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Apr 13, 2017 at 12:39 history edited CommunityBot
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Jul 1, 2015 at 5:07 vote accept Physics_Plasma
Jun 27, 2015 at 19:05 history edited Qmechanic CC BY-SA 3.0
Added explanation
Jun 27, 2015 at 18:45 comment added Physics_Plasma Oh, I might see it now. If the energy E that is remaining is not including the rest energy, then $E = \frac{p^2}{2m}$. Since we are investigating the case where $E\ll m$, then given E is just kinetic energy, then we get $E \ll \frac{p^2}{2E}$, or $E \ll p$.
Jun 27, 2015 at 18:10 comment added Qmechanic Only indirectly.
Jun 27, 2015 at 17:48 comment added Physics_Plasma Ok, so if I plug the plane wave expansion into the Lagrangian and keep the terms in the same order as they are in my second equation from the bottom, then I get $\mathcal{L} = E\Psi^{\dagger}\Psi + \frac{E^2}{2m}\Psi^{\dagger}\Psi - \frac{p^2}{2m}\Psi^{\dagger}\Psi -\frac{\lambda}{4m^2}(\Psi^{\dagger}\Psi)^2. $ I can clearly see how the second term is negligible compared to the first. Is there a reason why the second term is negligible compared to the third term, with the square of the momentum?
Jun 27, 2015 at 9:02 history edited Qmechanic CC BY-SA 3.0
added 19 characters in body
Jun 27, 2015 at 8:49 history answered Qmechanic CC BY-SA 3.0