《理论化学导论》（英文版）Solutions

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

First derive the general formulas for ox, ay, oz in terms of r, e, and and Or, 80 and ao in terms of x,y, and z. The general relationships are as follows x=r Sine Coso r2=x2+y2+z y=r Sine Sino sine= z=r Cose x2+y2+z2 First ax, ay, and az from the chain rule a ar a a0 a az =az 0 第+a Evaluation of the many "coefficients"gives the following Cose cose(φ Sine cosφ Cose Sind ap Sine sinφ Cose and 6

6 3. First derive the general formulas for ¶ ¶x , ¶ ¶y , ¶ ¶z in terms of r,q, and f, and ¶ ¶r , ¶ ¶q , and ¶ ¶f in terms of x,y, and z. The general relationships are as follows: x = r Sinq Cosf r 2 = x2 + y2 + z2 y = r Sinq Sinf sinq = x2 + y2 x2 + y2 + z2 z = r Cosq cosq = z x2 + y2 + z2 tanf = y x First ¶ ¶x , ¶ ¶y , and ¶ ¶z from the chain rule: ¶ ¶x = è ç æ ø ÷ ¶rö ¶x y,z ¶ ¶r + è ç æ ø ÷ ¶qö ¶x y,z ¶ ¶q + è ç æ ø ÷ ¶fö ¶x y,z ¶ ¶f , ¶ ¶y = è ç æ ø ÷ ¶rö ¶y x,z ¶ ¶r + è ç æ ø ÷ ¶qö ¶y x,z ¶ ¶q + è ç æ ø ÷ ¶fö ¶y x,z ¶ ¶f , ¶ ¶z = è ç æ ø ÷ ¶rö ¶z x,y ¶ ¶r + è ç æ ø ÷ ¶qö ¶z x,y ¶ ¶q + è ç æ ø ÷ ¶fö ¶z x,y ¶ ¶f . Evaluation of the many "coefficients" gives the following: è ç æ ø ÷ ¶rö ¶x y,z = Sinq Cosf , è ç æ ø ÷ ¶qö ¶x y,z = Cosq Cosf r , è ç æ ø ÷ ¶fö ¶x y,z = - Sinf r Sinq , è ç æ ø ÷ ¶rö ¶y x,z = Sinq Sinf , è ç æ ø ÷ ¶qö ¶y x,z = Cosq Sinf r , è ç æ ø ÷ ¶fö ¶y x,z = Cosf r Sinq , è ç æ ø ÷ ¶rö ¶z x,y = Cosq , è ç æ ø ÷ ¶qö ¶z x,y = - Sinq r , and è ç æ ø ÷ ¶fö ¶z x,y = 0

Upon substitution of these"coefficients Cos6CosφaSinφ = Sine Coso+ r a0r Sine師 Cose Sind a Cosφa ay=Sine Sin ar + r 80+r Sine ao, and az cose Sine a a +0 Next or, 80, and ao from the chain rule ar (, )最剧 gain evaluation of the the many"coefficients"results in 0φVx2+y2+ 0,Vx2+y2+z2 r√x2 φ Upon substitution of these"coefficients r ay

7 Upon substitution of these "coefficients": ¶ ¶x = Sinq Cosf ¶ ¶r + Cosq Cosf r ¶ ¶q - Sinf r Sinq ¶ ¶f , ¶ ¶y = Sinq Sinf ¶ ¶r + Cosq Sinf r ¶ ¶q + Cosf r Sinq ¶ ¶f , and ¶ ¶z = Cosq ¶ ¶r - Sinq r ¶ ¶q + 0 ¶ ¶f . Next ¶ ¶r , ¶ ¶q , and ¶ ¶f from the chain rule: ¶ ¶r = è ç æ ø ÷ ¶xö ¶r q,f ¶ ¶x + è ç æ ø ÷ ¶yö ¶r q,f ¶ ¶y + è ç æ ø ÷ ¶zö ¶r q,f ¶ ¶z , ¶ ¶q = è ç æ ø ÷ ¶xö ¶q r,f ¶ ¶x + è ç æ ø ÷ ¶yö ¶q r,f ¶ ¶y + è ç æ ø ÷ ¶zö ¶q r,f ¶ ¶z , and ¶ ¶f = è ç æ ø ÷ ¶xö ¶f r,q ¶ ¶x + è ç æ ø ÷ ¶yö ¶f r,q ¶ ¶y + è ç æ ø ÷ ¶zö ¶f r,q ¶ ¶z . Again evaluation of the the many "coefficients" results in: è ç æ ø ÷ ¶xö ¶r q,f = x x2 + y2 + z2 , è ç æ ø ÷ ¶yö ¶r q,f = y x2 + y2 + z2 , è ç æ ø ÷ ¶zö ¶r q,f = z x2 + y2 + z2 , è ç æ ø ÷ ¶xö ¶q r,f = x z x2 + y2 , è ç æ ø ÷ ¶yö ¶q r,f = y z x2 + y2 , è ç æ ø ÷ ¶zö ¶q r,f = - x2 + y2 , è ç æ ø ÷ ¶xö ¶f r,q = -y , è ç æ ø ÷ ¶yö ¶f r,q = x , and è ç æ ø ÷ ¶zö ¶f r,q = 0 Upon substitution of these "coefficients": ¶ ¶r = x x2 + y2 + z2 ¶ ¶x + y x2 + y2 + z2 ¶ ¶y + z x2 + y2 + z2 ¶ ¶z

xz a yz a Ox dy 西=y+xy+0花z Note, these many"coefficients"are the elements which make up the Jacobian matrix used whenever one wishes to transform a function from one coordinate representation to another. One very familiar result should be in transforming the volume element dxdydz to r2Sinedrdedd. For example f(x,y, z)dxdydz f(x(:.4)y(r,)z(r(.)p、( drddφ r tr a a ily az-zay a Sine i rSineSind(Cose a CoseSino a Cosφo rOse(SineSinp or+-p Smn6+CoCo$西 c -it 8

