As the semester goes on, I find myself to be quite often in a pickle to find spare time. Therefore, what I will be writing about is mostly what I need to know for my exams. Tedious as it may sound, there are still absolutely stunning proofs and derivations that can bring a smile to my face, and that is what I shall be posting about. Having come to terms that probably no one will ever read this, I do this more as a mental exercise to engage with the materials than to teach my supposedly non-existing audience about physics and maths.

Recently, in my taking of the Fourier Analysis course as part of my degree prorgramme, I came across the reciprocal relation: where a function is narrow, its Fourier transform will be wide, and vice versa. This is directly responsible for the Heisenberg’s Uncertainty principle. In fact, one can derive this arguably fundamental relation with just some algebra, doing so also highlights why the Gaussian is a really awesome function. In introductory QM courses, it is quite common for the ones teaching to shy away from the maths and sweep it all under the rug. I personally think this is a rather poor pedagogical choice, as students will most likely take several QM courses throughout their years in uni, this just results in time wasted covering material that would have otherwise stayed with the students had they been formally shown the derivation.

Suppose we start with a non-normalised Gaussian of the form:

\[f(x) = \exp\left[\frac{-x^2}{2\sigma^2}\right]\]

The function is rather easily normalisable by evaluating its integral for all of space:

\[\int_{-\infty}^{\infty}\dd{x} f(x) = \sqrt{2 \pi \sigma^2}\]

Despite the fact that our Gaussian never really touches the $x$ axis, but we can still rigorously define a so-called ‘width’ of our function, and this is just the standard deviation $\sigma$.

We can put our Gaussian into our newly acquired ‘machine’ that is the Fourier transformation and turn the cranks:

\[\tilde{f}(k) = \int_{-\infty}^{\infty} \dd{x} f(x) e^{-ikx}\]

This results in a daunting integral when we write in the full definition of $f(x)$ in all its glory, but in fact, it can be trivially solved with a highschooler’s maths trick.

\[\int_{-\infty}^\infty\dd{x} \exp\left[ \frac{-x^2}{2\sigma^2} – ikx\right]\]

From binomial expansion, we know:

\[\left( x+ \frac{a}{2} \right ) ^2 = x^2 + ax + \frac{a^2}{4}\]

which we can rearrange to give:

\[x^2 + ax = \left( x + \frac{a}{2} \right )^2 – \frac{a^2}{4}\]

By direct comparison:

\[a = 2 ik\sigma\]

Resubstituting into our integral will lead to a litany of cancellations:

\[\int_{-\infty}^{\infty} \dd{x} \exp \left[ -\frac{1}{2\sigma^2} (x+ik\sigma^2)^2 – \frac{\cancel{4}k^2\sigma^{\cancel{4}}}{\cancel{4}\cdot 2 \cancel{\sigma^2}}\right ]\]

Which leads to the gratifying result that the Fourier transform of a Gaussian is still a Gaussian, but with a catch:

\[\tilde{f}(p) = \exp \left [ -\frac{k^2\sigma^2}{2}\right ]\sqrt{2\pi \sigma^2}\]

The $\sigma$ is now in the numerator of the exponential! This is precisely what is referred to as the reciprocal relation. It is this change in the location of $\sigma$ that leads to the observation of a broad function in $x$ space being narrow in Fourier space. In short, it can be written as:

\[\tilde{\sigma} = \frac{1}{\sigma}\]

It is only a few short steps away from obtaining Heisenberg’s Uncertainty Principle!

If we take the de Broglie relation as an axiomatic starting point *i.e.*:

\[p = \hbar k\]

and think in terms of wave functions, we can subsititute in to what we have derived!

For the wave function in momentum space, we have:

\[\tilde{\psi}(p) = \sqrt{2\pi \sigma^2} \exp\left[-\frac{\hbar^2\sigma^2p^2}{2}\right]\]

And its probability amplitude is just the same function but squared, as $\tilde{\psi}(p) \in \mathbb{R}$:

\[|\tilde{\psi}(p)|^2 = 2\pi\sigma^2\exp\left[-\hbar^2\sigma^2p^2\right]\]

By virtue of $\tilde{\psi}(p)$ being a Gaussian, we know its width to be $\frac{\hbar}{\sqrt{2}\sigma}$. Comparing that to what we know for our wave function in $\mathbb{R}$:

\[|\psi(x)|^2 = e^{-\frac{x^2}{\sigma^2}}\]

We also know its width to be just $\frac{\sigma}{\sqrt{2}}$.

If we take the width of our corresponding Gaussians in momentum space and real space, and refer to them with convenient labels, I don’t know, say $\Delta x$ and $\Delta p$, and multiply them together:

\[\Delta x \Delta p = \frac{\cancel{\sigma}}{\sqrt{2}}\frac{\hbar}{\sqrt{2}\cancel{\sigma}} = \frac{\hbar}{2}\]

Not only do we recover the Canonical Commutation Relation, we have also developed a keen intuition of what this physically represents. In addition, the elimination of the dependence $\sigma$ in our final expression signifies that this hold true for all particles in the universe ( as $\sigma$ is unique to each particle), quite a fulfilling result indeed.

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