This is it.

All of my previous articles about Fourier Transform has been made, bit by bit, step by step, to build logical foundation for this last article of the Fourier Series posts.

Historically, Heisenberg’s famous Uncertainty Principle (HUP) were deducted from a statistical facts/observation from Quantum Mechanics. In the beginning, it is such a bizarre concept, and people wondering why it was there at the hearts of Quantum Mechanics.

In some early understanding of HUP, it was thought that QM uses wave function to represent quantum state evolution, then the HUP is the next logical consequences. Because any wave equation will inherently exhibit HUP inequality. From the field of waves and electromagnetic, it was already understood that you can’t locally sample frequency at the same time with arbitrary accuracy. Fourier Transform/Fourier Analysis forces a trade-off. If you try to measure precise frequency, you need to sample with a very long time window. The converse is also true. If you use a short time window, then the frequency will be blurry or less precise.

But then, an alternative explanation exists from Information theory.

We can measure entropy from probability distribution

The act of “observing” can be interpreted as gaining information. If you can gain information, then you can measure a change in entropy.

Naturally, any statistical observation will correspond to evolution of probability distribution, and thus an evolution of entropy.

For classical mechanics, an “observable” is almost always be measured deterministically. Meaning its “expectation” value is very precise. It has a very low variance. Two common observables related with classical mechanic theory (like Newton’s law), are position and momentum.

In the classical observation, both of measurements can have significantly low entropy, because it can be very precise. But why in Quantum Mechanics it is not the case?

Before answering that, let us take a look at how probability distribution works.

Essentially, probability distribution were derived after collecting observation of events. The state events will slowly build up a distribution. With more and more data, we can guess if the data collection approaches a certain “kind” of distribution. To determine if a distribution is the same “kind”, we can generate a “characteristic function”. If the characteristic function is of the same kind, then it has the same kind of distribution.

The key catch is, the way to compute characteristic function (even though the original idea came from a completely different motivation and methods), is exactly the same with how you compute a Fourier Transform. So both “waves” and “distribution” can be thought of from the same fundamental ideas.

This is what becomes the crucial point in Quantum Mechanics. In QM, we can only “observe” from the measurement result. For big objects, you measure it indirectly. For example, you measure “air pressure” not by calculating each individual molecule, but from the volume change exerted by the pressure. Similarly, you determine object position from the light signal originated from the object.

In QM, however, when you want to measure a photon, you need to capture the photon. There is just one photon per one measurement outcome. If you measure it by “locating” where the photon is captured, you measure its position. If you measure it by “colliding” it with a detector or change its trajectory, you measure its momentum. But you have to choose. Because there is only one photon to measure at any given moment. The statistical distribution is what you get if you repeat the measurements for some numbers of photons.

But just as any statistical distribution, there will be corresponding entropy.

Position and Momentum were Fourier Dual

The next key insight is that in QM, position is a Fourier dual of momentum. This means they are the same thing, just expressed in a different bases.

Supposes that you have position wave function $\psi(x)$, then the probability of observing the position is expressed as $\rho(x)=|\psi(x)|^2$.

The corresponding momentum is then $\phi(\xi)$ which were taken by Fourier transforming $\psi(x)$. The probability of observing the momentum is then $\gamma(\xi)= |\phi(\xi)|^2$.

Applying Entropic Uncertainty Principle

Because both position and momentum is just two representation of one single function, this is the same prerequisite with how we define Entropic Uncertainty Principle. Which I already explained in previous article.

$$H(\rho)+H(\gamma) \ge \operatorname{ln}(\frac{e}{2})$$

We have a little bit of problem, though. In the derivation from the previous article, we use Hausdorff-Young inequality. HY inequality, by definition uses standard unitary Fourier Transform convention, so that Plancherel Theorem can be applied easily.

However, in the physicist’s convention, the Fourier Transform is usually non-unitary. This is because they already uses existing unit to work with. But, no worries we can resolve it by changing the variable.

Recall that we define $\gamma(\xi)=|\phi(\xi)|^2$ and $\phi(\xi)$ a Fourier transform of $\psi(x)$.

That means $\xi$ is the unitary Fourier dual of $x$, not exactly the momentum in Physics. Rather, $\xi$ is usually called the spatial frequency. Or in another familiar term, it is the reciprocal of wave length, usually denoted by $\lambda$.

