Download Variational Method - Quantum Mechanics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Physics 137B: Midterm 1 Solutions September 27, 2011 Problem 1: Variational Method (5 points) Use a Gaussian trial function to obtain the lowest upper bound you can on the ground state energy of the linear potential V (x) = α|x|. Hint: The following integrals may be useful:∫ ∞ 0 dxe−yx 2 x2 = √ π 4 y− 3 2∫ ∞ 0 dxe−yx 2 = √ π 2 √ y (1) Solution 〈V 〉 = 2αA2 ∫ ∞ 0 xe−2bx 2 dx = 2αA2 ( − 1 4b e−2bx 2 )∞ 0 = αA2 2b (2) The normalization can be computed 1 = A2 ∫ ∞ −∞ e−2bx 2 dx = A2 √ π 2b (3) Therefore, A2 = √ 2b π , and so 〈V 〉 = α√ 2bπ . The expectation value of the double derivative 〈 ∂ 2 ∂x2 〉 = A2 ∫ ∞ −∞ e−bx 2 (4b2x2−2b)e−bx2dx = A24b22 √ π 4 (2b)−3/2−2b = √ 2b π 2b2 √ π(2b)−3/2−2b = −b (4) Therefore, the Hamiltonian expectation value reads 〈H〉 = ~ 2b 2m + α√ 2bπ . (5) Minimizing with respect to b, we obtain ∂H ∂b = ~2 2m − 1 2 α√ 2π b−3/2 → b3/2 = α√ 2π m ~2 ; b = ( αm√ 2π~2 )2/3 (6) Hmin = ~2 2m ( αm√ 2π~2 )2/3 + α √ 2π ( αm√ 2π~2 )−1/3 = 32 ( α2~2 2πm )1/3 (7) Problem 2: WKB Approximation (5 points) For spherically symmetrical potentials we can apply the WKB approximation to the radial part. In the case l = 0 the answer is∫ r0 0 p(r)dr = (n− 1/4)π~, (8) 1 where r0 is the turning point (in effect, we treat r = 0 as an infinite wall). Exploit this formula to estimate the allowed energies of a particle in the logarithmic potential V (r) = V0 ln(r/a) (9) (for constants V0 and a). Treat only the case l = 0. Show that the spacing between the levels is independent of mass. Hint: The following integral may be useful:∫ r0 0 √ log r0 r dr = r0 √ π 2 (10) Solution E = V0 ln r0 a defines r0. The quantization condition reads (n− 1/4)π~ = ∫ r0 0 √ 2m(V0 ln r0 a − V0 ln r/a)dr = √ 2mV0 ∫ r0 0 √ ln(r0/r) = √ 2mV0r0 √ π 2 (11) Solving for r0, we obtain r0 = (n− 1/4)~ √ 2π mV0 (12) Therefore, the energy levels are En = V0 ln[ (n− 1/4)~ a √ 2π mV0 ] = V0 ln(n− 1/4) + V0 ln[ ~ a √ 2π mV0 ] (13) The spacing is given by Em − En = V0 ln m− 1/4 n− 1/4 (14) which is independent of mass. Problem 3: Rotating Rigid Body (7 points) Consider an electrically neutral body that can rotate about a fixed axis, with moment of inertia I. The Hamiltonian for the system is H0 = L2 2I , (15) where L is the angular momentum. (a) What are the possible energy values and what is the degeneracy for each energy eigenvalue? (b) Now suppose that an electron is embedded in the rotating body at some fixed posi- tion off the axis, and an equal positive charge is placed on the axis. Correspondingly, we add a new term to our Hamiltonian, of the form H ′ = λS · L where λ is small and S is the electron spin operator. Use first order perturbation theory to determine the energies and degeneracies of the l = 0 and l = 1 eigenstates. 2
