# 8.2: Pertubation Theory and the Variational Method for Helium

Both perturbation theory and variation method (especially the linear variational method) provide good results in approximating the energy and wavefunctions of multi-electron atoms.Below we address both approximations with respect to the helium atom.

### Perturbation Theory of the Helium Atom

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, $$\hat {H} ^0$$, the zero order term, and the correction term $$\hat {H} ^1$$, the first order term. In the general expression below, there can be an infinite number of correction terms of increasingly higher order,

$\hat {H} = \hat {H} ^0 + \hat {H} ^1 + \hat {H} ^2 + \cdots \label {9-17}$

but usually it is not necessary to have more terms than $$\hat {H} ^0$$ and $$\hat {H} ^1$$. For the helium atom,

$\hat {H} ^0 = -\frac {\hbar ^2}{2m} \nabla ^2_1 - \frac {2e^2}{4 \pi \epsilon _0 r_1} - \frac {\hbar ^2}{2m} \nabla ^2_2 - \frac {2e^2}{4 \pi \epsilon _0 r_2} \label {9-18}$

$\hat {H} ^1 = \frac {2e^2}{4 \pi \epsilon _0 r_{12}} \label {9-19}$

The expression for the first-order correction to the energy is

$E^1 = \int \psi ^{0*} \hat {H} ^1 \psi ^0 d\tau \label {9-28}$

Equation $$\ref{9-28}$$ is a general expression for the first-order perturbation energy, which provides an improvement or correction to the zero-order energy we already obtained. For the helium atom, the integral in Equation $$\ref{9-28}$$ is

$E^1 = \int \int \varphi _{1s} (r_1) \varphi _{1s} (r_2) \frac {1}{r_{12}} \varphi _{1s} (r_1) \varphi _{1s} (r_2) d\tau _1 d\tau _2 \label {9-29}$

where the double integration symbol represents integration over all the spherical polar coordinates of both electrons $$r_1, \theta _1, \varphi _1 , r_2 , \theta _2 , \varphi _2$$. The evaluation of these six integrals is lengthy. When the integrals are done, the result is $$E^1$$ = +34.0 eV so that the total energy calculated using our second approximation method, first-order perturbation theory, is

$E_{appr ox2} = E^0 + E^1 = - 74.8 eV \label {9-30}$

$$E^1$$ is the average interaction energy of the two electrons calculated using wavefunctions that assume there is no interaction.

The new approximate value for the binding energy represents a substantial (~30%) improvement over the zero-order energy, 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 = -79.2 eV. So with two corrections to the energy, the calculated result is within 0.3% of the experimental value of -79.00 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 because we are left with the original zero-order wavefunctions. In the next section we will employ an approximation that modifies zero-order wavefunctions in order to address one of the ways that electrons are expected to interact with each other.The Variational Method of the Helium Atom

### Variational Method Applied to the Helium Method

Let's exame five trial wavefunctions for the helium atom within the variational method. The calculations begin with an uncorrelated wavefunction in which both electrons are placed in a hydrogenic orbital with scale factor $$\alpha$$. The next four trial functions use several methods to increase the amount of electron correlation in the wave function. As the summary of results that is appended shows this gives increasingly more favorable agreement with the experimentally determined value for the ground state energy of the species under study. The detailed calculations show that the reason for this improved agreement with experiment is due to a reduction in electron-electron repulsion.

Because of the electron-electron interaction 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 calculated using approximate methods. One of these is the variation method which requires the minimizing of the following variational integral.

$E = \dfrac{\int_0^{\infty} \Psi_{trial}^* \hat{H} \Psi_{trial} d\tau}{ \int_0^{\infty} \Psi_{trial}^2 d\tau} \label{7.3.1a}$

or via Dirac's Bra-Ket notion

$E = \dfrac{\langle \Psi_{trial}| \hat{H} | \Psi_{trial} \rangle }{\langle \Psi_{trial}| \Psi_{trial} \rangle}\label{7.3.1b}$

Five calculations that have been outlined below for the helium atom can be repeated for H-, Li+, Be2+, etc. The hydride anion is a particularly interesting case because the first two trial wavefunctions do not predict a stable ion. This indicates that electron correlation is an especially important issue for atoms and ions with small nuclear charge.

#### First Trial Wavefunction

As is clear from Equation $$\ref{7.3.1b}$$, the variational method approximation requires that a trial wavefunction with one or more adjustable parameters be chosen. A logical first choice for such a function would be to assume that the electrons in the helium atom occupy two identical, but scaled, hydrogen 1s orbitals.

$\Psi (1,2) = \Phi (1) \Phi (2) = \exp\left[- \alpha (r_1 +r_2)\right] \label{7.3.2}$

Results for the Helium Atom (Z=2)

$|\Psi \rangle_{trial} = \dfrac{\alpha^3}{\pi} \exp(-\alpha r_1) \exp(- \alpha r_2 )$

The variational method energy obtained after minimizing Equation $$\ref{7.3.1b}$$ and optimizing for $$\alpha$$ is

$E = -2.84766 \;E_h$

and the experimentally determined ground state energy is

$E_{\exp}= -2.90372 \;E_h$

The deviation of energy for the optimized trial wavefunction from the experimental value is

$\left| \dfrac{E(\alpha)-E_{\exp}}{E_{\exp}} \right| = \left| \dfrac{-2.84766 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| = 1.93 \%$

The value of -2.8477 hartrees is within 2% of the known ground state energy of the helium atom. The error in the calculation is attributed to the fact that the wavefunction is based on the orbital approximation and, therefore, does not adequately take electron correlation into account. In other words, this wavefunction gives the electrons too much independence, given that they have like charges and tend to avoid one another.

