Consider a particle of mass m constrained to move in the xy-plane on a circular ring...

 

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Consider a particle of mass m constrained to move in the xy-plane on a circular ring of radius a. The only variable of the system is the azimuthal angle o. The state of the system is described by a wave function () that must be periodic with period 27. The energy eigenstates are Un (0) = (on) = En = ħ²n² 2ma² (2) E 1 2π eing Ein². The ground state n = 0 has zero energy and is non-degenerate. All other states are doubly-degenerate because the states with in have the same energy level. = The particle now has a charge q and is placed in a uniform electric field & in the x-direction. We describe this by adding to the Hamiltonian the perturbation 3 E(2) F1-m nez E₁ = H' = -qEa cos Consider the degenerate ±1) states. (a) (4 points) Consider the first order energy shifts. Show that even with this perturbation the energy eigen- values are degenerate. Hint: Express coso as 1/2(ei + e-i) and consider what it does to each state |±1) in the basis, i.e. (o| H' |±1) Σ k#+1 (b) (4 points) Since the first order energy shift is zero, find the second order energy shift for the system in the 1) basis. Use the formula: ħ² 2ma² (±1| H' |k) (k| H' |±1) Et1 - Ek And compute both E(²), E(2). (c) (10 points) You should find that the second order energy shift is still degenerate! Time to do "degenerate perturbation theory". Consider the cross terms, that is write down the action of H' to second order in the |+1)-subspace. Do this by computing (1| H' |k) (k| H' \m) E₁ - Ek Σ k#+1 for the two remaining combinations of I and m, 1 = 1, m = -1 and 1= -1, m = 1. Consider a particle of mass m constrained to move in the xy-plane on a circular ring of radius a. The only variable of the system is the azimuthal angle o. The state of the system is described by a wave function () that must be periodic with period 27. The energy eigenstates are Un (0) = (on) = En = ħ²n² 2ma² (2) E 1 2π eing Ein². The ground state n = 0 has zero energy and is non-degenerate. All other states are doubly-degenerate because the states with in have the same energy level. = The particle now has a charge q and is placed in a uniform electric field & in the x-direction. We describe this by adding to the Hamiltonian the perturbation 3 E(2) F1-m nez E₁ = H' = -qEa cos Consider the degenerate ±1) states. (a) (4 points) Consider the first order energy shifts. Show that even with this perturbation the energy eigen- values are degenerate. Hint: Express coso as 1/2(ei + e-i) and consider what it does to each state |±1) in the basis, i.e. (o| H' |±1) Σ k#+1 (b) (4 points) Since the first order energy shift is zero, find the second order energy shift for the system in the 1) basis. Use the formula: ħ² 2ma² (±1| H' |k) (k| H' |±1) Et1 - Ek And compute both E(²), E(2). (c) (10 points) You should find that the second order energy shift is still degenerate! Time to do "degenerate perturbation theory". Consider the cross terms, that is write down the action of H' to second order in the |+1)-subspace. Do this by computing (1| H' |k) (k| H' \m) E₁ - Ek Σ k#+1 for the two remaining combinations of I and m, 1 = 1, m = -1 and 1= -1, m = 1.

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Part A Using first order perturbation theory EI 12 ffdd u where we have ... View the full answer