Introduction to Quantum Mechanics Physics Thursday February 21, Problem # 1 (10pts) We are given the operator U(m, n) defined by

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1 Department of Physics Introduction to Quantum Mechanics Physics 5701 Temple University Z.-E. Meziani Thursday February 1, 017 Problem # 1 10pts We are given the operator Um, n defined by Ûm, n φ m >< φ n 1 where φ n > are the eigenstates of the Hermitian operator Ĥ which could be the Hamiltonian of a system. We also can assume that the states φ n > form an orthonormal basis. Ûm, n following the rules of hermitian conju- a Here we calculate the adjoint Û m, n of gation. Û m, n φ m >< φ n φ n >< φ m ] b We evaluate the commutator [Ĥ, Ûm, n [Ĥ, Ûm, n ] [ ] Ĥ, φ m >< φ n Ĥ φ m >< φ n φ m >< φ n Ĥ since Ĥ is hermitian it can act on a ket or a bra m φ m >< φ n n φ m >< φ n c We want to prove the relation: m n φ m >< φ n m nûm, n Ûm, nû p, q δ nq Ûm, p 4 We start by expanding the product Ûm, nû p, q following the definition of Ûm, n. Ûm, nû p, q φ m >< φ n φ q >< φ p φ m > δ np < φ p δ nq φ m >< φ m where we have used the fact that the φ n > are orthonormal. δ nq Um, p 5

2 d The definition of the trace of an operator A is T r{â} i Let s evaluate the trace of Ûm, n T r{ûm, n} i i < φ i  φ i > 6 < φ i φ m >< φ n φ i > < φ i φ m >< φ n φ i > i δ im δ ni δ nm 7 e Assume  to be an operator, with matrix elements A mn < φ m  φ n > we want to prove the relation  m,n A mn Ûm, n 8 We start with A mn Um, n < φ m  φ n > φ m >< φ n 9 m,n m,n since A mn is a matrix element it is a scalar, therefore it can be positioned anywhere A mn Um, n φ m >< φ m  φ n >< φ n m,n m,n m,n φ m >< φ m  φ n >< φ n Using the closure relation on the basis { φ n >} we have φ m >< φ m 1 and φ n >< φ n 1 11 m we now replace each of the sums by unity. A mn Um, n φ m >< φ m A φ n >< φ n m,n m,n A mn Um, n ˆ1ˆ1  m,n m 10 1 f Finally we want to show that A pq T r{âû p, q}. Following the definition of the trace. T r{âû p, q} i i < φ i  φ p >< φ q φ i > < φ i  φ p >< φ q φ i > i A ip δ qi A pq 1

3 Problem # 10pts Consider a physical system whose two dimensional state space is spanned by the orthonormal basis formed by the two kets u 1 >, u >. In this basis the operator ˆσ y is defined by: 0 i 14 i 0 a The matrix ˆσ y is Hermitian because it is a square matrix with real diagonal elements and the off-diagonal elements are complex conjugate of each other. The eigenvalues of this operator are found through solving the secular equation det ˆσ x λî 0 15 This corresponds to λ i i λ λ 1 0 λ +1 and λ 1 16 We find two distinct eigenvalues thus non degenerate. We now solve for the eigenvectors for each eigenvalue. We start with the eigenvalue +1 and and solve for the eigenvector ψ + using the equation 0 i +1 ic i 0 and i i 17 Therefore the eigenvector of ˆσ y corresponding to the eigenvalue +1 is ψ + > u 1 > +i u > 18 < ψ + ψ + > + i We follow with the eigenvalue 1 and and solve for the eigenvector ψ using the equation 0 i 1 ic i 0 and i i 0 Therefore the eigenvector of Ŝx correponding to the eigenvalue 1 is ψ > u 1 > i u > 1 In summary the two eigenvectors are < ψ ψ > + i 1 1 ψ + > 1 u 1 > +i u > for + 1 ψ > 1 u 1 > i u > for 1 4

4 b Lets first write the matrix which represent the projector P + ψ + >< ψ + P + 1/ i/ 1/ i/ 1/ i/ i/ 1/ Second we write the matrix which represent the projector P ψ >< ψ P 1/ i/ 1/ i/ 1/ i/ i/ 1/ 5 6 Adding the matrices gives P + + P 1/ i/ i/ 1/ + 1/ i/ i/ 1/ ˆ1 7 Thus satisfies the closure relation. c The matrix M is Hermitian because it is a square matrix with real diagonal elements and the off-diagonal elements are complex conjugate of each other. The eigenvalues of this operator are found through solving the secular equation det ˆM λî 0 8 This corresponds to λ i i λ λ λ 0 6 λ λ + λ 0 λ 5λ λ 1, 5 ± 5 16 λ 1 +4 and λ +1 0 We find two distinct eigenvalues thus non degenerate. We now solve for the eigenvectors for each eigenvalue. We start with the eigenvalue +1 and and solve for the eigenvector ψ + using the equation i i +4 c +i 4 and i + 4 i 1 Therefore the eigenvector of ˆM corresponding to the eigenvalue +4 is ψ 1 > u 1 > i u > < ψ 1 ψ 1 > We follow with the eigenvalue +1 and and solve for the eigenvector ψ > using the equation i i 1 + i and i + 4

