Download Appropriate Wave Functions - Quantum Mechanics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Physics 137A: Second Midterm Closed Book and Closed Notes: 50 Minutes 1) (20 pts) Review problem: A particle of mass m is confined in the half-infinite, half-finite square well of depth of V o = |V o| and width a: V (x) = ∞ x < 0 −V o 0 < x < a 0 a < x V=0 V=-Vo x=0 x=a a) (4 pts) Assuming a bound state (E <0), write down appropriate wave functions for the in- terior (0 < x < a) and exterior (a < x) regions, taking into account the behavior of the wave function at x = 0 and at x = ∞. Please denote the wave number for the region 0 < x < a and a < x by k and κ, respectively, defining these wave numbers in terms of E, m, h̄, and V o. Since the wave function must vanish at the origin and at infinity, the only possibilities are φ(x) = { A sin kx, 0 < x < a Be−κx x > a where κ = √ −2mE/h̄ and k = √ 2m(E + V0)/h̄. 1 b) (4 pts) By matching the interior and exterior wave functions and their derivatives at the boundary x = a, determine the wave function up to one overall normalization constant and determine the eigenvalue condition. A sin (ka) = B exp (−κa) → B = A sin (ka) exp (κa) Ak cos (ka) = −Bκ exp (−κa) = −Aκ sin (ka) → − k cot (ka) = κ so the wave function : φ(x) = { A sin kx, 0 < x < a A sin ka eκ(a−x) x > a c) (4 pts) By solving the eigenvalue equation for the case where there is only one bound state – which you should place at zero binding energy – determine the condition on the potential parameters (V o and a) that will guarantee that at least one bound state exists. κ→ 0 k → √ 2mV0/h̄ so cot ( a √ 2mV0 h̄ ) = 0 ⇒ √ 2mV0 a h̄ = π 2 for exactly one zero− energy bound state so ⇒ a2V0 > π2h̄2 8m for bound states to exist d) (4 pts) What is the relationship of this problem to the fully finite well problem of depth V o and width 2a, centered on the origin (no calculations necessary here)? The solutions of this problem are equivalent to the odd solutions of the square well problem: the eigenvalues are the same, and if the solution is merely extended as an odd function to negative −x, the wave functions would be the same. e) (no calculations needed here, either) (4 pts) If we had kept the infinite potential for x < 0, but used the potential V (x) = 1 2 mω2x2 for x > 0, what would be the resulting spectrum of allowed eigenvalues En? As in 4d), the solutions would satisfy the harmonic oscillator potential Hamiltonian for x > 0 and would vanish at x = 0. Thus the solutions would correspond to the odd solutions of the harmonic oscillator problem. The spectrum of all solutions is En = h̄ω(n+ 1/2). This the odd solutions would have the spectrum En = h̄ω(n + 1/2), n=1,3,5,... or equivalently En = h̄ω(2n+ 3/2), n=0,1,2,... 2
