Energy of a Two Electron System

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:

H^=2me122me22Ze24πϵ0r1Ze24πϵ0r2+e24πϵ0r1r2.\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 | }.

2me12\frac{\hbar}{2 m_e} \nabla^2_1 is the kinetic energy for the first electron, same goes for the second. Ze24πϵ0r1\frac{Ze^2}{4 \pi \epsilon_0 r_1 } is Coulomb's law between the atom core's charge ZeZe and the first electron's charge ee. Lastly we have e24πϵ0r1r2\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.

Two Independent Electrons Estimation

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:

ψ+(r1,r2)=ψ100(r1)ψ100(r2),\psi_+ (\vec{r}_1, \vec{r}_2) = \psi_{100}(r_1) \psi_{100}(r_2),


ψ100(r)=Z3/2πa03/2eZr/a0\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, a0a_0 is the Bohr radius. So our wave function of the two electron system is

ψ+(r1,r2)=Z3πa03eZ(r1+r2)/a0,\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=2Z2(13.6 eV).E = 2Z^2 \cdot (-13.6 \text{ eV}).

Perturbation Theory Estimation

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)=ψ+e24πϵ01r1r2ψ+=54Z13.6 eV.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=2Z2(13.6 eV)+54Z13.6 eV.E = 2Z^2 \cdot (-13.6 \text{ eV}) + \frac{5}{4} Z \cdot 13.6 \text{ eV}.

Variational Method Estimation

Lets try getting an even better estimation with the variational method. If we change the atom number ZZ to a parameter α\alpha we get:

ψ0(r1,r2)=α3πa03eα(r1+r2)/a0.\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(α)=ψ0(r1,r2)H^ψ0(r1,r2)=(2α2+4Zα54α)(13.6 eV).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 α=Z\alpha = Z we get the same result as for the first order correction perturbation theory. The variational method says EgroundE(α)E_\text{ground} \leq E(\alpha) so lets check where the minimum of E(α)E(\alpha) is:

E(α)α=(4α+4Z54)=0α=Z516.\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

Eground=(Z516)2(13.6 eV).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=2Z=2) is 77.5-77.5 eV which is close to the experimental value of 78.975-78.975 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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