On Quantum Operators as Statistical Random Variables Defined on a Spinning Roulette

A quantum operator is a classical physical variable except that it assumes values in a probability distribution, i.e., a random variable, solved by the Schrödinger equation in the form of a wavefunction. To date, the logical foundation of this construct remains an open question. This paper casts the basic components of quantum mechanics in the framework of general statistics. Our starting point is to define the domain of the wavefunction to be the complex unit circle, thus faithfully observing the periodicity of a wave. We seek to crystallize the meanings of the fundamental elements of quantum mechanics, which often are lost in its formalism, by certain common random variables in ordinary applications, such as random walks, inventory cycles, and human healthy temperatures, all defined on the sample space of a unit circle.

Introduction

This paper casts the fundamental construct of the wavefunction $\psi$ with its underlying operator $H$ in quantum mechanics [1] into the framework of a random variable in statistics with a view to applying it to other disciplines [2]. A common property of many measurable variables is their periodicities, i.e., as waves. By definition, a wave repeats itself, thus like a spinning roulette, $e^{i\theta }$, cycle after cycle, such as an inventory cycle in business.

In the following Section 2, we will begin with a solution of the wavefunction $\psi$ for the appearance of a free photon in a $2\pi$-cycle as from its underlying electromagnetic wave, where we will translate all the involved quantum terms, such as the Schrödinger equation with its associated Hamiltonian, into their corresponding statistical constructs. We then proceed to additional illustrations of these quantum concepts by examples of a binary random variable as in a random walk, a uniform distribution of a declining inventory, and a normal distribution of human temperatures limited to within six sigmas as centered at their mean. In Section 3, we will present a summary exhibiting an explicit correspondence between all the quantum foundational terms and their statistical counterparts by the above binary random variable.

Analysis

As all elementary material particles can be produced by pair-creations of photons, we will base our analysis on a free photon. In Light [3], the photon is created by an electromagnetic wave (“EMW”), which has the following composition as from the Maxwell equations, as (1) shows: where $\mathbf{\text{E}}=$ the electric field, $\mathbf{\text{B}}=$ the associated magnetic field, and $S=\mathbf{\text{E}}\times \mathbf{\text{B}}\equiv$ the Poynting vector.

(1) $$\begin{array}{rl}EMW& =\mathbf{\text{E}}\oplus \mathbf{\text{B}}\oplus S\\ & =({\Vert \mathbf{\text{E}}\Vert }_{max}\cos (kx-\omega t),\frac{1}{c}{\Vert \mathbf{\text{E}}\Vert }_{max}\cos (kx-\omega t),c{\epsilon }_o{\Vert \mathbf{\text{E}}\Vert }_{max}^2{\cos }^2(kx-\omega t)),\end{array}$$

To observe relativistic invariance, we will define the unit of time to be the duration of the spin of EMW by exactly one radian, i.e., as in with $\omega :=radian/radian\equiv 1$ so that time $t\equiv$ distance $x$ (in the unit of the radius of EMW by whichever the reference frame happens to be).

(2) $$\begin{array}{r}(\cos t,\sin t)\equiv \cos t+i\sin t=e^{it}\end{array}$$

As the above (1) shows, the Poynting vector $S$, containing ${\cos }^2$, renders two energy cycles in any $2\pi$-cycles. Hence, we set up a Hamiltonian $H\equiv$ energy $E=2\hslash \omega \equiv 2\hslash$ for the photon (cf. [3], [4]) that manifests its presence at some $t\equiv x\in [-\pi ,\pi ]$, and solve the following Schrödinger equation, as (3) shows: where the factor $\pm i$ accounts for the two spin directions $\pm z$ of EMW due to $\cos (kx-\omega t)=\cos (\omega t-kx)$ as contained in the Poynting vector $S$ of EMW. Then where ${\displaystyle \frac{1}{\sqrt{\pi }}}\cos t$ is the probability amplitude for the photon to appear at $t\in [-\pi ,\pi ]$ in consistency with the Poynting vector $S$, as (1) shows, i.e., ${\int }_{-\pi }^{\pi }1/\pi {\cos }^{2}tdt=\left(1/\pi \right)\cdot \left({1/2t+1/4\sin 2t]}_{-\pi }^{\pi }\right)=1$.

