Helium Atom

The helium atom is represented by the following diagram,

This diagram represents two electrons with charge $-e$, and a nucleus with charge +2.

The Hamiltonian of the Helium atom is,

$$\hat{H}= -\frac{\hbar^2}{2\mu} \nabla_{r_1}^2 - \frac{2e^2}{r_1} -\frac{\hbar^2}{2\mu} \nabla_{r_2}^2 - \frac{2e^2}{r_2} +\frac{e^2}{r_{12}}.$$

Note that the term $\frac{e^2}{r_{12}}$ couples two one-electron hydrogenlike Hamiltonians. In order to find a solution to the eigenvalue problem,

$$\hat{H}\psi=E\psi,$$

we implement an approximate method. We first solve the problem by neglecting the coupling term. Then we consider such term to be a small perturbation, and we correct the initially zeroth-order eigenfunctions and eigenvalues by using perturbation theory.

Neglecting the coupling term, the Hamiltonian becomes,

$$\hat{H}^{(0)}= -\frac{\hbar^2}{2\mu} \nabla_{r_1}^2 - -\frac{\hbar^2}{2\mu} \nabla_{r_2}^2 - \frac{2e^2}{r_1} - \frac{2e^2}{r_2},$$

the sum of two independent one-electron Hamiltonians. The eigenfunctions of such Hamiltonian are,

$$\psi=R_{nl}(r_1) P_l^m(\theta_1) \frac{1}{\sqrt{2\pi}} e^{im\phi_1} R_{nl}(r_2) P_l^m(\theta_2) \frac{1}{\sqrt{2\pi}} e^{im\phi_2},$$

and the eigenvalues are,

$$E_{n_1n_2}^{(0)} = - \frac{z^2\mu e^4}{2 \hbar^2 n_1^2}- \frac{z^2\mu e^4}{2 \hbar^2 n_2}.$$

Exercise 43: Prove that,

$$<\psi_{100}\vert\frac{e^2}{r_{12}}\vert \psi_{100}> = \frac{5}{8} e^2\frac{z}{a}.$$

In order to illustrate how to correct the zeroth order solutions by implementing perturbation theory, we compute the first order correction to the ground state energy as follows,

$$E=E_{11}^{(0)}+ <\psi_{100}\vert\frac{e^2}{r_{12}}\vert \psi_{100}> = - \frac{z^2\mu e^4}{\hbar^2} + \frac{5}{8} e^2\frac{z}{a}.$$

Alternatively, the variational method could be implemented to obtain better results with simple functions $\tilde{\psi}$, e.g., products of hydrogenlike orbitals with an effective nuclear charge $z'$:

$$\tilde{\psi}= A^2 e^{-\frac{z'}{a}(r_1+r_2)}.$$

According to the variational theorem, the expectation value $<\tilde{\psi}\vert\hat{H}\vert \tilde{\psi}>$ is always higher than the ground state energy. Therefore, the optimum coefficient $z'$ minimizes the expectation value,

$$\tilde{E}(z') =<\tilde{\psi}\vert\hat{H}\vert \tilde{\psi}>,$$   where

$$\hat{H}=-\frac{\hbar^2}{2\mu} \nabla_{r_1} -\frac{z'e^2}{r_1}-\fr... ...z'e^2}{r_2}- \frac{(2-z')e^2}{r_1} - \frac{(2-z')e^2}{r_2} +\frac{e^2}{r_{12}}.$$

Computing the expectation value of $\hat{H}$ analytically we obtain,

$$<\tilde{\psi}\vert\hat{H}\vert \tilde{\psi}>= -\frac{z'^2 e^2}{a}... ...nt dr_1 \int dr_2 \frac{e^{-\frac{2 z'}{a}(r_1+r_2)} r_2^2 e^2 r_1^2}{r_1-r_2},$$

$$f(z')=-\frac{z'^2 e^2}{a} - 2 z' \frac{(2-z')}{a} e^2 + \frac{5}{8} z' \frac{e^2}{a}.$$

Therefore, the optimum parameter $z'$ is obtained as follows,

$$\frac{\partial f}{\partial z'}=0, \rightarrow z'=2-\frac{5}{16}, ... ...) 2 \frac{e^2}{a} + \frac{5}{8} \left(2 - \frac{5}{16} \right)^2 \frac{e^2}{a}.$$