I am trying to look into atomism possibly relation to modern quantum mechanics. I am new to wave functions stuff, but trying to learn. I started looking more closely at Heisenberg uncertainty principle.

When including my max-velocity that is a consequence of atomism I get that the momentum and position operator commute at the Planck scale, but not before that. In other words no uncertainty at the Planck scale. Same with the energy and time operator. If correct this again opens up for hidden variable theories at the Planck scale, and since all particles are made of Planck mass particles under atomism then there is likely no spooky action at distance. Could Albert have been right?

For example using the standard wave function (plane wave solution to Klein--Gordon)

\begin{equation}

\Psi=e^{i\left(\frac{p}{\hbar}x-\frac{E}{\hbar} t\right)}

\end{equation}

I get the standard operators when working with non-Planck mass particle, at the Planck mass particle the momentum and energy operator becomes 0, and yes the momentum and position operator then commute, the same with the energy and time operator. Otherwise not commute.

Also I get an upper bound on uncertainty from my max velocity formula for non-Planck particles (for Planck mass particles zero uncertainty)

\begin{equation}

\frac{\hbar}{2} \leq \sigma_E\sigma_t \leq \hbar \left(1-\frac{l_p}{\bar{\lambda}} \right)

\end{equation}

Revisiting the Derivation of Heisenberg's Uncertainty Principle: The Collapse of Uncertainty at the Planck Scale. (first draft, comments and critics welcome, likely some typos, and possibly even something formally wrong, but my concept I think is right. typo first line of my equation 24, delete 2pi and add a t)

Renormalization not needed

Uncertainty collapse at the Planck scale

Bells inequality collapses at Planck scale, spooky action no more.

Superposition collapses at Planck scale

Lorentz symmetry collapses at Planck scale

We get upper boundary on uncertainty in addition to lower

Negative energy solutions should possibly be interpreted as we also have upper boundary?

\begin{equation}

\sigma_E\sigma_t\leq \hbar c \left(1-\frac{l_p}{\bar{\lambda}}\right)

\end{equation}

Assume that we now multiply both sides with minus one and we get

\begin{equation}

-\sigma_E\sigma_t\geq -\hbar c \left(1-\frac{l_p}{\bar{\lambda}}\right)

\end{equation}

so mathematical negative energy and probabilities is just a hint of upper boundary? (of course negative energy and negative probabilities are Fake, but do they hint at simply upper boundary, flip of inequality sign.?)

but yes could be something formally wrong with my derivations, I am a farm boy, not a mathematician.