will not go through this story. Instead interested readers can see the bookby Frank Close called Neutrinos.Neutrinos first appeared in physics in 1900 in β decay. There was a transition:AZN0 AZ 1 N e(2.1)the energy of the electron theoretically was expected to beEe (mN mN 0 ) c2 Q(2.2)which suggested that the energy of the outgoing electron should have a Gaussian distribution:NEewhere the dashed distribution is the expected distribution while the experimental distribution is the solid line1 .A solution to this puzzle was suggested by Pauli which was that there was an extraparticle being emitted (Pauli’s “dark matter”),Ee (mN mN 0 ) c2 Q ν1(2.3)The spectrum was expected to be a Gaussian and not a delta function due to potential kinetic energyof the nuclei.4
2.1. NEUTRINO EXPERIMENTS5This particle was called neutrino and this idea was taken seriously by Fermi whocreated the famous four-fermi interaction:pe nνThis reaction implied the existence of a related reaction:νnpe Bethe-Peierls estimated this cross-section at the small value, σν 10 44 cm2 . It wascalculated that with a detector mass of about a ton and a neutrino flux of Φ 1011 ν’scm2 /s you can get a few events per day using:σν Φν few eventsNA Z Mdet Aday(2.4)Reines and Cowen in 1956 first discovered the neutrinos in an experiment using areactor:e νnThe reactor produced neutrinos which hit a target made of protons. Through the fourFermi interaction an electron and neutron were produced. The electron quickly annhilatedand the neutron underwent neutron capture2 . The delay between these two events is avery powerful discriminator which enables you to essentially eliminate the background.Muons were discovered in cosmic rays in the 1930’s in decays:π µ νν(2.5)To find if this neutrino was the same one as discovered from β decay, in 1962 Lederman,Schwarz, and Steinberger performed the first accelerator neutrino experiment:2Neutron capture is when a neutron enters a nucleus and remains there to form a heavier nucleus.
6CHAPTER 2. INTRODUCTIONShieldµ νpDetectorµT argetπ’sThe protons were accelerated into a target which produced pions. The pions decayedinto muons and neutrinos. The muons were then stopped by a shield and the neutrinospassed through the shield and hit the detector. The detector only produced muons andnot electrons. This is how flavor was discovered. If neutrinos from beta decay was thesame as the ones arising from muons then they would have discovered both electrons andmuons.In all these experiments neutrinos could have had a mass. If they did they therewould be kinematical effects. However, in all the experiments these kinematical effectswere very small. The best current bounds on the neutrino masses areH 3 He e νe “mνe ” 2.2eVπ µ νµ “mνµ ” 170keVτ 5π ντ “ mντ ” 182MeV32.2(2.6)(2.7)(2.8)Basic PropertiesNeutrinos in the SM have the following properties1. Form a SU (2)L doublet with the leptons: νLi Li(2.9)where i e, µ, τ with SU (3) SU (2) U (1) charges of(1, 2, 1)(2.10)Since the electric charge is T3 Y2 , the upper component of the doublet (which hasT3 1/2) has an electric charge, Qν 0.2. Concerning Lorentz transformations neutrinos form left Weyl spinors:νL PL ν(2.11)where1 γ52The equation of motion for a free neutrino isPL p/νL 0(2.12)(2.13)
