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Next: Probing energy levels and Up: Many-electron atoms Previous: Many-electron atoms

Hartree-Fock theory

For the helium atom, what happens if we try to use a product form as a ``guess'' wave function? Ignoring spin for the moment, we take as a guess of the solution for the ground-state wave function

\begin{displaymath}
\Psi_g({\bf r}_1,{\bf r}_2) =
{1 \over \sqrt{2}}
\left[\psi...
..._1)\psi_2({\bf r}_2)-\psi_1({\bf r}_2)\psi_2({\bf r}_1)\right]
\end{displaymath}

Then we know from problem set #4 that the corresponding guess of the ground-state energy

\begin{displaymath}
E_g = \int \Psi_g \hat{H} \Psi_g dV_1 dV_2 > E_0
\end{displaymath}

where $E_0$ is the true ground-state energy. Here, $dV_1$ and $dV_2$ are the volume elements of electron 1 and electron 2, respectively, and $\hat{H}$ is the Hamiltonian

\begin{displaymath}
\hat{H} = \hat{K}_1 + \hat{K}_2 +
+ {e^2 \over 4\pi \epsilo...
...r \vert{\bf r}_1\vert}
- {2 \over \vert{\bf r}_2\vert}\right]
\end{displaymath}

That is, we know that this guess is not correct, however, we can try to optimize the form of the functions $\psi_1({\bf r})$ and $\psi_2({\bf r})$ so as to make $E_g$ as small as possible, thereby making it approach $E_0$.



If we do this, we find that $\psi_1({\bf r})$ and $\psi_2({\bf r})$ satisfy a set of 2 coupled Schrödinger-like equations and that the potential energy $V$ in each equation corresponds to what the electron would experience if we averaged over all possible positions of the other electron (think back to the classical shell model!).



If we solve these equations, we find that both $\psi_1({\bf r})$ and $\psi_2({\bf r})$ bear a strikingly resemblance to the hydrogenic function $\psi_{100}(r,\phi,\theta) = \psi_{1s}(r,\theta,\phi)$, the 1$s$ wave function. The difference, however, is that it decays a little faster, as if its $Z$ value were between 1 and 2. In fact, the value one obtains from an actual calculation is $Z \approx 1.69$. We will denote this as $\psi_{1s(Z)}(r,\theta,\phi)$. We can imagine one of the electrons as ``occupying'' the state $\psi_1({\bf r})$ and the other as ``occupying'' the state $\psi_2({\bf r})$, but of course, we cannot say which electron is in which state, which is why we need the above form of the guess wave function. We also obtain two energies $\varepsilon_1$ and $\varepsilon_2$ that are close in energy but that the energies are lower than the energy $E_1$ of the hydrogen atom.



Now, we see the problem with ignoring spin. If $\psi_1$ and $\psi_2$ are both $\psi_{1s(Z)}$, then the guess wave function becomes

\begin{displaymath}
\Psi_g({\bf r}_1,{\bf r}_2) =
{1 \over \sqrt{2}}\left[\psi_...
...) -
\psi_{1s(Z)}({\bf r}_2)\psi_{1s(Z)}({\bf r}_1)\right] = 0
\end{displaymath}

However, if we multiply the spatial wave functions by spin wave functions, then the two orbitals we obtain are

\begin{displaymath}
\psi_1({\bf x}) = \psi_{1s(Z)}({\bf r})\psi_{\uparrow}(s_z),...
...
\psi_2({\bf x}) = \psi_{1s(Z)}({\bf r})\psi_{\downarrow}(s_z)
\end{displaymath}

Thus, if we substitute this into the expression for $\Psi_g({\bf x}_1,{\bf x}_2)$, we find that we can factor out $\psi_{1s(Z)}({\bf r}_1)\psi_{1s(Z)}({\bf r}_2)$ and we are left with the antisymmetric combination of spin wave functions. The only possibility for the combination of spin wave functions leads to

\begin{displaymath}
\Psi_g({\bf x}_1,{\bf x}_2) = {1 \over \sqrt{2}}
\psi_{1s(Z)...
... -
\psi_{\uparrow}(s_{z,2})\psi_{\downarrow}(s_{z,1})
\right]
\end{displaymath}

Note that another possible spin wave function would be $[\psi_{\uparrow}(s_{z,2})\psi_{\downarrow}(s_{z,1}) -
\psi_{\uparrow}(s_{z,1})\psi_{\downarrow}(s_{z,2})]/\sqrt{2}$ however, this is different from the above wave function by an overall minus sign and is, therefore, not actually different.

