# Triatomic molecule

1. Nov 11, 2007

### ultimateguy

1. The problem statement, all variables and given/known data
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

$$H_0 = E_0 $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$$

$$W = -a$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)$$$

$$H = H_0 + W = E_0 $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$$ $$-a$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)$$$

$$H = $\left( \begin{array}{ccc} E_0 & -a & 0 \\ -a & E_0 & -a \\ 0 & -a & E_0 \end{array} \right)$$$

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.

2. Nov 11, 2007

### Dr Transport

You're on the right track, you have to solve the cubic function in $$\lambda$$ to get the eigenenergies and their associated wave functions.