Quantum Mechanics I / Quantum theory I. 02 October, 2015 Assignment 2: Solution 1. Consider the following two kets: 0 1 0 1 −3i 2 B C B C B C B C j = B 2 + i C ; jφ = B −i C : @ A @ A 4 2 − 3i (a) Find the bra φj. (b) Evaluate the scalar product φj . (c) Why the products j jφ and φj j do not make sense? Answer (a) j = 2 + i 2 + 3i : (b) 0 1 −3i B C B C φj = 2 + i 2 + 3i B 2 + i C : @ A 4 = −6i + 2i − 1 + 8 + 12i; = 7 + 8i: (c) The j and jφ are column vectors and φj and j are row vectors. And we know that matrix multiplication is performed only when we have the number of columns of first matrix equal to the number of rows to the second matrix. Therefore the multiplication of the form j jφ and φj j does not make sense. Neither do these objects make nay physical sense. 2. Consider the states j = 3ijφ1 −7ijφ2 jχ = −|φ1 +2ijφ2 ; where jφ1 and jφ2 are orthonormal. Calculate the scalar products jχ and χj . Are they equal? 1 Quantum Mechanics I / Quantum theory I. 02 October, 2015 Answer jχ = −3i φ1j + 7i φ2j −|φ1 +2ijφ2 ; jχ = −14 + 3i: Similarly, χj = − φ1j − 2i φ2j 3ijφ1 −7ijφ2 ; = −14 − 3i: Hence jχ = χj ∗ as expected. 3. (a) Discuss the hermiticity of operators (A^ + A^y), i(A^ + A^y) and i(A^ − A^y). (b) Find the Hermitian adjoint of (1 + iA^ + 3A^2)(1 − 2iA^ − 9A^2) f(A^) = · 5 + 7A^ (c) Show that the expectation value of a Hermitian operator is real and that of an anti-Hermitian operator is imaginary. Answer (a)(A^ + A^y): A^ + A^yy= A^y + A:^ So this is hermitian. iA^ + A^y: iA^ + A^yy= −iA^ + A^y: So this is not hermitian, rather anti-Hermitian. iA^ − A^y: iA^ − A^yy = −iA^y − A^; = iA^ − A^y 2 Quantum Mechanics I / Quantum theory I. 02 October, 2015 So, this is hermitian. (b) (1 + iA^ + 3A^2)(1 − 2iA^ − 9A^2) f(A^) = · 5 + 7A^ (1 + iA^ + 3A^2)(1 − 2iA^ − 9A^2)y f(A^)y = · 5 + 7A^ (1 − 2iA^ − 9A^2)y(1 + iA^ + 3A^2)y = : (5 + 7A^)y (1 + 2iA^y − 9A^2y)(1 − iA^y + 3A^2y) = : 5 + 7A^y y This can be simplified further if A^ is hermitian, then A^ = A^y and A^2 = A^2 . (c) ^ Suppose A is a hermitian operator with real eigenvalues λi and eigenvector jλi , i.e ^ Ajλi = λij : The expectation value is, ^ ^ A = λijOjλi ; X ^ = jAjλi λij ; i X = λi jλi λij ; n X = λi j ; n X = λi: n The expectation value is the sum of eigenvalues which must be real. Hence A^ = ^ P Tr A = n λi. Now consider an anti-Hermitian operator B^, i.e.,B^y = B^. Suppose B^ has eigenvalues bi with eigenvectors jbi . ^ Bjbi = bijbi (1) ^y ∗ ^ bijB = bi bij = − bijB (2) From 1, we obtain ^ bijBjbi = bi bijbi = bi (3) 3 Quantum Mechanics I / Quantum theory I. 02 October, 2015 ^ ∗ But from 2, bijB = −bi bij: Hence the LHS of can be written as, ^ ∗ bijBjbi = −bi bijbi = −bi (4) ∗ Comparing 3 and 4, bi = −bi which is only possible if bi is imaginary. Now B^ = jB^j X ^ = jBjbi bij i X = bi jbi bij i X = bi Which must be imaginary i Since all the bi's are imaginary. 4. What conditions must the parameter " and the operator G^ satisfy so that the operator U^ = ei"G^ is unitary? Answer U^U^ y = ei"G^ei"G^y; = ei"G^ei"G^y ; ^ ^y = ei" G−G = 1^: This means that G^ = G^y or G^ must be hermitian to ensure that ei"G^ is unitary. Remember: i^2 i^3 ei"G^ = 1^ + i^+ + + :::::: 2! 3! ^ ^ and ei"A^ei"B^ = ei" A+B only if [A;^ B^] = 0. 5. Show that if A^−1 exists, the eigenvalues of A^−1 are just the inverse of those of A^. Answer If A^−1 exists then we have A^A^−1 = 1^ = A^−1A;^ 4 Quantum Mechanics I / Quantum theory I. 02 October, 2015 So, 1^j = j ; = A^−1A^j ; = A^−1aj ; = aA^−1j : 1 A^−1aj = j ; a Hence the eigenvalues of A^−1 are the inverse of the eigenvalues of A^. 