As another shorthand notation, we can use symbols ``1'' and ``2'' to represent any of the variables of electrons 1 and 2. Thus, the guess wave functions can be represented simply as

$\displaystyle \psi_g(1,2)$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}
\left[\psi_1(1)\psi_2(2)-\psi_1(2)\psi_2(1)\right]$  
       
  $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}\psi_{1s(Z)}(1)\psi_{1s(Z)}(2)
\left[\psi_{\uparrow}(1)\psi_{\downarrow}(2)-\psi_{\uparrow}(2)\psi_{\downarrow}(1)\right]$  

Given that the spin wave functions are rather simple, we can focus on the spatial parts of the wave functions in the proceeding discussion. For short, we denote the electronic configuration as $1s^2$.



The lowest HF energy obtained for helium, $\varepsilon_1 \approx - 2.83$Ry. Note how close this is to the formula $-Z^2/n^2$ for $n=1$ with $Z=1.69$! The next energy $\varepsilon_2$ is only slightly large than this value, so we see that $\varepsilon_1 \approx \varepsilon_2$, but strictly $\varepsilon_1 < \varepsilon_2$. However, the energies are so close that they can be considered as degenerate. In fact, there is a class of Hartree-Fock calculations (called ``closed-shell'' calculations), in which this degeneracy is assumed at the outset, and the pair of electrons is treated as a single unit or single quantum state.



Now, if we followed the same thing for lithium, we would need three functions to construct the guess wave function, which we could take as

\begin{displaymath}
\psi_1({\bf x}) = \phi_1({\bf r})\psi_{\uparrow}(S_z),\;\;\;...
...\;\;\;\;
\psi_3({\bf x}) = \phi_3({\bf r})\psi_{\uparrow}(S_z)
\end{displaymath}

where $\phi_1({\bf r})$, $\phi_2({\bf r})$, and $\phi_3({\bf r})$ are arbitrary spatial wave functions whose shape we optimize so as to make $E_g$ as close to the true ground state energy $E_0$ as possible. For lithium, the Hamiltonian is

\begin{displaymath}
\hat{H} = \hat{K}_1 + \hat{K}_2 + \hat{K}_3 + {e^2 \over 4\p...
...r \vert{\bf r}_2\vert} - {3 \over
\vert{\bf r}_3\vert}\right]
\end{displaymath}

The guess wave function we would need is
$\displaystyle \Psi_g({\bf x}_1,{\bf x}_2,{\bf x}_3)$ $\textstyle =$ $\displaystyle {1 \over \sqrt{6}}
\left[\psi_1({\bf x}_1)\psi_2({\bf x}_2)\psi_3...
...\psi_3({\bf x}_2) +
\psi_1({\bf x}_2)\psi_2({\bf x}_3)\psi_3({\bf x}_1)
\right.$  
  $\textstyle -$ $\displaystyle \left.\psi_1({\bf x}_1)\psi_2({\bf x}_3)\psi_3({\bf x}_2) -
\psi_...
...)\psi_3({\bf x}_1) -
\psi_1({\bf x}_2)\psi_2({\bf x}_1)\psi_3({\bf x}_3)\right]$  

and the guess energy is computed using

\begin{displaymath}
E_g = \int \Psi_g^*\hat{H}\Psi_g d{\bf x}_1 d{\bf x}_2 d{\bf x}_3
\end{displaymath}

where the notation $\int d{\bf x}$ means

\begin{displaymath}
\int d{\bf x}= \sum_{S_z=-\hbar/2}^{\hbar/2} \int dV
\end{displaymath}

That is, we sum over the two spin values and integrate over all space using either Cartesian ($dV = dxdydz$) or spherical ( $dV = r^2\sin\theta
drd\theta d\phi$) coordinates or any other coordinates we wish to use. Then, minimizing the guess energy by optimizing the shapes of the orbitals $\phi_1$, $\phi_2$, and $\phi_3$, we obtain two orbitals $\phi_1({\bf r})$ and $\phi_2({\bf r})$ that resemble 100 or 1s hydrogen-like orbitals $\psi_{100(Z)}(r,\phi,\theta)$ and $\phi_3$ resembles $\psi_{200(Z)}(r,\phi,\theta)$ but again, all of these orbitals decay faster than their counterparts for the hydrogen atom. The calculation gives an effective value of $Z \approx 2.64$. Thus, multiplying these by spin wave functions, we would obtain the three HF orbitals as
$\displaystyle \psi_1({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_2({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  
$\displaystyle \psi_3({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  

We would denote the electronic configuration as $1s^2 2s^1$. Here, we find $\varepsilon_1 \approx \varepsilon_2$, and $\varepsilon_3 > \varepsilon_1,\varepsilon_2$. Strictly, $\varepsilon_1 < \varepsilon_2 < \varepsilon_3$. The total HF energy for the ground state is -14.86 Ry, while the experimental value is -14.96 Ry. Note that if we neglected the electron-electron repulsion, the energy would be

\begin{displaymath}
E_{n_1 n_2 n_3} = -{9 \over n_1^2} - {9 \over n_2^2} - {9 \over n_3^2}
\end{displaymath}

with $n_1=n_2=n_3=1$, which gives -27 Ry.



