Arfken 3.4.6 Matrix C is not Hermition. But which is Hermitian. Likewise, Assignment 11 (C + C ) = (C + C ) = (C + C) [ i(c C ) ] = i(c C ) = i(c C) = i ( C C ) Arfken 3.4.9 The matrices A and B are both Hermitian: A = A and B = B. The adjoint of their product is (AB) = (AB) = B A = BA For the product then to be Hermitian, we must have AB = (AB) ger = BA, i.e. A and B must commute. Thus, this shows that if AB = (AB), then [A, B] = 0. To go the other way, and AB = (AB) and is Hermitian. 0 = [A, B] = [A, B] = (AB BA) = (AB) A B = (AB) AB 1
Arfken 3.4.12 (a) Two matrices U and H are related by U = e iah 1 + iah + (ia)2 2 H2 + (ia)3 H 3 + 3! First assume H = H and take the adjoint of the above relation U = 1 iah + ( ia)2 2! ( H 2 ) ( ia) 3 ( + H 3 ) + 3! It should be clear that we need to show (H n ) = H n. Briefly, for n = 1 and n = 2, this is straightforward to show. By induction we can then demonstrate the general case. Taking it as a result, we have U = 1 iah + (ia)2 H 2 (ia)3 H 3 + 2! 3! = e iah By now multiplying on the left by U = e iah, we can see that U must equal U 1 and therefore U is unitary. (b) Now assume U = U 1. We know that U = e iah. Its inverse is U 1 which we might guess is e iah. But we need to show this: U U 1 = e iah e iah and we have verified it. Being unitary, we have = e iah iah = 1 U = e iah = ( e iah) = e iah where we have two (matrix-valued) Taylor series which are equal: e iah = e iah. If the Taylor series are to be equal, we must have, in general, H n = (H n ) = (H ) n. This will be true provided, H = H, i.e. H is Hermitian. Arfken 3.5.4 Assume the matrix A is not symmetric but that it can be diagonalized by an orthogonal similarity transformation. We then have A il = R ij A jk (R T ) kl = R ij A jk R lk = R lk A jk R ij where R is the appropriate orthogonal matrix. Since A is diagonal, it is symmetric: A il = A li. This implies A il = R ik A jk R lj = R ik A jk (R T ) jl Relabeling indices so that k j and j k above, it become clear that this can only equal the first line if A jk = A kj, i.e. A is symmetric. This is a contradiction and we conclude that a non-symmetric matrix cannot be diagonalized. 2
Arfken 3.5.8 Two matrices, A and B, are diagonalized by the same transformation: These two diagonal matrices now commute: which will be true if and only if AB = BA. Arfken 3.5.12 0 = A B B A A = RAR T B = RBR T = RAR T RBR T RBR T RAR T = RABR T RBAR T = R(AB BA)R T For a rigid body defined by m 1 = 1 at (1, 1, 2), m 2 at ( 1, 1, 0), and m 3 at (1, 1, 2), the components of the inertia matrix are 3 I xx = m i (ri 2 x 2 i ) i=1 = m 1 (r 2 1 x 2 1) + m 2 (r 2 2 x 2 2) + m 3 (r 2 3 x 2 3) = 1 (6 1) + 2 (2 1) + 1 (6 1) = 12 with similar calculations leading to I yy = 12, I zz = 8, I xy = 4, and I xz = I yz = 0. Putting it together, 12 4 0 I = 4 12 0 0 0 8 (b) Getting the eigenvalues and eigenvectors requires the secular equation 0 = det(i λ1) = (8 λ) ( (12 λ) 2 16 ) = (λ 8) 2 (λ 16) Solving the eigenvalue equations for λ = 16 gives the equations x = y and z = 0 so we pick a normalized eigenvector of (1, 1, 0)/ 2. The degenerate eigenvalue λ = 8 gives the equations x = y and z can be anything. So one eigenvector associated with λ = 8 is (1, 1, 1)/ 3. Another eigenvector which would go with the λ = 8 eigenvalue is ( 1, 1, 2)/ 6 which, one can readily check, is orthogonal to both the other eigenvectors. Arfken 3.5.20 Diagonalize The secular equation is A = 1 0 0 0 1 1 0 1 1 0 = det(a λ1) = (1 λ) ( (1 λ) 2 1 ) = λ(λ 1)(λ 2) 3
