MT_26-31

MT_26-31 - Tutorial MT: Matrix Theory = [(1 !) + 1]2 [(1 !)...

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Unformatted text preview: Tutorial MT: Matrix Theory = [(1 !) + 1]2 [(1 !) 2] = 0 =) !1 = 1; ! 2 = !3 = 2: The eigenvector for ! = 1 is obtained from 0 10 1 2 1 1 v1 @ 1 2 1 A @ v2 A = 0 =) v1 = v2 = v3: 1 1 2 v3 Thus, 1 1 1 ^ V(1) = p @ 1 A : 3 1 0 10 1 1 1 1 v1 @ 1 1 1 A @ v2 A = 0 =) v3 = v1 v2 1 1 1 v3 0 For ! = 2, we have with v1 and v2 undetermined. This means that any values for v1 and v2 can be taken. W can, for e example, set v2 = 0: Then, v3 = v1, and the normalized eigenvector is 0 1 1 1 @ ^ V(2) = p 0 A: 2 1 Or, we can choose v1 = 0 =) 1 0 1 ^ V(3) = p @ 1 A : 2 1 Exercise 26. Following the suggested steps, we write Hui = i ui ; Huj = j uj : followed by uyHui = i uy ui ; j j uy Huj = j uy uj : i i Taking the complex conjugate of the second of these equations, we have y y y ui Huj = j ui uj = uj ui j = uy Hyui = uy Hui; j j 20 0 Tutorial MT: Matrix Theory or the pair, uyHui = i uy ui; j j uyHui = uy ui : j j j Subtracting the second of these from the first, we get, as desired, y 0 = i uj ui : j (i ) uy ui = 0: i i Since uy ui > 0 for any vector other than the null vector, we must have i = 0, i.e., i is real. i i Exercise 28. Since we have already shown that the i are real, we can write Eqn. 46 as (i j ) uy ui = 0: j Now, for i 6= j , we must have (uj ; ui ) = 0. Exercise 29 H (a1ui1 + a2ui2 + + amuim) = a1 Hui1 + a2Hui2 + + amHuim = a1 i ui1 + a2 i ui2 + + am i uim = i (a1ui1 + a2ui2 + + am uim ) : Assuming the uir are normalized, we can follow the Schmidt process by taking linear combinations of them to form a set of m orthogonal vectors. For example, take v1 = ui1. Then, let v2 = v1 + ui2 where and are chosen such that (v2; v1 ) = 0 and v2 is normalized. Continue in this fashion until all m linearly independent uir are combined into linear combinations to yield a set of m orthonormal vi . Exercise 30. The two degenerate eigenvectors (which share the eigenvalue ! = 2 ) are (as chosen by us) 0 1 0 1 1 0 1 1 ^ ^ V(2) = p @ 0 A and V(3) = p @ 1 A : 2 2 1 1 They are clearly not orthogonal: ^ ^ V(2) ; V(3) = 1 : 2 Exercise 27. If we set i = j in Eqn. 46, we have The matrix is hermitian, however, so we should be able to choose two degenerate eigenvectors 21 Tutorial MT: Matrix Theory ^ which are orthogonal. W take v1 = V(2) . Then, taking e ^ v2 = v1 + V(3) : ^ Requiring (v1 ; v2 ) = 0 yields + v1; V(3) = 0 or = =2: Thus, 0 1 1 0 1 @ 2 A @ 1 A v2 = p 1 = p 2 : 2 1 2 2 1 2 0 p Normalizing gives = 2= 3 and 1 1 1 v2 = p @ 2 A 6 1 where we have also adjusted the phase to make the first non-zero element positive. It is straight^ forward to check that v2 is an eigenvector of M with eigenvalue 2, and that it is orthogonal to V(1) . Exercise 31. W write U = (u1 u2 : : : un ). Then, e B B Uy = B @ and UyU = n o uyuj i 0 uy 1 uy 2 . . . uy n 1 C C C; A = fij g = I: From Problem 3 of T utorial V AM: Vector Algebra and an Introduction to Matrices, we then have, UUy = I: Now, B B Uy HU = B @ 0 u1 uy 2 . . . uy n y 1 C C C H (u1 u2 : : : un) A 22 Tutorial MT: Matrix Theory 0 0 0 uy 1 uy 2 . . . uy 1 uy 2 . . . uy n 1 1 B B = B @ B B = B @ uy n C C C (Hu1 Hu2 : : : Hun ) A C C C (1u1 2u2 : : : 2un) A 1uy u1 2uy u2 1 1 B 1uy u1 2uy u2 B 2 2 = @ 1 uy u1 2 uy u2 n n 0 1 0 0 B 0 2 0 = B @ 0 0 n 1 n uy un 1 n uy un C C 2 A nuy un n 1 C C: A 23 ...
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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.

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