Midterm Exam 1 Solutions - Quantum Mechanics | PHYSICS 137A, Exams of Quantum Mechanics

Download Midterm Exam 1 Solutions - Quantum Mechanics | PHYSICS 137A and more Exams Quantum Mechanics in PDF only on Docsity! Physics 137A: Quantum Mechanics I, Spring 2007 Midterm I solutions by JEM. Exam date 2/22/07 1. a. The maximum probability density for ψ1 is at the center x = a/2, and for ψ2 is at a/4, 3a/4. The maxima of the two wavefunctions have the same height, 2/a. b. ψ2 is an eigenstate of total energy with energy E2 = 4π2h̄2 2ma2 = 2π2h̄2 ma2 . (1) Since it is an eigenstate, it has zero variance, and its expectation value is the same as the energy above. c. The time evolution is ψ(x, t) = 1√ 2 [ e−iE1t/h̄ψ1(x) + ie−iE2t/h̄ψ2(x) ] , (2) where E2 is above and E1 = E2/4. 2. a. The normalization integral is, using that the wavefunction is even about the origin, 1 = 2|A|2 ∫ a/2 0 (a/2− x)2 dx = 2|A|2 1 3 (a/2)3, ψ(x) = A(a/2− |x|). (3) so a normalized wavefunction is ψ(x) = 2 √ 3 a3/2 (a/2− |x|). (4) b. The inner product 〈ψ1|ψ〉 is 4 √ 6/π2 (for the details of the integral, see the solution to Griffiths problem 2.7 on problem set 2). The inner product 〈ψ2|ψ〉 vanishes because ψ is even and ψ2 is odd. c. The probability is (4 √ 6/π2)2 = 96/π4. d. This is somewhat tricky because, in the simplest way to write the expectation value, the only nonvanishing contribution is a singular part at the origin: 〈E〉 = ∫ a/2 −a/2 ψ∗(x) ( − h̄ 2 2m ∂2ψ ∂x2 ) ψ(x) dx. (5) The second derivative vanishes except at the origin. At the origin, the first derivative jumps by −2A, which means that the integral of the second derivative over a small interval around the origin gives −2A. Then 〈E〉 = |A|2(a/2)2h̄ 2 2m = 6h̄2 ma2 (6) (An alternate way is to first integrate by parts to move one derivative to ψ∗, which will give the same result.) We can see that this makes sense if we note that it is slightly larger than E1. e. The expectation value of energy is constant in any state, including this one. Expand in the basis