6. Consider the following two kets: 0 1 0 1 5i 3 B C B C B C B C j = B 2 C ; jφ = B 8i C : @ A @ A −i −9i (a) Find j ∗ and j. (b) Is j normalized? If not, normalize it. (c) Are j and jφ orthogonal? Answer (a) 0 1 −5i B C ∗ B C j = B 2 C ; @ A i j = −5i 2 i : 0 1 5i B C B C j = −5i 2 i B 2 C ; @ A −i = 10 + 4 + 1; = 15: 5 Quantum Mechanics I / Quantum theory I. 02 October, 2015 j So, normalized ket will be ; r j 0 1 5i 1 B C B C = p B 2 C : 15 @ A −i (c) 0 1 5i B C B C φj = 3 − 8i 9i B 2 C ; @ A −i = 15i − 16i + 9; = 9 − i: As φj is not zero, so these are not orthonormal. 7. (a) Show that Tr(A^B^) = Tr(B^A^). (b) Show that the trace of a commutator is always zero. Answer (a) X TrA^B^ = ija^B^ji ; i X X = ijA^jj jjB^ji ; i j X X ^ ^ = AijBji; i j Similarly, ^ ^ X X ^ ^ ^ ^ Tr BA = BjiAij = Tr AB i j 6 Quantum Mechanics I / Quantum theory I. 02 October, 2015 (b) Tr[A;^ B^] = TrA^B^ − B^A^; = TrA^B^−TrB^A^; = TrA^B^−TrA^B^; = 0: 8. Show that the trace of an operator does not depend on the basis in which it is expressed. Answer Let we take two basis ji and j~j , with j~j = S^ji where S^ is a similarity transform ^ with elements Sji = ~jji . X TrA^= ~jjA^j~j ; j Since j~j = S^ji , X TrA^ = ijS^yA^S^ji ; i = TrS^yA^S^; ~ = TrA^; ~ Where A^ is the operator in the transformed bases. 9. Consider the two states, j = ijφ1 +3ijφ2 −|φ3 ; and jχ = jφ1 −ijφ2 +5ijφ3 ; where jφ1 , jφ2 and jφ3 are orthogonal and normalized. (a) Calculate j χj and jχ j. Are they equal? Calculate their traces and compare them. 7 Quantum Mechanics I / Quantum theory I. 02 October, 2015 (b) Find the Hermitian conjugates of j χj. Answer (a) These states in the jφ1 ; jφ2 ; jφ3 basis are given by, 0 1 i B C B C j = B 3i C ; @ A −1 0 1 1 B C B C jχ = B −i C ; @ A −5i Now 0 1 i B C B C j χj = B 3i C 1 i 5i ; @ A −1 0 1 i − 1 − 5 B C B C = B 3i − 3 − 15 C ; @ A −1 − i − 5i and similarly, 0 1 1 B C B C jχ j = B −i C −i − 3i − 1 ; @ A −5i 0 1 −i − 3i − 1 B C B C = B −1 − 3 i C ; @ A −5 − 15 5i (b) Hermitian Conjugate, also called the adjoint is, 8 Quantum Mechanics I / Quantum theory I. 02 October, 2015 0 1y i − 1 − 5 B C y B C j χj = B 3i − 3 − 15 C ; @ A −1 − i − 5i 0 1 −i − 3i − 1 B C B C = B −1 − 3 i C : @ A −5 − 15 5i 10. Consider a state which is given in terms of three orthonormal vectors jφ1 , jφ2 , and jφ3 as follows: 1 1 1 j = p jφ1 +p jφ2 +p jφ3 ; 15 3 5 ^ ^ 2 where jφn are eigenstates to an operator B such that Bjφn = (3n − 1)jφn with n = 1; 2; 3. (a) Find the norm of the state j . (b) Find the expectation value of B^ for the state j . (c) Find the expectation value of B^2 for the state j . Answer (a) The norm is, 2 1 1 1 1 1 1 j = j = p φ1j + p φ2j + p φ3j p jφ1 +p jφ2 +p jφ3 ; 15 3 5 15 3 5 1 1 1 = φ jφ + φ jφ + φ jφ ; 15 1 1 3 2 2 5 3 3 1 1 1 = + + ; 15 3 5 1 + 5 + 3 = ; 15 3 = : 5 9 Quantum Mechanics I / Quantum theory I. 02 October, 2015 (b) ^ ^ 1 1 1 Bj = B p jφ1 +p jφ2 +p jφ3 ; 15 3 5 2 11 26 = p jφ1 +p jφ2 +p jφ3 ; 15 3 5 ^ 1 1 1 2 11 26 jBj = p φ1j + p φ2j + p φ3j : p jφ1 +p jφ2 +p jφ3 ; 15 3 5 15 3 5 2 11 26 = p φ1jφ1 +p φ2jφ2 + φ3jφ3 ; 15 3 5 2 11 26 = p + p + p ; 15 3 5 2 + 55 + 78 = ; 15 135 = = 9: 15 (c). ^2 ^ ^ 1 1 1 B j = BB p jφ1 +p jφ2 +p jφ3 ; 15 3 5 ^ 2 11 26 = B p jφ1 +p jφ2 +p jφ3 ; 15 3 5 4 121 676 = p jφ1 + p jφ2 + p jφ3 ; 15 3 5 1 1 1 4 121 676 = p φ1j + p φ2j + p φ3j : p jφ1 + p jφ2 + p jφ3 ; 15 3 5 15 3 5 4 121 676 = φ jφ + φ jφ + φ jφ ; 15 1 1 3 2 2 5 3 3 4 121 676 = + + ; 15 3 5 2637 = = 175:8: 15 11.