(3) $$\begin{array}{r}\pm i\hslash \frac{d}{dt}\psi \left(t\right)=H\psi \left(t\right)=2\hslash \psi \left(t\right)\end{array}$$

(4) $$\begin{array}{rl}& \frac{d}{dt}\ln \psi \left(t\right)=\mp 2i,\text{ or}\\ & \psi \left(t\right)=c_0{e}^{\mp 2it}=\frac{1}{\sqrt{\pi }}{e}^{\mp it}(\frac{1}{2}{e}^{-it}+\frac{1}{2}{e}^{it})\\ & (\text{where }{c}_0=\frac{1}{\sqrt{\pi }}\text{as derived below},\text{ and}{e}^{\mp it}\equiv \frac{1}{2}{e}^{-it}+\frac{1}{2}{e}^{it}\text{as composed via }\text{a}\text{superposition})\\ & =\frac{1}{\sqrt{\pi }}\cos t\cdot e^{\mp it}\end{array}$$

Here we note that the common rendition of the Schrödinger equation, as (5) shows, does not apply to a free photon, which has $m=0$ (cf. [5]), with $V=0$ (cf. [6]).

(5) $$\begin{array}{r}i\hslash \frac{\mathrm{\partial }\psi }{\mathrm{\partial }t}=-\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}^2\psi +V\psi \end{array}$$

Cast in the general setting of a statistical random variable, the above calculation renders the following identification:

The sample space $=[-\pi ,\pi ]\equiv S^1$ (a unit circle) with probability measure $p\left(\theta \right)=1/\pi {\cos }^2\theta$ for the random variable photon’s energy appearances,

$$\begin{array}{rl}{E}_{\theta }& :=\left\{2\hslash |appearingat\theta \in [-\pi ,\pi ]asthephoton\gamma \right\}\\ & \equiv positionof\gamma at\mathrm{\Theta }:=\left\{\theta |\theta \in [-\pi ,\pi ]\right\}\end{array}$$

The above general form of the Schrödinger equation, as (3) shows, has the advantage of its applicability to diverse fields, but $H$ without the algebraic form [7] cannot yield a solution for the probability distribution of the random variable in question (our earlier solution for the free photon’s positions, as (4) shows, was only guided by a priori knowledge of the Poynting vector $S$). In this connection, consider $\psi \left(\theta \right)=f\left(\theta \right)e^{i\theta },$ where $\theta \in [0,2\pi ]$ and $f\left(\theta \right)=$ the probability amplitude at $\theta$ of an underlying random variable $X.$ Then

$$\begin{array}{rl}-i\frac{d}{d\theta }\psi \left(\theta \right)& =-i\cdot f^{{ }^{\prime }}\left(\theta \right)e^{i\theta }+f\left(\theta \right)e^{i\theta }\\ & =(-i\cdot \frac{f^{{ }^{\prime }}\left(\theta \right)}{f\left(\theta \right)}+1)\psi \left(\theta \right),f\left(\theta \right)\ne 0\\ & \equiv H\psi \left(\theta \right)\end{array}$$

That is, without any information on $f^{{ }^{\prime }}\left(\theta \right)/f\left(\theta \right)$, the equation cannot be solved. As such, in the following we will illustrate the Schrödinger equation by simply imposing a given Hamiltonian $H$.

