Problem Set 5 for Quantum Mechanics I - Fall 2007 | PHYS 580, Assignments of Quantum Mechanics

Download Problem Set 5 for Quantum Mechanics I - Fall 2007 | PHYS 580 and more Assignments Quantum Mechanics in PDF only on Docsity! PHYSICS 580 – FALL 2007 PROBLEM SET 5 – DUE: TUESDAY, OCT. 30 1. (LQM 4-6) Consider a potential step VI(x) =    0, x < a V (x), a < x < b ∞, x > b, where V (x) is a function that we do not know and a > 0. Suppose however, that we know the reflection amplitude N(E) = −e2iδ(E) at x = a. a) Show that δ(E) must be real. b) Consider a particle in the potential well defined by VII(x) =    ∞, x < 0 VI(x), x > 0. Show that to first order in δ(E) (assumed small), the energy levels of this well are given in terms of the energy levels of the square well VIII(x) =    ∞, x < 0 0, 0 < x < a ∞, x > a by E = E0 − p0h̄ ma δ(E0) where E0 is the energy level of the square well and p0 = √ 2mE0. c) Give a qualitative argument (at least) that δ(E) must be positive. 2. (LQM 4-7) a) Show that the energy levels of a double square well VS(x) =    ∞, | x |> b 0, a <| x | < b ∞, | x |< a are doubly degenerate. b) Now suppose that the barrier between −a and a is very high, but finite. Assume that the potential between −a and a is symmetric about the origin. There is now the possibility of tunneling from one well to the other, and this possibility has the effect of splitting the degeneracy of the double well in part (a). Let E0 be an energy level of the well in part (a), and assume that E0 is reasonably less than the barrier height. Assume that in the neighborhood of E0 the reflection amplitude of the barrier at −a is of the form −e2iδ(E) where δ(E) is real, positive and ¿ 1. (Can δ(E) be otherwise under these conditions?) Also assume S(E) to be of the form iJ(E) where J is small and positive. (Must the leading term in S(E as the barrier height becomes infinite be of this form?) Show that to lowest order in δ and J the well with the finite barrier has two levels E corresponding to each degenerate level of the double square well in (a), given by E± = E0 − p0h̄ m(b− a) [ δ(E0) ∓ J(E0) 2 ] . What are the parities of the split levels? c) If δ(E) is positive, as it generally must be, then one level of the double well with tunneling is lower in energy than E0. This effect provides a qualitative explanation of the “one electron bond” between two ions, e.g., the bond in H+2 . The ions when far apart, correspond to the double square well of part (a). When they are close together there is a possibility of an electron tunneling from one ion to the other. Due to the tunneling splitting the electron, and hence the two ions plus electron, can have a lower energy when the two ions are close together; hence they bind. Estimate the binding energy of two ions due to this mechanism. 3. (LQM 4-11) A particle of mass m is in a potential V (x) = vδ(x − a) + vδ(x + a) where v < 0. a) What is the wave function for a bound state with even parity? b) Find the expression that determines the bound state energies for even parity states, and deter- mine graphically how many even parity bound states there are. c) Solve for the even parity bound state energy analytically in the case that m|v|a/h̄2 ¿ 1. d) Repeat parts (a) and (b) for odd parity. For what values of |v| are there bound states? e) Find the even and odd parity state binding energies for a À h̄2/m|v|. Explain physically why these energies move closer and closer together as a →∞. 4. (LQM 4-12) The effective mass m∗ of an electron in a solid with energy E(k) is defined in one dimension by 1 m∗ = 1 h̄2 d2E dk2 . Calculate m∗, in the Kronig-Penney model with delta functions, at the bottom and at the top of the lowest energy band. Explain the physical significance of the results. 6. (LQM 4-16) [Basic structure of the Josephson effect] A box, containing a particle, is divided in two into a right and a left compartment by a thin partition. Suppose that the amplitude for the particle being on the left side of the box is ψ1 and the amplitude for it being on the right side of the box is ψ2. Neglect spatial variations of these amplitudes within the halves of the box. Suppose that the particle can tunnel through the partition and that the rate of change of the amplitude on the right is K/ih̄ times the amplitude on the left, where K is real: ih̄ ∂ψ2 ∂t = Kψ1. Assume in the absence of tunneling, i.e., an impermeable partition, that ∂ψ1/∂t = 0.