Momentum itself can be related by wavelength like this $p=\frac{h}{\lambda}$, where $h$ is the Planck’s constant. Which can then be re-substituted as $p=h\xi$. This comes from de Broglie proposals.

Because we want to express the relation in terms of statistical distribution of $p$, the momentum, then we need to transform $\gamma(\xi)$ into $\gamma(p)$. We use the fact that the probability distribution should both be summed up to 1. Then the distribution were just rescaled by some factors.

$$\begin{align*} \int \gamma(\xi) \, d\xi&=\int \gamma(p) \, dp = 1 \\ \gamma(\xi) &= \gamma(p) \, \frac{dp}{d\xi} = \gamma(p) \, h \\ \end{align*}$$

We now calculate the entropy of the momentum.

$$\begin{align*} H_\xi(\gamma) &= - \int \gamma(\xi) \operatorname{ln}(\gamma(\xi)) d\xi \\ &= - \int h \gamma(p) \operatorname{ln}(h \gamma(p)) \frac{dp}{h} \\ &= - \int \gamma(p) dp \operatorname{ln}(h) - \int \gamma(p) \operatorname{ln}(\gamma(p)) dp \\ &= - \operatorname{ln}(h) + H_p(\gamma) \\ \end{align*}$$

Substituting back to the to the Entropic Uncertainty inequality:

$$\begin{align*} H_x(\rho)+H_\xi(\gamma) &\ge \operatorname{ln}(\frac{e}{2}) \\ H_x(\rho)+H_p(\gamma) - \operatorname{ln}(h) &\ge \operatorname{ln}(\frac{e}{2}) \\ H_x(\rho)+H_p(\gamma) &\ge \operatorname{ln}(\frac{e}{2}) + \operatorname{ln}(h) \\ &\ge \operatorname{ln}(\frac{eh}{2}) \\ \end{align*}$$

Variance of position and momentum

The original Heisenberg Uncertainty Principle, actually express it using a multiplication for variance between position and momentum. But our current Uncertainty inequality uses addition. How come it will change into a product?

First. Let’s define what a variance is. Given repeated observation of position and momentum, the variance is that how far apart each outcomes ends up in the distribution.

If we have no other information, by Maximum Entropy Principle, the best possible distribution to match this is just the Normal distribution (again).

Suppose the Normal distribution for each position and momentum distribution has corresponding variance $\sigma_x$ and $\sigma_p$.

The entropy of Normal distribution is completely determined by its variance.

$$H_x = \frac{1}{2} \operatorname{ln}(2\pi e \sigma_x^2)$$

Combining all together

If the position/momentum uses any distribution other than the Normal distribution, its entropy can’t be higher than the Normal distribution.

So we have

$$\begin{align*} H_x &\le\frac{1}{2} \operatorname{ln}(2\pi e \sigma_x^2) \\ H_p &\le\frac{1}{2} \operatorname{ln}(2\pi e \sigma_p^2) \\ \end{align*}$$

Adding it together to get the upper bound:

$$\begin{align*} H_x(\rho) + H_p(\gamma) &\le \frac{1}{2} \operatorname{ln}(2\pi e \sigma_x^2) + \frac{1}{2} \operatorname{ln}(2\pi e \sigma_p^2) \\ &\le \frac{1}{2} \operatorname{ln}(2^2 \pi^2 e^2 \sigma_x^2 \sigma_p^2) \\ &\le \operatorname{ln}(2\pi e \sigma_x \sigma_p) \\ \end{align*}$$

Then combine with the lower bound from Entropic Uncertainty Principle

$$\begin{align*} \operatorname{ln}(2\pi e \sigma_x \sigma_p) &\ge H_x(\rho) + H_p(\gamma) \ge \operatorname{ln}(\frac{eh}{2}) \\ \operatorname{ln}(2\pi e \sigma_x \sigma_p) \ge \operatorname{ln}(\frac{eh}{2}) \\ \sigma_x \sigma_p \ge \frac{h}{4\pi} \\ \end{align*}$$

So that’s the Heisenberg Uncertainty Principle. As we can see, it turns out that the product between the variance came from the logarithmic nature of the expression, that changes addition into multiplication.

Other alternative famous notation is to use $\Delta$ to express the standard deviation and uses reduced Planck’s constant (h-bar) $\hbar=\frac{h}{2\pi}$. The expression becomes:

$$\Delta{x} \cdot \Delta p \ge \frac{\hbar}{2}$$