Fine structure of H, momentum-potential commutation

[Pengwuino]

I'm going through Maggiore's QFT text and he is doing the fine structure constant for the Hydrogen atom. While trying to get to the Schrodinger-like Hamiltonian, he comes to a point where he has a term proportional to $\epsilon * p^2$ where $\epsilon = (E - m)$ which eventually is the energy eigenvalues. So he has this part that i'll pretty much write down in its entirety:

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$$\left(\epsilon - {{p^2} \over {2m}} - V +\epsilon {{p^2} \over {8m^2}} + {{p^4} \over {16m^3}} + {{Vp^2}\over{8m^2}} - {{1}\over{4m^2}}(\bf{\sigma} \cdot \bf{p})V(\bf{\sigma} \cdot \bf{\sigma}) \right) \phi_S = 0$$

At lowest order, we have of course $\epsilon \phi_S \approx \left( {{p^2} \over {2m}} + V \right)\phi_S$. Therefore the term $\epsilon {{p^2}\over{8m^2}}$ in the above equation can be rewritten as

$$\epsilon {{p^2}\over{8m^2}} = {{p^2}\over{8m^2}} \epsilon \approx {{p^2}\over{8m^2}}\left( {{p^2}\over{2m}} + V \right)$$

Of course $\epsilon$ is a c-number and we can write it both to the left or to the right of $p^2$. When we substitute it with $p^2/(2m) + V$ the difference between writing it to the left or to the right is $\cal{O}(p^6/m^4)$ and therefore can be neglected, at the order at which we are working.

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The order we are working at is keeping terms up to $\cal{O}(p^4/m^3)$ and $\cal{O}(Vp^2/m^2)$.

The bolded statement does not make sense to me. My initial reaction was to say that it means $[p^2,V] = \cal{O}(p^4)$ which means it would introduce terms $\cal{O}(p^6)$ which are beyond what we want, but I have no idea how this could be or if this is even the correct interpretation. So what does this mean?

Also, at the beginning he makes the approximation of the eigenvalue being approximately the kinetic plus potential "at lowest order". Now, is this because we know that the relativistic energy to 2nd order is proportional to $p^4$ which means resulting terms would again be $\cal{O}(p^6)$ and we don't have to include them?

 

[mark726]

The bolded statement means that the difference between writing \epsilon to the left or right of p^2 is of order \cal{O}(p^6/m^4). This means that when we substitute \epsilon with p^2/(2m) + V, the resulting difference in terms will be of order \cal{O}(p^6/m^4), which is beyond the order we are working at (up to \cal{O}(p^4/m^3) and \cal{O}(Vp^2/m^2)). Therefore, this difference can be neglected.

The reason for making the approximation of the eigenvalue being approximately the kinetic plus potential "at lowest order" is because we are working at a certain order (up to \cal{O}(p^4/m^3) and \cal{O}(Vp^2/m^2)) and including higher order terms (such as \cal{O}(p^6)) would go beyond this order and be negligible. This is why we can neglect the terms \cal{O}(p^6) and only consider the terms up to \cal{O}(p^4/m^3) and \cal{O}(Vp^2/m^2).