Butadiene

Chemistry Level 5

Butadiene (H 2 _2 C = = CH − - CH = = CH 2 _2 ) is a four-carbon molecule arranged in a chain. In this molecule, there are π \pi -bonds that result from the interactions of the electrons in the 2 p z 2 p_z -orbitals of carbon. The wave function ψ \psi of the π \pi -electrons can be approximated as a linear combination of the atomic orbitals: ψ ≈ ∑ i = 1 4 λ i ϕ i \psi \approx \sum_{i=1}^4 \lambda_i \phi_i where ϕ i \phi_i denotes the 2 p z 2 p_z -orbital at the i i -th carbon atom ( i = 1 , 2 , 3 , 4 i = 1,2,3,4 ). The λ i ∈ R \lambda_i \in \mathbb {R} are the coefficients, which form a vector λ = ( λ 1 , λ 2 , λ 3 , λ 4 ) \lambda = (\lambda_1, \lambda_2, \lambda_3, \lambda_4) . The π \pi -states result as a solution of the matrix equation ( α β 0 0 β α β 0 0 β α β 0 0 β α ) ⏟ = H ⋅ ( λ 1 λ 2 λ 3 λ 4 ) = ε ⋅ ( λ 1 λ 2 λ 3 λ 4 ) \underbrace{\left( \begin{array}{cccc} \alpha & \beta & 0 & 0 \\ \beta & \alpha & \beta & 0 \\ 0 & \beta & \alpha & \beta \\ 0 & 0 & \beta & \alpha \end{array} \right)}_{= H} \cdot \left( \begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{array} \right) = \varepsilon \cdot \left( \begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{array} \right) with energy parameters α , β ∈ R \alpha, \beta \in \mathbb {R} . The quantity ε \varepsilon denotes an energy eigenvalue of the Hamiltonian matrix H H . There are a total of four energy eigenvalues ε a < ε b < ε c < ε d \varepsilon_\text {a} <\varepsilon_\text {b} <\varepsilon_\text {c} <\varepsilon_\text {d} as shown above in the energy diagram. In the ground state, the π \pi -electrons have the total energy E = 2 ( ε a + ε b ) = 4 α − ξ ⋅ ∣ β ∣ E = 2 (\varepsilon_a + \varepsilon_b) = 4 \alpha - \xi \cdot | \beta | What is the numerical value of ξ \xi ?

Hint: The eigenvalues ε \varepsilon of H H results as solutions of the characteristic polynomial det ( H − ε E ) = 0 \text{det} (H- \varepsilon E) = 0 , where E E denotes the unit matrix ,


The answer is 4.47213595499958.

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Jun 16, 2018

This problem is an example of the application of the Hückel method .

We look for solutions of the characteristic polynomial det ( H − ε E ) = ∣ α − ε β 0 0 β α − ε β 0 0 β α − ε β 0 0 β α − ε ∣ = β 4 ∣ x 1 0 0 1 x 1 0 0 1 x 1 0 0 1 x ∣ = 0 , x = α − ε β \text{det}(H - \varepsilon E) = \left| \begin{array}{cccc} \alpha - \varepsilon & \beta & 0 & 0 \\ \beta & \alpha - \varepsilon & \beta & 0 \\ 0 & \beta & \alpha - \varepsilon & \beta \\ 0 & 0 & \beta & \alpha - \varepsilon \end{array} \right| = \beta^4 \left| \begin{array}{cccc} x & 1 & 0 & 0 \\ 1 & x & 1 & 0 \\ 0 & 1 & x & 1 \\ 0 & 0 & 1 & x \end{array} \right| = 0, \quad x = \frac{\alpha - \varepsilon}{\beta} The determinant can be calculated using the Laplace formula ∣ x 1 0 0 1 x 1 0 0 1 x 1 0 0 1 x ∣ = x ⋅ ∣ x 1 0 1 x 1 0 1 x ∣ − 1 ⋅ ∣ 1 0 0 1 x 1 0 1 x ∣ = x ⋅ ( x ⋅ ∣ x 1 1 x ∣ − 1 ⋅ ∣ 1 0 1 x ∣ ) − 1 ⋅ ( 1 ⋅ ∣ x 1 1 x ∣ − 0 ⋅ ∣ 1 1 0 x ∣ ) = x ⋅ ( x ⋅ ( x 2 − 1 ) − 1 ⋅ ( x − 0 ) ) − 1 ⋅ ( 1 ⋅ ( x 2 − 1 ) − 0 ) = x 4 − 3 x 2 + 1 = ! 0 ⇒ x 2 = 3 2 ± 9 4 − 1 = 3 ± 5 2 = ( 1 ± 5 ) 2 4 ⇒ x a,b,c,d = 1 + 5 2 , 5 − 1 2 , − 5 − 1 2 , − 1 + 5 2 \begin{aligned} \left| \begin{array}{cccc} x & 1 & 0 & 0 \\ 1 & x & 1 & 0 \\ 0 & 1 & x & 1 \\ 0 & 0 & 1 & x \end{array} \right| &= x \cdot \left| \begin{array}{ccc} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{array} \right| - 1 \cdot \left| \begin{array}{ccc} 1 & 0 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{array} \right| \\ &= x \cdot \left( x \cdot \left| \begin{array}{cc} x & 1 \\ 1 & x \end{array} \right| - 1 \cdot \left| \begin{array}{cc} 1 & 0 \\ 1 & x \end{array} \right| \right) - 1 \cdot \left( 1 \cdot \left| \begin{array}{cc} x & 1 \\ 1 & x \end{array} \right| - 0 \cdot \left| \begin{array}{cc} 1 & 1 \\ 0 & x \end{array} \right| \right) \\ &= x \cdot ( x \cdot (x^2 - 1) - 1 \cdot (x - 0)) - 1 \cdot (1 \cdot (x^2 - 1) - 0)\\ &= x^4 - 3 x^2 + 1 \stackrel{!}{=} 0 \\ \Rightarrow \quad x^2 &= \frac{3}{2} \pm \sqrt{\frac{9}{4} - 1} = \frac{3 \pm \sqrt{5}}{2} = \frac{(1 \pm \sqrt{5})^2}{4}\\ \Rightarrow \quad x_\text{a,b,c,d} &= \frac{1 + \sqrt{5}}{2} , \quad \frac{\sqrt{5}-1}{2}, \quad -\frac{\sqrt{5}-1}{2}, \quad -\frac{1 + \sqrt{5}}{2} \end{aligned} The results correspond to the golden ratio ϕ = 1 + 5 2 \phi = \frac{1 + \sqrt{5}}{2} and his "little brother" ϕ ‾ = 1 − 5 2 \overline{\phi} = \frac{1 - \sqrt{5}}{2} . The ground state energy results to E = 2 ε a + 2 ε b = 4 α − 2 β ( x a + x b ) = 4 α − 2 5 β E = 2 \varepsilon_\text{a} + 2 \varepsilon_\text{b} = 4 \alpha - 2 \beta (x_\text{a} + x_\text{b}) = 4 \alpha - 2 \sqrt{5} \beta so that ξ = 2 5 \xi = 2 \sqrt{5} .

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