## Motivation:

The textbook answer to the question of why Quantum Mechanics has a probabilistic description goes along the lines of the Heisenberg Uncertainty Principle. This is true even in Landau’s famous account of Quantum Mechanics . In essence, we can’t simultaneously know both the position and momentum of quanta.

However, it may be demonstrated that the HUP is nothing more than a mathematical consequence of the wave description which is in itself not more than a useful approximation for engineering applications. This suggests that we may need to analyse actual quantum mechanical phenomena in order to understand the origins of these uncertainties.

## Defining uncertainties within the context of wave equations:

Given the De Broglie hypothesis and the Everettian interpretation of quantum mechanics, the physical behaviour of any object is described by a wave function and therefore the universe itself may be identified with a wave function $$\Psi (x,t)$$.

If $$x$$ denotes the position of an object and $$p$$ its momentum, its associated wave function may generally be expressed as the sum of many orthogonal waves:

\begin{equation} \Psi(x) \propto \sum_n a_n \cdot \Psi_n = \sum_n a_n \cdot e^{i p_n x/\hbar} \end{equation}

where $$\lvert \Psi(x) \rvert^2 = \sum_n \lvert a_n \rvert^2$$ may be interpreted as the probability that an object is localised at $$x$$.

Now, if we bring (1) to the continuum limit:

\begin{equation} \Psi(x) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} \phi(p) \cdot e^{i p x/\hbar} dp \end{equation}

so $$\phi(p)$$ is the Fourier transform of $$\Psi(x)$$.

Since $$\lvert \Psi(x) \rvert^2$$ is the probability density for position, we may calculate its standard deviation and use this as a representation of our statistical uncertainty. In fact, we shall demonstrate that the uncertainty principle is a direct consequence of basic facts in Fourier analysis.

## Deriving the uncertainty principle via Fourier analysis:

If we subtract the mean from each variable, we may define the variances for position and momentum as follows:

\begin{equation} \sigma_x^2 = \int_{-\infty}^{\infty} x^2 \cdot \lvert \Psi(x) \rvert^2 dx \end{equation}

\begin{equation} \sigma_p^2 = \int_{-\infty}^{\infty} p^2 \cdot \lvert \phi(p) \rvert^2 dp \end{equation}

Now, if we define the function:

\begin{equation} f(x) = x \cdot \Psi(x) \end{equation}

which may be interpreted as a vector in a certain function space, we may define the variances as inner-products:

\begin{equation} \sigma_x^2 = \int_{-\infty}^{\infty} \lvert f(x) \rvert^2 dx = \langle f,f \rangle \end{equation}

Given that $$\Psi(x)$$ and $$\Phi(p)$$ are Fourier transforms of each other, it may be shown that:

\begin{equation} g(x) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} p \cdot \phi(p) e^{-ipx/\hbar} = -i \hbar \frac{d}{dx} \cdot \Psi(x) \end{equation}

By applying Parseval’s theorem, we have:

\begin{equation} \sigma_p^2 = \int_{-\infty}^{\infty} \lvert p \cdot \phi(p) \rvert^2 dp = \int_{-\infty}^{\infty} \lvert g(x) \rvert^2 dx = \langle g,g \rangle \end{equation}

Now, given that for any $$z \in \mathbb{C}$$:

\begin{equation} \lvert z \rvert^2 = \text{Re}(z)^2 + \text{Im}(z)^2 \geq \text{Im}(z)^2 = \Big(\frac{z -\bar{z}}{2i}\Big)^2 \end{equation}

if we let $$z = \langle f,g \rangle$$ and $$\bar{z} = \langle g,f \rangle$$ we have:

\begin{equation} \lvert \langle f,g \rangle \rvert^2 \geq \Big(\frac{\langle f,g \rangle -\langle g,f \rangle}{2i}\Big)^2 \end{equation}

By evaluating the inner products, we find:

\begin{equation} \langle f,g \rangle -\langle g,f \rangle = i \hbar \end{equation}

Finally, from (9) and (10) it follows that:

\begin{equation} \sigma_x^2 \cdot \sigma_p^2 \geq \lvert \langle f,g \rangle -\langle g,f \rangle \rvert^2 \geq \Big(\frac{\langle f,g \rangle -\langle g,f \rangle}{2i}\Big)^2 = \frac{\hbar^2}{4} \end{equation}

and so we may deduce that the product of standard deviations yields:

\begin{equation} \sigma_x \cdot \sigma_p \geq \frac{\hbar}{2} = \frac{h}{4 \pi} \end{equation}

## Discussion:

As demonstrated here, the Heisenberg Uncertainty Principle is a useful mathematical observation but it does not provide any profound epistemological insights into the fundamental source of uncertainty in quantum mechanics. That said, I believe it is possible to make important progress in this direction by carefully analysing what Feynman considered the essential mystery in quantum mechanics. The double-slit experiment.

I plan to proceed with such an analysis using the tools of information theory as any serious account of uncertainty is of an information-theoretic nature. Moreover, such an information-theoretic investigation into the foundations of quantum mechanics was originally proposed by John Wheeler.

1. L. D. Landau & L. M. Lifshitz. Quantum Mechanics-Nonrelativistic Theory. Butterworth-Heinemann. 1981.
2. Wheeler. INFORMATION, PHYSICS, QUANTUM: THE SEARCH FOR LINKS. 1989.
3. Feynman. The Feynman Lectures in Physics. 1963.
4. Wikipedia. The Uncertainty Principle. 2021.