Question

A molecule is made up of three identical atoms at the corners of an equilateral triangle as shown in Fig. We consider its ion to be made by adding one electron with some amplitude on each site. Suppose the matrix element of the Hamiltonian for the electron on two adjacent sites i, j is \(\langle i|H|j\rangle\) = -a for i \(\ne\) j.

(a) Calculate the energy splittings.

(b) Suppose an electric field in the .z direction is applied, so that the potential energy for the electron on top is lowered by b with |b| <<|a|. Now calculate the levels.

(c) Suppose the electron is in the ground state. Suddenly the field is rotated by 120° and points toward site 2. Calculate the probability for the electron to remain in the ground state.

Answer 1

(a) Denote the basis vectors by \(|1\rangle,|2\rangle,|3\rangle\) and let \(\langle i|H|i\rangle = E_0\), i = 1,2,3. Then

The solution gives energy levels E_1,2 = E₀ + a (two-fold degenerate} and E₃ = E₀ - 2a.

(b) The H matrix is now

E₂ has the lowest energy and thus corresponds to the ground state, with wave function

(c) After the rotation of the field the system has the same configuration as before but the sites are renamed: 

1 → 2, 2 → 3, 3 → 1.

Hence the new ground state is

Hence the probability for the electron to remain in the ground state is