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V25.0109: General Chemistry I (Honors)
Notes for Lecture 8


Now that we have introduced the basic concepts of quantum mechanics, we can start to apply these concepts to build up matter, starting from its most elementary constituents, namely atoms, up to molecules, supramolecular complexes (complexes built from weak interactions such as hydrogen bonds and van der Waals interactions), networks, and bulk condensed phases, including liquids, glasses, solids,... As explore the structure of matter, itself, we should stand back and wonder at how the mathematically elegant, yet somehow not quite tangible, structure of quantum theory is able to describe so accurately all phases of matter and types of substances, ranging from metallic and semiconducting crystals to biological macromolecules, to morphologically complex polymeric materials.



We begin with the simplest system, the hydrogen atom (and hydrogen-like single-electron cations), which is the only atomic system (thus far) for which the Schrödinger equation can be solved exactly for the energy levels and wave functions. To this end, we consider a nucleus of charge $+Ze$ at the origin and a single electron a distance $r$ away from it. If we just naïvely start by writing the classical energy

\begin{displaymath}
{p_x^2 \over 2m} + {p_x^2 \over 2m} + {p_x^2 \over 2m} - {Ze^2 \over
4\pi\epsilon_0\sqrt{x^2 + y^2 + z^2}} = E
\end{displaymath}

where we have written $r = \sqrt{x^2 + y^2 + z^2}$ in terms of its Cartesian components, then we see immediately that the energy is not a simple sum of terms involving $(x,p_x)$, $(y,p_y)$, and $(z,p_z)$, and therefore, the wave function is not a simple product of a function of $x$, a function of $y$ and a function of $z$. Thus, it would seem that we have already hit a mathematical wall. Fortunately, we are not confined to work in Cartesian coordinates. In fact, there are numerous ways to locate a point ${\bf r}$ in space, and for this problem, there is a coordinate system, known as spherical coordinates, that is particularly convenient because it uses the distance $r$ of a point from the origin as one of its explicit coordinates.




next up previous
Next: Spherical coordinates
Mark E. Tuckerman 2008-10-08