Principles of Quantum Mechanics · Problem 12.3.7

Rotational Invariance and Angular Momentum — R. Shankar

Physics — Quantum Mechanics · isotropic-oscillator-2d

Problem

*(The Isotropic Oscillator).* Consider the Hamiltonian H=Px2+Py22μ+12μω2(X2+Y2)H = \frac{P_x^2 + P_y^2}{2\mu} + \frac{1}{2}\mu\omega^2(X^2 + Y^2) (1) Convince yourself [H,Lz]=0[H, L_z] = 0 and reduce the eigenvalue problem of HH to the radial differential equation for REm(ρ)R_{Em}(\rho). (2) Examine the equation as ρ→0\rho \to 0 and show that REm(ρ)⟶ρ→0ρ∣m∣R_{Em}(\rho) \underset{\rho \to 0}{\longrightarrow} \rho^{|m|} (3) Show likewise that up to powers of ρ\rho REm(ρ)⟶ρ→∞e−μωρ2/2ℏR_{Em}(\rho) \underset{\rho \to \infty}{\longrightarrow} e^{-\mu\omega\rho^2/2\hbar} So assume that REm(ρ)=ρ∣m∣e−μωρ2/2ℏUEm(ρ)R_{Em}(\rho) = \rho^{|m|} e^{-\mu\omega\rho^2/2\hbar} U_{Em}(\rho). (4) Switch to dimensionless variables ε=E/ℏω\varepsilon = E/\hbar\omega, y=(μω/ℏ)1/2ρy = (\mu\omega/\hbar)^{1/2}\rho. (5) Convert the equation for RR into an equation for UU. (I suggest proceeding in two stages: R=y∣m∣fR = y^{|m|}f, f=e−y2/2Uf = e^{-y^2/2}U.) You should end up with U′′+[(2∣m∣+1y)−2y]U′+(2ε−2∣m∣−2)U=0U'' + \left[\left(\frac{2|m|+1}{y}\right) - 2y\right]U' + (2\varepsilon - 2|m| - 2)U = 0 (6) Argue that a power series for UU of the form U(y)=∑r=0∞CryrU(y) = \sum_{r=0}^{\infty} C_r y^r will lead to a two-term recursion relation. (7) Find the relation between Cr+2C_{r+2} and CrC_r. Argue that the series must terminate at some finite rr if the y→∞y \to \infty behavior of the solution is to be acceptable. Show ε=r+∣m∣+1\varepsilon = r + |m| + 1 leads to termination after rr terms. Now argue that rr is necessarily even—i.e., r=2kr = 2k. (Show that if rr is odd, the behavior of RR as ρ→0\rho \to 0 is not ρ∣m∣\rho^{|m|}.) So finally you must end up with E=(2k+∣m∣+1)ℏω,k=0,1,2,…E = (2k + |m| + 1)\hbar\omega, \quad k = 0, 1, 2, \ldots Define n=2k+∣m∣n = 2k + |m|, so that En=(n+1)ℏωE_n = (n+1)\hbar\omega. (8) For a given nn, what are the allowed values of ∣m∣|m|? Given this information show that for a given nn, the degeneracy is n+1n+1. Compare this to what you found in Cartesian coordinates (Exercise 10.2.2). (9) Write down all the normalized eigenfunctions corresponding to n=0,1n = 0, 1. (10) Argue that the n=0n = 0 function must equal the corresponding one found in Cartesian coordinates. Show that the two n=2n = 2 solutions are linear combinations of their counterparts in Cartesian coordinates. Verify that the parity of the nn states is (−1)n(-1)^n as you found in Cartesian coordinates.

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