- #1
ultimateguy
- 122
- 1
Homework Statement
Consider an electron of a linear triatomic molecule formed by three equidistant atoms. We use |\phi_A>, |\phi_B>, |\phi_C> to denote three orthonormal staes of this electron, corresponding respectively to three wave functions localized about the nuclei of atoms A, B and C. We shall confine ourselves to the subspace of the state space spanned by |\phi_A>, |\phi_B>, |\phi_C>.
When we neglect the possibility of the electron jumping from one nucleus to another, its energy is described by the Hamiltonian H_0 whose eigenstates are the three states |\phi_A>, |\phi_B>, |\phi_C> with the same eigenvalue E_0. The coupling between states |\phi_A>, |\phi_B>, |\phi_C> is described by an additional Hamiltonian W defined by:
W|\phi_A> = -a|\phi_B>
W|\phi_B> = -a|\phi_A> - a|\phi_C>
W|\phi_C> = -a|\phi_B>
where a is a real positive constant.
a) Calculate the energies and stationary states of the Hamiltonian H = H_0 + W.
2. The attempt at a solution
H_0 = E_0 \[ \left( \begin{array}{ccc}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \end{array} \right)\]
W = -a\[ \left( \begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> 1 & 0 & 1 \\<br /> 0 & 1 & 0 \end{array} \right)\]
H = H_0 + W = E_0 \[ \left( \begin{array}{ccc}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \end{array} \right)\] -a\[ \left( \begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> 1 & 0 & 1 \\<br /> 0 & 1 & 0 \end{array} \right)\]
H = \[ \left( \begin{array}{ccc}<br /> E_0 & -a & 0 \\<br /> -a & E_0 & -a \\<br /> 0 & -a & E_0 \end{array} \right)\]
When I try to calculate the eigenvalues of this matrix to get the energies, I end up with an algebraic mess that involves a cubic function for \lambda and I'm not sure how to solve, so I think I'm probably on the wrong track.
