Square Root of Klein-Gordon

Replacing $E$ and $p$ by $i\hbar\frac{\partial}{\partial t}$ and $-i\hbar\nabla$, respectively, in the energy-momentum relation
we obtain the Klein-Gordon operator
$$\square=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2$$
$\square$ 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
we obtain
i\hbar\frac{\partial}{\partial t}=\sqrt{-\hbar^2c^2\nabla^2+m_0^2c^4}
\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
Since $v<c$,
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
So, $\sqrt{p^2c^2+m_0^2c^4}$ can be understood as an operator as
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.

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