By John Walsh

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The probability to find this neutrino in the state |νe at time t is Pe (t) = | νe |ν(t) |2 = cos2 θ e−iE1 t/¯h + sin2 θ e−iE2 t/¯h 2 , which gives, after a simple calculation: Pe (t) = 1 − sin2 (2θ) sin2 (E1 − E2 )t 2¯h . We have E1 − E2 = (m21 − m22 )c4 /(2pc). Defining the oscillation length by L = 4π¯ hp/(|∆m2 | c2 ), we obtain Pe (t) = 1 − sin2 (2θ) sin2 πct L . 3. For an energy E = pc = 4 MeV and a mass difference ∆m2 c4 = 10−4 eV2 , we obtain an oscillation length L = 100 km. 4. The time of flight is t = /c.

1. 2 Oscillations of Three Species; Atmospheric Neutrinos We now consider the general formalism with three neutrino species. We denote |να , α = e, µ, τ the “flavor” neutrinos and |νi , i = 1, 2, 3 the mass eigenstates. 8) ⎝ Uµ1 Uµ2 Uµ3 ⎠ i=1 Uτ 1 Uτ 2 Uτ 3 ∗ Uαi = δαβ ) and it can be written as: This matrix is unitary ( i Uβi ⎞ ⎞⎛ ⎞⎛ ⎛ 0 s13 e−iδ c13 c12 s12 0 1 0 0 ˆ = ⎝ 0 c23 s23 ⎠ ⎝ ⎠ ⎝ −s12 c12 0 ⎠ 0 1 0 U iδ 0 −s23 c23 0 0 1 −s13 e 0 c13 where cij = cos θij and sij = sin θij . The complete experimental solution of the problem would consist in measuring the three mixing angles θ12 , θ23 , θ13 , the phase δ, and the three masses m1 , m2 , m3 .

We want to know the minimal distance necessary in order to observe oscillations. We assume that both mixing angles θ12 and θ23 are equal to π/4, which corresponds to maximum mixing. 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. e. ij ≥ Lij /10. This corresponds to 12 ≥ 14000 km for the oscillation resulting from the superposition 1 ↔ 2, and 23 ≥ 400 km for the oscillation resulting from the superposition 2 ↔ 3.

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