Is Perturbation Theory Enough for Relativistic Corrections?
| Main Author: | Francesco R. Ruggeri |
|---|---|
| Format: | info publication-preprint eJournal |
| Terbitan: |
, 2019
|
| Subjects: | |
| Online Access: |
https://zenodo.org/record/3459745 |
Daftar Isi:
- In some recent articles in the literature (1),(2) the Schrodinger equation: p*p/2m W(x) + V(x) W(x) = E W(x) ((1)) is replaced with an equation using a relativistic momentum operator: [sqrt(p*p + m*m) - m] W(x) + V(x)W(x) = E W(x) ((2)) The first operator on the LHS of ((2)) is then expanded in powers of 1/(m*m). Next, p is replaced with hbar/i d/dx (or gradient in 3D) and perturbation theory used. In this note, we consider p*p as being a conditional average given by: p*p= [Integral dp f(p) exp(ipx) p*p ] / [Integral dp f(p) exp(ipx)] ((3)) The denominator is the wavefunction W(x). This result should hold for the nonrelativistic case. In the relativistic case, p is replaced everywhere with prel where prel= p/ sqrt(1-p*p/(m*m)) prel*prel = Integral dprel p*p (1+ p*p/(m*m)) f(prel) exp(iprel x) ((4)) And W(x) = Integral dprel f(prel) exp(iprel x) ((5)) To first order, we argue that Integral dprel f(prel) exp(iprel x) = Integral dp f(p) exp(i p x), but Integral p*p f(prel) dprel exp(i prel x) contributes a term . For higher orders, even Integral dprel f(prel) exp(i prel x) contributes. Thus, we try to argue that one does not only have a change in the prel*prel operator, but one must consider changes of prel within Integral dprel f(prel) exp(iprel x) itself. Thus, perturbation theory may not be enough in such cases.