- The expectation value of the energy in this state is greater than or equal to the ground state energy E 0 : ψ | H ˆ | ψ ≥ E 0 . - All particles in nature belong to one of the following classes:. - The state vector of N identical bosons is totally symmetric with respect to the exchange of any two of these particles.. - The state vector of N identical fermions is totally antisymmetric with respect to the exchange of any two of these particles.. - of the one particle Hilbert space.. - (a) If the particles are bosons, the state vector of the system with N 1 particles in the state | n 1 , N 2 particles in the state | n 2 , etc., is:. - (b) If the particles are fermions, the state corresponding to one parti- cle in the state | n 1 , another in the state | n 2 , etc., is given by the Slater determinant:. - Since the state vector is antisymmetric, two fermions cannot be in the same quantum state (Pauli’s exclusion principle). - The above states form a basis of the N − fermion Hilbert space.. - We assume that the state of the system is. - at time t = 0. - The probability to find the system in the state. - at time t is:. - At time t = 0, we assume that the system is in the eigenstate | i of ˆ H 0 . - To first order in ˆ H 1 , the probability amplitude to find the system in another eigenstate. - f at time t is:. - In the case of a time-independent perturbation H 1 , the probability is:. - For simplicity, we assume that the matrix elements f | V ˆ | i only depend on the energies E f of the states | f. - To lowest order in ˆ V , this coupling results in a finite lifetime τ of the state. - i : the probability to find the system in the state | i at time t >. - The function ρ(E) is the density of final states. - When the spin degree of freedom of the particle comes into play, this density of state must be multiplied by the number of possible spin states 2s + 1, where s is the spin of the particle. - The quantity L 3 represents the normalization volume (and cancels identically with the normalization factors of the states. - The lifetime τ of the excited state due to this spontaneous emission is given by:. - where ˆ D is the electric dipole operator.. - To second order in V , the elastic scattering cross-section for an incident particle in the initial momentum state p and the final momentum state p is given by:. - We consider. - where θ is the scattering angle between p and p . - In the case where the range a of the potential tends to infinity, we recover the Coulomb cross section:. - In general, interference effects can be observed between the various q con- tributing to the sum which defines V ( q. - In the case of a charge distribution, V ˜ is the Rutherford amplitude, and the form factor F is the Fourier transform of the charge density.. - In order to study the general problem of the scattering of a particle of mass m by a potential V ( r. - This corresponds to the superposition of an incident plane wave e i k·r and a scattered wave. - When the wavelength of the incident particle λ ∼ k − 1 is large compared to the range of the potential, the amplitude f does not depend on u and u (at least if the potential decreases faster than r − 3 at infinity). - There exists in nature another particle, the µ lepton, or muon, whose physical properties seem com- pletely analogous to those of the electron, except for its mass m µ 200 m e . - A neutrino beam produced in an accelerator can interact with a neutron (n) in a nucleus and give rise to the reactions. - ν e + n → p + e and ν µ + n → p + µ , (1.1) whereas the reactions ν e + n → p + µ or ν µ + n → p + e are never observed.. - In (1.2) we have introduced the antiparticles ν ¯ µ et ¯ ν e . - There is a (quasi) strict symmetry between particles and their antiparticles, so that, in the same way as the electron is associated with the neutrino ν e , the antielectron, or positron, e + is associated with the antineutrino ¯ ν e . - reactions of (1.1) and (1.2). - For a long time, physicists believed that neutrinos were zero-mass particles, as is the photon. - The proof that neutrino masses are not all zero is a great discovery of the last ten years.. - In the present study, we show how the mass differences of neutrinos can be measured by a quantum oscillation effect. - neutrinos ν e , ν µ and ν τ , which are produced or detected experimentally are not eigenstates of the mass, but rather linear combinations of mass eigenstates ν 1 , ν 2 , ν 3 , with masses m 1 , m 2 , m 3. - They can be pro- duced in accelerators, in nuclear reactors, and also in the atmosphere by cos- mic rays, or in thermonuclear reaction inside stars, in particular the core of the sun, and in supernovae explosions.. - 1.1 Mechanism of the Oscillations. - This simple case will allow us to understand the underlying physics of the general case. - The average energy of the (anti-)neutrinos produced in reactors is E = 4 MeV, with a dispersion of the same order.. - In all what follows, we will assume that if m is the neutrino mass and p and E its momentum and energy, the mass is so small that the energy of a neutrino of mass m and momentum p is. - m 1 and m 2 are the respective masses of the states | ν 1 and | ν 2 , and we assume m 1 = m 2. - If the physical states of the neutrinos which are produced (reactions (1.2)) or detected (reactions (1.1)) are not | ν 1 and | ν 2 , but linear combinations of these:. - At time t = 0, one produces a neutrino of momentum p in the state. - Calculate the state | ν (t) at time t in terms of | ν 1 and | ν 2. - What is the probability P e for this neutrino to be detected in the state. - ν e at time t? The result will be expressed in terms of the mixing angle θ and of the oscillation length L. - Express the probability P e as a function of the distance = ct.. - The mass of the muon satisfies m µ c 2 = 106 MeV. - Putting together that data and the results of numerous experiments performed on solar neutrinos, the physicists of Kamland come to the following results:. - Ratio between the numbers of observed electron neutrinos and those expected in the absence of oscillations as a function of the distance to the reactor. - The complete experimental solution of the problem would consist in measuring the three mixing angles θ 12 , θ 23 , θ 13 , the phase δ, and the three masses m 1 , m 2 , m 3 . - At time t = 0 a neutrino is produced with momentum p in the state. - Express, in terms of the matrix elements U αi , its state at a
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