Calculate ground state energy of helium atom using variational method
Answers
We use perturbation theory to approach the analytically unsolvable helium atom Schrödinger equation by focusing on the Coulomb repulsion term that makes it different from the simplified Schrödinger equation that we have just solved analytically. The electron-electron repulsion term is conceptualized as a correction, or perturbation, to the Hamiltonian that can be solved exactly, which is called a zero-order Hamiltonian. The perturbation term corrects the previous Hamiltonian to make it fit the new problem. In this way the Hamiltonian is built as a sum of terms, and each term is given a name. For example, we call the simplified or starting Hamiltonian, H^0 , the zero order term, and the correction term H^1 .
H^=H^0+H^1(8.2.1)
The Hamilonian for the helium atom (in atomic units) is:
H^0H^1=−12∇21−2r1H atom Hamiltonian−12∇22−2r2H atom Hamiltonian=1r12=1|r1−r2|(8.2.2)(8.2.3)
The expression for the first-order correction to the energy is
E1=⟨ψ0|H^1|ψ0⟩=∫ψ0∗H^1ψ0dτ(8.2.4)
Equation 8.2.4 is a general expression for the first-order perturbation energy, which provides an improvement or correction to the zero-order energy we already obtained. Hence, E1 is the average interaction energy of the two electrons calculated using wavefunctions that assume there is no interaction.
The solution to H^0 (Equation 8.2.2 ) is the product of two single-electron hydrogen wavefunctions (scaled by the increased nuclear charge) since H^0 can be separated into independent functions of each electron (i.e., Separation of Variables).
|ψ0⟩=|φ1s(r1)φ1s(r2)⟩(8.2.5)
So the integral in Equation 8.2.4 is
E1=∬φ1s(r1)φ1s(r2)1r12φ1s(r1)φ1s(r2)dτ1dτ2(8.2.6)
where the double integration symbol represents integration over all the spherical polar coordinates of both electrons r1,θ1,φ1,r2,θ2,φ2 . The evaluation of these six integrals is lengthy. When the integrals are done, the result is E1 = +34.0 eV so that the total energy calculated using our second approximation method, first-order perturbation theory, is
Eapprox2=E0+E1=−74.8eV(8.2.7)
The new approximate value for the binding energy represents a substantial (~30%) improvement over the zero-order energy:
E0=2n2+2n2=4Ehhartrees=108.8eV(8.2.8)
so the interaction of the two electrons is an important part of the total energy of the helium atom. We can continue with perturbation theory and find the additional corrections, E2 , E3 , etc. For example,
E0+E1+E2=−79.2eV.(8.2.9)
So with two corrections to the energy, the calculated result is within 0.3% of the experimental value of -79.01 eV. It takes thirteenth-order perturbation theory (adding E1 through E13 to E0 ) to compute an energy for helium that agrees with experiment to within the experimental uncertainty. Interestingly, while we have improved the calculated energy so that it is much closer to the experimental value, we learn nothing new about the helium atom wavefunction by applying the first-order perturbation theory to the energy above. He need to expand the wavefunctions to first order perturbation theory, which requires more effort. Below, we will employ the variational method approximation to modify zero-order wavefunctions to address one of the ways that electrons are expected to interact with each other.
THE HARTREE UNIT OF ENERGY
The hartree is the atomic unit of energy (named after the British physicist Douglas Hartree) and is defined as
Eh=2RHhc
where RH is the Rydberg constant, h is the Planck constant and c is the speed of light.
Eh=4.359×10−18J=27.21eV.
The hartree is usually used as a unit of energy in atomic physics and computational chemistry. As discussed before for hydrogen emission, IR, and microwave spectroscopies, experimental measurements prefer the electronvolt ( eV ) or the wavenumber ( cm−1 ).
Variational Method Applied to the Helium Method
As discussed in Section 6.7, because of the electron-electron interactions, the Schrödinger's Equation cannot be solved exactly for the helium atom or more complicated atomic or ionic species. However, the ground-state energy of the helium atom can be estimated using approximate methods. One of these is the variational method which requires the minimizing of the following variational integral.
Etrial=⟨ψtrial|H^|ψtrial⟩⟨ψtrial|ψtrial⟩=∫∞0ψ∗trialH^ψtrialdτ∫∞0ψ2trialdτ(8.2.10)(8.2.11)
The five trial wavefunctions discussions below are equally "valid" trial wavefunctions that describe the probability of finding each electron (technically the wavefunction squared). What separates the "poor" approximations from the "good" approximation is whether the trial wavefunction predicts experimental results. Consequently, for all the approximations used for the rest of this TextMap, it is important to compare the theoretical results to the "true" (i.e., experimental) results. No matter how complicated an approximation is, it is only as good as the accuracy of its predicted values to experimental values.