The quantum harmonic oscillator with potential leads

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Ordinary differential equations · Special functions and orthogonal polynomials

Conditions of Validity, Ch Series solutions, singular points and where the special functions come from

Problem

The quantum harmonic oscillator with potential V(x)=12mω2x2V(x)=\tfrac12 m\omega^2x^2 leads, after the substitution ξ=mω/ℏ x\xi=\sqrt{m\omega/\hbar}\,x and ψ(x)=e−ξ2/2h(ξ)\psi(x)=e^{-\xi^2/2}h(\xi), to Hermite's equation h′′−2ξh′+(2ε−1)h=0h''-2\xi h' + (2\varepsilon-1)h=0, where ε=E/(ℏω)\varepsilon=E/(\hbar\omega). The coefficients of the power series h(ξ)=∑kakξkh(\xi)=\sum_k a_k\xi^k about ξ=0\xi=0 obey ak+2=ak 2k+1−2ε(k+1)(k+2),a_{k+2}=a_k\,\frac{2k+1-2\varepsilon}{(k+1)(k+2)}, splitting into an even series (a0a_0) and an odd series (a1a_1). ((a)) Show that if neither series terminates, h(ξ)h(\xi) grows like eξ2e^{\xi^2} as ∣ξ∣→∞|\xi|\to\infty, so that ψ=e−ξ2/2h\psi=e^{-\xi^2/2}h is unbounded and cannot be normalised. ((b)) Show that termination of a series at index k=nk=n requires ε=n+12\varepsilon=n+\tfrac12, and hence that the allowed energies are En=(n+12)ℏωE_n=(n+\tfrac12)\hbar\omega, n=0,1,2,…n=0,1,2,\ldots. ((c)) An oscillator has angular frequency ω=4.00×1014 rad s−1\omega=4.00\times10^{14}\ \mathrm{rad\,s^{-1}}. Using the reduced Planck constant ℏ=1.0546×10−34 J s\hbar=1.0546\times10^{-34}\ \mathrm{J\,s}, find the energy of the n=3n=3 state in electronvolts (1 eV=1.6022×10−19 J1\ \mathrm{eV}=1.6022\times10^{-19}\ \mathrm{J}), and state the number of nodes of its wavefunction, given that a Hermite polynomial of degree nn has exactly nn real zeros.

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