# Lecture 19: Variational Method, Effective Charge, and Matrix Representation

Recap of Lecture 18

Last Lecture addressed the angular moment of an electron revolving around the nucleus. This is described by the $$l$$ quantum number and an electron in any non-spherical orbital (i.e., an s orbital) will have an angular moment (you should know formula). The $$m_l$$ quantum number designates the orientation of that angular moment wrt the z-axis (and there is a formula for that too) and the degeneracy can be partial broken by magnetic fields. We discussed that we cannot always do a one-to-one correspondence between quantum numbers to orbitals. We discussed basic spectroscopy of H-like system and specifically selection rules (for n,l, and ml). We introduced the He system, discussed we cannot solve it and need to do approximations. We discussed the first approximation, which is the worst possible one.

## The "ignorance is bliss" Approximation (redux)

If we ignore the $$\dfrac {e^2}{4\pi\epsilon_0 r_{12}}$$ electron-electron repulsion term, then the electron 1 coordinates are separable from electron 2 coordinates, so that

$\Psi_{total} = \psi_{el_{1}}\psi_{el_{2}}$

or in braket notation

$| \Psi_{total} \rangle = | \psi_{el_1} \rangle | \psi_{el_2} \rangle$

With some operator algebra, something important arises - the one electron energies are additive:

$\hat{H} \Psi_{total} = (\hat{H}_{el_1} + \hat{H}_{el_2}) \psi_{n\ {el_1}} \psi_{n\ {el_2}} = (E_{n_1} + E_{n_2}) \psi_{n\ {el_1}} \psi_{n\ {el_2}}$

or in bra-ket notation

$\hat{H} | \Psi_{total} \rangle = \hat{H} | \psi_{el_1} \rangle | \psi_{el_2} \rangle = (E_{n_1} + E_{n_2}) | \psi_{1} \rangle | \psi_{1} \rangle$

The energy for a ground state Helium atom (both electrons in lowest state) is then

$E_{He_{1s}} = \underset{\text{energy of single electron in helium}}{E_{n_1}} + \underset{\text{energy of single electron in helium}}{E_{n_2}} = -R\left(\dfrac{Z^2}{1}\right) -R \left(\dfrac{Z^2}{1}\right) = -8R$

where $$R$$ is the Rydberg constant ($$13.6 \; eV$$ that also maps the lowest energy of the hydrogen atom and $$Z=2$$ for the helium nucleus. Experimentally, we find that the total energy for a ground state Helium atom $$E_{He_{1s}} = -5.8066\,R$$. The big difference in results is due to the approximation we used to get to the value of $$-8\;R$$. The "ignorance is bliss" approximation overestimates the energy of the helium atom greatly; this is a poor approximation and we need to address electron-electron repulsion properly (or better at least).

The general form of the "Ignorance is Bliss" approximation which almost as wrong as you can be for multi-electron atoms

$\underbrace{E_{z} = nZ^2E_{z=1}}_{\text{bad approximation}}$

## Electron-electron repulsion results in "Shielding"

One way to take electron-electron repulsion into account is to modify the form of the wavefunction. A logical modification is to change the nuclear charge, $$Z$$, in the wavefunctions to an effective nuclear charge, from +2 to a smaller value, $$Z_{eff}=\zeta$$. The rationale for making this modification is that one electron partially shields the nuclear charge from the other electron.

Figure: Electron-electron shielding leads to a reduced effective nuclear charge. The attractive force of the nucleus on electron 2,$$V(r_2)$$, is partially countered by the repulsive force between electron 1 and electron 2, $$V(r_{12})$$.

Electron-electron shielding leads to a reduced effective nuclear charge. $$Z_{eff}$$ is used instead of $$Z$$ in radial part of wavefunctions.

