Solutions First determine the eigenvalue 1-2 (-1-)(2-)-22=0 2+λ-2λ+12-4=0 (-3)X+2)=0 Next, determine the eigenvectors. First the eigenvector associated with eigenvalue -2 12‖Cr 22LC21 C11+2C21=-2C11 C11=-2C21(Note: The second row offers no new information, e.g. 2C11 2C21=-2C21) C112+C212= 1(from normalization (-2C212+C212=1 C212=0.2 C21=V0. 2, and therefore C1=-2vo
1 Solutions 1. a. First determine the eigenvalues: det ë ê é û ú -1 - l 2 ù 2 2 - l = 0 (-1 - l)(2 - l) - 22 = 0 -2 + l - 2l + l2 - 4 = 0 l2 - l - 6 = 0 (l - 3)(l + 2) = 0 l = 3 or l = -2. Next, determine the eigenvectors. First, the eigenvector associated with eigenvalue -2: ë ê é û ú ù -1 2 2 2 ë ê é û ú C ù 11 C21 = -2 ë ê é û ú C ù 11 C21 -C11 + 2C21 = -2C11 C11 = -2C21 (Note: The second row offers no new information, e.g. 2C11 + 2C21 = -2C21) C112 + C212 = 1 (from normalization) (-2C21) 2 + C212 = 1 4C212 + C212 = 1 5C212 = 1 C212 = 0.2 C21 = 0.2 , and therefore C11 = -2 0.2
For the eigenvector associated with eigenvalue 3 12 12 22儿C C 12 4C12=-2C C12=0.5C22(again the second row offers no new information) C122+C222=1(from normalization) 0.25C22+C 1.25C C22=0.8 C22=V0.8=2v0.2, and therefore C12=v0.2 Therefore the eigenvector matrix becomes 202V02 √022V02 b first determine th -2-入0 0 2-λ From la. the solutions then become -2 -2 and 3. Next. determine the eigenvectors First the eigenvector associated with eigenvalue 3(the third root)
2 For the eigenvector associated with eigenvalue 3: ë ê é û ú ù -1 2 2 2 ë ê é û ú C ù 12 C22 = 3 ë ê é û ú C ù 12 C22 -C12 + 2C22 = 3C12 -4C12 = -2C22 C12 = 0.5C22 (again the second row offers no new information) C122 + C222 = 1 (from normalization) (0.5C22) 2 + C222 = 1 0.25C222 + C222 = 1 1.25C222 = 1 C222 = 0.8 C22 = 0.8 = 2 0.2 , and therefore C12 = 0.2 . Therefore the eigenvector matrix becomes: ë ê é û ú ù -2 0.2 0.2 0.2 2 0.2 b. First determine the eigenvalues: det ë ê ê é û ú ú ù -2 - l 0 0 0 -1 - l 2 0 2 2 - l = 0 det [ ] -2 - l det ë ê é û ú -1 - l 2 ù 2 2 - l = 0 From 1a, the solutions then become -2, -2, and 3. Next, determine the eigenvectors. First the eigenvector associated with eigenvalue 3 (the third root):
200 II C11 0-12|C C 31 C 2 C13=3C13(row one) C13=0 C23+ 2C33=3C23(row two) 33=2C23(again the third row offers no new information) C132+C232+C332=1(from normalization) 5C232=1 C23=v0. 2, and therefore C33=2v0.2 Next, find the pair of eigenvectors associated with the degenerate eigenvalue of-2. First, root one eigenvector one 2C11=-2Cl1(no new information from row one) 1+2C31=-2C21( row two) C21=-2C31(again the third row offers no new information) 11+ C212+ C312=1(from normalization C1 C C11 5C312 Ith three unknowns. Second root two eigenvector two
