Introduction to Quantum Mechanics · Problem 8.25

The Variational Principle — David J. Griffiths

Physics — Quantum Mechanics · variational_hydrogen_ion

Problem

In Problem 8.8 we found that the trial wave function with shielding (Equation 8.28), which worked well for helium, is inadequate to confirm the existence of a bound state for the negative hydrogen ion. Chandrasekhar used a trial wave function of the form ψ(r1,r2)≡A[ψ1(r1)ψ2(r2)+ψ2(r1)ψ1(r2)],(8.81)\psi(\mathbf{r}_1, \mathbf{r}_2) \equiv A\left[\psi_1(r_1)\psi_2(r_2) + \psi_2(r_1)\psi_1(r_2)\right], \qquad (8.81) where ψ1(r)≡Z13πa3e−Z1r/a,andψ2(r)≡Z23πa3e−Z2r/a.(8.82)\psi_1(r) \equiv \sqrt{\frac{Z_1^3}{\pi a^3}}e^{-Z_1 r/a}, \quad \text{and} \quad \psi_2(r) \equiv \sqrt{\frac{Z_2^3}{\pi a^3}}e^{-Z_2 r/a}. \qquad (8.82) In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetrized with respect to interchange. The spin state—which is irrelevant to the calculation—is evidently antisymmetric.) Show that by astute choice of the adjustable parameters Z1Z_1 and Z2Z_2 you can get ⟨H⟩\langle H\rangle less than −13.6-13.6 eV. *Answer:* ⟨H⟩=E1x6+y6(−x8+2x7+12x6y2−12x5y2−18x3y4+118xy6−12y8),\langle H\rangle = \frac{E_1}{x^6 + y^6}\left(-x^8 + 2x^7 + \frac{1}{2}x^6 y^2 - \frac{1}{2}x^5 y^2 - \frac{1}{8}x^3 y^4 + \frac{11}{8}xy^6 - \frac{1}{2}y^8\right), where x≡Z1+Z2x \equiv Z_1 + Z_2 and y≡2Z1Z2y \equiv 2\sqrt{Z_1 Z_2}. Chandrasekhar used Z1=1.039Z_1 = 1.039 (since this is larger than 1, the motivating interpretation as an effective nuclear charge cannot be sustained, but never mind—it's still an acceptable trial wave function) and Z2=0.283Z_2 = 0.283.

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