8 Solved Problems in Homework 4 on the Quantum Mechanics | PHY 389K, Assignments of Quantum Mechanics

Download 8 Solved Problems in Homework 4 on the Quantum Mechanics | PHY 389K and more Assignments Quantum Mechanics in PDF only on Docsity! PHY 389K QM1, Homework Set 5 Solutions Matthias Ihl 10/12/2006 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/PHY389K/ 1 Problem 1 We make use of the following identities: J1 = 1 2 (J+ + J−), (1) J+J− + J−J+ = 2( ~J 2 − J23 ). (2) Now, we have ∆ψJ1 = √ 〈j,m|J21 |j,m〉 − 〈j,m|J1|j,m〉2 = √ 〈j,m|1 4 (J2+ + J 2 − + J+J− + J−J+) |j,m〉 − 1 2 〈j,m|(J+ + J−)|j,m〉2 = √ 1 2 〈j,m| ( ~J2 − J23 ) |j,m〉 = √ 1 2 (j(j + 1) −m2) Therefore, with ψa = |j, 0〉 and ψb = |j, j〉, ∆ψaJ1 = √ j(j + 1) 2 , ∆ψbJ1 = √ j 2 . 1 This result is not very surprising. In quantum mechanics, we know that J1 and J3 do not commute and therefore you should expect that the uncertainty of J1 in an eigenstate of J3 is non-vanishing. 2 Problem 2 From [Si, Sj] = iǫijkSk, we find that 〈f j′m′ |Si|f jm〉 ∝ δjj ′ , (3) since we know that S3|f jm〉 = m|f jm〉, S1|f jm〉 = 1 2 (S+ + S−)|f jm〉 = 1 2 (αm+1|f jm+1〉 + αm|f jm−1〉), S2|f jm〉 = 1 2i ((S+ − S−)|f jm〉 = 1 2i (αm+1|f jm+1〉 − αm|f jm−1〉), where αm+1 = √ (j −m)(j +m+ 1) and αm = √ (j +m)(j −m+ 1). Also, we have ~S2|f jm〉 = j(j + 1)|f jm〉, (4) with ~S2 = ∑3 i=1 S 2 i . From {Si, Sj} = 12δij, we find that Si, Si = 2S 2 i = 1 2 ⇒ S2i = 1 4 ⇒ ~S2 = 3 4 . (5) Therefore we have ~S2|f jm〉 = 3 4 |f jm〉 = j(j + 1)|f jm〉, (6) from which we conclude that j = 1 2 is the only allowed j-value. We have found the corresponding spin operators Si in homework 4: Si = 1 2 σi, i ∈ {1, 2, 3}, (7) 5 Problem 5 (a) We use the result of problem 4 to calculate the matrix elements 〈er|Jij|es〉. 〈er|Jij|es〉 = ǫijk〈er|Jk|es〉 = ǫijkǫrksi~ = −i~ǫijkǫkrs = −i~(δirδjs − δisδjr) = i~(δisδjr − δirδjs), where use was made of relation (2.13) of the textbook. The constant c is thus c := i~. (12) (b) In three dimensions, using Cartesian coordinates, we have ~x = 3 ∑ i=1 ~eix i = 3 ∑ i=1 ~ei(~x · ~ei), (13) i.e., ~x · ~ei = xi, where the ~ei are the canonical Cartesian basis vectors in R3. Here, we want to find the relation between the |ei〉 defined above in part (a) and the basis vectors in cartesian coordinates. Hence, our aim is to check whether (in spherical coordinates) 〈~̂x|ei〉 = xi r . (14) We are only interested in the angular part, so we use ~̂x = ~x r and set r = 1. We have, 〈~̂x|e3〉 = 〈r = 1, θ, φ|l = 1, m = 0〉 = R(r = 1) · Yl=1,m=0(θ, φ) = R(r = 1) √ 3 4π cosθ ∝ x 3 r , as expected. R(r) := Rn=0,l=1(r) is given in terms of associated Laguerre polynomials at r = 1. Moreover, 〈~̂x|e1〉 = 〈r = 1, θ, φ|(|l = 1, m = 1〉 − |l = 1, m = −1〉)( −1√ 2 ) = R(r = 1) (Yl=1,m=1(θ, φ) − Yl=1,m=−1(θ, φ)) (−1√ 2 ) = R(r = 1) (−1√ 2 ) ( − √ 3 8π e−iφP |m|=1 l=1 (cosθ) − √ 3 8π eiφP |m|=1 l=1 (cosθ) ) = R(r = 1) √ 3 16π (eiφ + e−iφ)Pm=1l=1 (cosθ) = −R(r = 1) √ 3 4π cosφ sinθ ∝ x 1 r , and 〈~̂x|e2〉 = 〈r = 1, θ, φ|(|l = 1, m = 1〉 + |l = 1, m = −1〉)( i√ 2 ) = R(r = 1) (Yl=1,m=1(θ, φ) + Yl=1,m=−1(θ, φ)) ( i√ 2 ) = R(r = 1) ( i√ 2 ) ( − √ 3 8π e−iφP |m|=1 l=1 (cosθ) + √ 3 8π eiφP |m|=1 l=1 (cosθ) ) = R(r = 1) √ 3 4π i 2 (eiφ − e−iφ)Pm=1l=1 (cosθ) = R(r = 1) √ 3 4π sinφ sinθ ∝ x 2 r . Thus we have shown (up to proper normalization) that {|ei〉}i=1,2,3 are the basis vectors of three dimensional (euclidean) space in spherical coordinates. 6 Problem 6 I will use the following shorthand notation: |j = 1/2, m1〉 = α|1/2, 1/2〉+ β|1/2,−1/2〉. (15) We have J1|j = 1/2, m1〉 = 1 2 (H+ +H−)(α|1/2, 1/2〉+ β|1/2,−1/2〉) = 1 2 (αH−|1/2, 1/2〉+ βH+|1/2,−1/2〉) = 1 2 (α|1/2,−1/2〉+ β|1/2, 1/2〉) ! = m1|j = 1/2, m1〉, ⇐⇒ α = β ∨ α = −β. We choose |α| = |β| = 1√ 2 for normalization. We conclude |j = 1/2, m1 = +1/2〉 = 1√ 2 (|1/2, 1/2〉+ |1/2,−1/2〉), |j = 1/2, m1 = −1/2〉 = 1√ 2 (|1/2, 1/2〉 − |1/2,−1/2〉. We have two different values for m1 = ±12 . now, let us calculate ~J2|j = 1/2, m1 = ±1/2〉 = ( H+H− +H 2 3 −H3 ) 1√ 2 (|1/2, 1/2〉 ± |1/2,−1/2〉) = 1√ 2 ( H+H−|1/2, 1/2〉+ (H23 −H3)|1/2, 1/2〉 ±(H23 −H3)|1/2,−1/2〉 ) = 1√ 2 ( |1/2, 1/2〉+ (1 4 − 1 2 )|1/2, 1/2〉 ± (1 4 + 1 2 )|1/2,−1/2〉 ) = 3 4 |j = 1/2, m1 = ±1/2〉. This is precisely j(j + 1) for j = 1/2, as expected.