Old Test 2 with Resolution - Introduction to Quantum Mechanics II | PHY 4605, Exams of Physics

Download Old Test 2 with Resolution - Introduction to Quantum Mechanics II | PHY 4605 and more Exams Physics in PDF only on Docsity! PHY4605–Introduction to Quantum Mechanics II Spring 2005 Test 2 SOLUTIONS March 9, 2005 Short Answer. Must attempt (only) 3 of 6. Circle answers to be graded. 1. Identify or short answer: (a) Born-Oppenheimer approx. Discuss the philosophy and its implementation in the case of the H2 molecule or ion. The B-O approx. is a method useful for problems where one has fast and slow degrees of freedom. In the H2 molecule, the protons are slow and the electrons are fast due to the difference in the masses. Thus the elec- trons may be assumed to a good accuracy to instantaneously adopt their configuration of lowest energy around the 2 protons “placed” in space at some separation rab. The protons feel an effective potential determined by this electron cloud, and try to minimize this potential at their equilibrium separation. The B-O method then fixes rab, factorizes the wave function into an electron and proton part φψ, then keeps only the terms in the proton motion which involve the large spatial gradients of the proton wave function ψ. This gives an equation for the proton motion involving the eigenvalue of the electronic part E(0) playing the role of an effective po- tential for the protons, (T + E(0))ψ = Eψ, where T is the proton KE and E(0) is the solution to H0φ = E (0)φ. E(0) can be found variationally for a given rab and therefore depends on rab. It has a minimum for the ground state of the electron system around r = 0.74 ◦ A. Expanding E(0) around its minimum gives the frequency of small oscillations, i.e. the vibrational modes of the molecule. (b) For Lithium, Z = 3, electron configuration 1s22s, standard tables give the radius of the atom as 1.5 ◦ A . If, on the other hand, one looks at the formula for the radius of the n = 2 Bohr orbit for a hydrogenic 1-electron atom with nuclear charge Ze, we find the Bohr radius a0 = 4π²0h̄ 2/(mZe2), and the 2s radial function goes like e−r/2a0 , so we might have guessed that the “size” of the atom would be 2a0 = 2 ∗ (0.529 ◦ A)/3 ' 0.35◦A (where 0.529◦A is the Bohr radius for Hydrogen with Z = 1). What physical phenomenon was neglected in our estimate which might lead to a larger orbit of the valence electron if included? Explain. The argument neglects screening, sometimes called shielding. The valence (2s) electron is further from the nucleus, and therefore effectively “sees” a nuclear charge closer to 3− 2 = +1e rather than +3e because the two 1s electrons are also with high probability inside the orbit. The orbit is thus 1 a factor of three or so larger, not quite the “right” answer, but in the right direction. (c) Hyperfine structure of hydrogen. State its origin, give the magnitude of the splitting of the H ground state either in terms of fraction of a Rydberg using the fine structure constant, or the wavelength of the photon emitted in the hyperfine transition. Hyperfine structure arises from the dipole-dipole interaction of the pro- ton and the electron spin. The Bohr energy of the ground state doesn’t depend on electron spin, but including hyperfine coupling splits it into a proton-electron spin triplet and spin singlet, with energy separation of or- der (m/mp)α 2R, where α is fine structure constant and R = 13.6eV . The famous photon emitted has wavelength 21 cm. (d) Give the electronic configuration à la chemistry for the Boron atom, with Z = 5. Then enumerate the allowed term symbols for the ground state from the point of view of allowed angular momenta alone. Boron is 1s22s21p. Since the two 1s and two 2s electrons are in a singlet state (S = 0) and have zero orbital angular momentum, the spin and angular momentum are determine entirely by the 1p electron. The term diagrams are therefore 2S+1Lj = 2P1/2 and 2P3/2, with j = 1/2 and 3/2 being the two values of j you can get from adding s = 1/2 and ` = 1. The 2P1/2 turns out to be the ground state, but you can’t tell that from the angular momentum arguments alone. (e) You are told that an excited state ψn, n 6= 0 of a quantum system you cannot solve is an eigenstate of Hermitian operator Q, Qψn = qnψn, of lowest possible energy (but not the ground state, which has a different quantum number q0). You also know that [Q,H]=0. Show that any trial wavefunction φ which obeys Qφ = qnφ can be used to provide an upper bound to En. Take a basis for Hilbert space χm,q, where q is the eigenvalue of Q, and m represents all other eigenvalues. A general function may be expanded u = ∑ m,q cmqχm,q. A function with definite eigenvalue qn is however φ =∑ m cmqnχmqn . 〈φ|H|φ〉 = ∑ mm′ c∗mqncm′qn〈χmqn |H|χm′qn〉 = ∑ m,m′ c∗mqncm′qn〈χmqn |Em′qn |χm′qn〉 = ∑ m,m′ c∗mqncm′qnδmm′Em′qn = ∑ m |cmqn |2Emqn ≥ En ∑ m |cmqn |2 = En since En is by definition the lowest of the energies corresponding to qn, and assuming φ is normalized. 2 Eigenfunctions are ψn = e inφ, n integer, e’values En = h̄ 2n2/(2ma2), n = 0,±1,±2.... (b) A potential is now applied to the ring of form V (φ) = A cos φ sin φ, with A a constant. Calculate the matrix Vnn′ = 〈n|V |n′〉, n, n′ = ±1. 〈1|V |1〉 = A ∫ dφ2π0 cos φ sin φ = 0 〈−1|V | − 1〉 = A ∫ dφ2π0 cos φ sin φ = 0 〈1|V | − 1〉 = A ∫ dφ2π0 e −2imφ cos φ sin φ = −iπ 2 〈−1|V |1〉 = A ∫ dφ2π0 e 2imφ cos φ sin φ = i π 2 , so V =   0 −iA/4 iA/4 0   . (c) Calculate the first-order energy shifts of the state(s) with n = ±1 using degenerate perturbation theory, and find the “good” eigenstates in which V is diagonal. The two states | ± 1〉 are degenerate for A = 0, and the matrix V is not diagonal, so we have to use full-fledged degenerate perturbation theory, and find the basis in which the matrix becomes diagonal. The eigenvalues are found from the characteristic polynomial, δE(1) = ±A/4. To find the eigenvectors you can use any method you like, e.g. consider the equations V | ± 1′〉 = ±A/4| ± 1′〉, and put |1′〉 = a|1〉 + b| − 1〉, etc., then solve for coefficients, remembering normalization of wave function. Find |1′〉 = 1√ 2 (| − 1〉+ i|1〉) | − 1′〉 = 1√ 2 (|1〉+ i| − 1〉) (d) The potential is now removed (A = 0), and a second identical particle added to the ring. Both particles have spin 1. Write down i) the Hamilto- nian, ii) a valid wavefunction for the ground and first excited states of the two particles, and iii) the energies of these states. If you could not do part a), you can still express your answer in terms of the ψn which are solutions to that part. Hamiltonian is H = p21/(2m) + p 2 2/(2m). Ground state is constant, Ψ0(φ1, φ2) = 1 · χt, E = 0, where χt is triplet spin state of 2 spin-1 5 particles. For 1st excited state we can put one particle in the n = ±1 state, Ψ±1,+(φ1, φ2) = ( 1 · e±iφ2 + e±iφ1 · 1 ) χt Ψ±1,−(φ1, φ2) = ( 1 · e±iφ2 − e±iφ1 · 1 ) χs overall symmetric under 1 ↔ 2 since spin-1 particles are bosons. 6