Linear Triatomic Molecule Hamiltonian Eigenvalues

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SUMMARY

The discussion focuses on calculating the eigenvalues and stationary states of the Hamiltonian for a linear triatomic molecule with three equidistant atoms, represented by the states |\phi_A>, |\phi_B>, and |\phi_C>. The Hamiltonian is defined as H = H_0 + W, where H_0 is a diagonal matrix with eigenvalue E_0 and W is a coupling matrix involving a real positive constant 'a'. The resulting Hamiltonian matrix is H = [[E_0, -a, 0], [-a, E_0, -a], [0, -a, E_0]]. The eigenvalues can be determined by solving the cubic equation derived from this matrix.

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Homework Statement


Consider an electron of a linear triatomic molecule formed by three equidistant atoms. We use [tex]|\phi_A>, |\phi_B>, |\phi_C>[/tex] 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 [tex]|\phi_A>, |\phi_B>, |\phi_C>[/tex].

When we neglect the possibility of the electron jumping from one nucleus to another, its energy is described by the Hamiltonian [tex]H_0[/tex] whose eigenstates are the three states [tex]|\phi_A>, |\phi_B>, |\phi_C>[/tex] with the same eigenvalue [tex]E_0[/tex]. The coupling between states [tex]|\phi_A>, |\phi_B>, |\phi_C>[/tex] is described by an additional Hamiltonian W defined by:

[tex]W|\phi_A> = -a|\phi_B>[/tex]
[tex]W|\phi_B> = -a|\phi_A> - a|\phi_C>[/tex]
[tex]W|\phi_C> = -a|\phi_B>[/tex]

where a is a real positive constant.

a) Calculate the energies and stationary states of the Hamiltonian [tex]H = H_0 + W[/tex].

2. The attempt at a solution

[tex]H_0 = E_0 \[ \left( \begin{array}{ccc}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \end{array} \right)\][/tex]

[tex]W = -a\[ \left( \begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> 1 & 0 & 1 \\<br /> 0 & 1 & 0 \end{array} \right)\][/tex]

[tex]H = H_0 + W = E_0 \[ \left( \begin{array}{ccc}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \end{array} \right)\][/tex] [tex]-a\[ \left( \begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> 1 & 0 & 1 \\<br /> 0 & 1 & 0 \end{array} \right)\][/tex]

[tex]H = \[ \left( \begin{array}{ccc}<br /> E_0 & -a & 0 \\<br /> -a & E_0 & -a \\<br /> 0 & -a & E_0 \end{array} \right)\][/tex]

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 [tex]\lambda[/tex] and I'm not sure how to solve, so I think I'm probably on the wrong track.
 
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You're on the right track, you have to solve the cubic function in [tex]\lambda[/tex] to get the eigenenergies and their associated wave functions.
 

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