Wong, Introduction to Mathematical Physics, Oxford UP, 19913Wong, Introduction to Mathematical Physics, Oxford UP, 1991Answers and hints to selected problemsCopyright 2003 by Chun Wa WongTypos:Please report typos and errors to: cwong@physics.ucla.edu Thank you.Notations:boldface = vector: a, A, nunit vectors:e, i, j, k, nmatrix = {1st row, 2nd row, ...}:{{1,2}, {3,4}} = 22 matrixChapter 1:1.2.1.(-i + 8j - 5k)/(90).1.2.4.If xy = diagonal on the xy plane, xyz = diagonal to the opposite corner:xy,yz = yz,zx = zx,xy = 60 , xy,xyz = 35.3 .1.2.5.Hint:Distance from a plane = r.n, where r = vector to any point on the plane, n = unit vector normal to the plane.1.2.7.A sphere of radius b centered at a.1.2.9.Hint:Use Example 1.2.7, or decompose into components X = x e(a) + y e(b) + z e(c), where a = b c.1.2.11.A.(BC) = 0.1.3.1.1jk as a 33 matrix = {{0, 0, 0}, {0, 0, 1}, {0, -1, 0}}.1.3.3.Hints:There are 4! = 24 permutations of a, b, c, d, and 4!/2! = 12 permutations of a, a, b, c.1.4.2.(a) e(r)/r, (b) -2xi - 2yj + 4zk, (c) 1/r2, (d) 0.1.4.3.(r =(1, 1, 1)) = - (1/6) [2i + (1 + 3)j + (1 + 3)k].1.4.4.They have common equipotentials and lines of force.1.4.5.Hint:Show that the radius vector to the contact point between sphere and plane is also normal to the plane.1.4.6.30 .1.5.1.Hint:Use permutation symbols. 1.5.2.Hint:Use # 1.5.1 (a).1.5.3.Hint:Use # 1.5.1 (a), Eq. (1.57), and # 1.5.1 (c), respectively.1.5.5.Hint:Use Eq. (1.60).1.6.3.(a) 0, (b) /384.1.7.2.(a) 4/5, (b) 0.1.7.3.(a) 0,(b) 4.1.7.6.Hints:CS B.d = 0, CS D.d = Q (total charge); CS = closed surface.1.7.7.p > . 1.8.1.0.1.8.3.Force in (b) is not conservative if n =/= 0.1.8.4.(a) 0,(b) 2 x, where x = .i. 1.10.1.See Appendix 1.4 on p.70 of Wong.1.10.4.v = (d/dt)e + (d/dt)e + (dz/dt)ez; a not given.1.10.6Hints:Derive and use the following expressions:v = hi (dui/dt) ei,a = { [(dhi/dt) (dui/dt) + hi (d2ui/dt2)] ei + hi (dui/dt) (dei/dt) }.1.11.1.Hint:First express in spherical coordinates.1.11.3.(a) e, 1/; (b) 2r, 6;(c) Not given.1.11.5.(a) 3, (0, 0, 0);(b) 0, - e /;(c) 0, ez(1 + ln )/; (d), (e) Not given.1.11.9.(a) (.V) = ei (/si) [(1/pj) (/sj) (pj Vj)]; (b) .(V) = (1/pi) (/si) [(pi/hk) (/sj) (hk Vk)]; (c), (d) Not given. Answers for Chapter 2:2.2.2.Use the identity: R(-) R() = R(-).2.2.3.A special case of the general result R() R() = R(+).2.2.13. = -30, = 60.2.2.14.Use the definition ij = einew . ejold to get = {{0,1,0},{0,0,1},{1,0,0}}.2.3.2.Use the identity: det(AB) = det(A) det(B).2.3.4.When k i, the resulting expression is a determinant with two identical rows or columns.2.3.6.{{-43,22,-3},{38,-20,6},{-3,6,-3}}/24.2.3.7.(a) Do it by long hand.(b) Prove and use the identity lmn det A = i,j,k ijk Ali Amj Ank .2.3.8.x = (-1, 2, 0)/3.2.4.1.x = const a1 a2; e(x) = (1, -2, 1)/6.2.4.2.Rank = 2 for all three cases. For (c), first show that a3 = -a1 + 2a2 .2.5.1. (c) The eigenvectors are e1 = (1, -1, 0)/2, e2 = (1, 1, -2)/2, e3 = (1, 1, .2)/2.(d) The eigenvectors are e1 = (-1, -1, 1, 1)/4, e2 = (0, 0, -1, 1)/2, e3 = (-1, 1, 0, 0)/2, e4 = (1, 1, 1, 1)/4.(e) The eigenvectors are e1 = (-1, 0, 0, 1)/2, e2 = (-1, 0, 1, 0)/2, e3 = (-1, 1, 0, 0)/2, e4 = (1, 1, 1, 1)/4. 2.6.1.(a) 1 = 3.12, e1 = (8.36, 1)/8.42; 2 = 0.21, e2 = (1, -2.79)/2.96.2.6.2.Hint:Show that det (K-M) is a polynomial of degree n-m in with the help of a unitary transformation that diagonalizes M.2.7.1.Hint:Have AB and then BA multiply an eigenvector ei of A. Discuss separately the cases when the eigenvalues are non-degenerate and when one group of eigenvalues are degenerate.2.7.6.Hint:Use the result det Q 0.2.7.7.There are only n-1 eigenvalues and eigenvectors.2.7.8.Hint:First relate eH to eD.2.9.1.