Our first example is that of a binary random variable $X\left(\theta \right)\in \{0,1\},\theta \in [0,2\pi ],$ denoting any two alternative quantitative or qualitative outcomes in a Bernoulli trial, such as a random walk, with probability $P\left(0\right)=w_1$ and $P\left(1\right)=1-w_1\equiv w_2$ over the sample space $\left\{\theta |\theta \in [0,2\pi ]\right\}$, where $w_1=$ the probability measure $\nu$ of $[0,\pi )$ and $w_2=\nu \left([\pi ,2\pi )\right)$. Then the Schrödinger equation is:

so that

$$\begin{array}{r}-i\frac{d}{d\theta }\psi \left(\theta \right)=\left(\begin{array}{ll}{w}_1& 0\\ 0& w_2\end{array}\right)\psi \left(\theta \right)\end{array}$$

here in particular if $w_1=w_2=1/2$, then

$$\begin{array}{rl}& \frac{d}{d\theta }{\psi }_j\left(\theta \right)=i{w}_j{\psi }_j\left(\theta \right),j=1,2,and\\ & {\psi }_j\left(\theta \right)=c_j{e}^{i{w}_j\theta }=\sqrt{w_j}\cdot e^{i{w}_j\theta }\end{array}$$

is the wavefunction of a fair coin of “H” or “T,” or a quantum spin of up or down (as implied by the Pauli matrix ${\sigma }_z$ incidentally). Note that $X:\theta \in [0,2\pi ]\mapsto x\in \{0,1\}$ corresponds to $X:\omega t\in [0,2\pi ]\mapsto \hslash \omega t=E\cdot t$. As such, $X^{-1}\left(\left\{x\right\}\right)$ corresponds to $Et/\hslash$; the exponent $i\cdot \left(1/2\right)\cdot X^{-1}\left(\left\{{\delta }_{ij}\right\}\right)$, as (6) shows, corresponds to $i\omega t$, where $w_j=1/2\equiv$ the relative frequency or probability $\equiv$ the angular frequency of the EMW $\equiv \omega$. A destructive wave interference [8] may be illustrated by two investment returns of their coefficient of correlation $\rho =-1$ [4], so that the net profit of this portfolio has $\psi \left(\theta \right)=f\left(\theta \right)e^{i\theta }-f\left(\theta \right)e^{i\theta }=0$.

(6) $$\begin{array}{r}{\psi }_j\left(X^{-1}\left(\left\{{\delta }_{ij}\right\}\right)\right)=\frac{\sqrt{2}}{2}{e}^{i\cdot \left(\frac{1}{2}\right)\cdot X^{-1}\left(\left\{{\delta }_{ij}\right\}\right)}\end{array}$$

As a second example, we consider a discrete uniform distribution of a declining inventory over a cycle of 30 days from 30 units to 1 unit. Then we define $X:[0,2\pi )\to \{30,29,\cdots ,1\}$ by $X:\left(2\pi /30\right)\cdot [k,k+1)\mapsto 30-k,k=0,1,\cdots ,29$, so that $\psi \left(X^{-1}\left(\left\{x\right\}\right)\right)={\displaystyle \frac{1}{\sqrt{30}}}{e}^{i{X}^{-1}\left(\left\{x\right\}\right)}$, with the Hamiltonian $H={\left(1/30\right)}_{diag,30\times 30}$.

Lastly, we consider a truncated normal distribution for $X=$ healthy human temperatures of range $=\mu \pm 3\sigma ={98}^{\circ }\pm 3\times {0.2}^{\circ }$, of six standard deviations centered at the mean. Then set $X:(-\pi ,\pi )\to ({97.4}^{\circ },{98.6}^{\circ })$, so that $X\left(0\right)={98}^{\circ }=\mu$ and $X\left([0,\pi /3]\right)={0.2}^{\circ }=\sigma$. Since

we have

$$\begin{array}{r}{\left({\int }_{-3}^3\frac{1}{\sqrt{2\pi }}{e}^{-0.5z^2}dz\right)}^{-1}\approx \frac{1}{0.9975}\approx 1.0025\end{array}$$

and

$$\begin{array}{r}{\left|\psi \left(X^{-1}\left(\left\{x\right\}\right)\right)\right|}^2=1.0025\times \left(\frac{1}{0.2\sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-98}{0.2}\right)}^2}\right)\end{array}$$

with the Hamiltonian

$$\begin{array}{r}\psi \left(X^{-1}\left(\left\{x\right\}\right)\right)=\left|\psi \left(X^{-1}\left(\left\{x\right\}\right)\right)\right|\cdot e^{i\cdot \left(\frac{x-98}{0.2}\right)\cdot \left(\frac{\pi }{3}\right)}\end{array}$$