For Beryllium, we would obtain four orbitals, two of which resemble 1s orbitals and two of which resemble 2s orbitals. Again, these decay as if they have an effective $Z$ value between 3 and 4. We can, therefore, represent these orbitals as

$\displaystyle \psi_1({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_2({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  
$\displaystyle \psi_3({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_4({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  

We would denote the electronic configuration as $1s^2 2s^2$.



For Boron, we would obtain 5 orbitals, two of which would resemble $1s$, two would resemble $2s$, and the fifth would resemble a $2p_x$ orbital. Again, all of these would decay as if they have an effective $Z$ between 4 and 5. We can thus represent the 5 orbitals as

$\displaystyle \psi_1({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_2({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  
$\displaystyle \psi_3({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_4({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  
$\displaystyle \psi_5({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2p_x(Z)}({\bf r})\psi_{\uparrow}(s_z)$  

and denote the electronic configuration as $1s^2 2s^2 2p_x^1$. Here the HF energies would follow the pattern $\varepsilon_1 \approx \varepsilon_2$, $\varepsilon_3 \approx \varepsilon_4$, $\varepsilon_3,\varepsilon_4 > \varepsilon_1,\varepsilon_2$, and $\varepsilon_5 > \varepsilon_3,\varepsilon_4$. Strictly, $\varepsilon_1 < \varepsilon_2 < \varepsilon_3 < \varepsilon_4 <
\varepsilon_5$.



For carbon, we obtain 6 orbitals, and this time, the 6th orbital does not resemble $2p_x$ but rather $2p_y$. Hence, we would represent the 6 orbitals as

$\displaystyle \psi_1({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_2({\bf x})$ $\textstyle =$ $\displaystyle \psi_{1s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  
$\displaystyle \psi_3({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_4({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2s(Z)}({\bf r})\psi_{\downarrow}(s_z)$  
$\displaystyle \psi_5({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2p_x(Z)}({\bf r})\psi_{\uparrow}(s_z)$  
$\displaystyle \psi_6({\bf x})$ $\textstyle =$ $\displaystyle \psi_{2p_y(Z)}({\bf r})\psi_{\uparrow}(s_z)$  

These all decay as if they had an effective $Z$ value between 5 and 6. The HF energies follow the pattern $\varepsilon_1 \approx \varepsilon_2$, $\varepsilon_3 \approx \varepsilon_4$, $\varepsilon_5 \approx \varepsilon_6$, $\varepsilon_3,\varepsilon_4 > \varepsilon_1,\varepsilon_2$, and $\varepsilon_5,\varepsilon_6 > \varepsilon_3,\varepsilon_4$. Note that for Boron and Carbon, the energies of the $2p$ orbitals are larger than for the $2s$ orbitals, which is not the case for hydrogen. The electronic configuration would be $1s^2 2s^2 2px_1 2p_y^1$.



For Nitrogen, the 7th orbital would resemble a $2p_z$ orbital, and we would just add $\psi_7({\bf x}) = \psi_{2p_z(Z)}\psi_{\uparrow}(s_z)$. Now, for oxygen, the 8th orbital again resembles a $2p_x$ orbital, and we would add this to the list but with a down spin wave function $\psi_8({\bf x}) = \psi_{2p_x(Z)}\psi_{\downarrow}(s_z)$. For oxygen, we find that the 5-8 HF energies are approximately equal, $\varepsilon_5\approx \varepsilon_6 \approx \varepsilon_7 \approx
\varepsilon_8$, although strictly $\varepsilon_5 < \varepsilon_6 < \varepsilon_7 < \varepsilon_8$.



These examples illustrate a more general procedure known as Hartree-Fock theory or the Hartree-Fock approximation (named after Douglas Hartree and Vladimir Fock) developed in 1930. Hartree-Fock (HF) theory makes several important assumptions:

1.
The guess wave function is always composed of combinations of products of single-electron functions $\psi_1({\bf r})$, $\psi_2({\bf r})$,... that properly account for the identical nature of the electrons.

