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1.8: N-dimensional cyclic systems

  • Page ID
    221676
  • 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:

    clipboard_e663f445cb323ef610c93dec07ba41fe7.png

    There are two derivations to this problem.

    Polynomial Derivation

    The Hückel determinant is given by,

     

     

    $$
    D_{N}(x)=\left|\begin{array}{ccccccccc}
    x & 1 & & & & & & & \\
    1 & x & 1 & & & & & & \\
    & 1 & x & \ddots & & & & & \\
    & & 1 & \ddots & \ddots & & & & \\
    & & & \ddots & \ddots & \ddots & & & \\
    & & & & \ddots & \ddots & \ddots & & \\
    & & & & & \ddots & \ddots & 1 & \\
    & & & & & & \ddots & x & 1 \\
    & & & & & & 1 & x
    \end{array}\right|=0 \quad \text { where } \quad x=\frac{\alpha-E}{\beta}
    $$

    From a Laplace expansion one finds,

    DN(x) = xDn-1(x) - DN-2(x)

    Where

    clipboard_e8df885d6f2729c34ffbc37f7ea708c91.png

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

    D3(x) = xD2(x) – D1(x) = x(x2–1) – x = x(x2–2)

    D4(x) = xD3(x) – D2(x) = x2(x2–2) – (x2–1)

    \[ \vdots \nonumber \]

    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

    clipboard_ed3d9af7c123326eb19a86d6b823977e6.png

    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,

    clipboard_ecae7fe1444ce0da502bb815247e75b3b.png

    Thus the amplitude of ψj 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,

    clipboard_ea2aec33a3557d8665c244b626e703286.png

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

    clipboard_e4383b4c2ed9b73a4fcf1d7190b2e4fa0.png

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

    clipboard_eeb6a87f7e8ad64dabfa00891976d59af.png

    Evaluating the integrals,

    clipboard_ebab174e6c9157628039ca86704446754.png

    Substituting for cjm,

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

    clipboard_e63dd30e66aa2e29b9b8d09c626bfa968.png

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

    clipboard_e7d799f952e061cf6a06e17b69510e19c.png

    clipboard_ef7bf60442035b6e32c9e718d0f71df1f.png

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

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

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

    clipboard_e833d8f09c94f75a81848b80e0d383c47.png

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

    clipboard_e627b8a3833d79aecc1cf43e070c35c3b.png

    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,

    clipboard_ea958414a38b7968b1b7b2d5df73eddeb.png

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