with eigenvalues λ = 0, 1, 2. The eigenvector associated with the first eigenvalue can be found from the equations x = 0 and y + z = 0. It is (0, 1, 1)/ 2. For the second eigenvalue, the eigenvector can be determined from the equations z = 0 and y = 0 with x anything. The second eigenvector is thus (1, 0, 0). For λ = 2, the equations for the eigenvector are x = 0 and y = z. Thus we have (0, 1, 1)/ 2. Arfken 3.5.27 Diagonalize The secular equation is 0 = det(a λ1) A = 5 0 2 0 1 0 2 0 2 = (5 λ)(1 λ)(2 λ) + 2(1 λ) 2 = (1 λ) [ λ 2 7λ + 6 ] = (λ 1) 2 (λ 6) with eigenvalues λ = 1, 1, 6. The eigenvector associated with the last eigenvalue λ = 6 can be found from the equations y = 0 and x = 2z. It is (2, 0, 1)/ 5. For the degenerate eigenvalue, the eigenvectors can be determined from the equations 2x = z and with y anything. Thus one eigenvector is (1, 0, 2)/ 6. To get another, just notice that (0, 1, 0) satisfies the equations and is orthogonal to the other two eigenvectors. Arfken 3.6.3 is The secular equation for the matrix a b A = c d 0 = det(a λ1) = (a λ)(d λ) bc = λ 2 λ(a + d) + ad bc = λ 2 λtra + det A 4
Arfken 3.6.7 In bra-ket notation A r i = λ i r i (1) A r j = λ j r j (2) Taking the adjoint of Eq. (2), we get (A r j ) = (λ j r j ) r j A = λ r j (3) Now multiply by r j on the left of Eq. (1) and by r i on the right of Eq.(3). Finally, subtract the two r j A A r i = (λ i λ j ) r j r i The left hand side is zero for a Hermitian matrix (A = A ). For i j (and no degeneracy) the eigenvectors are orthogonal. For i = j, the eigenvalues must be real: λ i = λ i. Now take Eq. (1) and multiply by A 1 A 1 A r i = λ i A 1 r i which can be re-written A 1 r i = 1 λ i r i Multiply this by r j on the left and subtract from this Eq. (3) r j A 1 A r i = ( 1 λ i λ j ) r j r i For a unitary matrix, the left hand side is zero and for i = j, we must have λ i λ i = 1 Thus if a matrix is both Hermitian and unitary, λ i λ i = 1 and the eigenvalues can only be ±1. Arfken 3.6.14 We have A = 1 2 2 5 1 4 The transpose, à together with Aà and ÃA are à = 1 2 1 5 2 4 8 6 Aà = 1 5 6 17 5 0 ÃA = 1 5 0 20 1 0 = 0 4 5
(b) The eigenvalues of AÃ come out of the secular equation: 0 = det(aã λ1) = ( 8 5 λ)(17 36 λ) 5 25 = (λ 4)(λ 1) Thus, λ 2 n = 1, 4. The eigenvectors, g n, associated with these are (2, 1)/ 5 and (1, 2)/ 5, respectively. (c) The eigenvalues of ÃA are simple since it is a diagonal matrix. They are as before: λ2 n = 1, 4. However, the eigenvectors, f n, are (1, 0) and (0, 1). (d) Note that (e) By construction, we find A f 1 = 1 2 2 1 = 1 2 = λ 5 1 4 0 5 1 1 g 1 A f 2 = 1 2 2 0 = 1 2 = λ 5 1 4 1 5 4 2 g 2 Ã g 1 = 1 2 1 1 2 = 1 5 = λ 5 2 4 5 1 5 0 1 f 1 Ã g 2 = 1 2 1 1 1 = 1 0 = λ 5 2 4 5 2 5 10 2 f 2 A = n λ n g n f n = λ 1 g 1 f 1 + λ 2 g 2 f 2 1 2 1 2 = 1 ( 1 0 ) + 2 ( 0 1 ) 5 1 5 4 = 1 2 2 5 1 4 6