4. Practice with Pauli matrix operators: Consider the three 2 x 2 Pauli matrices of, oy, of. These matrices have the following properties: they are unitary, Hermitian, square to identity, and satisfy o"oy = 10%, 636% = io", 0 0 = ios. Using these properties, show the following: (a) eldon = cos(0) 1 + isin(0) on where n is a three-component unit-vector, and O is a real number. (b) (8 . a) (6 . b) = a . b1 + io . (a x b), where a, b are three-component vectors. 2 (c) tr (8 . a) (8 . b) (8 . c) = 2ia . (bx c), where a, b, care three-component vectors, and 'tr' denotes trace. (d) Lizx,y,2 08075 = 20a,868,y - da,Boy,8 where o'. denotes the (a, B) element of matrix o', e.g., of.1 = 1, 0 2 =0 etc. Hint: Consider a 2 x 2 matrix A and express it in terms of Pauli matrices as A = c1 +a . o, where c is a constant, and a is a vector, both of which you can determine in terms of matrix A using Pauli matrices' aforementioned properties. Since this equality holds for any 2 x 2 matrix A, it leads to the identity that you are asked to prove.Expanding &simplifying each term separately we find. = 8( B , 8 ) 8 ( x, 8 ) - " (ap ) ( x , 8 )we have a" 0 pauli Matrices of 6 2 So 6 x B 6 y can be written as Expanding the above equation gives us uterms 1) OXO = 0 2 ) OXI = 0 3 ) 1 X0 = 0 9 ) 1X1 = 1 5 98 5 X8 - X = 0+0+0+ 1 =(1)if d=s and B= & whileitis 0 otherwise This is what the ( 1 ) seepresent it just a scales multiple Now to expand Ex Boys, we can express each element as a combination of kronecker dellals ) 6 x 5 ( 11 ) 5 ( 11 ) 8 ( 2 , 8 ) 8 3 , 8 ) + 6 1 1 6 1 2 8 (x , 8 ) (BIS ) 5 12 5 2 , 8 ( 6, 8 ) 5 ( Bid ) + 6 7 25 2 2 8 ( of, 8 ) $ ( B, s )2 1 8 / 2 , 8 ) 8 ( B , S ) + 6 7 6 21 0 12 6 , 8 1 813 , 8 ) - 6 21 15x 21 5 ( or , * ) = ( Bid ) + 6 21 2 6 22 , S ( x , 8 ) S( B ,8 ) + 6 22 6 11 1 6 (x, x ) s ( Bid ) + 6 2 2 6 12 2 8 ( 2 1 8 ) ( B , S ) + 6 ,,6 8 (21 8) 8( B1 8)7 6 27 6 7 8 ( N , 8 ) S ( B , S ) Using the Values of the Pauli Matrix element, 11=0 67 12 = 1, 6 21 2 = 01 ; 6,2 2 = 0 . Substituenthese into the expansion .& simply we get OxDS x x 2 8 @, 8 ) ( B, s ) + as ( x , 8 ) 8 ( B, S ) = 2 8 (, 8 ) S(B,8 ) 6 X B 5 8 8 = R S ( x , 8 ) S ( B , 8 ) rely solve 6 4x B 698 y 2 Z & OXB and Add them