8 ¶ ¶q = x z x2 + y2 ¶ ¶x + y z x2 + y2 ¶ ¶y - x2 + y2 ¶ ¶z ¶ ¶f = -y ¶ ¶x + x ¶ ¶y + 0 ¶ ¶z . Note, these many "coefficients" are the elements which make up the Jacobian matrix used whenever one wishes to transform a function from one coordinate representation to another. One very familiar result should be in transforming the volume element dxdydz to r2Sinqdrdqdf. For example: õóf(x,y,z)dxdydz = õ ô ô ó f(x(r,q,f),y(r,q,f),z(r,q,f)) ï ï ï ï ï ï ï è ç ï æ ø ÷ ¶xö ¶r qf è ç æ ø ÷ ¶xö ¶q rf è ç æ ø ÷ ¶xö ¶f rq è ç æ ø ÷ ¶yö ¶r qf è ç æ ø ÷ ¶yö ¶q rf è ç æ ø ÷ ¶yö ¶f rq è ç æ ø ÷ ¶zö ¶r qf è ç æ ø ÷ ¶zö ¶q rf è ç æ ø ÷ ¶zö ¶f rq drdqdf a. Lx = -h i îï í ïì þï ý ïü y ¶ ¶z - z ¶ ¶y Lx = -h i è ç æ ø ÷ ö rSinqSinf è ç æ ø ÷ ö Cosq ¶ ¶r - Sinq r ¶ ¶q - -h i è ç æ ø ÷ ö rCosq è ç æ ø ÷ ö SinqSinf ¶ ¶r + CosqSinf r ¶ ¶q + Cosf rSinq ¶ ¶f Lx = - -h i è ç æ ø ÷ ö Sinf ¶ ¶q + CotqCosf ¶ ¶f b. Lz = -h i ¶ ¶f = - ih- ¶ ¶f Lz = -h i è ç æ ø ÷ ö -y ¶ ¶x + x ¶ ¶y

4 d-B 4x4-12x2+3 16x3-24x 48x2-24 lI 20 60X iii. e3x+e-3x 3(e3x-e-3x)9( -4x+2 B(v)is an eigenfunction of A(i) (1-x2) dx b(v 24x-24x3-12x3+3x 36x3+27x 9(4x3-3x)(eigenvalue is-9) B(iii )is an eigenfunction of A(ii x)(eigenvalue is 9) B(ii )is an eigenfunction of A(iii) d

9 4. B dB/dx d 2B/dx2 i. 4x4 - 12x2 + 3 16x3 - 24x 48x2 - 24 ii. 5x4 20x3 60x2 iii. e 3x + e-3x 3(e3x - e-3x) 9(e3x + e-3x) iv. x2 - 4x + 2 2x - 4 2 v. 4x3 - 3x 12x2 - 3 24x B(v.) is an eigenfunction of A(i.): (1-x2) d2 dx2 - x d dx B(v.) = (1-x2) (24x) - x (12x2 - 3) 24x - 24x3 - 12x3 + 3x -36x3 + 27x -9(4x3 -3x) (eigenvalue is -9) B(iii.) is an eigenfunction of A(ii.): d2 dx2 B(iii.) = 9(e3x + e-3x) (eigenvalue is 9) B(ii.) is an eigenfunction of A(iii.): x d dx B(ii.) = x (20x3)

20 4(5x4)(eigenvalue is 4) B() is an eigenfunction of A(vi dx2- 2x dx b(= (48x2-24)-2x(16x3-24x) 48x2-24-32x4+48x2 -32x4+96x2-24 8(4x4-12x2+3)(eigenvalue is-8) B(iv )is an eigenfunction of A(v. x dx2 +(I-x)dx B(iv) x(2)+(1-x)(2x-4) 2x+2x-4-2x2+4x 2x2+8x-4 -2(x2-4x +2)(eigenvalue is-2)

10 20x4 4(5x4) (eigenvalue is 4) B(i.) is an eigenfunction of A(vi.): d2 dx2 - 2x d dx B(i) = (48x2 - 24) - 2x (16x3 - 24x) 48x2 - 24 - 32x4 + 48x2 -32x4 + 96x2 - 24 -8(4x4 - 12x2 + 3) (eigenvalue is -8) B(iv.) is an eigenfunction of A(v.): x d2 dx2 + (1-x) d dx B(iv.) = x (2) + (1-x) (2x - 4) 2x + 2x - 4 - 2x2 + 4x -2x2 + 8x - 4 -2(x2 - 4x +2) (eigenvalue is -2) 5