which gives the probability density at $x\in (97.4,98.6)\iff \theta \in (-\pi ,\pi )$, i.e., the familiar bell curve but only erected above a circle as a 3-D graph. As such, we see that the essence of the Schrödinger equation, $i\hslash (d/dt)\psi (t\equiv \theta )=H\psi (t\equiv \theta )$, is to solve for $\pm \sqrt{p\left(\theta \right)}\cdot e^{ip\left(\theta \right)\theta }$, with $H={(p\left(\theta \right)\equiv \omega \left(\theta \right)\equiv {\displaystyle \frac{E}{\hslash }})}_{\theta \in S^1}$, and ${\int }_{-\pi }^{\pi }p\left(\theta \right)d\theta =1$; i.e., the exponent $ip\left(\theta \right)\theta$ in $e^{ip\left(\theta \right)\theta }$ depicts a histogram $(\theta ,p\left(\theta \right))$ erected above a unit circle.

$$\begin{array}{r}H={(1.0025\times \left(\frac{1}{0.2\sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-98}{0.2}\right)}^2}\right))}_{x\in (97.4,98.6)}\end{array}$$

Summary

We summarize the above analysis by a direct correspondence between (a) the foundational constructs of quantum mechanics: {state, observable, operator, its eigenvalues and eigenvectors, the wavefunction $\psi$, the Schrödinger equation, the Hamiltonian, the superposition principle} and (b) our earlier example of a binary random variable: $X\left(\theta \right)\in \{0,1\}$ with $P\left(0\right)=w_1=1/2=P\left(1\right)=w_2$.

The state vector [9], also known as the wavefunction $\psi$, is

with eigenvalues $0$ and $1$, the probabilities of which are $1/2$ and $1/2$.

$$\begin{array}{rl}& \psi (\theta \in [0,2\pi ])\equiv \psi \left(X^{-1}\left(\{0,1\}\right)\right)\\ & =\frac{\sqrt{2}}{2}{e}^{i{X}^{-1}\left(\left\{0\right\}\right)}+\frac{\sqrt{2}}{2}{e}^{i{X}^{-1}\left(\left\{1\right\}\right)}\\ & =\frac{\sqrt{2}}{2}{e}^{i\theta }|{ }_{\theta \in [0,\pi )}+\frac{\sqrt{2}}{2}{e}^{i\theta }|{ }_{\theta \in [\pi ,2\pi )}\\ & =\text{superposition of eigenvectors }\left(\begin{array}{c}{e}^{i{X}^{-1}\left(\left\{0\right\}\right)}\\ 0\end{array}\right)\text{ and }\left(\begin{array}{c}0\\ e^{i{X}^{-1}\left(\left\{1\right\}\right)}\end{array}\right)\end{array}$$

The Hamiltonian $H={(P\left(0\right)=1/2,P\left(1\right)=1/2)}_{diag,2\times 2}$ is the underlying operator, of its own eigenvalues, $1/2$ and $1/2$ for the probability distribution of the observable $0$ or $1$.

An eigenstate is a phase $\left\{\theta |\theta \in [0,\pi )\right\}$ or $\left\{\theta |\theta \in [\pi ,2\pi )\right\}$, and a general state is a linear combination of the eigenstates, which also satisfies the Schrödinger equation $i(d/d\theta )\psi \left(\theta \right)=H\psi \left(\theta \right)$; its general solution $\psi \left(\theta \right)=f\left(\theta \right)e^{i\theta }$ can be considered as a spinning roulette - - when it stops (correspondingly, after one energy cycle of $\pi -radian$ in ${\cos }^2(kx-\omega t)$ the particle must manifests itself), $\psi$ collapses to a particular eigenstate with the outcome of the “Schrödinger cat” [10], [11] determined whether being observed or not [12]. The Schrödinger equation is based on Planck’s energy formula $E=\hslash \omega$, where $\omega \equiv$ angular frequency $=$ relative frequency or probability; its remarkableness lies in its service to solve for the probability distribution of a random variable.