2.2. BASIC PROPERTIES7This has a non-trivial solution if and only if the determinate of p/ is equal to zero.Finding the determinate gives:p20 p2 0 E p The helicity operator isĥ (2.14)(2.15)p0 p/σ·p p p (2.16)Acting on the neutrino gives,ĥνL EνL p (2.17)Inserting in the possible energies we found above we have,(E 0 ĥ 1E 0 ĥ 1(2.18)So νL can either be a particle with negative helicity or an antiparticle with positivehelicity. The simple existence of neutrinos breaks Parity and charge conjugationsince Parity sends left neutrinos to right neutrinons which don’t exist and chargeconjugation to left anti-neutrinos which again don’t exist.3. Neutrinos interact in the SM through: ggµν L,a γµ Z µ νL,aLSM ν L,a γµ W L,a h.c. 2 cos θW2(2.19)where the sum over a runs over the generations.It is due to the the neutral current interactions that we know there are three flavorsof neutrinos. At LEP they did the experiment:νae Ze νaThey then the measured the ratio of the invisible width of the Z boson to theoreticalwidth for a single neutrino. Since the Z boson couples equally to each flavor, theresult should give the number of neutrino flavors. They measured:Γinv (Z) 2.984(8)Γνa ν aThus the SM must have 3 families of neutrinos.(2.20)
Chapter 3Massive ν’s3.1Working in free spaceWe begin by studying massive neutrinos but ignoring any gauge interactions. In the SMthe massive fermions are Dirac fermions which means that they are a four-componentspinor which is the smallest representation of the Lorentz group Parity. A free Diracfermion has the Lagrangian: (3.1)LD Ψ i / m ΨwhereΨ ψL ψR PL Ψ PR Ψ(3.2)so the mass term is of the form, mΨΨ m ψ L ψR ψ R ψL (3.3)This implies immediately that for neutrinos in the SM, which only have the left component, we cannot write down such a mass term. If you want to write a Dirac mass forneutrinos you need to have a four-component representation of the neutrino and rqeuirea right-handed component for the neutrinos, νR . Since these neutrinos haven’t beendetected anywhere they can’t interact with anything and are called sterile neutrinos.While one can introduce a Dirac mass, this is not the only way that one could addmass to the neutrinos. Majorana realized that you can give mass to particles in a moreeconomical way. Majorana fermions are two component Weyl fermions that have themass term: mψ L ψL c h.c.(3.4)LM ajorana 2whereTψL c Cψ L Cγ0 ψL (3.5)C iγ0 γ2(3.6)where8
3.1. WORKING IN FREE SPACE9A Dirac mass is a coupling between a fermion and its conjugate since T TTψ L C Cψ L (ψL c )T C(3.7)(3.8)and thus a Dirac mass is of the form,ih c Tc T(3.9) m ψ L ψR ψ R ψL m (ψL ) CψR (ψR ) CψLhim (ψL c )T CψR (ψR c )T CψL ψR T CψL c ψL T CψR c2(3.10) 10 mψLC h.c.(3.11) ψLT (ψR c )Tm 0ψR c2Similarly one can rearrangemLM ajorana 2m 2m 2the Majorana mass for left handed particles as: ψ L ψLc h.c. ψLT CψL h.c. m 0ψLTc TC h.cψL (ψR )0 0ψRc(3.12)(3.13)(3.14)Exercise 1.a) Show that the Majorana mass does not vanish identically in a trivial way. In otherwords that PR ψLc ψLc .b) Show that the mass term in Lorentz invariant. Recall thatiµνψ e 4 ωµν σ ψ(3.15)i µ ν[γ , γ ]4(3.16)whereσµν c) Show that the equations of motion derived from this Lagrangian in momentum spacegive solutions that have the equationE 2 p2 m2(3.17)Exercise 2. Write down a four-component spinor asΨM ψL ψLc(3.18)and show that this Lagrangian can be written as a Dirac Lagrangian 1LM ajorana ΨM i / m ΨM2(3.19)
10CHAPTER 3. MASSIVE ν’SExercise 3. Show that a Dirac fermion is equivalent to two degenerate Majorana fermions.To do this you need to start with a Dirac field:Ψ ψL ψR(3.20)You then define Majorana fields the same way we above,ΨM1 ψL ψLcΨM2 ψR ψRc(3.21)(3.22)The Dirac mass term can then written in terms of these new fields as 0 mψM1 mΨΨ ψ M1 ψ M2m 0ψM23.2(3.23)Interacting CaseWe know that for any fermion in the SM that carries charge we can’t write down massterms since the left and right handed fermions have different charges. Furthermore, wecan’t write down Majorana mass terms due to the conjugation in ψLc .3.2.1Dirac mass In the SM we have a Higgs field which has the quantum numbers: 1, 2, 12 . We also have Φ̃ iσ2 Φ with quantum numbers: 1, 2 21 . The Higgs acquires a VEV: Φ 0 v/ 2 ,Φ̃ v/ 20(3.24)To add a Dirac mass we need right-handed neutrinos such that we can write a Yukawainteraction of the form,(1,1,0) LΦ̃ νR h.c.(3.25)LY uk λν {z} ( 1,1,0 )Upon spontaneous symmetry breaking (SSB) this term gives a Dirac mass:with (ν L mν νR h.c.)(3.26)vλν 2(3.27)
