Test 2 Solutions for Introduction to Quantum Mechanics II | PHY 4605, Exams of Physics

Download Test 2 Solutions for Introduction to Quantum Mechanics II | PHY 4605 and more Exams Physics in PDF only on Docsity! PHY4605–Introduction to Quantum Mechanics II Spring 2004 Test 2 Solutions March 17, 2004 1. Short Answer. Must attempt (only) 3 of 6. Circle answers to be graded. (a) Given the exact normalized eigenfunctions ψn of a quantum system, Hψn = Enψn, prove the variational principle, i.e. show that for any normalized φ, 〈φ|H|φ〉 ≥ E0, where E0 is the exact ground state energy. |φ〉 = ∑ n cn|n〉, |n〉 normalized 〈n|n〉 = 1 (1) 〈φ|H|φ〉 = ( ∑ n c∗n〈n| ) H ( ∑ n′ cn′|n′〉 ) (2) = ∑ nn′ c∗ncn′〈n|H|n′〉 = ∑ nn′ c∗ncn′En′ 〈n|n′〉︸ ︷︷ ︸ δnn′ (3) = ∑ n |cn|2En ≥ ( ∑ n |cn|2 ) ︸ ︷︷ ︸ 1 E0 = E0 (4) (b) A chemistry book lists the electronic configuration of an N atom in its ground state as 1s22s22p3 and states that the wave function for this state is simply the product of hydrogenic orbitals corresponding to the 7 electrons in the levels designated. Criticize this statement. i) product must be antisymmetrized with respect to all particles ii) “hydrogenic” orbitals - not really hydrogenic since z = 7 iii) Even antisymmetrized product of correct one-electron orbitals won’t be eigenstate of N-atom Hamiltonian since there is an inter-electron Coulomb interaction. (c) An atomic state is referred to by its spectroscopic term, 3P2. What are its quantum numbers S, L, J? Using hydrogenic wavefunctions ψn`m and 1 2-particle spin states |SMS〉, write down an appropriate wave function for 2 electrons which might correspond to this term. “term” 2s+1Lj } S = 1, L = 1, J = 2. Note J is maximum value, given S = 1, L = 1. Therefore must have |2, 2 J,MJ 〉 = |1, 1, 1, 1〉 L,S,ML,MS , ML + MS = MJ . (d) Explain the physical origin of fine structure and hyperfine structure in Hydrogen. Fine structure: relativistic effects/spin-orbit coupling. Hyperfine structure: spin-spin interaction between electron and proton. (e) Angular momentum. Given two angular momenta L1 and L2, with eigen- values of the operators L21 and L 2 2 given by 6h̄ 2 and 15h̄2/4, what are the allowed possible eigenvalues of the total angular momentum operator squared L2 = (L1 + L2) 2? L̂21 = 6h̄ 2 = L1(L1 + 1)h̄ 2 ⇒ L1 = 2 (5) L̂22 = 15h̄ 2 = L2(L2 + 1)h̄ 2 ⇒ L2 = 3/2 (6) Allowed values of L: 2− 3/2 ≤ L ≤ 2 + 3/2 ⇒ 1 2 , 3 2 , 5 2 , 7 2 . Eigenvalues of L2 : 3 4 h̄2, 15 4 h̄2, 35 4 h̄2, 63 4 h̄2. (f) H2 molecule. Sketch the effective potential for the two protons as a func- tion of the distance between them for the ground state and first excited state. In the same picture, sketch the vibrational levels and explain (do not calculate) how you would estimate them. Approximate E(r) near potential minimum in 1Σg state by simple har- monic oscillator around r0. Calculate h̄ω0(n + 1/2), ω0 = √ k m , k = from E ′′ (r0), m = mP . 2 3. H− Ion. Consider an H− ion, which is a hydrogen atom with two electrons. (a) Write down the Hamiltonian for this system. You may assume that the proton is infinitely heavy. H = p̂21 2m + p̂22 2m + e2 |r1 − r2| − e2 r1 − e 2 r2 (13) (b) Show that, if you neglect the Coulomb interaction between the electrons, the Hamiltonian separates into equations for each electron. H = H1 + H2, Hα = p̂2α 2m − e 2 rα (14) “Additive Hamiltonian theorem”⇒ E = E1 + E2, ψ = ψ1(r1)ψ2(r2) is a solution. (15) (c) Continue to neglect the Coulomb interaction between the electrons, and assume that both electrons are in the ground (n = 1) state. Write a valid wavefunction Ψ1s1s for the electrons, including specification of the possible spin states. Do the same for one electron in a 1s state and the other in a 2s state, Ψ1s2s. Ψ1s1s = symmetric ψ1s(r1)ψ1s(r2) antisymmetric χsinglet , χsinglet = 1√ 2 (↑↓ − ↓↑) (16) Ψ1s2s = [ ψ1s(r1)ψ2s(r2)± ψ1s(r2)ψ2s(r1) ] χsinglettriplet (17) χtriplet =    ↑↑ 1√ 2 (↑↓ + ↓↑) ↓↓ (18) (d) Including the Coulomb interaction between electrons, would you expect spin triplet states to be lower or higher in energy than spin singlet states with the same orbital quantum numbers? Why? 5 Triplet states spatially antisymmetric⇒ two electrons spend less “time” close to each other⇒ lower energy since 〈 e2 r12 〉 is minimized. 6 4. Angular momentum. (a) Sketch the angular and radial forms of the hydrogenic wavefunctions φn`m for the cases (n = 2, ` = 1,m) and (n = 3, ` = 1, m). Make sure you make a separate sketch for each of the 2l + 1 degenerate cases. (b) Given eigenstates of angular two independent angular momenta L1 and L2 designated by |`1,m1, `2,m2〉, consider the state |Ψ〉 = 1√ 3 |2, 2, 1,−1〉+ 1√ 6 |2, 1, 1, 0〉 − 1√ 2 |2, 0, 1, 1〉. (19) Show that this state is an eigenstate of total Lz ≡ L1z + L2z with m = 1. Each state in sum has m1+m2 = 1 ⇒ eigenstate of total Lz with eigenvalue h̄. (c) Show that |ψ〉 is an eigenstate of total L ≡ L1 + L2 with L = 2. Use: L2 = L21 + L 2 2 + 2L1 · L2 (20) = L21 + L 2 2 + 2 ( L1zL2z ) + L1+L2− + L1−L2+ (21) Check: 2L1zL2z|Ψ〉 = −4√ 3 |221− 1〉+ 0 · |2110〉+ 0|2011〉 (22) L1+L2−|Ψ〉 = 1√ 3 · 0 + 1√ 6 √ 6− 2 √ 2|221− 1〉 − 1√ 2 √ 6 √ 2|2110〉 (23) L1−L2+|Ψ〉 = 1√ 3 √ 6− 2 √ 2|2110〉+ 1√ 6 √ 6 √ 2|2011〉 − 1√ 2 · 0 (24) L2 h̄2 |Ψ〉 = (L1(L1 + 1) + L2(L2 + 1) )|Ψ〉+ (− 4√ 3 + 2√ 3 )|221− 1〉 (25) + (− √ 6 + 2 √ 2√ 3 )|2110〉+ √ 2|2011〉 (26) = ( 2(2 + 1) + 1(1 + 1) )|Ψ〉+ (−2) 1√ 3 |221− 1〉 (27) + (−2) 1√ 6 |2110〉+ 2 1√ 2 |2011〉 (28) = (6 + 2− 2)|Ψ〉 = 6|Ψ〉 = 2(2 + 1)|Ψ〉 (29) 7