# Coulomb Integral in Helium Variational Approximation

$$\def\braket#1{\langle{#1}\rangle}$$I've been trying to get the ground state energy of the Helium atom using the trial wave function $$\psi(r_1,r_2)=Ne^{-\alpha (r_1^2+r_2^2)}$$ Where $$N$$ is the normalization constant and $$\alpha$$ is the variational parameter.

Here is my progress.

• Normalization:

Writing the trial function as $$\psi(r_1,r_2)=Ne^{-\alpha r_1^2}e^{-\alpha r_2^2}=N\phi(1)\phi(2)$$ We can get:

$$\braket{\psi|\psi}=(4\pi)^2\int_0^\infty{r_1^2e^{-\alpha r_1^2}dr_1}\int_0^\infty{r_2^2e^{-\alpha r_2^2}dr_2}=(\dfrac{\pi}{\alpha})^{3/2}$$ Where $$\int_0^\infty{r^2e^{-\alpha r^2}dr}=\dfrac{1}{4}\sqrt{\dfrac{\pi}{\alpha^3}}$$

so $$N=(\dfrac{\alpha}{\pi})^{3/2}$$

• Variational Method

Having the Helium hamiltonian

$$\braket{H}=\braket{h_1}+\braket{h_2}+\braket{H´}$$

we added and subtracted a term for making the trial wave function a solution for each electron:

$$\braket{h_1}=\braket{-\dfrac{\hbar^2}{2m}\nabla^2-\dfrac{k\alpha}{r_1}+\dfrac{k\alpha}{r_1}-\dfrac{Zk}{r_1}}$$ $$=\braket{-\dfrac{\hbar^2}{2m}\nabla^2-\dfrac{k\alpha}{r_1}}+\braket{\dfrac{k\alpha}{r_1}}-\braket{\dfrac{Zk}{r_1}}$$ $$=E_1^{He}(\alpha)+k(\alpha-Z)\braket{\dfrac{1}{r_1}}$$

Where the integral is calculated: $$\braket{\dfrac{1}{r_1}}=\dfrac{\alpha^3}{\pi^3}(4\pi)\int_0^\infty {\phi(1)^2 \dfrac{1}{r_1} r_1^2 dr_1}$$ $$=\dfrac{\alpha^3}{\pi^3}(4\pi)\int_0^\infty {e^{-2\alpha} r_1 dr_1}=\dfrac{\alpha^3}{\pi^3}(4\pi)\dfrac{1}{4\alpha} =\dfrac{\alpha^2}{\pi^2}$$ So: $$\braket{h_1}=E_1^{He}(\alpha)+k(\alpha-Z)\dfrac{\alpha^2}{\pi^2}$$ Similarly $$\braket{h_2}=E_1^{He}(\alpha)+k(\alpha-Z)\dfrac{\alpha^2}{\pi^2}$$

We now have to calculate the expectation value of the interaction of the electrons: $$\braket{H´}=\braket{\dfrac{k}{r_{12}}}$$

Where: $$\braket{\dfrac{1}{r_1}}=\int \int \dfrac{\psi_is^*(1,2) \psi_is^(1,2)}{r_{12}} d\tau_1 d\tau_2$$ $$=N^2\int \int \dfrac{e^{-2\alpha (r_1^2+r_2^2)}}{r_{12}} d\tau_1 d\tau_2$$

I've no idea how to evaluate it, I can´t even google it.