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Unformatted text preview: Tutorial MT: Matrix Theory 2 4 3 = I 1 + + ii + 2! 4! 3! = I cos + ii sin : Exercise 16. We have ^ = (sin cos ') 1 + (sin sin ') 2 + (cos ) 3 ; so that r h i (^ )2 = (sin cos ')2 + (sin sin ')2 + cos2 I r + sin2 cos ' sin ' [12 + 21] + (sin cos cos ') [13 + 31] + (sin cos sin ') [23 + 32 ] : Since the i anticommute, the last three terms vanish and (^ )2 = sin2 cos2 ' + sin2 ' + cos2 I = I: r Exercise 17. We repeat the steps in Exercise 15 with i replaced by ^ . Since (^ )2 = I, r r the same development obtains. Exercise 18. The term ``secular'' comes from early work in the study of planetary motions, and means ``lasting a very long (but not eternal) time.'' That this apt description of planetary orbits would become a part of the phraseology relating to linear equations is not difficult to imagine. Exercise 19. ! 1 = !2 1 = 0 =) ! = 1; j1 !Ij = 1 ! ! i = !2 1 = 0 =) ! = 1; j2 !Ij = i ! 1 ! 0 = 1 + !2 = 0 =) ! = 1: j3 !Ij = 0 1 ! Thus, all three Pauli matrices have the same eigenvalues. Exercise 20. Let V(1) and V(2) both be eigenvectors of  with different eigenvalues !1 and ! 2, respectively. Now suppose V(1) and V(2) are not linearly independent, i.e., that we can write V(1) = V(2) for some . Then V(1) = !1 V(1) = ! 1V(2) = V(2) = ! 2V(2) ; or ! 1 = !2 ; in contradiction to our assumption. The supposition must therefore be false. Exercise 21. a. 1: 17 Tutorial MT: Matrix Theory ! = +1
(+) 1 V1 = (+) (+1) V1 =) v1 v1 =) v2 = v1 =) ! = 1
() 1 V1 (+) V1 = : 0 1 1 0 v1 v2 = v1 v2 = () (1) V1 0 1 v1 v1 =) = 1 0 v2 v2 =) v2 = v1 v1 =) V() = : 1 v1 b. 2 : ! = +1 2V(+) 2 = (+1) V(+) 2 0 i v1 v1 = ) = i 0 v2 v2 = ) iv2 = v1 or v2 = iv1 v1 (+) = ) V2 = : iv1 0 i v1 v1 =) = i 0 v2 v2 =) iv2 = v1 iv2 () =) V2 = : v2 ! = 1
() 2 V2 = () (1) V2 c. 3 : ! = +1
(+) 3 V3 1 0 = =) 0 1 = ) v1 = v1 and v2 = v2 v1 (+) = ) V3 = : 0
(+) (+1) V3 v1 v2 = v1 v2 18 Tutorial MT: Matrix Theory ! = 1
() 3 V3 () (1) V3 1 0 = =) 0 1 =) v1 = v1 and v2 = v2 0 () =) V3 = : v2 v1 v2 = v1 v2 Exercise 22. If  is diagonal, its elements can be written j k = j jk (no summation) and assume there are no repetitions among the j : Eqn. 42 can be written
n X k=1 (i) (j jk ! i jk ) Vk = 0 = This has as a nontrivial solution, for given i; n X k=1 (i) (j ! i) j kVk = (j !i ) Vj(i) : Vj = 0 for j 6= i; ! i = i for j = i: ^ (i) ^ Normalizing gives Vj = ij as the jth element of V(i) . (i) Exercise 23. If V(i) = !i V(i) , then  V = V(i) = !i V(i) = ! i V(i) . Exercise 24. The normalization of these twodimensional vectors is straightforward and yields 1 1 1 1 (+) () ^ ^ V1 = p ; V1 = p ; 2 1 2 1 1 1 1 1 ^ (+) = p ^ () = p V2 ; V2 ; 2 i 2 i 1 0 ^ (+) = ^ () V3 ; V3 = : 0 1 Exercise 25. We first find the eigenvalues from the secular equation: 1 ! 1 1 1 1 ! 1 jM !Ij = 1 1 1 ! (i) = (1 !)3 (1 !) (1 !) 1 1 (1 !) = (1 !)3 3 (1 !) 2 h i = (1 !)2 + 2 (1 !) + 1 [(1 !) 2] 19 ...
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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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