3.2. INTERACTING CASE3.2.211Majorana MassesA Majorana mass term for a left-handed neutrino is of the form, mνLT CνL . In the SMprior to spontaneous symmetry breaking such a term is forbidden such it breaks U (1)Ysymmetry.The next suggestion would be to form such a term through spontaneous symmetrybreaking. However, there are some important requirements1. The Majorana mass term has to come from a term with two SU (2) doublets2. Both SU (2) doublets need to be either unbarred or both barred to have a Majoranamass3. The term must conserve hyperchargeThese are very strong requirements. One can go ahead and try different terms and itseasy to see that most naive combinations are forbidden. To try to better understandthe type of allowed terms consider the charges of the Majorana term. It has an isospin,T3 1, and hypercharge 2. This suggests that it can only arise from a term thattransfers as a vector under isospin rotations. The lowest order term that transforms as avector under isospin and has a neutrino-neutrino coupling is given by, LT iσ2 σL(3.28)Exercise 4. Show that this term transforms as a vectorHowever, we have yet to find a isospin singlet. To do this we have to take theproduct of this term with another term which transforms as a vector under SU (2) andhas hypercharge of 2. If we had such a particle, , we could write: T L (3.29)Y uk f L iσ2 σL · h.c.Then when the electrically neutral component of the Higgs (the third component) gets aVEV we would get a Majorana mass. However, in the SM where we don’t have such amass we can still get a Majorana mass by taking a product of Higges: LY uk α LT iσ2 σL · H T iσ2 σH h.c.(3.30)Note that this term cannot arise from loop in the SM due to conservation of leptonnumber.Matching dimensions we see that we have[α] 1We can write(3.31)λ(3.32)Λwhere Λ is a scale of new physics. A big question of new physics is if neutrinos areMajorana, what the hell is this scale? If neutrinos are Majorana then we are very closethe situation that Fermi was in when he was describing β decay. The four-point Fermiinteractions had dimensionful coupling which was an effective theory:α
12CHAPTER 3. MASSIVE ν’SppSMe nne WννeIn this way you can undertand that the Fermi coupling in the limit of low energy isjust1GF 2(3.33)MWNow with neutrinos we are in a very similar situation where we have the couplingΦΦνSM?νcνThe good thing about having this possibility is that neutrinos are very light comparedto the other particles in the SM and masses arising from non-renormalizable operatorsare naturally suppressed. Plotting fermion masses on a Logrithmic scale gives roughly,quarksν’sleptonseVkeVMeVGeVTeVThere is a gap of at least 6 orders of separating the neutrino masses from the remainingfermions. If there was a Dirac mass then its hard to understand how this would comeabout because after all neutrinos are in the same doublet as the leptons. However, in thecase of Majorana neutrinos if Λ v then the neutrino mass would bemν mFvΛ(3.34)Having Dirac or Majorana masses has very different implications in terms of globalsymmetries. If neutrinos are Dirac fermions then Lepton number (Li eiα Li ) is stilla good symmetry. However, if neutrinos are Majorana then no such symmetry remainssince a Majorana mass term breaks all possible symmetries.Deciding whether we have Dirac or Majorana neutrinos is a fundamental question inneutrino physics. In principle Dirac and Majorana neutrinos leave different traces. Takefor example a neutrino beam from pion decay. The following will always produce a muonat the detector,
