MathPhys

There is a new TV series titled Constellation on Apple TV+. The plot so far (I don’t intend to be a spoiler and I don’t actually know much about the plot either because there have been only three episodes) is that a Swedish astronaut Jo Ericsson (portraited by Noomi Rapace) has returned to Earth after a catastrophic disaster led by a weird circumstance hit the ISS. But somethings in Jo’s life are amiss after her return home …

The show is entertaining, but more interestingly it appears to embrace Hugh Everett’s many-worlds interpretation of quantum mechanics. This perspective of quantum mechanics assumes the existence of the universal wavefunction which is the wave function of the entire universe and it never collapses unlike wave functions in Copenhagen interpretation. In this perspective, all the possible quantum states as different realities exist simultaneously, hence the name many-worlds interpretation. Another peculiar aspect of this perspective is that the observers are part of the wavefunction, i.e. the observers and the observed are all mixed together. What this implies is that there is no measurement in many-worlds interpretation. The proponents of many-worlds interpretation seem to support the idea of philosophical realism regarding the universal wavefunction, that it exists even in the absence of any mind perceiving it. It is an interesting concept but I am not buying it. Perhaps no one cares whether I believe it and this is merely my own opinion. First, my impression is that often physicists don’t seem to distinguish hypotheses and facts, and mathematical entities and physical entities (that we can actually measure and/or observe). Wave functions are not physical entities but merely convenient mathematical entities to represent the wave nature of particles and their quantum states. Extending the wave nature of matter (consequently, the probabilistic nature of quantum mechanics) to the macroscopic world or to the entire universe is too far-fetched to be even remotely true. More importantly, you cannot have a viable physical theory without considering measurement.  If there is no measurement, the theory is not even falsifiable, quoting Wolfgang Pauli, it is not even wrong.

If you don’t care about all this, I think you can enjoy the show.

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.