Question

A particle of mass m moves one-dimensionally in the oscillator potential V(x) = 1/2 mω²x². In the nonrelativistic limit, where the kinetic energy T and momentum p are related by T = p²/2m, the ground state energy is well known to be 1/2 hω.

Allow for relativistic corrections in the relation between T and p and compute the ground state level shift \(\Delta E\) to order \(\frac 1{c^2}\) (c =speed of light).

Answer 1

In relativistic motion, the kinetic energy T is

to order \(\frac 1{c^2}\). The \(\frac{-p^4}{8m^3c^2}\) term may be considered as a perturbation. Then the energy shift of the ground state is