Download Problems 5 Solutions - Quantum Mechanics | PHYS 324 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 324 Solutions to Selected Problems #5 Autumn 2003 1. Prob. 3.4 in your text: To keep things easy to type, I will use vector notation familiar from R3: |a〉 = (ax, ay, az); 〈a| = (a∗x, a∗y, a∗z) and 〈a|b〉 = a∗xbx + a∗yby + a∗zbz. Then: |e1〉 = (1 + i, 1, i); |e2〉 = (i, 3, 1); |e3〉 = (0, 28, 0) 〈e1| = (1 − i, 1,−i); 〈e2| = (−i, 3, 1); 〈e3| = (0, 28, 0) 〈e1|e1〉 = (1 − i)(1 + 1) + 1 + (−i)i = 4; 〈e1|e2〉 = (1 − i)i + (1)3 + (−i)1 = 4; 〈e1|e3〉 = 28 We follow the steps outlined in the problem: |e′1〉 = |e1〉 ||e1|| = |e1〉√ 〈e1|e1〉 = 1 2 |e1〉 = 1 2 (1 + i, 1, i) |f2〉 = |e2〉 − 〈e′1|e2〉|e′1〉 = |e2〉 − 〈e1|e2〉 2 |e1〉 2 = |e2〉 − |e1〉 = (−1, 2, 1− i) |e′2〉 = |f2〉 ||f2|| = |f2〉√ 1 + 4 + 2 = 1√ 7 (−1, 2, 1− i) |f3〉 = |e3〉 − 〈e′1|e3〉|e′1〉 − 〈e′2|e3〉|e′2〉 = |e3〉 − 〈e1|e3〉 2 |e′1〉 − 1√ 7 (−1, 2, 1 + i) · (0, 28, 0)|e′2〉 = (0, 28, 0)− 7(1 + i, 1, i)− 8(−1, 2, 1− i) = (0− 7− 7i + 8, 28− 7− 16, 0− 7i− 8 + 8i) = (1− 7i, 5, i− 8) |e′3〉 = |f3〉 ||f3|| = 1√ 50 + 25 + 9 (1 − 7i, 5, i− 8) = 1 2 √ 21 (1 − 7i, 5, i− 8) 2. Prob. 3.9 in your textbook: A = −1 1 i 2 0 3 2i −2i 2 ; B = 2 0 −i 0 1 0 i 3 2 a.) A + B = −1 + 2 1 + 0 i + −i 2 + 0 0 + 1 3 + 0 21 + i −2i + 3 2 + 2 = 1 1 0 2 1 3 3i 3 − 2i 4 b.) AB = −2 + 0 − 1 0 + 1 + 3i i + 0 + 2i 4 + 0 + 3i 0 + 0 + 9 −2i + 0 + 6 4i + 0 + 2i 0 − 2i + 6 2 + 0 + 4 = −3 1 + 3i 3i 4 + 3i 9 6 − 2i 6i 6 − 2i 6 2 c.) [A,B] = AB−BA; BA = −2 + 0 + 2 2 + 0 − 2 2i + 0 − 2i 0 + 2 + 0 0 + 0 + 0 0 + 3 + 0 −i + 6 + 4i i + 0 − 4i −1 + 9 + 4 = 0 0 0 2 0 3 6 + 3i −3i 12 [A,B] = −3 1 + 3i 3i 4 + 3i 9 6 − 2i 6i 6 − 2i 6 − 0 0 0 2 0 3 6 + 3i −3i 12 = −3 1 + 3i 3i 2 + 3i 9 3 − 2i 3i − 6 6 + i −6 d.) Ã = −1 2 2i 1 0 −2i i 3 2 e.) A∗ = −1 1 −i 2 0 3 −2i 2i 2 f.) A† = Ã∗ = −1 2 −2i 1 0 2i −i 3 2 g.) Tr(B) = 2 + 1 + 2 = 5 h.) det(B) = 2(2 − 0) − 0(0 − 0) + (−i)(0 − i) = 4 − 1 = 3 i.) B−1 = C̃ det(B) ; quad C = 2 − 0 0 0 − i 0 − 3i 4 − 1 6 − 0 0 + i 0 2 − 0 = 2 0 −i −3i 3 −6 i 0 2 where C is the matrix of cofactors of B. B−1 = 1 3 C̃ = 1 3 2 −3i i 0 3 0 −i −6 2 BB−1 = 1 3 2 0 −i 0 1 0 i 3 2 2 −3i i 0 3 0 −i −6 2 = 1 3 4 + 0 − 1 −6i + 6i 2i− 2i 0 3 0 2i − 2i 3 + 9 − 12 −1 + 4 = I det(A) = −1(0 + 6i) − 1(4− 6i) + i(−4i− 0) = 0. A does not have an inverse. 3 Prob. 3.10 in your textbook: a.) Aa = −1 1 i 2 0 3 2i −2i 2 i 2i 2 = −i + 2i + 2i 2i + 0 + 6 −2 + 4 + 4 = 3i 6 + 2i 6 b.) a†b = (−i −2i 2 ) 2 1 − i 0 = −2i + (−2i)(1− i) + 2(0) = −4i− 2 c.) ãBb = ( i 2i 2 ) 2 0 −i 0 1 0 i 3 2 2 1 − i 0 = ( i 2i 2 ) 4 1 − i 2i + 3(1 − i)
