Question

Diagonalization of 2 X 2 Hermitian matrices. Usually we are in need of diagonalizing 2 X 2 matrices. The most general form of such matrices in a base { 1), 2)} is 11) = ( 6 ). 12) = (i). A = ( Aust) -(8%). where a and a' are real numbers and b is complex. However, we have seen in quantum mechanics that it is always possible to redefine the phase of the base vectors 11) + |1') = pia 1), 12) + |2') = eiß 12). Show that in the new base the new elements of matrix A12 of A are A12 = e'(B-a) b. Consider that b= |ble, and then choose (a-B) = 8 to eliminate the phase of b, where b can be considered as real.

Answer 1

10) = (6), 17=(8) = AE A, 1X11 + 42,127[2] + A2 113(2) + 4, 12X1|| Now define li> 11'>= ela y = 117 = e r> 127 12'>= e( 127 127 = 1/2') > A = A la l< le tice & A 3 127 <2 1 &'p + Aeo 13 (2) eP + A 2/27 <plea = A, 11'><il + A22/2²>(211 B-a) + 412 e"X+1 + 12X61 » A, 2 = A 2 é (B-d - betod)

A la bé (B-2) A lg-2) a b=1bles Now define - A = ( a lblee- (5=(x-p) a 6 choose 8 = x-ß we can -> É = I a lbl .. = Libl a . ab) I , where lbal bis real * Let me know if you have any questions !