Space Discretization in Quantum Physics and Precision Principle

Abstract

This paper explores the consequences of discretization of space in quantum mechanics. Specifically, I show that under normal conditions in space where the wave function is uniform across a quantum of space, Heisenberg’s uncertainty principle is identically satisfied, where the “normal conditions” are defined as $p_x\Delta x/\hbar \ll 1$ with $\Delta x$ characterizing the size of a quantum of position space or equivalently $\Delta p_x x/\hbar \ll 1$ with $\Delta p_x$ characterizing the size of a quantum of momentum space (this limiting behavior is also at the core of Nyquist sampling theorem; see Jerri (1977); Vaidyanathan (IEEE Trans. Circ. Syst. I Fundam. Theory Appl. 48(9), 1094–1109, 2001); Ozaktas et al. (2001)). Moreover, I derive an equation similar to Heisenberg’s uncertainty principle which imposes a constraint on the minimum values of “precision” of space (i.e. the maximum values for $\Delta x$ and $\Delta p_x$). Additionally, I show that in discretized space the simultaneous minimum position and minimum momentum uncertainties are unphysical.

Appendix

In this section I will prove two results I previously used.

1) Here a method of Lagrange multiplier is used to show that the minimum uncertainty wave function is

$$\Psi _{ijk}= \frac{1}{\sqrt{\Delta x\Delta y\Delta z}}\delta _{ii'}\delta _{jj'}\delta _{kk'}$$

And the maximum uncertainty wave function is

$$\Psi _{ijk}= \lim _{n\rightarrow \infty }\frac{1}{\sqrt{\Delta x\Delta y\Delta z}}\left[ \frac{1}{\sqrt{2}}\delta _{i(-n)}\delta _{j(-n)}\delta _{k(-n)}+\frac{1}{\sqrt{2}}\delta _{in}\delta _{jn}\delta _{kn}\right] $$

Suppose the coordinate system is chosen such that $\langle {x}\rangle =0$. The constrain function is $g(|\Psi _{-N-1}|,\dots ,|\Psi _N|)=(|\Psi _{-N-1}|^2+\dots +|\Psi _N|^2)\Delta x-1$ and the objective function is $f(|\Psi _{-N-1}|,\dots ,|\Psi _N|)=\sum _{i}((i+\frac{1}{2})^2 +\frac{1}{12})\Delta x^2|\Psi _i|^2\Delta x$. Then $\nabla g = [|\Psi _{-N-1}|,\dots ,|\Psi _N|]2\Delta x$ and $\nabla f = 2\Delta x^3[(-N-\frac{1}{2})^2|\Psi _{-N-1}|,\dots ,((N+\frac{1}{2})^2+\frac{1}{12})|\Psi _N|]$. One $\Psi$ that maximizes f is $\Psi _i=\frac{1}{\sqrt{\Delta x}}\left[ \frac{1}{\sqrt{2}}\delta _{i(-N-1)}+\frac{1}{\sqrt{2}}\delta _{iN}\right]$ because it satisfies g and satisfies $\lambda \nabla g = \nabla f$ $\forall i$. First I will show $\lambda \nabla g = \nabla f$ is satisfied for $i=-N-1$

$$\begin{aligned} 2\lambda |\Psi _{-N-1}|&=((-N-\frac{1}{2})^2+\frac{1}{12})|\Psi _{-N-1}| 2\Delta x^3 \end{aligned}$$

So, $\lambda =((-N-\frac{1}{2})^2+\frac{1}{12}) \Delta x^3$. In order for $\lambda \nabla g = \nabla f$ to be satisfied for $i=N+1$ this lambda has to also equal to $((N+\frac{1}{2})^2+\frac{1}{12})\Delta x^3$ (and it does).

$$\begin{aligned} \lambda&=((-N-\frac{1}{2})^2+\frac{1}{12})\Delta x^3\\&= ((N+\frac{1}{2})^2+\frac{1}{12})\Delta x^3 \end{aligned}$$

The rest are satisfied because both sides of $\lambda \nabla g = \nabla f$ are zero. Other $\Psi$ of similar form that satisfy $\lambda \nabla g = \nabla f$ are local maxima because they will be distributed closer to $\langle {x}\rangle =0$ and thus have lower variance.

Similarly, one $\Psi$ that minimizes f is $\Psi _i = \frac{1}{\sqrt{\Delta x}}\delta _{i0}$ because it satisfies g and $\lambda \nabla g = \nabla f$ if $\lambda =\Delta x^2(\frac{1}{12})$.

$$\lambda \nabla g =2\lambda \Delta x[0,\dots ,1,\dots ,0]=2\frac{\Delta x^3}{12}[0,\dots ,1,\dots ,0]=\nabla f$$

Other $\Psi$ that may satisfy Lagrange equation will be distributed further from $\langle {x}\rangle$ and thus have a higher variance.

2) A method of Lagrange multiplier is used to show that $\sum _k|\Psi _{k11}|\sqrt{\Delta x}\le \sum _k\frac{1}{\sqrt{N}}=\frac{N}{\sqrt{N}}$. Suppose the objective function to be maximized is $f(|\Psi _1|,\dots ,|\Psi _1|)=|\Psi _1|+\dots +|\Psi _N|$. The constraint function is $g(|\Psi _1|,\dots ,|\Psi _1|)=|\Psi _1|^2+\dots +|\Psi _N|^2-1$. Then $\nabla f=[1,\dots ,1]=\frac{\sqrt{N}}{2}[2|\Psi _1|,\dots ,2|\Psi _N|]=\frac{\sqrt{N}}{2}\nabla g$, which implies $|\Psi _i|=\frac{1}{\sqrt{N}}$