Appendix A Diagonalization of a Hermitian Matrix. TheCase2× 2 Let us consider a hermitian operator H. Eigenstates and eigenvalues of the operator H are given by the equation H |i=Ei |i . (A.1) We will assume that the states |i are normalized. We have i|k=δik. (A.2) Let |α be a full system of normalized and orthogonal states: α |α=δα α. (A.3) In the α-representation we have α |H|αα|i=Ei α |i (A.4) α Equation (A.4) has nonzero solution if the condition Det(H − E) = 0(A.5) is satisfied. This equation determines the eigenvalues of the matrix H. It is obvious from (A.2) and (A.3) that i|αα|k=δik (A.6) α and α |ii|α=δα α . (A.7) i From (A.7) and (A.4) we find Bilenky, S.: Appendix. Lect. Notes Phys. 817, 237–256 (2010) DOI 10.1007/978-3-642-14043-3 c Springer-Verlag Berlin Heidelberg 2010 238 Appendix A α |H|α= α |i Ei i|α. (A.8) i Let us determine a matrix U by the following relation Uαi =α|i. (A.9) From (A.6) and (A.7) follows that U is a unitary matrix: UU† = 1, U † U = 1. (A.10) Equation (A.4) in the matrix form can be written as follows H = UEU†, (A.11) where Eik = Ei δik. Two sets of basic vectors |α and |i are connected by the unitary transformation. In fact we have |α= | |α= ∗ | , | = |αα| = |α . i i Uαi i i i Uαi (A.12) i i α α Let us now consider 2 × 2 real, symmetrical matrix H with trace equal to zero. We have −ab H = , (A.13) ba where a and b are real quantities. For the eigenstates and the eigenvalues of the matrix H we have the following equation H ui = Ei ui . (A.14) For the eigenvalues Ei from (A.5) we find from (A.5) 2 2 E1,2 =∓ a + b . (A.15) Further we have H = OEOT , (A.16) where O is a real orthogonal 2 × 2matrix(OT O = 1) and −10 E = a2 + b2 . (A.17) 01 Appendix A 239 The matrix O has the following general form cos θ sin θ O = . (A.18) − sin θ cos θ From (A.16), (A.17) and (A.18) we find the following equations for the angle θ a = a2 + b2 cos 2θ, b = a2 + b2 sin 2θ. (A.19) From these relations we find b a tan 2 θ = , cos 2 θ = √ . (A.20) a a2 + b2 Let us notice two extreme cases 1. The nondiagonal element b is equal to zero. In this case θ = 0 (there is no mixing) and E1,2 =∓a 2. The diagonal element a is equal to zero. In this case θ = π/4 (maximal mixing) and E1,2 =∓b Appendix B Diagonalization of a Complex Matrix Let us consider a complex n × n matrix M. It is obvious that MM† is the hermitian matrix with positive eigenvalues. In fact, we have † MM |i=ai |i, (B.1) where † 2 ai =i| MM |i= |i| M |k| > 0. (B.2) k = 2 Let us put ai mi .Wehave MM† = Um2 U †, (B.3) ( 2) = 2 δ U is a unitary matrix and m ik mi ik. The matrix M can be presented in the form M = UmV†, (B.4) √ where m =+ m2 and1 − V † = m 1 U † M (B.5) The matrix V is an unitary matrix. In fact, from (B.5) we find − V = M† Um 1. (B.6) From (B.3), (B.5) and (B.6)wehave 1 We assumed that all eigenvalues of the matrix MM† are different from zero. Thus, the diagonal matrix m−1 does exists. 241 242 Appendix B − − − − V † V = m 1 U † MM† Um 1 = m 1 U † Um2 U † Um 1 = 1(B.7) Thus, we have shown that a complex n × n nonsingular matrix M can be diagonal- ized by the bi-unitary transformation (B.4). Equation (B.4) is used in the procedures for the diagonalization of quark, lepton and neutrino mass terms. Appendix C Diagonalization of a Complex Symmetrical Matrix Let us consider a complex matrix M which satisfies the condition M = M T . (C.1) We have shown in Appendix B that any complex matrix M can be presented in the form M = VmW†, (C.2) where V and W are unitary matrices and mik = mi δik, mi > 0. From (C.2)we obtain the relation M T = W † T mVT . (C.3) From (C.2) and (C.3)wehave MM† = Vm2 V †, M T M T † = W † T m2 W T . (C.4) Taking into account that M is a symmetrical matrix, we obtain the following relation Vm2 V † = W † T m2 W T . (C.5) From this relation we find W T Vm2 = m2 W T V. 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