Midterm I Questions on Quantum Mechanics - Fall 2007 | PHYSICS 137B, Exams of Quantum Mechanics

Download Midterm I Questions on Quantum Mechanics - Fall 2007 | PHYSICS 137B and more Exams Quantum Mechanics in PDF only on Docsity! Physics 137B, Fall 2007 Quantum Mechanics II Midterm I 10/9/2007, 9:40-11:00 a.m. No books, notes, or calculators are allowed. Please start each of the 4 problems on a fresh side. You should make an effort to answer every problem. I. Consider a one-dimensional harmonic oscillator: H = px 2 2m + kx2 2 . (1) Some useful facts: the energy levels of the one-dimensional harmonic oscillator are E = (n+1/2)h̄ω, n = 0, 1, . . ., and the lowest two wavefunctions are of the form ψ0 = α1 exp(−βx2), ψ1 = α2x exp(−βx2) (2) where α1, α2, β are positive constants. All the following questions are for a single spinless particle. (a) Using the exact values for the energy levels and ω = √ k/m, find the change in the ground state energy induced by a small increase in the mass m→ m+ dm, to first order in dm. The ground state energy is h̄ω/2 = h̄ √ k/(m+ dm)/2 ≈ h̄ √ (k/m)(1− dm/m)/2 ≈ (h̄/2) √ k/m(1− dm/2m), where≈means that we have kept only linear order in dm. So the change is h̄ √ k/m(−dm/4m). (b) Explain how you would use first-order perturbation theory for the change in (a) by writing the change induced by dm as a perturbation. Write, but do not calculate, the integral that will give the first-order perturbation theory result for the energy change. The perturbation Hamiltonian is −(p2/2m)(dm/m). The first-order energy correction is then 〈ψ0| − (p2/2m)(dm/m)|ψ0〉 (3) . You can check that this is indeed equal to the above result, since the expected kinetic energy in the harmonic oscillator is half the total energy. (c) Is the second order correction to the ground state energy positive, negative, or zero? The second-order correction from the ground state is always either negative or zero. Since all even states are connected by the perturbation Hamiltonian in this case, the second-order shift is negative. II. (a) Consider the 2p levels of the hydrogen atom and ignore fine structure corrections. Including spin, how many levels are there (for a single electron)? There are six levels with ` = 1 and s = 1/2, which we can label by two quantum numbers m` = −1, 0, 1 and ms = −1/2, 1/2. Without fine structure, these are all degenerate. (b) Suppose that some interaction generates the perturbation H ′ = AL · S. (4) What are the energy shifts induced by this perturbation, to first order? What are the degeneracies? 1 Ignoring fine structure the six levels are initially degenerate. We can rewrite H ′ = AL · S. = A(L+ S) 2 − L2 − S2 2 = A(J2 − L2 − S2)/2. (5) Then the angular momentum addition rule tells us that the total momentum quantum number j takes the values j = 3/2 (4 states) or j = 1/2 (2 states). Since J2 has the eigenvalue j(j+1)h̄2, and for all these states L2 has eigenvalue `(`+ 1)h̄2 = 2h̄2 and S2 has eigenvalue s(s+ 1)h̄2 = (3/4)h̄2, the j = 3/2 states are eigenstates of H ′ with eigenvalue Ah̄2((3/2)(5/2) − 2 − 3/4)/2) = Ah̄2/2, and the j = 1/2 states are eigenstates of H ′ with eigenvalue Ah̄2((1/2)(3/2)− 2− 3/4)/2 = −Ah̄2. Working in the total angular momentum basis has diagonalized the perturbation Hamiltonian. So the final energies are E = E0 +Ah̄2/2 (6) for four levels, and E = E0 −Ah̄2 (7) for two levels, where E0 is the unperturbed energy (numerical value of E0 for hydrogen n = 2 levels is -13.6 eV / 4). III. Consider a s = 1/2 fermion in a box: the particle is confined to move in one dimension between x = 0 and x = L. The potential is zero in that region and infinite elsewhere. The one-particle energy eigenfunctions are ψn(x) = √ 2 L sin( nπx L ) (8) and the energy levels are En = π2n2h̄2 2mL2 . (9) (a) Consider the first-order energy shifts from a small sloping of the bottom of the box, i.e., a correction to the potential V ′(x) = α(x− L/2). (10) Which (if any) energy levels will shift under this perturbation at first order? (Note that every level is still two-fold degenerate from spin.) The first-order energy shift for orbital state n is∫ L 0 |ψn(x)|2V ′(x) dx. (11) Since ψn(x) is either odd or even around x = a/2, |ψn(x)|2 is even around x = a/2. The perturba- tion V ′(x) is odd around x = a/2. Hence all these integrals are zero and there are no energy shifts. The two-fold spin degeneracy is unaffected by the perturbation, which does not involve spin. (b) Write an expression for the energy shift of the (spin-degenerate) ground state to second order. Explain if any terms in the infinite sum must be zero by the symmetry of the original potential. You do not have to compute the nonzero terms in the energy shift. The second-order perturbation theory result for the ground state energy is E (2) 1 = ∑ j 6=1 |H ′j1|2 E (0) 1 − E (0) j . (12) 2