3.2. INTERACTING CASE13µ µ π νµregardless of whether the neutrino is Dirac or Majorana. For a Dirac fermion the muonhas to be of negative sign to conserve lepton number. However if a neutrino has aMajorana mass then it may spontaneously turn into an antineutrino and subsequentlydecay into a antimuon.Unfortunately the rate for getting these negative muons is approximately equal to therate of getting a position muons. The rate of getting a positive (R ) or negative muon(R ) are related by: m 2ν(3.35)R R Ebecause to make the transition from neutrino to antineutrino you need a helicity flip.Since mν is so small such a measurement is typically not practical.The best method to look for Majorana neutrinos is in what’s known as neutrinolessdouble beta decay (ββ0ν):A, Zββ2νA, Z 2e νeνee This process is very high order in the weak interaction and is very rare. However it hasbeen measured to have rateTββ2ν 1018 1020 y(3.36)If neutrinos are Majorana then such a process can take place without any outgoingantineutrinos since the two antineutrinos can annhilate one another. The rate has beenmeasured to be roughly,Tββ0ν 1025 y(3.37)The ββ0ν decay has a larger phase space so such a measurement is not hopeless. Onecan then go ahead and measure the energy distribution. Depending on the type of massthe neutrinos have we have two possible distributions (notice the analogy to our earlierdiscussion):NEe Ee
14CHAPTER 3. MASSIVE ν’Swhere now the dashed distibution holds if the neutrinos are Majorana while the soliddistribution holds if neutrinos are Dirac.3.3Neutrino MassesWhat we have learned from experiments was thatmν m , mq(3.38)Neutrinos have a very different mixing pattern to the CKM mixing. There have been alot of works in the literature to try to understand the mixing pattern however we haven’tgotten very far in this aspect. On the other hand we do have a compelling explanation forwhy the neutrinos have small masses. This explanation is that neutrinos are Majoranaand their masses arise from a non-renormalizable operator. This non-renormalizableoperator represents some new physics, but what in particular is not yet understood,Φ̃Φ̃νSM?LcLThere are essentially three possibilities:The particles could exchange a fermion,Φ̃Φ̃singlet ortripletLLcThe fermion could be a singlet fermion (type 1), triplet fermion (type 2). Alternatively,you could have a t-channel process through a triplet scalar boson in the t channel (type3):Φ̃Φ̃tripletLLc
3.3. NEUTRINO MASSES15Of course one can consider the term coming from loop level. In this case there willbe the non-renormalizability suppressed as well as loop suppression.We focus only on type 1. Since the intermediate particles are singlets they are oftencalled right-handed neutrinos. This gives the terms:1 cLtype 1 LSM λLΦ̃NR N R mR NR2(3.39)where NR is the new singlet fermion. At low energies we have,Φ̃λΦ̃ p MN Φ̃Φ̃LcLcλip/ mRLLThe coupling in the four point interaction is:λT1λMR(3.40)The neutrino mass in this model is then,mν λTv2λMR(3.41)If we measure neutrino masses then how can measure the scale MR ? The Yukawa couplings in the SM are:tope 10 61λIf we expect the neutrino Yukawas to be of similar order as those in the SM then whatare then if we take MR O(GUT) then the neutrino Yukawas will be of the order of thetop yukawa. If you take MR O(TeV) then you get Yukawas of order of the electronYukawa. So we have,MR [TeV, GU T ](3.42)The energy scale can be anything!To constrain the couplings you can use natuarlness1 . The corrections to the Higgsmass due to these new fermions are given by,LHHN1This arguement fails if you have SUSY.
