Connection to chemistry

These examples have been kindly provided by Dr. Felix Schumacher from Essen.

Periodic boundary conditions

The six p-electrons in a benzene ring (C₆H₆) form a system of three conjugated double bonds that are ultimately completely delocalized, i.e., in the language of chemists, smeared out over the whole molecule. The wave functions of these electrons have a nodal plane in the plane of the molecule, and in what follows we shall only consider the behaviour along the benzene ring; this then genuinely gives a one-dimensional problem. This yields the following eigenfunctions ( $L$ is the as yet unknown circumference of the benzene ring):

States of even parity:

$\varphi_n (x) = \sqrt{\dfrac{2}{L}} \cdot \cos\biggl(2n\pi\dfrac{x}{L}\biggr) \quad n = 0,1,2,3,...$

States of odd parity:

$\Psi_n (x) = \sqrt{\dfrac{2}{L}} \cdot \sin\biggl(2n\pi\dfrac{x}{L}\biggr) \quad n = 0,1,2,3,...$

Ground state:

$\varphi_0 (x) = \dfrac{1}{\sqrt{L}} \quad \text{(i. e. a constant function)}$

For definiteness, one can imagine that $x$ is the distance of the electron from a particular C atom, measured along the benzene ring. The wave function must then have the period $L$ .

In what follows, we shall estimate the circumference of the benzene ring, i. e. the quantity $L$ . Apart from the ground state, all states are doubly degenerate. Taking the Pauli principle into account and including spin, the electron state with $n = 0$ is occupied by two electrons and the state with $n = 1$ by four. The first excitation occurs from $n = 1$ to $n = 2$ .

Because of the periodic boundary condition, it must additionally hold that:

$j (x) = j (x + L) \quad \text{and} \quad \psi (x) = \psi (x + L) \, \text{.}$

The energy eigenvalues are as follows:

$E_n = \dfrac{h^2 \cdot n^2}{2mL^2} \, \text{.}$

The corresponding energy difference is therefore:

$\Delta E = \dfrac{h^2}{2mL^2} \cdot (2^2 -1 ) = \dfrac{3 \cdot h^2}{2mL^2} \, \text{.}$

Optically, this transition lies in the near UV at $\lambda = 200$ nm; setting

$\Delta E = \dfrac{h \cdot c}{\lambda} \quad \text{, one obtains} \quad L = \sqrt{\dfrac{3h\lambda}{2mc}} \, \text{.}$

Under the square root there are only natural constants and the measured quantity $\lambda = 200$ nm. This allows the circumference of the benzene ring to be determined; one obtains $L = 0{,}832 \text{nm}$ . The distance between two C atoms in the benzene ring is therefore

$d = \dfrac{L}{6} = 0{,}142 \, \text{nm.}$

The literature gives $d = 0{,}139 \text{nm}$ . This is a deviation of only 2%.

For the spatial distribution of the p-electrons along the benzene ring, the model yields a constant probability density. This is typical of an aromatic compound.

An application example for the linear potential well with infinitely high walls are the polyenes, linear carbon-hydrogen compounds with conjugated double bonds. If one considers the p-electrons along the molecule (e. g. β-carotene), it turns out that the probability of finding an electron is greatest at the ends of the molecule. In addition reactions, atoms therefore always attach at the ends of the molecule.

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