#### Second Trial Wavefunction

Some electron correlation can be built into the wavefunction by assuming that each electron is in an orbital which is a linear combination of two different and scaled hydrogen 1s orbitals.

$\Phi = \exp(- \alpha r_1) + \exp(- \beta r_2) \label{7.3.3}$

Under the orbital approximation this assumption gives a trial wavefunction of the form

$\Psi (1,2)= \Phi (1) \Phi (2) \label{7.3.4a}$

$= {\exp(- \alpha r_1 )\exp(- \alpha r_2)}+\exp(- \alpha r_1 )\exp(- \beta r_2)+\exp(- \beta r_1 )\exp(- \alpha r_2 )+\exp(- \beta r_1 )\exp(- \beta r_2 ) \label{7.3.4b}$

Inspection of this trial wavefunction indicates that 50% of the time the electrons are in different orbitals, while for the first trial wavefunction the electrons were in the same orbital 100% of the time. Notice the enormous increase in the complexity of the variational expression for the energy for this trial wavefunction (Equation $$\ref{7.3.1}$$). However, the calculation is very similar to that using the previous trial wavefunction. The differences are that in this case the expression for the energy is more complex and that it is being minimized simultaneously with respect to two parameters ($$\alpha$$ and $$\beta$$) rather than just one ($$\alpha$$).

Results for the Helium Atom (Z=2)

$|\Psi \rangle_{trial} = \exp (-\alpha r_1) \exp (-\alpha r_2) + \exp (\alpha r_1) \exp (-\beta r_2) + \exp (-\beta r_1) \exp (-\alpha r_2) + \exp (-\beta r_1) \exp (-\beta r_2)$

The variational method energy obtained after minimizing equation $$\ref{7.3.1a}$$ and optimizing for $$\alpha$$ and $$\beta$$ is

$E =-2.86035 \;E_h$

Deviation from experimental value:

$\left| \dfrac{E(\alpha)-E_{\exp}}{E_{\exp}} \right| = \left| \dfrac{-2.86035 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| = 1.49 \%$

Clearly introducing some electron correlation into the trial wavefunction has improved the agreement between theory and experiment.

#### Third Trial Wavefunction

The extent of electron correlation can be increased further by eliminating the first and last term in the second wavefunction (Equation $$\ref{7.3.4b}$$). This yields a wavefunction of the form,

$\Psi (1,2) = \exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \label{7.3.5}$

This trial wavefunction places the electrons in different scaled hydrogen 1s orbitals 100% of the time this adds further improvement in the agreement with the literature value of the ground state energy is obtained. This result is within 1% of the actual ground state energy of the helium atom.

Results for the Helium Atom (Z=2)

$|\Psi \rangle_{trial} = \exp (-\alpha r_1 ) \exp (-\beta r_2) + \exp (-\beta r_1 ) \exp (-\alpha r_2)$

The variational method energy obtained after minimizing Equation $$\ref{7.3.1a}$$ and optimizing for $$\alpha$$ and $$\beta$$ is

$E = -2.87566 \;E_h$

Deviation from experimental value:

$\left| \dfrac{E(\alpha)-E_{\exp}}{E_{\exp}} \right| = \left| \dfrac{-2.87566 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| = 0.97 \%$

#### Fourth Trial Wavefunction

The third trial wavefunction, however, still rests on the orbital approximation and, therefore, does not treat electron correlation adequately. Hylleraas took the calculation a step further by introducing electron correlation directly into the first trial wavefunction by adding a term, $$r_{12}$$, involving the inter-electron separation.

$\Psi(1,2) = \left(\exp[- \alpha ( r_1 + r_2 )]\right) \left(1 + \beta r_{12} \right) \label{7.3.6}$

In the trial wavefunction shown above, if the electrons are far apart, then $$r_{12}$$ is large then the magnitude of the wave function increases favors that configuration. This modification of the trial wavefunction has further improved the agreement between theory and experiment to within 0.5%.

Results for the Helium Atom (Z=2)

$|\Psi \rangle_{trial} = [\exp(- \alpha (r_1 +r_2) ] (1 + \beta r_{12})$

The variational method energy obtained after minimizing equation $$\ref{7.3.1a}$$ and optimizing for $$\alpha$$ and $$\beta$$ is

$E = - 2.89112\; E_h$

Deviation from experimental value:

$\left| \dfrac{E(\alpha)-E_{\exp}}{E_{\exp}} \right| = \left| \dfrac{- 2.89112 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| = 0.43 \%$

#### Fifth Trial Wavefunction

Chandrasakar brought about further improvement by adding Hylleraas's $$r_{12}$$ term to the third trial wave function (Equation $$\ref{7.3.5}$$) as shown here.

$\Psi (1,2) = \left[\exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \right][1 + \gamma r _{12} ] \label{7.3.7}$

Chandrasakar's three parameter wavefunction gives rise to a fairly complicated variational expression for ground state energy. However it also gives a result for helium that is within 0.07% of the experimental value for the ground state energy.

Results for the Helium Atom (Z=2)

$|\Psi \rangle_{trial} =[\exp (- \alpha r_1 ) \exp (-\beta r_2) + \exp (-\beta r_1) \exp (-\alpha r_2 )] (1 + \gamma r_{12})$

The variational method energy obtained after minimizing Equation $$\ref{7.3.1a}$$ and optimizing for $$\alpha$$, $$\beta$$ and $$\gamma$$ is

$E = -2.90143 \;E_h$

Deviation from experimental value is

$\left| \dfrac{E(\alpha)-E_{\exp}}{E_{\exp}} \right| = \left| \dfrac{E(\alpha)--2.90372 \;E_h}{-2.90372 \;E_h} \right| = 0.0789 \%$