16CHAPTER 3. MASSIVE ν’SThe corrections areδm2H λ2MRMR2 log216πµ(3.43)Exercise 5. Show that the corrections indeed take this formThe predictions of type 1 seasaw models are the following:1. ββ0ν provided MR . 100MeV2. Matter-antimatter asymmetry in the evolution of the universe which might explainthe one we observe. However, this is only true if there is CP violation in the leptonsector which we haven’t found yet.3. New states which have not yet been observed.The type 1 Lagrangian mass terms can be written as (see Eqs 3.14 and 3.11),111 Ltype 1 νLT CmL νL νRT Cm R νR ν L m D νR h.c. nTL MCnL h.c.222where νL (νL1 , νL2 , ., νLn ), νR (νR1 , νR2 , ., νRn ), nL (νL , νR c ), mL mDM mTD mR(3.44)(3.45)where mR , mL are complex symmetric n n matrices and mD is a complex n n matrix.Exercise 6. Show that the matrix form shown above is consistent with the explicit form.We first consider the special case of a single generation. Diagonalization in this caseis straightforward. Furthermore, for simplicity we consider the mass parameters to bereal. In this case the matrix can be diagonalized by an orthogonal matrix: cos θ sin θU (3.46) sin θ cos θWe denote the diagonalized fields by χL : nL Uχ1Lχ2L We now find the angle that diagonalizes the mass matrix: m1 0 U T MU0 m2 cθ sθmL mDcθ s θ s θ cθmD mR sθ cθ 22sθ mR cθ mL s2θ mD c2θ mD 12 s2θ (mL mR ) c2θ mD 21 s2θ (mL mR ) s2θ mL c2θ mR s2θ mD(3.47)(3.48)(3.49)(3.50)
3.3. NEUTRINO MASSES17This matrix is diagonal iftan 2θ 2mDmR mL(3.51)To calculate the mass eigenvalues we use the following relationsmR mLc2θ p 2,4mD (mR mL )2c2θ 1 1mR mL p 2,2 2 4mD (mR mL )22mDs2θ p 24mD (mR mL )2s2θ 1 1mR mL p 22 2 4mD (mR mL )2(3.52)(3.53)which gives,m1/2mL mR 2s mL mR2 2 m2D(3.54)The masses are real but can be positive or negative. In terms of the mass eigenstates theLagrangian takes the form,1 Lm χTL Md χL h.c.2 1 m1 χ1LT Cχ1L m2 χ2LT Cχ2L h.c.2(3.55)(3.56)Consider the Dirac spinors,χ1 χ1L η1 χ1Lc ,χ2 χ2L η2 χ2Lc(3.57)where η1/2 is positive if m1/2 0.With these we have, χχ χL ηχLc (χL ηχ1Lc ) χLcT ηχLT C (χL ηχ1Lc ) η χLT CχL χLcT Cχ1Lc η χLT CχL h.c.(3.58)(3.59)(3.60)(3.61)Thus we can alternatively write the mass term as two products of Majorana spinors: 1 m1m2 Lm χ1 χ1 χ2 χ22 η1η21 ( m1 χ1 χ1 m2 χ2 χ2 )2(3.62)(3.63)
18CHAPTER 3. MASSIVE ν’S3.3.1Dirac SpinorConsider now the special case of a Dirac spinor. In this case mD m and mL mR 0.This gives mass eigenvalues of m and a mixing angle of θ π4 . The mass eigenstatesare,1νL (χ1L χ2L ) ,21 χ1L (νL νR c ) ,21νR c ( χ1L χ2L )21χ2L (νL νL c )2(3.64)(3.65)This gives,1χ1 (νL νR c νL c νR ) ,21χ2 (νL νR c νL c νR )2(3.66)which gives,χ̄1 χ1 νL νR νR c νL c νL c νR c νR νL (.)(3.67)where we omit the negative terms as they will cancel with the strictly positive χ̄2 χ2contribution. Thus we have, mνL νR νR c νL c νL c νR c νR νL2 m (νL νR νR νL ) Lm (3.68)(3.69)where we have used the identity, νL c νR c . This finally be rewritten in terms of a Diracspinor: Lm mν D νD(3.70)where νD νL νR .Under a global U (1) phase transformation we have,ν D νD ei(LL LR ) νL νR e i(LL LR ) νR νL(3.71)Theres a conserved Lepton number only ifLL LR(3.72)The same can be shown for χ 2 χ2 .For the Dirac case, the mass term is also particularly simple: Lm m νL νL νR c νR c νL c νL c νR νR m (3.73)(3.74)
3.3. NEUTRINO MASSES3.3.219See-sawNow consider another type-1 see-saw mechanism limit, mL mD mR . In this casem1 mL 2m2D,mRm2 mR(3.75)Thus we have one eigenstate that is very heavy with a mass of approximately, mR . Thesecond eigenstate has a very small mass suppressed by the scale mR . Furthermore, in thisD1. The left handed components of the mass eigenstates are approximatelylimit θ mmRgiven by,χ1L νL mD cν ,mR Rχ2L νR c mDνLmR(3.76)which gives full mass eigenstates,mD(ν c η1 νR )mR RmDχ2 νR c η2 νR (ν η2 νR c )mR Rχ 1 νL η 1 νL c (3.77)(3.78)The eigenstates are approximately just the left and right handed neutrinos but with aD.small admixture of order mmRWe now consider the same limit but for the full n-generation case. We want to blockdiagonalize the matrix M: m̃L 0mL