(a) (r) = kQ (s.r)/r3, k = 1/(40).(b) (r) = kQ (3 cos2 - 1)/(4r3).2.9.2.Direction of propagation is e(k) for u1 and u4, and - e(k) for u2 and u3 .2.9.3.The identity group element is the rectangular zero matrix Z whose matrix elements are zeros. The inverse of a matrix A and -A.2.9.4.Hint:Products of nonsingular matricies are nonsingular.2.9.5.Hint:All rotation matrices are nonsingular and made up of direction cosines.2.9.6.Hint:These groups all have the same group table, i.e., all 16 matrix products MiMj of the four group elements Mi.2.10.1.J = (Jx + Jy + Jz)/3.2.10.2.J3 can also be conisdered a 22 (sub)matrix (in the xy-plane). Then J32 = I. Use this to show that exp(i J3) = cos + i J3 sin in this 2-dim subspace before going back to 33 matrices.2.10.5.(a) Try proof by induction.(b) First show that [B, An] = nAn-1 = (/A) An .(c) Show that both the left-hand side and the right-hand side satisfy the differential equation (d/dx) g(x;A,B) = (A+B) g(x;A,B). Since the two sides agree at x = 0, they must agree at all values of x.2.10.7.Hint:First show that for any vector A: (d/dt)fixed A = (d/dt)rot A + A .(b) The first line of the equation should read afixed = (d/dt)rot vfixed + vfixed . 2.10.8.There are 12 terms, 6 terms with positive signs, and six terms with negative signs.These terms cancel pairwise.2.11.1.Ther are n(n-1)/2 conditions corresponding to the off-diagaonal matrix elements of I above the main diagonal, and n conditions from the diagonal matrix elements. The conditions from the off-diagonal matrix elements of I below the main diagonal are the same as those above the main diagonal for the orthogonality relations.2.11.2.Conditions corresponding to off-diagonal matrix elements of I above and below the main diagonal are independent of each other for the unitarity relations.Answers for Chapter 3:3.3.1.(a)a0 = /2, an = [(-1)n - 1]/(n2); bn = (-1)n+1/n.(b)a0 = 1, an = 0; bn = [1 - (-1)n]/(n).(c)an = 0; bn = { [1/(a-n)] sin(a-n) - [1/(a+n)] sin(a+n) }/. (d)an = { [1/(a-n)] sin(a-n) + [1/(a+n)] sin(a+n) }/; bn = 0.3.3.2.(a)an = (-1)n (eL - e-L)/{L[1 + (n/L)2 ]; bn = an(- n/L).(b)a0 = L, an = [(-1)n - 1] 2L/(n)2 ; bn = 0.(c)a0 = L/2, aodd n = 0; aeven n = [(-1)n/2 - 1] 4L/(n)2 ; bn = 0.(d)an = 0; bn = (2/n) [cos(n/3) - (-1)n].(e)an = 0; bodd n = (-1)(n-1)/2 4L/(n)2 , beven n = 0.3.3.3.All bn = 0.(a)a0 = (eL - 1) (2/L), an = [eL(-1)n - 1] 2/{L[1 + (n/L)2 ].(b)a0 = L, a odd n = - 4L/(n)2 , aeven n = 0.(c) & (e)a0 = L/2, a n 0 only when n = 2, 6, 10, ... when a n = - 8L/(n)2 .(d)a0 = 4/3; an = - (2/n) sin(n/3).3.3.4.All an = 0.(a)bn = [eL(-1)n - 1] 2/{L[1 + (n/L)2 ] (-n/L).(b)bn = (2L/n) (-1)n+1 .(c) & (e)beven n = 0, bodd n = (-1)(n-1)/2 4L/(n)2 .(d)bn = (2/n) [cos(n/3) - (-1)n].3.3.5.Use the Fourier-series expansion of the Dirac -function to get directly the resultf(x) = a0 /2 + [ cn cos(nx/L) + dn sin(nx/L) ], wherecn = (1/L), dn = (1/L). 3.3.6.Answers are all of the form f(x) = a0 /2 + [ an cos(2nx/L) + bn sin(2nx/L) ], where a0 = (2/L), an = (2/L), bn = (2/L). (a)a0 = (eL - 1) (2/L), an = (eL - 1) 2/{L[1 + (n/L)2 ]; bn = an (-2n/L).(b)a0 = L, an = 0; b n = - L/(n) ,(c) & (e)a0 = L/2, a n = [(-1)n - 1] L/(n)2 ; bn = 0 .(d)a0 = 4/3; an = - (1/n) sin(2n/3); bn = (1/n) [cos(2n/3) - 1].3.3.7.f(x) = (/4) [(/2) - |x|]. 3.3.8.g1(x) = odd part of f(x) = f odd (x); g2(x) = f odd, even (x);g3(x) = f (x), if x>0; = - f(|x|), if x0): F{1/(x-ia)} = i (2) e ak (-k); F{1/(x+ia)} = - i (2) e -ak (k). 3.6.1.