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I'm learning how to use variational method to calculate the energy of ground state helium. I understand how we can set up a trial wavefunction that is a product of two hydrogen-like wavefunctions; however, when finding the integral $\int_{a}^{b} \phi H \phi dr_1dr_2$ (for $\phi=\phi_1(r_1)\phi_2(r_2)$), what would the volume element be? Would it be $16\pi^2 r_1^2 r_2^2$? I'm assuming the denominator is one for normalized wavefunction of hydrogen-like atom.

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If you are working in spherical coordinates, and if the integrand has no dependence on the angular coordinates, then that volume element would be correct ($4\pi$ from each integration over $\theta_i \phi_i$ and $r_i^2$ for each spherical shell); however, the interelectronic potential $\frac{1}{r_{12}}$ depends not only on the radial coordinates, but also on the angular coordinates of each electron, so you can't just integrate those out.

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  • $\begingroup$ I see. So the $r^2$ terms would be correct, but the $4\pi$ term wouldn't be? $\endgroup$ – John Smith Nov 20 '16 at 6:24
  • $\begingroup$ The volume element $\mathrm{d}V$ (or commonly denoted $\mathrm{d}\tau$ in quantum chemistry) is equal to $r^2 \sin\theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$. This is the most general form and the one you should use. In the special case where your integrand is only dependent on $r$, then the multiple integral can be separated out, and the integrals over $\theta$ and $\phi$ work out to be equal to $4\pi$. $\endgroup$ – orthocresol Nov 21 '16 at 1:55

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