3 ë ê ê é û ú ú ù -2 0 0 0 -1 2 0 2 2 ë ê ê é û ú ú C ù 11 C21 C31 = 3 ë ê ê é û ú ú C ù 11 C21 C31 -2 C13 = 3C13 (row one) C13 = 0 -C23 + 2C33 = 3C23 (row two) 2C33 = 4C23 C33 = 2C23 (again the third row offers no new information) C132 + C232 + C332 = 1 (from normalization) 0 + C232 + (2C23) 2 = 1 5C232 = 1 C23 = 0.2 , and therefore C33 = 2 0.2 . Next, find the pair of eigenvectors associated with the degenerate eigenvalue of -2. First, root one eigenvector one: -2C11 = -2C11 (no new information from row one) -C21 + 2C31 = -2C21 (row two) C21 = -2C31 (again the third row offers no new information) C112 + C212 + C312 = 1 (from normalization) C112 + (-2C31) 2 + C312 = 1 C112 + 5C312 = 1 C11 = 1 - 5C312 (Note: There are now two equations with three unknowns.) Second, root two eigenvector two:
2C12=-2C12(no new information from row one) C22+ 2C32=-2C22(row two) C22=-2C32(again the third row offers no new information) C122+C222+C322=1(from normalization) C12+(-2C32)2+ C122+5C 12=(1-5C32)(Note: again, two equations in three unknowns C11C12+ C21 C22+ C3 C32=0( from orthogonalization) Now there are five equations with six unknowns Arbitrarily choose C11=0 (whenever there are degenerate eigenvalues, there are not unique eigenvectors because he degenerate eigenvectors span a 2-or more-dimensional space, not two unique directions. One al ways is then forced to choose one of the coefficients and then determine all the rest; different choices lead to different final eigenvectors but to identical spaces spanned by these eigenvectors) C11=0=y1-5C C31=V02 C21=-2V0.2 C1l C12+ C21 C22+C31 C32=0(from orthogonalization) 0+-2V0.2(-2C32)+V0.2C32=0 5C32=0
4 -2C12 = -2C12 (no new information from row one) -C22 + 2C32 = -2C22 (row two) C22 = -2C32 (again the third row offers no new information) C122 + C222 + C322 = 1 (from normalization) C122 + (-2C32) 2 + C322 = 1 C122 + 5C322 = 1 C12 = (1- 5C32 2 ) 1/2 (Note: again, two equations in three unknowns) C11C12 + C21C22 + C31C32 = 0 (from orthogonalization) Now there are five equations with six unknowns. Arbitrarily choose C11 = 0 (whenever there are degenerate eigenvalues, there are not unique eigenvectors because the degenerate eigenvectors span a 2- or more- dimensional space, not two unique directions. One always is then forced to choose one of the coefficients and then determine all the rest; different choices lead to different final eigenvectors but to identical spaces spanned by these eigenvectors). C11 = 0 = 1 - 5C312 5C312 = 1 C31 = 0.2 C21 = -2 0.2 C11C12 + C21C22 + C31C32 = 0 (from orthogonalization) 0 + -2 0.2(-2C32) + 0.2 C32 = 0 5C32 = 0
C32=0,C2=0,andC12=1 Therefore the eigenvector matrix becomes 0 0 -2V020V02 √020202 KE 吗2-(m2-m2 KE.=加m(Px2+p2+p2) 1hak hana KE Oy)T(iaz K.E.=2mlax2+ b. p=mv= ipx+ jpy kp p=o/×/tk/ba h a ioy +kia where i,j, and k are unit vectors along the x, y, and z axes
5 C32 = 0, C22 = 0, and C12 = 1 Therefore the eigenvector matrix becomes: ë ê ê é û ú ú 0 1 0 ù -2 0.2 0 0.2 0.2 0 2 0.2 2. a. K.E. = mv2 2 = è ç æ ø ÷ mö m mv2 2 = (mv)2 2m = p2 2m K.E. = 1 2m(px 2 + py 2 + pz 2) K.E. = 1 2mî í ì þ ý ü è ç æ ø ÷ ö -h i ¶ ¶x 2 + è ç æ ø ÷ ö -h i ¶ ¶y 2 + è ç æ ø ÷ ö -h i ¶ ¶z 2 K.E. = -h-2 2mî í ì þ ý ü ¶ 2 ¶x2 + ¶ 2 ¶y2 + ¶ 2 ¶z 2 b. p = mv = ipx + jpy + kpz p = î í ì þ ý ü i è ç æ ø ÷ ö -h i ¶ ¶x + j è ç æ ø ÷ ö -h i ¶ ¶y + kè ç æ ø ÷ ö -h i ¶ ¶z where i, j, and k are unit vectors along the x, y, and z axes. c. Ly = zpx - xpz Ly = z è ç æ ø ÷ ö -h i ¶ ¶x - x è ç æ ø ÷ ö -h i ¶ ¶z