Introduction to Quantum Mechanics · Problem 8.24

The Variational Principle — David J. Griffiths

Physics — Quantum Mechanics · rubber_band_helium

Problem

Although the Schrödinger equation for helium itself cannot be solved exactly, there exist "helium-like" systems that do admit exact solutions. A simple example is "rubber-band helium," in which the Coulomb forces are replaced by Hooke's law forces: H=−ℏ22m(∇12+∇22)+12mω2(r12+r22)−λ4mω2∣r1−r2∣2.(8.78)H = -\frac{\hbar^2}{2m}\left(\nabla_1^2 + \nabla_2^2\right) + \frac{1}{2}m\omega^2\left(r_1^2 + r_2^2\right) - \frac{\lambda}{4}m\omega^2|\mathbf{r}_1 - \mathbf{r}_2|^2. \qquad (8.78) (a) Show that the change of variables from r1\mathbf{r}_1, r2\mathbf{r}_2, to u≡12(r1+r2),v≡12(r1−r2),(8.79)\mathbf{u} \equiv \frac{1}{\sqrt{2}}(\mathbf{r}_1 + \mathbf{r}_2), \quad \mathbf{v} \equiv \frac{1}{\sqrt{2}}(\mathbf{r}_1 - \mathbf{r}_2), \qquad (8.79) turns the Hamiltonian into two independent three-dimensional harmonic oscillators: H=[−ℏ22m∇u2+12mω2u2]+[−ℏ22m∇v2+12(1−λ)mω2v2].(8.80)H = \left[-\frac{\hbar^2}{2m}\nabla_u^2 + \frac{1}{2}m\omega^2 u^2\right] + \left[-\frac{\hbar^2}{2m}\nabla_v^2 + \frac{1}{2}(1-\lambda)m\omega^2 v^2\right]. \qquad (8.80) (b) What is the exact ground state energy for this system? (c) If we didn't know the exact solution, we might be inclined to apply the method of Section 8.2 to the Hamiltonian in its original form (Equation 8.78). Do so (but don't bother with shielding). How does your result compare with the exact answer? *Answer:* ⟨H⟩=3ℏω(1−λ/4)\langle H\rangle = 3\hbar\omega(1 - \lambda/4).

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