# 1.8: N-dimensional cyclic systems

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This lecture will provide a derivation of the LCAO eigenfunctions and eigenvalues of N total number of orbitals in a cyclic arrangement. The problem is illustrated below:

There are two derivations to this problem.

## Polynomial Derivation

The Hückel determinant is given by,

$D_{N}(x)=\left|\begin{array}{ccccccccc} x & 1 & & & & & & & \\ 1 & x & 1 & & & & & & \\ & 1 & x & \ddots & & & & & \\ & & 1 & \ddots & \ddots & & & & \\ & & & \ddots & \ddots & \ddots & & & \\ & & & & \ddots & \ddots & \ddots & & \\ & & & & & \ddots & \ddots & 1 & \\ & & & & & & \ddots & x & 1 \\ & & & & & & 1 & x \end{array}\right|=0$

where

$x=\frac{\alpha-E}{\beta}$

From a Laplace expansion one finds,

$D_N(x)=x D_{n-1}(x)-D_{N-2}(x)$

Where

$\begin{array}{l} D_1(x)=x \\ D_2(x)=\left|\begin{array}{ll} x & 1 \\ 1 & x \end{array}\right|=x^2-1 \end{array} \notag$

With these parameters defined, the polynomial form of DN(x) for any value of N can be obtained,

$\begin{array}{c} D_3(x)=x D_2(x)-D_1(x)=x\left(x^2-1\right)-x=x\left(x^2-2\right) \\ D_4(x)=x D_3(x)-D_2(x)=x^2\left(x^2-2\right)-\left(x^2-1\right) \end{array} \notag$

$\vdots \nonumber$

and so on

The expansion of DN(x) has as its solution,

$x={-2}\cos \dfrac{2\pi}{N}j (j= 0, 1, 2, 3...N-1) \nonumber$

and substituting for x,

$E = \alpha + 2\beta\cos \dfrac{2\pi}{N}j (j= 0, 1, 2, 3...N-1) \nonumber$

## Standing Wave Derivation

An alternative approach to solving this problem is to express the wavefunction directly in an angular coordinate, θ

For a standing wave of λ about the perimeter of a circle of circumference c,

$\psi_j = \sin \dfrac{c}{\lambda} \theta \nonumber$

The solution to the wave function must be single valued ∴ a single solution must be obtained for ψ at every 2nπ or in analytical terms,

Thus the amplitude of $$ψ_j$$ at atom m is, (where c/λ = j and θ = (2π/N)m)

$\psi_{j}(m) = \sin{2m\pi}{N}j (j= 0, 1, 2, 3...N-1) \nonumber$

Within the context of the LCAO method, ψj may be rewritten as a linear combination in φm with coefficients cjm. Thus the amplitude of ψj at m is equivalent to the coefficient of φm in the LCAO expansion,

$\psi_{j} = \displaystyle \sum_{k=1}^N C_{jm\phi m}$

Where

$C_{jm} = \sin{2\pi m}{N}j (j= 0, 1, 2, 3...N-1) \nonumber$

The energy of each MO, ψj, may be determined from a solution of Schrödinger’s equation,

$\begin{array}{l} \mathrm{H} \psi_j=\mathrm{E}_j \psi_j \\ \left.\left|\mathrm{H}-\mathrm{E}_{\mathrm{j}}\right| \psi_j\right\rangle=0 \\ \left.\left|\mathrm{H}-\mathrm{E}_j\right| \sum_m^{\mathrm{N}} \mathrm{c}_{j m} \phi_m\right\rangle=0 \end{array} \notag$

The energy of the φm orbital is obtained by left–multiplying by φm,

$\left\langle\phi_m\left|H-E_j\right| \sum_m^N c_{j m} \phi_m\right\rangle=0 \notag$

but the Hückel condition is imposed; the only terms that are retained are those involving φm, φm+1, and φm-1. Expanding,