mDTTU (3.79)uL U χL , U MU UmTD MR0 M̃RU is an orthogonal 2n 2n matrix. Since we take the eigenvalues of MR to be large thismatrix is almost diagonal. We take as an ansatz U to be of the form, 1ρU U T U 1 O(ρT ρ)(3.80) ρT 1and we furthermore, take ρ to be small.Multiplying out the matrices gives, mL ρmTD mD ρTmL ρ mD ρMRTU MU ρT mL mTD MR ρT mTD ρ ρT mD MR(3.81)This matrix is diagonal if ρ mD m 1R and gives the mass matrices:m̃L mL mD MR 1 mTDm̃R MR(3.82)(3.83)Diagonalization of the effective mass martrix m̃L gives n light Majorana neutrinos madeup primarily of νL with a small admixture of order of mD MR 1 of the “sterile” right handed
20CHAPTER 3. MASSIVE ν’Sneutrinos. Diagonalization of the matrix m̃R gives the masses for the heavy eigenstatescomposed primarily of the right handed neutrinos with a small admixture of the lefthanded neutrinos.The neutrino masses as a function of the right handed neutrino Majorana mass canbe calculated. At very small right handed neutrino Majorana masses, the left and rightneutrinos just have a Dirac mass. However, as the right neutrino mass increases, there issplitting and we get 3 very heavy neutrinos and 3 very light:Neutrino massesm3m2m1MR
Chapter 4Lepton MixingWhether neutrinos are Dirac or Majorana the Yukawa coupling will still not be diagonalin flavor. The Lagrangian will include one of the two terms:Majorana: L c v2v)ν L,a λ(MνL,b h.c. Dirac: L ν L,a λ(D)ν h.c.ννabab R,b2Λ2(4.1)Its not hard to show that the Majorana mass matrix must be symmetric n n complexmatrix. In the case of Dirac mass λ(D) is a symmetric n nR complex matrix (the numberof right handed neutrinos need not be equal to the number of left handed neutrinos). To(M )go from the flavor basis to the mass basis you can always write λν as a diagonal matrix:)λ(M Vν† diag (mνL ) Vνν(4.2)Vν† diag (mνL ) Uν(4.3)λ(D)ν In the Dirac case the unitary matrix on the right and on the left are different.First we consider the Dirac mass case. You can always go to the mass basis and rotateyour fields:νR0 Uν νR 0R U RνL0 Vν νL 0L V L(4.4)(4.5)(4.6)(4.7)Now in terms of the primed fields the masses are diagonal, however the gauge interactionsare no longer diagonal. The only interaction which remains non-diagonal is the chargecurrent interactions which in terms of neutrinos looks likegLSM ν 0L Vν† V γµ 0L Wµ h.c. {z }2UP M N S21(4.8)
22CHAPTER 4. LEPTON MIXINGThe UP M N S matrix is completely analogous to the CKM matrix. An interesting exerciseis to compute how many physical parameters are in the PMNS matrix. A typical unitarymatrix hasn(n 1)(4.9)2physical parameters where the plus is for phases and the minus is for angles. Howeverthis is not the correct answer.Exercise 7. Perform this exercise for both Dirac and Majorana neutrinos.When you produce neutrinos they are produced in a flavor eigenstate, say ν forexample. The neutrino is a linear combination of mass eigenstates: ν i XUi νi i(4.10)iwhere Ui are entries from the PMNS matrix. When neutrinos propagate in space, themass eigenstates are the ones that propagate. When the flavor is again measured aftera while there is a probability to measure it as any of the flavor eigenstates. Thus leptonnumber for each generation is broken, though total lepton number is still conserved.Pictorically we have, β ανsourcetargetIn the beginning we have, νeν1 νµ UP M N S ν2 ντν3(4.11)At some point farther away these states will pick up phases: νi (p)i e iEi t νi (p)i(4.12)pwhere Ei p2 m2i . Due to this interference between states we have a non-vanishingprobability to find a new type of neutrino. This gives rise to the phenomena of neutrinooscillations. We now answer the question “if you produce a flavor α what is the probabilitythat you will detect a flavor β a given distance away?” We label this probability P ( α β ).