A region of negative charge density between one of the electrons and the +2 nucleus makes the potential energy between them more positive (decreases the attraction between them). We can effect this change mathematically by using $$Z_{eff} < 2$$ in the wavefunction expression. If the shielding were complete, then $$Z_{eff}$$ would equal 1. If there is no shielding, then $$Z_{eff} = 2$$. So a way to take into account the electron-electron interaction is by saying it produces a shielding effect. The shielding is not zero, and it isn't complete, so the effective nuclear charge is between one and two.

The lower the energy level, the higher the probability of finding the electron close to the nucleus. Also, the lower momentum quantum number gets, the closer it is to the nucleus.

The presence of $$Z_{eff}$$ in the radial portions of the wavefunctions also means that the electron probability distributions associated with hydrogenic atomic orbitals in multi-electron systems are different from the exact atomic orbitals for hydrogen. The Figure above compares the radial distribution functions for an electron in a 1s orbital of hydrogen (the ground state), a 2s orbital in hydrogen (an excited configuration of hydrogen) and a 1s orbital in helium that is described by the best variational value of $$Z_{eff}$$. Our use of hydrogen-like orbitals in quantum mechanical calculations for multi-electron atoms helps us to interpret our results for multi-electron atoms in terms of the properties of a system we can solve exactly.

Shielding is like a Nickelback concert

More electrons create the shielding or screening effect. Shielding or screening means the electrons closer to the nucleus block the outer valence electrons from getting close to the nucleus.

Imagine being in a crowded auditorium in a Nickelback concert. The enthusiastic fans are going to surround the auditorium, trying to get as close to the celebrity on the stage as possible. They are going to prevent people in the back from seeing the celebrity or even the stage. This is shielding. The stage is the nucleus and the celebrity is the protons. The fans are the electrons. Electrons closest to the nucleus will try to be as close to the nucleus as possible. The outer/valence electrons that are farther away from the nucleus will be shielded by the inner electrons. Therefore, the inner electrons prevent the outer electrons from forming a strong attraction to the nucleus. The degree to which the electrons are being screened by inner electrons can be shown by ns<np<nd<nf where n is the energy level. The inner electrons will be attracted to the nucleus much more than the outer electrons. Thus, the attractive forces of the valence electrons to the nucleus are reduced due to the shielding effects. That is why it is easier to remove valence electrons than the inner electrons. It also reduces the nuclear charge of an atom.

## Penetration

Penetration is the ability of an electron to get close to the nucleus. Thus, the closer the electron is to the nucleus, the higher the penetration. Electrons with higher penetration will shield outer electrons from the nucleus more effectively. Electrons with higher penetration are closer to the nucleus than electrons with lower penetration. Electrons with lower penetration are being shielded from the nucleus more.

## Slater's Rules Crudely Approximate Shielding

Slater's rules are an old and not-unreasonable way to estimate the effective nuclear charge $$Z_{eff}$$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell" (e.g., to compare the effective nuclear charge and shielding 3d and 4s in transition metals). The proper approach requires solving the Schrödinger equations (e.g., in the variational method), but this is a good rule of thumb that illustrates the gist.

• Step 1: Write the electron configuration of the atom in the following form:

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . .

• Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups
• Step 3: Slater's Rules is now broken into two cases:
• the shielding experienced by an s- or p- electron, and
• the shielding experienced by a d- or f- electron.

Graphical depiction of Slater's rules with shielding constants indicated.

Group Other electrons in the same group Electrons in group(s) with principal quantum number n and azimuthal quantum number < l Electrons in group(s) with principal quantum number n-1 Electrons in all group(s) with principal quantum number < n-1
[1s] 0.30 - - -
[ns,np] 0.35 - 0.85 1
[nd] or [nf] 0.35 1 1 1

Sum together the contributions as described in the appropriate rule above to obtain an estimate of the shielding constant, $$S$$. S is found by totaling the screening by all electrons except the one in question.

$S = \sum n_i S_i$

We can quantitatively represent this difference between $$Z$$ and $$Z_{eff}$$ as follows:

$S=Z-Z_{eff}$

Rearranging this formula to solve for $$Z_{eff}$$ we obtain:

$Z_{eff}=Z-S$

For a more basic overview of shielding and orbital penetration, check this module out.

