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Next: Hartree-Fock theory Up: lecture_9 Previous: Spin wave functions

Many-electron atoms

The hydrogen atom is the only atom for which exact solutions of the Schrödinger equation exist. For any atom that contains two or more electrons, no solution has yet been discovered (so no solution for the helium atom exists!) and we need to introduce approximation schemes.



Let us consider the helium atom. The nucleus has a charge of $+2e$, and if we place the nucleus at the origin, there will be an electron at a position ${\bf r}_1$ with spin $s_{z,1}$ and an electron at position ${\bf r}_2$ and spin $s_{z,2}$. As usual, we consider the nucleus to be fixed. The classical energy is then

\begin{displaymath}
{\vert{\bf p}_1\vert^2 \over 2m_e} + {\vert{\bf p}_2\vert^2 ...
...ert{\bf r}_1\vert}
- {2 \over \vert{\bf r}_2\vert}\right] = E
\end{displaymath}

Note that $E$ is not simply a sum of terms for electron 1 and electron 2, $E \neq \varepsilon_1 + \varepsilon_2$. Therefore, it is not possible to write the wave function $\Psi({\bf x}_1,{\bf x}_2)$ as a simple product of the form $\psi_1({\bf x}_1)\psi_2({\bf x}_2)$, nor is it even possible to use the special form we introduced for identical particles

\begin{displaymath}
{1 \over \sqrt{2}}
\left[\psi_1({\bf x}_1)\psi_2({\bf x}_2)-\psi_1({\bf x}_2)\psi_2({\bf x}_1)\right]
\end{displaymath}

because these simple products are not correct solutions to the Schrödinger equation.



This means that the wave function $\Psi({\bf x}_1,{\bf x}_2)$ depends on the full set of 6 coordinates $x_1,y_1,z_1,x_2,y_2,z_2$ or $r_1,\theta_1,\phi_1,r_2,\theta_2,\phi_2$ if spherical coordinates are used, and 2 spin coordinates $s_{z,1}$, $s_{z,2}$, and that this dependence is not simple! In fact, as the number of electrons increases, the number of variables on which $\Psi$ depends increases as well. For an atom with $M$ electrons, the wave function $\Psi$ depends on $3M$ coordinates! Thus, it is clear that the wave function for a many-electron atom is a very unwieldy object!



As a side bar, we note that the 1998 Nobel prize in chemistry was awarded to Walter Kohn for the development of an extremely elegant theory of electronic structure known as density functional theory. In this theory, it is shown that the wave function $\Psi$, which depends on $3M$ coordinates, can be replaced by a much simpler object called the electron density denoted $\rho({\bf r}) = \rho(x,y,z)$. This object depends on only three variables for a system of any number of electrons. In density functional theory, it is shown that any physical quantity can be computed from this electron density $\rho({\bf r})$.



Consider now an imaginary form of helium in which the two electrons do not interact. For this simplfied case, the energy is simply

$\displaystyle E$ $\textstyle =$ $\displaystyle {\vert{\bf p}_1\vert^2 \over 2m_e} + {\vert{\bf p}_2\vert^2 \over...
...n_0}
\left[{2 \over \vert{\bf r}_1\vert}
+ {2 \over \vert{\bf r}_2\vert}\right]$  
       
  $\textstyle =$ $\displaystyle \left[{\vert{\bf p}_1\vert^2 \over 2m_e} - {2e^2 \over 4\pi\epsil...
...}_2\vert^2 \over 2m_e} - {2e^2 \over 4\pi\epsilon_0
\vert{\bf r}_2\vert}\right]$  
       
  $\textstyle =$ $\displaystyle \varepsilon_1 + \varepsilon_2$  

For this imaginary helium atom, the wave function can be expressed as an antisymmetric product, and because we have included spin, in the ground state, both electrons can be in the 1s($Z=2$) spatial orbital without having the wave function vanish. Recall that, when $Z=2$, the 1s orbital is

\begin{displaymath}
\psi_{100}(r,\theta,\phi) = {1 \over \sqrt{\pi}}
\left({2 \over a_0}\right)^{3/2}e^{-2r/a_0} \equiv \psi_{1s(Z=2)}({\bf r})
\end{displaymath}

Multiplying this by a spin wave function, the wave function for one of the electrons is

\begin{displaymath}
\psi({\bf x}) = \psi_{1s(Z=2)}({\bf r})\psi_{m_s}(S_z)
\end{displaymath}

Given this form, the antisymmetrized two-electron wave function becomes
$\displaystyle \Psi({\bf x}_1,{\bf x}_2)$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}
\left[\psi_{1s(Z=2)}({\bf r}_1)\psi_{\uparrow}...
..._{\uparrow}(S_{z_2})
\psi_{1s(Z=2)}({\bf r}_1)\psi_{\downarrow}(S_{z_1})\right]$  
       
  $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}\psi_{1s(Z=2)}({\bf r}_1)\psi_{1s(Z=2)}({\bf r}...
...downarrow}(S_{z_2}) -
\psi_{\uparrow}(S_{z_2})\psi_{\downarrow}(S_{z_1})\right]$  

which, by virtue of the spin wave functions, does not vanish.

Given this wave function, the energy would just be the sum of the energies of two electrons interacting with a nucleus of charge $+2e$. We would need two quantum numbers $n_1$ and $n_2$ for this, and from our study of hydrogen-like atoms, the energy would be

\begin{displaymath}
E_{n_1 n_2} = -{4 \over n_1^2} - {4 \over n_2^2}
\end{displaymath}

in Rydbergs. This comes from the fact that the energy of one electron interacting with a nucleus of charge $+Ze$ is $E_n = -Z^2/n^2$ in Rydbergs. So, the ground-state energy $(n_1=1, n_2=1)$ would be -8 Ry. In real helium, the electron-electron Coulomb repulsion $e^2/(4\pi\epsilon_0 r_{12})$ increases the ground state energy above this value (the experimentally measured ground-state energy is -5.8 Ry). The tendency is for the electrons to arrange themselves such that the Coulomb repulsion is as small as possible.



Subsections
next up previous
Next: Hartree-Fock theory Up: lecture_9 Previous: Spin wave functions
Mark E. Tuckerman 2011-11-03