Evaluating the integrals,

$\begin{array}{l} \alpha \mathrm{c}_{j m}-\mathrm{c}_{j m} \mathrm{E}_{\mathrm{j}}+\beta\left\lfloor\mathrm{c}_{\mathrm{j}(\mathrm{m}+1)}+\mathrm{c}_{j(\mathrm{~m}-1)}\right\rfloor=0 \\ \alpha \mathrm{c}_{j m}+\beta\left\lfloor\mathrm{c}_{j(m+1)}+\mathrm{c}_{\mathrm{j}(\mathrm{m}-1)}\right\rfloor=\mathrm{c}_{j m} \mathrm{E}_{\mathrm{j}} \end{array} \no tag$

Substituting for cjm,

$\alpha \sin \dfrac{2\pi m}{N}j + \beta \left( \sin \dfrac{2\pi (m+1)}{N}j + \sin \dfrac{2\pi (m-1)}{N}j \right) = E_{j} \sin \dfrac{2\pi m}{N}j \nonumber$

$\alpha + \dfrac{ \beta \left( \sin \dfrac{2\pi (m+1)}{N}j + \sin \dfrac{2\pi (m-1)}{N}j \right)}{ \sin \dfrac{2\pi m}{N}j} = E_{j} \nonumber$

Making the simplifying substitution, $$\kappa=\frac{2 \pi}{N} j$$
$\mathrm{E}_{\mathrm{j}}=\alpha+\beta\left(\frac{\sin \kappa \mathrm{m} \cdot \cos \kappa+\sin \kappa \cdot \cos \kappa \mathrm{m}+\sin \kappa \mathrm{m} \cdot \cos \kappa-\sin \kappa \cdot \cos \kappa \mathrm{m}}{\sin \kappa \mathrm{m}}\right) \notag$

$E_{j} = \alpha + 2\beta \cos k \nonumber$

$E_{j} = \alpha + 2\beta \cos \dfrac{2\pi}{N}j (j= 0, 1, 2, 3...N-1) \nonumber$

Let’s look at the simplest cyclic system, N = 3

Continuing with our approach (LCAO) and using Ej to solve for the eigenfunction, we find…

$\psi_j=\sum_m e^{i j \theta} \phi_m \quad \text { for } j=0, \pm 1, \pm 2 \ldots\left\{\begin{array}{l} \pm \frac{N}{2} \text { for } N \text { even } \\ \pm \frac{(N-1)}{2} \text { for } N \text { odd } \end{array}\right.$

Using the general expression for ψj, the eigenfunctions are:

$\psi_{0} = e^{i(0)0} \phi_{1} + e^{i(0) \dfrac{2\pi}{3}} \phi_{2} + e^{i(0) \dfrac{4\pi}{3}} \phi_{3} \nonumber$

$\psi_{1} = e^{i(1)0} \phi_{1} + e^{i(1) \dfrac{2\pi}{3}} \phi_{2} + e^{i(1) \dfrac{4\pi}{3}} \phi_{3} \nonumber$

$\psi_{-1} = e^{i(-1)0} \phi_{1} + e^{i(-1) \dfrac{2\pi}{3}} \phi_{2} + e^{i(-1) \dfrac{4\pi}{3}} \phi_{3} \nonumber$

Obtaining real components of the wavefunctions and normalizing,

$\begin{array}{ll} \psi_{0}=\phi_{1}+\phi_{2}+\phi_{3} \rightarrow & \psi_{0}=\frac{1}{\sqrt{3}}\left(\phi_{1}+\phi_{2}+\phi_{3}\right) \\ \psi_{+1}+\psi_{-1}=2 \phi_{1}-\phi_{2}-\phi_{3} \rightarrow & \psi_{1}=\frac{1}{\sqrt{6}}\left(2 \phi_{1}-\phi_{2}-\phi_{3}\right) \\ \psi_{+1}-\psi_{-1}=\phi_{2}-\phi_{3} \rightarrow & \psi_{2}=\frac{1}{\sqrt{2}}\left(\phi_{2}-\phi_{3}\right) \end{array}$

Summarizing on a MO diagram where α is set equal to 0,

This page titled 1.8: N-dimensional cyclic systems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Daniel Nocera via source content that was edited to the style and standards of the LibreTexts platform.