- The neutrino counters have an accuracy of the order of 10% and the energy is E = 4 GeV. - Above which distances 12 and 23 of the production point of the neutrinos can one hope to detect oscillations coming from the superpositions 1 ↔ 2 and 2 ↔ 3?. - Such neutrinos are produced in the collision of high energy cosmic rays with nuclei in the atmosphere at high altitudes.. - To simplify things, we assume that all muons decay before reaching the surface of the Earth. - Deduce that, in the absence of neutrino oscillations, the expected ratio between electron and muon neutrinos. - The corrections to the ratio R µ/e due to the fact that part of the muons reach the ground can be calculated accurately. - In order to explain this relative decrease of the number of ν µ ’s, one can think of oscillations of the types ν µ ν e and ν µ ν τ . - The Super-Kamiokande ex- periment consists in varying the time of flight of the neutrinos by measuring selectively the direction where they come from, as indicated on Fig. - The underground detector measures the flux of electron and muon neutrinos as a function of the zenithal angle α. - Right: number of atmospheric neutrinos detected in the Super-Kamiokande experiment as a func- tion of the zenithal angle (this picture is drawn after K. - neutrinos coming from above (cos α ∼ 1) have traveled a distance equal to the atmospheric height plus the depth of the detector, while those coming from the bottom (cos α. - 1) have crossed the diameter of the Earth (13 400 km).. - Given the weakness of the interaction of neutrinos with matter, one can con- sider that the neutrinos propagate freely on a measurable distance between a few tens of km and 13 400 km.. - Can one observe a ν e ν µ oscillation of the type studied in the first part?. - The angular distributions of the ν e and the ν µ are represented on Fig. - 1.2, together with the distributions one would observe in the absence of oscillations. - In view of the above results, we assume that there is only a two- neutrino oscillation phenomenon: ν µ ν τ in such an observation. - We there- fore use the same formalism as in the first part, except that we change the names of particles.. - By comparing the muon neutrino flux coming from above and from below, give an estimate of the mixing angle θ 23 . - Section 1.1: Mechanism of the Oscillations: Reactor Neutrinos 1.1.1. - The probability to find this neutrino in the state | ν e at time t is P e (t. - We have E 1 − E 2 = (m 2 1 − m 2 2 )c 4 /(2pc). - Defining the oscillation length by L = 4π¯ hp. - 1 − sin 2 (2θ) sin 2 πct. - A ν µ energy of only 4 MeV is below the threshold of the reaction ν µ +n → p +µ. - In order to detect a significant decrease in the neutrino flux ν e , we must have. - The typical distances necessary to observe this phenomenon are of the order of a fraction of the oscillation length.. - We notice that the KamLAND data point corresponds to the second oscillation of the curve. - We conclude that the probability P α → β to observe a neutrino of flavor β at time t is. - The oscillation lengths are propor- tional to the energy. - We saw in the first part that if this mixing is not maximum, the visibility of the oscillations is reduced and that the distance which is necessary to observe the oscillation phenomenon is increased.. - By resuming the argument of the first part, we find that the modification of the neutrino flux of a given species is detectable beyond a distance ij such that sin 2 (π ij /L ij. - In practice, part of the muons reach the ground before decaying, which modifies this ratio. - Naturally, this effect is taken into account in an accurate treatment of the data.. - For an energy of 4 GeV, we have found that the minimum distance to observe the oscillation resulting from the 1 ↔ 2 superposition is 14000 km. - We therefore remark that the oscillations ν e ν µ , corresponding to the mixing 1 ↔ 2 which we studied in the first part cannot be observed at terrestrial distances. - most to the diameter of the Earth (0.04 s), the energy difference E 1 − E 2 and the oscillations that it induces can be neglected.. - The deficit in muon neutrinos is not due to the oscillation ν e ν µ of the first part. - Indeed, we have seen in the previous question that this oscillation is negligible at time scales of interest. - No oscillation ν e ν τ appears in the data. - In the framework of the present model, this is interpreted as the signature of a very small (if not zero) θ 13. - where the averaging is performed on the energy distribution of the neutrino.. - If we measure the neutrino flux coming from the top, we have L 23 , which gives P top = 1. - half of the top value (200 events).. - The difficulty of such experiments comes from the smallness of the neutrino in- teraction cross sections with matter. - In 1998, the first undoubted observation of the oscillation ν τ ν µ was announced in Japan by the Super-Kamiokande experiment Fukuda Y. - This experiment uses a detector containing 50 000 tons of water, inside which 11 500 photomutipliers detect the Cherenkov light of the electrons or muons produced. - In 1992, the GALLEX experiment, using a Gallium target in the Gran Sasso, also confirmed the solar neutrino deficit (P. - We are interested in the ground state of the external electron of an alkali atom (rubidium, cesium. - As in the case of atomic hydrogen, the ground state is split by the hyperfine interaction between the electron magnetic moment and the nuclear magnetic moment µ n . - This splitting of the ground state is used to devise atomic clocks of high accuracy, which have numerous applications such as flight control in aircrafts, the G.P.S. - 2.1 The Hyperfine Splitting of the Ground State. - Give the degeneracy of the ground state if one neglects the magnetic interaction between the nucleus and the external electron. - nucleus: s n , m n a basis of the total spin states (external electron + nucleus).. - As in the hydrogen atom, one can write the corresponding Hamiltonian (restricted to the spin subspace) as:. - where A is a characteristic energy, and where ˆ S e and ˆ S n are the spin operators of the electron and the nucleus, respectively. - (b) Show that the two states. - by diagonalizing 2 × 2 matrices of the type:. - Recover the particular case of the hydrogen atom.. - What are the degeneracies of the two sublevels E 1 and E 2. - Show that the states of energies E 1 and E 2 are eigenstates of the square of the total spin ˆ S 2. - Give the corresponding value s of the spin.. - Sketch of the principle of an atomic clock with an atomic fountain, using laser-cooled atoms. - The atoms are initially prepared in the energy state E 1 , and are sent up- wards (Fig. - At the end of the descent, one detects the number of atoms which have flipped from the E 1 level to the E 2 level. - In all what fol- lows, the motion of the atoms in space (free fall) is treated classically. - In order to simplify things, we consider only one atom in the sub-level of energy E 1 . - This state (noted | 1 ) is coupled by the electromagnetic wave to only one state (noted | 2 ) of the sublevel of energy E 2 . - We assume that the time to cross the cavity is very brief and that this crossing results in an evolution of the state vector of the form:. - The initial state of the atom is | ψ(0. - Taking the limit → 0, show that the state of the atom after this round-trip is given by:. - Give the probability P (ω) to find an atom in the state | 2 at time T . - After the round-trip, each atom is in the state given by (2.1). - We measure separately the numbers of atoms in the states | 1 and | 2 , which we note N 1 and N 2 (with N 1 + N 2 = N. - What is the statistical distribution of the random variables N 1 and N 2 ? Give their mean values and their r.m.s. - ω − ω 0 | introduced by the random nature of the variable N 2 − N 1 . - 2.2 we have represented the precision of an atomic clock as a function of the number N of atoms per pulse
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