# Taylor expansion of $H = \sqrt{m^2 - \hbar^2 \nabla^2}$

$$H = \sqrt{m^2 - \hbar^2 \nabla^2}$$

Suppose that there is a equation like this. How do you taylor-expand this equation? I am extremely confused.

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Use the expansion of $\sqrt{1-x}$ about $x=0$, $$\begin{eqnarray*} H &=& \sqrt{m^2 - \hbar^2 \nabla^2} \\ &=& m\sqrt{1-\frac{\hbar^2\nabla^2}{m^2}} \\ &=& m\left(1-\frac{1}{2}\frac{\hbar^2\nabla^2}{m^2} - \frac{1}{8} \frac{\hbar^4\nabla^4}{m^4} + \ldots\right) \\ &=& m -\frac{\hbar^2\nabla^2}{2m} - \frac{\hbar^4\nabla^4}{8m^3} - \ldots \end{eqnarray*}$$ If we put back the factors of $c$ we'll find this is an expansion in small $\hbar \nabla/(m c)$, i.e., in large $c$. Keeping the first two terms we recover Schrödinger's Hamiltonian for a free particle (up to an additional, unphysical constant).