5 i 5 Therefore the eigenvector of M corresponding to the eigenvalue +1 is ψ > u 1 > + i u > 6 < ψ ψ > In summary the two eigenvectors are 1 ψ 1 > u 1 > i u > ψ > u 1 > + i u > Lets first write the matrix which represents the projector P 1 ψ 1 >< ψ 1 1/ 1/ P 1 i 1/ i / i / / i / / Second we write the matrix which represents the projector P ψ >< ψ / / P i / i / i 1/ 1/ i/ / 1/ Adding the matrices gives 1/ i / P 1 + P i + / / Thus satisfies the closure relation. e Finally we consider the matrix: / i / i/ / 1/ 0 h /i 0 h /i 0 h /i 0 h /i 0 7 for for ˆ1 4 4 To find the eigenvalues we need to solve the secular equation: λ h /i 0 ˆL y i h /i λ h /i 0 i h λ i h/ λ i h/ i h λ i h/ i h/ /i λ 0 λ i h/ λ i h/ λ λ h / + i h i h λ λ λ h / h / λ λ h 0 44

6 therefore the eigenvalues are: λ 1 + h, λ 0 and λ h 45 Let start to solve for the eigenvector corresponding to the eigenvalue λ 1 + h 0 i h/ 0 i h/ 0 i h/ 0 i h/ 0 ˆL y ψ 1 > + h ψ 1 > 46 c + h c 47 i i i i c c i c 48 Therefore the eigenvector of L z corresponding to the eigenvalue + h is ψ 1 > u 1 > +i u > u > 49 < ψ 1 ψ 1 > ψ 1 > 1 u 1 > +i u > u > we follow by solving for the eigenvector corresponding to the eigenvalue λ 0 0 i h/ 0 i h/ 0 i h/ 0 i h/ 0 ˆL y ψ 0 > 0 5 c 0 c 5 i 0 i i c 0 i 0 c 0 c 54 Therefore the eigenvector of L z corresponding to the eigenvalue 0 is ψ 0 > u 1 > + u > 55 < ψ 0 ψ 0 >

7 ψ 0 > 1/ u 1 > + u > 57 Finally we solve for the eigenvector corresponding to the eigenvalue λ h 0 i h/ 0 i h/ 0 i h/ 0 i h/ 0 ˆL y ψ 1 > h ψ 1 > 58 c h c 59 i i i c i c c i c 60 Therefore the eigenvector of L z corresponding to the eigenvalue h is ψ 1 > u 1 > i u > u > 61 < ψ 1 ψ 1 > ψ 1 > 1 u 1 > i u > u > 6 6 Problem # 10pts a We consider the two vectors : ψ 0 > 1 u 1 > + i u > 1 u > ψ 1 > 1 u 1 > + i u > 64 First we check the normalization of these kets using the properties of the basis { u 1 >, u >, u >} 1 < ψ 0 ψ 0 > < u 1 i < u < u u 1 > + i u > 1 u > therefore ψ 0 > is normalized. Similarly < ψ 1 ψ 1 > 1/ < u 1 i/ < u 1/ u 1 > +i/ u > therefore ψ 1 > is not normalized.

8 b The projector ˆρ 0 onto the state ψ 0 > reads: In the { u 1 >, u >, u >} representation we can write: ˆρ 0 1/ i/ 1/ i/ 1/ 1/ The projector ˆρ 1 onto the state ψ 1 > reads: ˆρ 0 ψ 0 >< ψ 0 67 /4 i /4 /4 i/ /4 1/4 i/4 /4 i/4 1/4 68 ˆρ 1 ψ 1 >< ψ 1 < ψ 1 ψ 1 > 69 In the { u 1 >, u >, u >} representation we can write: ˆρ i 1 0 i 1 0 i i i i Inspecting the matrices of ˆρ 0 and ˆρ 1 we notice that the diagonal matrix elements are real and that with respect to the diagonal the matrix elements are conjugate of each other. Therefore we conclude that these matrices are Hermitian.