2.
When the shapes of these functions are optimized, each electron is subject to an effective potential $V$ that is generated by averaging over the positions of all the other electrons. For the $i$th electron, $V$ depends only on ${\bf r}_i$.

3.
For an atom, $V$ depends only on the magnitude $\vert{\bf r}_i\vert$, which means that the effective potential is spherically symmetric.
If an atom has $M$ electrons, then HF theory yields $M$ functions $\psi_1({\bf r}),...,\psi_M({\bf r})$ and $M$ energies $\varepsilon_1,...,\varepsilon_M$. These energies are all negative, $\varepsilon_i < 0$, and they are ordered such that

\begin{displaymath}
\varepsilon_1 < \varepsilon_2 < \cdots < \varepsilon_M
\end{displaymath}

We will give a physical interpretation of these energies shortly.



So how do we calculate probabilities in HF theory? What we are interested in is the probability of finding one of the $M$ electrons in a small volume $dV$ about the point ${\bf r}$ independent of the positions of the remaining $M-1$ electrons. This probability is given in terms of the electron density introduced earlier. That is

\begin{displaymath}
P({\rm electron\ is\ in\ }dV{\ about\ }{\bf r}) = \rho({\bf r})dV
\end{displaymath}

and the electron density can be computed from the squares of the HF orbitals:

\begin{displaymath}
\rho({\bf r}) = \sum_{S_z = -\hbar/2}^{\hbar/2}\sum_{i=1}^M \vert\psi_i({\bf r},S_z)\vert^2
\end{displaymath}

This is the same density that lead to the chemistry Nobel prize!



To give an idea of how well HF theory can predict the ground state energies of several atoms, consider the table below (all energies are in Ry):

$\displaystyle \underline{\rm Atom}$ $\textstyle \;\;\;\;\; \underline{\rm HF\ Energy}\;\;\;\;\;$ $\displaystyle \underline{\rm Experiment}$  
$\displaystyle {\rm He}$ $\textstyle -5.72$ $\displaystyle -5.80$  
$\displaystyle {\rm Li}$ $\textstyle -14.86$ $\displaystyle -14.96$  
$\displaystyle {\rm Ne}$ $\textstyle -257.10$ $\displaystyle -257.88$  
$\displaystyle {\rm Ar}$ $\textstyle -1053.64$ $\displaystyle -1055.20$  



The HF theory motivates the so-called ``aufbau'' principle for expressing/building up electron configurations in atoms. ``Aufbau'' is German for ``building up'' (Aufbauen = ``to build up''). We see that the HF orbitals are ordered according to their energies and that as the energy increases, their shapes follow the ``ladder'' of wave functions of the hydrogen atom. Thus, as we consider atoms with more and more electrons, we simply have to ``occupy'' more of the HF orbitals in order of increasing energy. Since groups of these orbitals are close in energy, there are two rules we need to introduce for this procedure of occupying orbitals:

1.
Pauli exclusion principle: No two electrons in an atom can have the same set of quantum numbers.
2.
Hund's rule: When electrons are added to a set of HF orbitals of approximately equal energy, a single electron enters each orbital before a second one can enter any orbital. The lowest energy configuration is always the one with parallel spins.
Now, let us apply these rules to fill the orbitals in the first row elements (which are all we care about for the next two chapters). The filling will refer to the figure below:

figure=electron_configs.eps,height=4.0in,width=4.0in
Figure: Table of electronic configurations
Starting with hydrogen ($Z=1$), HF theory gives the exact answer because the H atom is exactly soluble. Thus, we generate the exact function $\psi_{100}(r,\phi,\theta)$, and we put the one electron into this orbital for the ground state. We write this as

\begin{displaymath}
1s^1
\end{displaymath}



For Helium ($Z=2$), HF theory is no longer exact, however as we discussed, HF theory gives two orbitals $\psi_1({\bf r})$ and $\psi_2({\bf r})$, both of which resemble $\psi_{100}$ (but decays a little faster). The energies $\epsilon_1$ and $\epsilon_2$ are also very similar. Thus, we regard these orbitals as having roughly the same energy, and we place one electron into $\psi_1$ and one into $\psi_2$. We regard this as placing two electrons into the single orbital $\psi_{100}$ and assigning the electrons the same set of H-like quantum numbers $n,l,m$. To be consistent with Pauli's exclusion principle, they need to have one quantum number that is different, so we assign one electron $m_s=1/2$ and the other $m_s=-1/2$ (see figure). We write the electron configuration as

\begin{displaymath}
1s^2
\end{displaymath}

signifying that two electrons occupy the $\psi_{100}$ orbital (although in reality, they occupy two different orbitals that happen to be very similar to each other).



