Introduction to Quantum Mechanics · Problem 5.15

Identical Particles — David J. Griffiths

Physics — Quantum Mechanics · helium_electron_repulsion

Problem

(a) Calculate ⟨1/∣r1−r2∣⟩\langle 1/|\mathbf{r}_1 - \mathbf{r}_2| \rangle for the state ψ0\psi_0 (Equation 5.41). Hint: Do the d3r2d^3\mathbf{r}_2 integral first, using spherical coordinates, and setting the polar axis along r1\mathbf{r}_1, so that ∣r1−r2∣=r12+r22−2r1r2cos⁡θ2.|\mathbf{r}_1 - \mathbf{r}_2| = \sqrt{r_1^2 + r_2^2 - 2r_1 r_2 \cos\theta_2}. The θ2\theta_2 integral is easy, but be careful to take the positive root. You'll have to break the r2r_2 integral into two pieces, one ranging from 0 to r1r_1, the other from r1r_1 to ∞\infty. Answer: 5/4a5/4a. (b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it to E0E_0 (Equation 5.42) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we're still working with an approximate wave function, so don't expect perfect agreement.)
Intuitive explanation

The key to this problem is recognizing that the electron–electron repulsion energy ⟨1/∣r1−r2∣⟩\langle 1/|\mathbf{r}_1 - \mathbf{r}_2| \rangle is a two-body expectation value you can evaluate by integrating over one electron's coordinate while the other stays fixed. The hint guides you to exploit spherical symmetry: by aligning your coordinate system with r1\mathbf{r}_1, the angular integrals become tractable, but you must split the radial integral at r1r_1 because the distance formula changes character there. Once you have this result, part (b) simply scales it by the Coulomb constant to get the actual interaction energy, then adds it to the unperturbed ground state energy to see how much electron repulsion shifts the total.

Intuitive explanation

The key to this problem is recognizing that the electron–electron repulsion energy ⟨1/∣r1−r2∣⟩\langle 1/|\mathbf{r}_1 - \mathbf{r}_2| \rangle is a two-body expectation value you can evaluate by integrating over one electron's coordinate while the other stays fixed. The hint guides you to exploit spherical symmetry: by aligning your coordinate system with r1\mathbf{r}_1, the angular integrals become tractable, but you must split the radial integral at r1r_1 because the distance formula changes character there. Once you have this result, part (b) simply scales it by the Coulomb constant to get the actual interaction energy, then adds it to the unperturbed ground state energy to see how much electron repulsion shifts the total.

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