4.1. NEUTRINO OSCILLATIONS IN VACUUM4.123Neutrino Oscillations in VacuumWe derive this result using simple quantum mechanics and treat the neutrinos as planewaves. At some initial time you produce a neutrino is a flavor state;X να (t0 )i Uα,i νi (p)i(4.13)iThese states are eigenstates of the Hamiltonian and are on shell:Ĥ νi (p)i Ei νi (p)iqEi m2i p2(4.14)(4.15)We evolve the initial state through the evolution operator:X iEi (t t0 ) να (t)i Uαie νi (p)i(4.16)iThe probability is simplyP (να νβ ) hνβ να (t)i 2(4.17)2 X iEi (t t0 )Uβ,i Uα,ei(4.18)iThe difference of the energies is,qq22Ej Ei p mj p 2 mi m2j m2i2 p (4.19)(4.20)Furthermore,L t t0Inputting in these results gives the master formula:X i m2ji L/2 p Uα,j Uβ,i Uα,ieP (να νβ ) Uβ,j(4.21)(4.22)ijThis equation assumes a few things. Firstly that the neutrino mass eigenstate eachstarts off with the same momenta. Further you may wonder how the neutrino statescan be a good approximation of plane waves if you initially say that these states arelocallized at the source and detector. This derivation also did not incorporate the twokey complications:1. The uncertainty in the momentum of the source and detector.
24CHAPTER 4. LEPTON MIXING2. Coherence over macroscopic distances.One can go ahead and make this more rigorous but that is beyond the scope of thesenotes.The master formula gives the “probability of appearance” for α 6 β and the “probilityof dissappearance” for α β. One can show that the master formula can be written ina slightly more famililar form using the Unitarity of the matrix:P (να νβ ) δαβ 4Xi j 2Xi j 2 m2ij L Re Uα,i Uβ,i Uα,j Uβ,j sin4E m2ij L Im Uα,iUβ,i Uα,j Uβ,jsin2E(4.23)Exercise 8. Prove equation 4.23Solution 1.To compute the corresponding formula for neutrinos we just need to take the complexconjugate of the product of matrix. This gives,P (ν α ν β ) δαβ 4Xi j 2Xi j 2 m2ij L Re Uα,iUβ,i Uα,j Uβ,jsin4E m2ij L Im Uα,iUβ,i Uα,j Uβ,jsin2E(4.24)Thus to the extent that the imaginary term does not vanish we will have a difference inthe oscillation probability for neutrinos and antineutrinos. In order to avoid this termthe phases in the imaginary parts need to add to zero. It is a simple exercise to showthat in the case that m 0 for one set of the neutrinos this term vanishes.For only two generations there is one angle in the mixing matrix and its given by, U cos θ sin θsin θ cos θ (4.25)Plugging in to the result above for dissappearance1 m2 LP (να νβ ) sin 2θ sin sin2 2θ sin24EP (να να ) 1 P (να νβ )212 1.27 m2 (eV )L(km)E(GeV )It helps to note that for i j we can only have i 1, j 2. (4.26)(4.27)
4.1. NEUTRINO OSCILLATIONS IN VACUUM25P (να νβ )LLosc π E(GeV )1.27 m2 (eV )In a real experiment we don’t have the pleasure the varying L but we can still vary theenergy:P (να νβ )EThe position of the first maxima is given byEmax 1.27 m2 (eV 2 )L(km)π/2(4.28)If you want to do a neutrino experiment for which you want to get the angle θ and m2then you should chooseE m2(4.29)LIfE/L m2(4.30)Then the probability is approximately:22P (να νβ ) sin 2θ m 1.27L(km)E(GeV ) (4.31)Any measurement of the probability of disspearance can’t disentangle the contributionof m2 and sin2 2θ. Typical measurement plot is given by,
26CHAPTER 4. LEPTON MIXINGlog m2dresueaMndbalog sin2 θ If E L m2 then the sin2 m2 oscillates very quickly and averages out to 1/2.Then the probability is roughly:P (να νβ ) 1 2sin 2θ2(4.32)In this limit the length is so large that you lose your coherence and all you get is themixing angle. This gives the following plot:log m2Measured bandlog sin2 θ Finally in the optical region where E/L m2 we have:log m2Measurement regionlog sin2 θ