For Lithium ($Z=3$), HF theory yields three orbitals $\psi_1$, $\psi_2$, and $\psi_3$. The first two closely resemble $\psi_{100}$, while the third resembles $\psi_{200}$. HF theory also yields three energies $\varepsilon_1$, $\varepsilon_2$ and $\varepsilon_3$ with $\varepsilon_1 \approx \varepsilon_2$ and $\varepsilon_3 > \varepsilon_2 > \varepsilon_1$. Each electron occupies on HF orbital, but we regard this as two electrons in a $\psi_{100}$ orbital and the third in a $\psi_{200}$ orbital (see figure). We write this as

\begin{displaymath}
1s^22s^1
\end{displaymath}

Note that it takes two electrons to fill the 1s orbital, after which we move to a 2s-like orbital. In this case, we say that the 1s electrons form a closed shell, since we have now moved on to a different value of $n$, i.e., $n=2$. The 2 1s electrons are called core electrons and the 2s electron is called a valence electron.



Coming to Beryllium ($Z=4$), we have 4 HF orbitals, two of which resemble $\psi_{100}$ and two of which resemble $\psi_{200}$, so we imagine that an electron in each HF orbital is equivalent to 2 in the $\psi_{100}$ and 2 in the $\psi_{200}$. In each pair, the electrons have opposite spins to be consistent with the Pauli exclusion principle. We write this as

\begin{displaymath}
1s^2 2s^2
\end{displaymath}

Now, we have two core electrons and two valence electrons.



For Boron ($Z=5$), there are 5 HF orbitals. $\psi_1$ and $\psi_2$ resemble $\psi_{100}$, $\psi_3$ and $\psi_4$ resemble $\psi_{200}$, and $\psi_5$ resembles $\psi_{21m}$, i.e. a generic $p$-orbital (we do not specify $m$ because the direction of the orbital does not matter). The energies are ordered such that $\varepsilon_1 \approx \varepsilon_2$, $\varepsilon_3 \approx \varepsilon_4$. Interestingly, $\varepsilon_5$ is larger than all of these and not close to $\varepsilon_4$. Unlike in hydrogen, where orbitals with the same $n$ but different values of $l$ have the same energy, here orbitals with the same $n$ but different values of $l$ have different energies. Thus, $\psi_5$ is at a higher energy. Hence, when we fill, we start with $1s$, then $2s$ and finally $2p$. We can choose any of the $2p$ orbitals, so we just start with $2p_x$. We thus write the electronic configuration as

\begin{displaymath}
1s^2 2s^2 2p_x^1
\end{displaymath}



For carbon ($Z=6$), the situation is much the same. However, here we have $\psi_5$ and $\psi_6$ both with $p$-like character and similar in energy. Here, however, we need Hund's rule. We first put a spin-up electron in the $2p_x$ orbital and then another spin-up electron in the $2p_y$ orbital. The electronic configuration is

\begin{displaymath}
1s^2 2s^2 2p_x^1 2p_y^1
\end{displaymath}



For nitrogen ($Z=7$), we would place a spin-up electron in the $2p_z$ orbital and write $1s^22s^2 2p_x^1 2p_y^1 2p_z^1$. When we get to oxygen ($Z=8$), we have now placed one electron in each of the three $p$-like orbitals. We now have to return to $p_x$ and place another electron in it, which will now be spin-down to be consistent with the Pauli exclusion principle. Its electronic configuration is $1s^22s^2 2p_x^2 2p_y^1 2p_z^1$. For fluorine ($Z=9$), it is $1s^22s^2 2p_x^2 2p_y^2 2p_z^1$. Finally, when we get to Neon ($Z=10$), we have $1s^22s^2 2p_x^2 2p_y^2 2p_z^2$, which we can write simply as $1s^22s^2 2p^6$. With Neon, as with Helium, we have completed a shell, and the next HF orbital will be like $\psi_{300}$ and have a higher energy, thereby starting the next shell. Thus, we see that for Neon, 8 electrons are needed to complete a shell, the 2 2s electrons and the 6 2p electrons. The fact that 8 electrons close a shell for Neon lend this atom special stability and make it highly unreactive.


next up previous
Next: Probing energy levels and Up: Many-electron atoms Previous: Many-electron atoms
Mark E. Tuckerman 2011-11-03