Vol 3 - Landau, Lifshitz - Quantum mechanics - non-relativistic theory (3ed., 1991) · Problem 5

Angular Momentum

Physics — Landau & Lifshitz · transmission-coefficient-low-energy-limit

Problem

Determine how the transmission coefficient tends to zero as E→0E \to 0, assuming that the potential energy U(x)U(x) decreases rapidly at distances ∣x∣≥a|x| \geq a, where aa is the dimension of the interaction region. SOLUTION. For distances k∣x∣≪1k|x| \ll 1, EE can be neglected in Schrödinger's equation. If also ∣x∣≫a|x| \gg a, the potential energy can also be neglected, and the equation becomes −(ℏ2/2m) d2ψ/dx2=0-(\hbar^2/2m)\,\mathrm{d}^2\psi/\mathrm{d}x^2 = 0; the solution of this may be written as ψ=a1+b1xfor x<0,ψ=a2+b2xfor x>0.(1)\psi = a_1+b_1 x \quad \text{for } x < 0, \qquad \psi = a_2+b_2 x \quad \text{for } x > 0. \tag{1} The relation between a1a_1, b1b_1 and a2a_2, b2b_2 can be found by solving the equation at distances ∣x∣∼a|x| \sim a. It is linear: a1=ρa2+μb2,b1=νa2+τb2.(2)a_1 = \rho a_2+\mu b_2, \qquad b_1 = \nu a_2+\tau b_2. \tag{2} The coefficients ρ\rho, μ\mu, ν\nu and τ\tau are real and independent of the energy, which does not appear in the equation.† The solution (1) must be the same as the first two terms in the expansion of (25.1) and (25.2) in powers of kk, so that a1=1+B,b1=ik1(1−B),a2=A,b2=ik1A.a_1 = 1+B, \quad b_1 = ik_1(1-B), \quad a_2 = A, \quad b_2 = ik_1 A. Substituting these in (2) and solving for AA, we get, for small kk, A≈2ik/νA \approx 2ik/\nu, whence D≈4k2/ν2∼ED \approx 4k^2/\nu^2 \sim E. The transmission coefficient thus tends to zero in proportion to the particle energy. This is of course true for the examples in Problems 2 and 4. † Since the flux is constant, ρτ−μν=1\rho\tau - \mu\nu = 1.

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