Question

1- 5. Two particles each of mass m are fixed at the end of a rigid rod of length 2a. This rod lies in the xy plane and is free to rotate in that plane about an axis passing through the midpoint of the rod and perpendicular to it (that is, parallel to the z-axis). Neglect the inertial properties of the rod in the rest of this question z-axis 1. Derive the classical expression for the kinetic energy of the system in terms of its angular momentum and moment of inertia. 2. Using your expression from Q1, show carefully and fully you can now deduce the Schrodinger equation for this system. To get any credit, you must show from first principles how you deduce the appropriate operator for the angular momentum 3. Solve the Schrodinger equation for this system, giving correctly normalized wavefunctions and showing that the energies vary as E, E where n is an integer. Make sure that you clearly express the constant E, in terms of m, a, and other constants 4. Now suppose the particle have charges of +q and -q, and the rotor is in a region of uniform electric field ε-ie. Derive the new Hamiltonian for the rotor (do NOT attempt to solve it.) 5. Treat the interaction of the charges with the electric field as a perturbation to the energies given in Q3, and find the perturbed energies correct to second order, explaining clearly why, although the unperturbed states are degenerate, non-degenerate perturbation theory can be used in this case.

Answer 1

I (system) = I(rod) +I( bobs)

=1/12 * ML²+ 2 MR²  

here let mass of rod= M

mass of bob = m

length of rod= 2a

I = 1/12* M * 2a² + 2 m r²  

again, rotational kinetic energy = $1/2 * I\omega 2$

omega in terms of angular momentum would be,

L = m v r =m r² w

w=l/ mr²

putting the value of I and w you can get the value of kinetic energy of the system