- Question 4 20 Points Let Us Compute The First Relativistic Correction To The Energies Of The Stationary States Of A Ha 1 (144.88 KiB) Viewed 13 times
Question 4 [20 points] Let us compute the first relativistic correction to the energies of the stationary states of a harmonic oscillator. Recall that in special relativity, the kinetic energy is given by T = √√p²c² + m²c4 - mc², which for small momenta (p << mc) can be expanded as T,= p² p4 2m 8m³c² E' = +... By considering only the first term p2/2m and the usual potential V = mwx², we found the eigenstates Un) of the harmonic oscillator with their corresponding energies En. Since the potential is the same in the relativistic case, we just need to compute the correction coming from the second term in T, proportional to p4. To first order, we can assume that the states n) do not change, so the correction to the energy is simply 1 8m³c2 (Vn/p²|yn) Compute En for general n. It will be very useful to use the expression of p in terms of raising and lowering operators. To save time, think carefully about which terms will contribute and only keep the nonzero ones.