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Unformatted text preview: Tutorial MT: Matrix Theory = I + R + R2 + R3 + + Rn + S = S + RS + R2S + : The value of this solution will depend upon its convergence, in a sense that relates to the matrices involved. Exercise 8. One can see by inspection that the Pauli matrices are hermitian and that their traces vanish. They all have determinant 1: That each is unitary can be seen by forming the products y i: i 0 1 0 1 1 0 y 1 = 1 y = 2 = = ; 1 1 1 1 0 1 0 0 1 0 i 0 i 1 0 y y 2 2 2 = 2 2 = 2 = = ; i 0 i 0 0 1 1 0 1 0 1 0 y 3 = 3 y = 2 = = : 3 3 3 0 1 0 1 0 1 Furthermore, 1 2 2 1 0 i = = i 0 0 i 0 1 = = i 0 1 0 0 1 1 0 1 0 =i = i3; 0 1 i 0 1 0 = i = i3 ; 0 i 0 1 i 0 0 i so that 1 2 21 = [1 ; 2 ] = 2i3. Similarly, 2 3 = 3 2 = i1, 3 1 = 1 3 = i2, [2 ; 3 ] = 2i1 and [3; 1] = 2i2. Finally, we can consider the anticommutators: fi; j g i j + j i. From Eqns. 28 and 29, we have fi; j g = 0 for i 6= j and fi; i g = 2: Exercise 9. You've already done the essential computations in the previous Exercise. Exercise 10. For i 6= j, from Eqn. 28, Trij = iijk Trk = 0. For i = j; from Eqn. 29, Tr2 =TrI. i Exercise 11. Writing m11 m12 M= = m11 + m22 + m33 + m0I m21 m22 0 1 0 i 1 0 1 0 = m1 + m2 + m3 + m0 1 0 i 0 0 1 0 1 m0 + m3 m1 im2 = =) m1 + im2 m0 m3 m0 + m3 = m11 ; m0 m3 = m22 =) m0 = (m11 + m22) =2; m3 = (m11 m22) =2; m1 + im2 = m21 ; m1 im2 = m12 =) m1 = (m12 + m21) =2; m2 = i (m12 m21) =2: 15 Tutorial MT: Matrix Theory Exercise 12. Using Eqn. 33 (and the summation convention,) Tr (i M) = = = Tr (M) = Exercise 13. Writing D= d11 d12 d21 d11 = d11 + d22 + d33 d3 d1 id2 = =) d1 + id2 d3 Tr (imj j) + Tr (im0I) mj 2ij + m0 Tri 2mi + 0 = 2mi : m11 + m22 = 2m0: d3 = d11; d1 = (d12 + d21 ) =2; d2 = i (d12 d21) =2: Exercise 14. If D is unitary, then D1 = Dy = d y + d y + dy 1 1 2 2 3 3 = d 1 + d 2 + d3 : 1 2 3 Now, D1D = I = (d 1 + d2 + d3) (d11 + d2 2 + d3 3 ) 1 2 3 2 2 2 = jd1j + jd2j + jd3j I + (d d2 dd1) 3 1 2 +(dd3 dd2) 1 + (dd1 dd3) 2 2 3 3 1 where we have used Eqns. 28 and 29. Thus, the di's must be relatively real (i.e., have the same complex phase) and jd1j2 + jd2j2 + jd3j2 = 1: Exercise 15. By definition (see Eqn. 23), exp (ii) =
1 X 1 (ii)n n! n=0 = I + ii + Since 2 = I, this can be written as i exp (ii ) = I + ii (ii )2 (ii)3 (ii)4 + + + : 2! 3! 4! 2 3 4 I i i + I + 2! 3! 4! 16 ...
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This note was uploaded on 07/13/2008 for the course PHY 201 taught by Professor Covatto during the Spring '08 term at ASU.
 Spring '08
 COVATTO

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