Question

2. Consider a particle of mass M attached to a rigid massless rod of fixed length R whose other end is fixed at the origin. The rod is free to rotate about its fixed point. (a) Give an argument why the Hamiltonian for the system may be written as 21 21 with/-MR2 (b) If the particle carries charge q, and the rotor is placed in a constant magnetic field B, what is the modified Hamiltonian? (e) What is the energy spectrum for small B?

Answer 1

The motion of the particle is in circular motion. The angular momentum is, L = R times P = R p sin 90 = Rp (1) = R Mv = R^2 Mv/R = MR^2 [v/R] = I omega Here, the total energy of the system is in the form of kinetic energy. E = 1/2 Mv^2 = 1/2 (Mv)^2/M = p^2/2M The Hamiltonian of the system is equal to the total energy of the system. From Classical mechanics the total energy of the system is given as, H = p^2/2M = (Mv)^2/2M = Mv^2/2 = M(R omega)^2/2 = 1/2 (MR^2) omega^2 = 1/2 I omega^2 = (1/2 I omega^20(I/I) = 1/2 (I omega)/I = L^2/2 I Hence, the Hamilton of the system is written as H = L^2/2 I = (R times p)^2/2 I The modified form of Hamiltonian is, H = (R rightarrow times(p rightarrow - q A rightarrow))^2/2 I When the particle having charge q and rotor is placed in