I solved for the bound state energies of a system using the Dirac equation, I was told that to compare my result to the energies obtained with the Schrodinger equation I need to do the non-relativistic limit of that result. It should be similar to this question but I didn't manage to do it. Here's the expression
$$ E^2 = \displaystyle\frac{n^2\pi^2\hbar^2c^2}{L^2} + m^2c^4 \; . $$
The relativistic energy squared is a multiple of mc^2 always. So you have
$$ E^2= m^2 \ c^4 \ \left(1+\frac{h^2 n^2}{c^2 L^2 m^2} \right) $$
The relativistic correction to the rest mass term is small, so use the expansion of the square root
$$ E = m c^2 \ \left(1+ \frac{h^2 n^2}{2 c^2 L^2 m^2}\right) = m c^2 +\frac{1}{2m } \left( \frac{n h}{L}\right)^2 $$
$$ p =\frac{n h}{L}$$
has the dimension of momentum, so the nonrelativistic kinetic energy is as usual $p^2/(2m)$, where p is the absolute value of momentum of a standing wave in an intervall of length $L$ with n nodes.
