Replacing $E$ and $p$ by $i\hbar\frac{\partial}{\partial t}$ and $-i\hbar\nabla$, respectively, in the energy-momentum relation

$$E^2=p^2c^2+m_0^2c^4$$

we obtain the *Klein-Gordon operator*

$$\square=\frac{m_0^2c^2}{\hbar^2},$$

where

$$\square=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2$$

$\square$ is called the *d’Alembertian* or the *wave operator*. This time, replacing $E$ and $p$ by $i\hbar\frac{\partial}{\partial t}$ and $-i\hbar\nabla$ in the square root of the energy-momentum relation

\begin{equation}

\label{eq:sqrte-m}

E=\sqrt{p^2c^2+m_0^2c^4}

\end{equation}

we obtain

\begin{equation}

\label{eq:sqrtkg}

i\hbar\frac{\partial}{\partial t}=\sqrt{-\hbar^2c^2\nabla^2+m_0^2c^4}

\end{equation}

\eqref{eq:sqrtkg} is called the square root of Klein-Gordon operator. I said mathematics is full of weird stuff here but so is physics. The RHS of \eqref{eq:sqrtkg} is pretty weird. It is supposed to be an operator, meaning supposed to act on a wave function. But how does it do that? Does it even exist? I read or heard somewhere (sorry I don’t remember details) that Schrödinger studied the equation

\begin{equation}\label{eq:sqrtkg2}i\hbar\frac{\partial\psi(\mathbf{r},t)}{\partial t}=\sqrt{-\hbar^2c^2\nabla^2+m_0^2c^4}\psi(\mathbf{r},t)+V(\mathbf{r})\psi(\mathbf{r},t)\end{equation}

before he discovered his celebrated equation, the one we now call the Schrödinger equation. What I vaguely remember is that the equation was too difficult or inconvenient to deal with so he eventually gave up on it. I am not really privy to the history of quantum mechanics let alone the history of physics in general, so I couldn’t find a reference to this. But that is not important anyway. What’s important is to make sense of the operator $\sqrt{-\hbar^2c^2\nabla^2+m_0^2c^4}$. Let us go back to the square root of the energy-momentum relation \eqref{eq:sqrte-m}. With $p^2=m^2v^2$ and $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$, we have

\begin{align*}

E&=\sqrt{p^2c^2+m_0^2c^4}\\

&=m_0c^2\sqrt{1+\frac{p^2}{m_0^2c^2}}\\

&=m_0^2c^2\sqrt{1+\left(\frac{m}{m_0}\right)^2\left(\frac{v}{c}\right)^2}

\end{align*}

Since $v<c$,

\begin{align*}

\left(\frac{m}{m_0}\right)^2\left(\frac{v}{c}\right)^2&=\frac{1}{1-\frac{v^2}{c^2}}\frac{v^2}{c^2}\\

&=\left(1+\frac{v^2}{c^2}+\frac{v^4}{c^2}+\frac{v^6}{c^6}+\cdots\right)\frac{v^2}{c^2}

\end{align*}

For a particle with speed $v$ slower than $c$, certain high-order terms of $\frac{v^2}{c^2}$ can be negligible. This means that, mathematically speaking, $\left(\frac{m}{m_0}\right)^2\left(\frac{v}{c}\right)^2=\frac{p^2}{m_0^2c^2}$ is nilpotent. Also, if any physical effect of certain high-order terms of $\frac{v^2}{c^2}$ is not observed from experiments, we can safely assume that $\left(\frac{m}{m_0}\right)^2\left(\frac{v}{c}\right)^2=\frac{p^2}{m_0^2c^2}$ is nilpotent. In here, we proved that if $N$ is a nilpotent operator, $I+N$ has a square root and it is given by

$$\sqrt{I+N}=I+\frac{1}{2}N-\frac{1}{8}N^2+\frac{1}{16}N^3-\cdots$$

So, $\sqrt{p^2c^2+m_0^2c^4}$ can be understood as an operator as

\begin{align*}

\sqrt{p^2c^2+m_0^2c^4}&=m_0c^2\sqrt{1+\frac{p^2}{m_0^2c^2}}\\

&=m_0c^2\left(1+\frac{p^2}{2m_0^2c^2}-\frac{1}{8}\frac{p^4}{m_0^4c^4}+\frac{1}{16}\frac{p^6}{m_0^6c^6}+\cdots\right)\\

&=m_0c^2+\frac{p^2}{2m_0}-\frac{1}{8}\frac{p^4}{m_0^3c^2}+\frac{1}{16}\frac{p^6}{m_0^5c^4}+\cdots

\end{align*}

For example, in one-dimensional case, since $\hat p=-i\hbar\frac{\partial }{\partial x}$, we have

$$\sqrt{-\hbar^2c^2\frac{\partial^2}{\partial x^2}+m_0^2c^4}=m_0c^2-\frac{\hbar^2}{2m_0}\frac{\partial^2}{\partial x^2}-\frac{1}{8}\frac{\hbar^4}{m_0^3c^2}\frac{\partial^4}{\partial x^4}-\frac{1}{16}\frac{\hbar^6}{m_0^5c^4}\frac{\partial^6}{\partial x^6}+\cdots$$

where $\frac{\partial^2}{\partial x^2}$ is assumed to be nilpotent.

*Update*: While \eqref{eq:sqrtkg} is Lorentz invariant, one cannot add a potential energy like \eqref{eq:sqrtkg2} in Lorentz invariant way. This is another reason square root of Klein-Gordon equation was not favored by physicists. The relativistically corrected $s$-wave equation in the paper *Relativistically corrected Schrödinger equation with Coulomb interaction* by J. L. Friar and E. L. Tomusiak, Physical Review C, Volume 29, Number 4, 1537, is not really relativistic as it is not Lorentz invariant.