Example $$\PageIndex{1}$$

What is the shielding constant experienced by a 2p electron in the nitrogen atom?

Strategy:
Determine the electron configuration of nitrogen (Z=7), then write it in the appropriate form. Use the Slater Rules to calculate the shielding constant for the electron.
Solution:

A N: 1s2 2s2 2p3

N: (1s2)(2s2,2p3)

B S[2p] = 1.00(0) + 0.85(2) + 0.35(4) = 3.10

Now, what is it effective charge?

Effective Charge:

$Z_{eff}=Z-S$

$Z_{eff} = 7 - 3.10 = 3.9$

Now, what is it effective energy of this electron if is followed the hydrogen atom energy (single electron system)?

Effective Energy:

$$E_{2p}=-R\dfrac{Z^2}{n^2}$$.

Let's substitute $$Z_{eff}$$ for $$Z$$ then

$E_{2p}=-R \dfrac{Z_{Eff}^2}{n^2}$

$E_{2p} = - R \dfrac{(3.10)^2}{2^2} \approx -2.4R$

Slaters rules provide an approximate guide to explain why certain orbitals fill before others since electrons in different orbitals will have a different effective nuclear charge due to different shielding resulting in differing energy levels.

(left) This energy levels for a hydrogen atom (the degeneracy per level is $$n^2$$). (right) The energy levels for a multi-electron atom (the $$\ell$$ degeneracy is broken, but not the $$m_l$$ degeneracy.

## The Variational Method Approximation

Start with the lowest energy state of a system $$\psi_0$$. For the hydrogen atom, this state is $$\psi_{100}$$, for the Harmonic Oscillator this state is $$v=0$$, and for the Rigid Rotor this state is $$J=0$$. Variational theorem states that if you substitute any function $$\phi$$ for the true wavefunction $$\psi_0$$, then calculate

$\color{red} E_{\phi} = \dfrac {\int \phi^* \hat{H} \phi d\tau}{\int \phi^* \phi d\tau} \ge E_o \label{1a}$

where $$E_0$$ is the true energy of the system. Equation $$\ref{1a}$$ in braket notation is

$\color{red} E_{\phi} = \dfrac { \langle \phi | \hat{H} | \phi \rangle} { \langle \phi | \phi \rangle} \ge E_o \label{1b}$

Note that $$E_{\phi} = E_0$$ when $$\psi_0 = \phi$$. This means that we can use any $$\phi$$ and we will find $$E_{\phi}$$, which is an upper bound for $$E_0$$. The closer $$\phi$$ is to $$\psi_0$$, the lower the value of $$E_{\phi}$$. The minimization of $$E_{\phi}$$ leads to a better estimate of $$E_0$$ and $$\psi_0$$. Set $$\phi(\alpha,\beta,\gamma)$$ with adjustable parameters $$\alpha, \beta, \gamma$$, and solve for the parameters by differentiating $$E_{\phi}$$ with respect to each parameter and setting the derivatives in equal to 0. In formulas,

$\dfrac {dE_{\phi}}{d\alpha} = 0,$

$\dfrac {dE_{\phi}}{d\beta}=0$

$\dfrac {dE_{\phi}}{d\gamma}=0\label{2}$

Equation $$\ref{1b}$$ is call the variational theorem and states that for a time-independent Hamiltonian operator, any trial wave function will have an variational energy expectation value that is greater than or equal to the true ground state wave function corresponding to the given Hamiltonian. Because of this, the variational energy is an upper bound to the true ground state energy of a given molecule.

The optimum parameters in a trial wavefunction are found by searching for minima in the potential landscape spanned by those parameters

The general approach of this method consists in choosing a "trial wavefunction" depending on one or more parameters, and finding the values of these parameters for which the expectation value of the energy is the lowest possible (Figure $$\PageIndex{2}$$).

Figure $$\PageIndex{2}$$: Simplified algorithmic flowchart of the Variational Method approximation

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.