Last edited: 2021-03-23 17:35:14

**Calculating the energy of the stationary state of a two electron system, like a helium atom or lithium ion, gets quiet a bit more tricky than for a single electron system. In fact, calculating the Schrödinger equation is impossible. Luckily we have methods like perturbation theory and the variational method to estimate the energy.**

First we need to establish the Hamiltonian for the system which is:

$\hat{H} = -\frac{\hbar}{2 m_e} \nabla^2_1 - \frac{\hbar}{2 m_e} \nabla^2_2 - \frac{Ze^2}{4 \pi \epsilon_0 r_1 } - \frac{Ze^2}{4 \pi \epsilon_0 r_2 } + \frac{e^2}{4 \pi \epsilon_0 | \vec{r}_1 - \vec{r}_2 | }.$$\frac{\hbar}{2 m_e} \nabla^2_1$ is the kinetic energy for the first electron, same goes for the second. $\frac{Ze^2}{4 \pi \epsilon_0 r_1 }$ is Coulomb's law between the atom core's charge $Ze$ and the first electron's charge $e$. Lastly we have $\frac{e^2}{4 \pi \epsilon_0 | \vec{r}_1 - \vec{r}_2 | }$ which is Coulomb's law between the two electrons, this term is positive because the electrons repel each other.

The first estimation we'll do is having the two electrons not react with each other which means that the Coulomb's law term between them is zero. Then the combined wave function of the system is the following:

$\psi_+ (\vec{r}_1, \vec{r}_2) = \psi_{100}(r_1) \psi_{100}(r_2),$where

$\psi_{100}(r) = \frac{Z^{3/2}}{\sqrt{\pi} a_0^{3/2}} e^{Zr/a_0}$is the wave function of a single electron system, $a_0$ is the Bohr radius. So our wave function of the two electron system is

$\psi_+ (r_1, r_2) = \frac{Z^{3}}{\sqrt{\pi} a_0^{3}} e^{Z(r_1+r_2)/a_0},$which leads the stationary state energy of:

$E = 2Z^2 \cdot (-13.6 \text{ eV}).$But that is not so satisfactory nor a good estimate, which leads to our second estimation method: perturbation theory. Lets use the first order correction to estimate the impact that Coulomb's law between the two electrons has on the energy:

$E^{(1)} = \langle \psi_+ | \frac{e^2}{4 \pi \epsilon_0} \frac{1}{|\vec{r}_1 - \vec{r}_2|} | \psi_+ \rangle = \frac{5}{4} Z \cdot 13.6 \text{ eV}.$So now our estimated stationary state energy is:

$E = 2Z^2 \cdot (-13.6 \text{ eV}) + \frac{5}{4} Z \cdot 13.6 \text{ eV}.$Lets try getting an even better estimation with the variational method. If we change the atom number $Z$ to a parameter $\alpha$ we get:

$\psi_0 (r_1, r_2) = \frac{\alpha^{3}}{\sqrt{\pi} a_0^{3}} e^{\alpha (r_1+r_2)/a_0}.$We now use this wave function as the base to calculate the expectation value of the Hamiltonian:

$E(\alpha) = \langle \psi_0 (r_1, r_2) | \hat{H} | \psi_0 (r_1, r_2) \rangle = (-2 \alpha^2 + 4 Z \alpha - \frac{5}{4} \alpha) \cdot (-13.6 \text{ eV}).$Here we can see that if we choose $\alpha = Z$ we get the same result as for the first order correction perturbation theory. The variational method says $E_\text{ground} \leq E(\alpha)$ so lets check where the minimum of $E(\alpha)$ is:

$\frac{\partial E(\alpha)}{\partial \alpha} = (-4 \alpha + 4Z - \frac{5}{4}) = 0 \Longrightarrow \alpha = Z - \frac{5}{16}.$So our estimated stationary state energy with the variational method is

$E_\text{ground} = (Z-\frac{5}{16})^2 \cdot (-13.6 \text{ eV}).$For example, the estimated stationary state energy with the variational method for helium ($Z=2$) is $-77.5$ eV which is close to the experimental value of $-78.975$ eV, in other words this method of estimation works very well! You now know a couple of methods to estimate the stationary state energy for electrons!

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