Solutions to Quantum Mechanics Homework Set 7: Calculating Eigenvalues and Eigenvectors, Assignments of Quantum Mechanics

Download Solutions to Quantum Mechanics Homework Set 7: Calculating Eigenvalues and Eigenvectors and more Assignments Quantum Mechanics in PDF only on Docsity! PHY 389K Quantum Mechanics, Homework Set 7 Solutions Matthias Ihl 04/01/2008 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/teaching.html 1 Problem 1 1. By Theorem (3.10.27) on p.237 of Sakurai, T (2) 2 = 〈11; 11|11; 22〉X+1X+1 = 12(x + iy)2 = 12(x2 − y2 + 2ixy) T (2) 1 = 〈11; 10|11; 21〉X+1X0 + 〈11; 01|11; 21〉X0X1 = −1 2 (x + iy)z − 1 2 (x + iy)z = −xz − iyz T (2) 0 = 〈11; 00|11; 20〉X0X0 + 〈11; 1 − 1|11; 20〉X+1X−1 + 〈11;−11|11; 20〉X−1X+1 = 2√ 6 z2 + 1√ 6 (−1 2 (x2 + y2) + 1√ 6 (−1 2 (x2 + y2) = 1√ 6 (3z2 − r2) T (2) −1 = 〈11; 0 − 1|11; 2− 1〉X0X−1 + 〈11;−10|11; 2− 1〉X−1X0 = 1 2 (x − iy)z + 1 2 (x − iy)z = xz − iyz T (2) −2 = 〈11;−1 − 1|112 − 2〉X−1X−1 = 12(x − iy)2 = 12(x2 − y2 − 2ixy) Therefore, xy = T (2) 2 −T (2) −2 2i = − i 2 (T (2) 2 − T (2) −2 ) xz = T (2) −1 −T (2) 1 2 = −1 2 (T (2) 1 − T (2) −1 ) x2 − y2 = T (2)2 + T (2) −2 1 2. Q ≡ e〈α, j, j| √ 6T (2) 0 |α, j, j〉 = √ 6e〈α, j, j|T (2)0 |α, j, j〉 = √ 6e〈j, 2; j, j|j, 2; j, 0〉a,j||T 2||α,j〉√ 2j+1 Therefore, e〈α, j, m′|x2 − y2|α, j, m = j〉 = e〈α, j, m|T (2)2 + T (2) −2 |α, j, m〉 = e〈α, j, m′|T (2)−2 |α, j, m〉 = e〈j, 2; j, m′|j, 2; j,−2〉 〈α,j||T 2||α,j〉√ 2j+1 = Q√ 6 〈j,2;j,m′|j,2;j,−2〉 〈j,2;j,j|j,2;j,0〉 2 Problem 2 For convenience, let α = eQ 2s(s−1)~2 .And assume H = α(φxxS 2 x + φyyS 2 y + φzzS 2 z ) = A(3S 2 z − S2) + B(S2+ + S2−) (1) holds. Then, H = α(φxxS 2 x + φyyS 2 y + φzzS 2 z ) = A(3S 2 z − S2) + B(S2+ + S2−) = A(2S2z − S2x − S2y + B(2S2x − 2S2y) = (−A + 2B)S2x + (−A − 2B)S2y + 2AS2z Comparing the coefficients of S2x and S 2 y , A = −α(φxx + φyy) 2 and B = α(φxx − φyy) 4 Because φxx + φyy + φzz = 0, αφzz = 2A = −α(φxx + φyy)(the coefficients of S2z ). So, indeed, the equality (1) holds with A = − eQ 4s(s − 1)(φxx + φyy) and B = eQ 8s(s − 1)(φxx − φyy) • Eigenvectors and eigenvalues H|3 2 〉 = 3A|3 2 〉 + 2 √ 3B| − 1 2 〉 H|1 2 〉 = −3A|1 2 〉 + 2 √ 3B| − 3 2 〉 H| − 1 2 〉 = −3A| − 1 2 〉 + 2 √ 3B|3 2 〉 H| − 3 2 〉 = 3A| − 3 2 〉 + 2 √ 3B|1 2 〉