(a) F{G(t)} = (2)-1/2 /( + + i),(b) f(t) = A e -(+)t (t). 3.6.2.Hints: 1/[( - + ) ( - - )] = [1/( - + ) - 1/( - - )] / (+ - - ). 3.7.1.(2)-1/2 .3.8.1.(b)|g(k)|2 = 0 at k = .3.8.2.Hint:Show that + is Hermitian, but - is anti-Hermitian. 3.8.3.Hint:Show that if vi = eigenvector of M, then so is Nvi .3.9.2.(a)c0 = (1/k) sin k, c1 = (3i /k)[(1/k) sin k - cos k], c2 = [15/(2k)]{ [(2/k2) - 1] sin k + 2 cos k } - [5/(2k)] sin k. (b) c0 = (1/) sinh , c1 = (3/)[(1/) sinh - cosh ], c2 = [15/(2)] { [(2/2) + 1] sinh - (2/) cosh } - [5/(2)] sinh . (c) c0 = (1/2) (1 + )/[1 - (1/n)], c1 = (3/2) (1 - )/[2 - (1/n)], c2 = (15/4) (1 + )/[3 - (1/n)] - (5/4) (1 + )/[1 - (1/n)],where = (-1)-1/n = e -i/n .3.9.3.F(x) = n Fn en(x), where en(x) = [(2n + 1)/2]1/2 Pn (x), Fn = ( en, F) = en(x) F(x) dx. (x-x) = n en(x)en(x) = [(2n + 1)/2] n Pn(x)Pn(x) .3.11.1.(a) Integrate entry 4 of Appendix 3A to get 4 /90 ( = 1.0823...).(b)First integrate entry 2 of Appendix 3B to get n (1/n6 ) = 6/945. By separating even and odd n terms, show that the final result is (31/32) 6 /945 = 0.98555...3.11.2.(/2) n (1/n2 ) = 3/12.3.12.1.Hint:Start with the Fourier series for x2 , Example 3.3.3.3.12.2.Hint:Start with the Fourier series for x , Example 3.3.1.3.12.3.Hint:The function x that appears at an intermediate step is eliminated by using the Fourier series given in Example 3.3.1.3.12.4.(a)(-x)/2 in (0,2),(b)(3x 2 - 6x + 2 2 )/12 in [0,2].Answers for Chapter 6:The answers are given in many of the problems in this chapter.6.2.4.(a)Only at x = (n + 1/2) on the x-axis.(b)Only at y = (n + 1/2) on the y-axis.6.2.10.Prove and use the sum formula: (n=0,1,...,N) eiNx = [ei(N+a)x - 1]/(eix-1).6.2.12.(a)An inversion transformation at z = 0.(b)An inversion transformation at z = -d/c plus a scaled change.(c)Translation plus a transformation (b).6.3.1.(a)Function is in the second of two branches: f2(2+) = f2(-2-) = -31/2, f2(2-) = f2(-2+) = 31/2.(b)Function is in the third of three branches: f3(2+) = f3(-2-) = k231/3, f3(2-) = f3(-2+) = k231/3 ei2/3; k = ei2/3.(c)Function is in the third of four branches: F4(2+) = f4(-2-) = -31/4, f4(2-) = f4(-2+) = -i31/4.6.3.2.(a)6 sheets(b)6 sheets(c)9 sheets(d)3 sheets(e)Same as the function ln(z+i) - ln(z-i): sheets(f) sheets on that sheet of z containing the branch point z = -1 of the logarithmic function.(g)6 sheets(h)6 sheets(i) The function (z + 1)1/3 has a Riemann surface of 6 sheets. The branch pointz = -1 and the branch line of the logarithmic function ln(z + 1)1/3 can be made to appear on only one of the two square-root branches. There are 3 log branches. 6.3.3.On the first branch for most choices of branch lines.6.4.3.(b)Simple poles at zn = (2/2n+1) and an essential singularity (where?).(c)A simple pole on one of the sheets of (z + i).(d)A simple pole and many double poles6.6.1.These are two separate problems:(a)f(z) = z2 + iC, where C = const.(b)f(z) = eiz + C, where C = const.6.8.1.(e)A Taylor series for |z-2| < 1, and a Laurent series for |z-2| > 1 with zero regular part.6.9.1.(h)Res[ f(0)] = 0; Res [f(n) ] = 1/(n)3 .6.10.1.(b)Use Eq. (6.56) to evaluate the residue.(c)RHS should read /18. Use Eq. (6.56) to evaluate the residues.6.13.3.(b)- ln t - , where , the Euler-Mascheroni constant, is conveniently defined by the integral .Note on Bromwich integrals: The trouble with many Bromwich integrals is that after simplification, one still has a difficult integral to evaluate. I think it is acceptable if the last integral is found by looking up a table of integrals, or equivalently by using a computer software such as Mathematica.