4.2. NEUTRINO OSCILLATIONS IN MATTER27This discussion referred to neutrinos propagating in vacuum. However, most neutrinoexperiments don’t pass through the vacuum. The effect of matter can be significant aswe will see.4.2Neutrino Oscillations in MatterA neutrino passing through the Earth can undergo the following processes:e, p, ne, p, nZνe,µ,τνee Wνe,µ,τνee where the second diagram only involves electrons since muons and taus aren’t for innormal matter. The neutral current interactions is harmless as it effects each neutrinoflavor equally. On the other hand since only electrons are found on Earth. The electronneutrino will be effected more strongly then the other flavors as the neutrino beam passesthrough matter.To calculate this effect we use the Fermi approximation:GFHCC heγµ (1 γ5 ) νe ν e γµ (1 γ5 ) eimatter2GF e γµ (1 γ5 ) e ν e γµ (1 γ5 )
LECTURE NOTES Lecture notes based in part on a lectures series given by Pilar Hernandez at TASI 2013, Neutrinos[1], and on notes written by Evgeny Akhmedov in 2000, Neutrino Physics[2
Introduction of Chemical Reaction Engineering Introduction about Chemical Engineering 0:31:15 0:31:09. Lecture 14 Lecture 15 Lecture 16 Lecture 17 Lecture 18 Lecture 19 Lecture 20 Lecture 21 Lecture 22 Lecture 23 Lecture 24 Lecture 25 Lecture 26 Lecture 27 Lecture 28 Lecture
Ina Sarcevic University of Arizona Models of Neutrino Mass with a Low Scale Symmetry Breaking New Interactions of Supernova Relic Neutrinos Probing Neutrino Properties with Supernova Neutrinos Experimental Detection of the New Interactions via Light Scalar (HyperK, GADZOOKS, UNO, MEMPHYS) BBN and SN1987A Constraints on the Parameter Space of the
Study of Acoustic Ultra-high-energy Neutrino Detection (SAUND) J. Vandenbroucke et al. ApJ 2005 7 hydrophones at undersea Navy array in Bahamas Searched for neutrino signals in 195-day exposure First acoustic neutrino limit, not competitive with radio but new technique SAUND-II now underway, 49 hydrophones (103 km2)
Project Report Yi Li Cornell University yl2326@cornell.edu Rudhir Gupta Cornell University rg495@cornell.edu Yoshiyuki Nagasaki Cornell University yn253@cornell.edu Tianhe Zhang Cornell University tz249@cornell.edu Abstract—For our project, we decided to experiment, desig
Atmospheric Neutrino Oscillation In the simplest, two-flavor, oscillation case, the survival probability depends on m2 L/E ν The various atmospheric neutrino sub-samples span a range of five decades in neutrino energy Eν From 100 MeV to 10 TeV Depending on the neutrino's arrival direction, L spans a range of three decades From 15 km .
GEOMETRY NOTES Lecture 1 Notes GEO001-01 GEO001-02 . 2 Lecture 2 Notes GEO002-01 GEO002-02 GEO002-03 GEO002-04 . 3 Lecture 3 Notes GEO003-01 GEO003-02 GEO003-03 GEO003-04 . 4 Lecture 4 Notes GEO004-01 GEO004-02 GEO004-03 GEO004-04 . 5 Lecture 4 Notes, Continued GEO004-05 . 6
Aman Agarwal Cornell University Ithaca, NY aa2398@cornell.edu Ivan Zaitsev Cornell University Ithaca, NY iz44@cornell.edu Xuanhui Wang, Cheng Li, Marc Najork Google Inc. Mountain View, CA {xuanhui,chgli,najork}@google.com Thorsten Joachims Cornell University Ithaca, NY tj@cs.cornell.edu AB
filter True for user-level API (default is False – admin API) persistent_auth True for using API REST sessions (default is False) . UI Plugin API (Demo) Scheduling API VDSM hooks. 51 UI Plugins Command Line Interface . 52 Web Admin user interface Extend oVirt Web Admin user interface. 53 Web Admin user interface. 54 Web Admin user interface . 55 Web Admin user interface. 56 Web Admin user .
