Fourier Transforms of Simple Functions

Introduction

Hello! In our previous lesson, we mastered the techniques for evaluating Gaussian integrals, which are essential for normalizing quantum wave functions. We saw how methods like the polar coordinate trick and completing the square allow us to solve integrals that don't have simple antiderivatives. I mentioned that this skill would be immediately useful for our next topic, and today you'll see exactly why.

This lesson addresses the learning outcome: Compute Fourier transforms and inverse Fourier transforms of simple functions, including Gaussians.

The Fourier transform is a cornerstone of quantum mechanics. It's the mathematical tool that allows us to translate between a particle's description in position space (its wave function, $ψ (x)$ ) and its description in momentum space (its momentum wave function, $ψ ~ (k)$ ). In essence, it decomposes a function into the spectrum of "wavenumbers" (which are directly proportional to momentum) that it contains. Today, we'll build an intuition for what the Fourier transform does, establish its formal definition, and then apply the integration skills you've just learned to compute the Fourier transform of the most important function for our course: the Gaussian.

What is the Fourier Transform? A Visual Intuition

Before diving into the formulas, it's incredibly helpful to build a conceptual model of what the Fourier transform actually does. The core idea is to take a signal or function and figure out which pure frequencies (or sine waves) make it up, and how strong each of those frequencies is.

To get a strong visual intuition for this process, the following video from 3Blue1Brown is unparalleled. It introduces the idea of "winding" a function around a circle and tracking the "center of mass" of the resulting shape to detect frequencies.

But what is the Fourier Transform? A visual introduction.

Please watch the following segments from the video 'But what is the Fourier Transform? A visual introduction.' by 3Blue1Brown. This will provide a beautiful and intuitive foundation for the entire concept.

Watch the video from the beginning until 17:02. Focus on these key ideas: Decomposition: How a complex signal can be seen as a sum of simple, pure frequencies. The Winding Machine: The core analogy of wrapping the function's graph around a circle at a specific 'winding frequency'. Center of Mass: How the center of mass of this wound-up graph spikes when the winding frequency matches a frequency present in the original signal. Complex Numbers: How complex exponentials (e^{-i \omega t}) provide a compact and elegant mathematical language for describing this winding process. The Integral: How the idea of averaging points to find the center of mass leads naturally to an integral formula.

The key takeaway from the video is that the Fourier transform is a machine that takes a function $g (t)$ and a "winding frequency" $f$ as inputs. It outputs a complex number whose magnitude tells you "how much" of that frequency is present in the original function. The integral we saw at the end, $\int g (t) e^{- i 2 π f t} d t$ , is the precise mathematical formulation of this "winding and averaging" process.

From Fourier Series to the Fourier Transform

Your interest in quantum mechanics provides a fantastic physical context for understanding the formal definition of the Fourier transform. We can think of it as an extension of the more familiar Fourier series. A Fourier series is used for periodic functions, like the allowed wave functions for a particle confined to a circle. The Fourier transform arises when we consider what happens in the limit that the circle's radius becomes infinite, and the particle is free to move on an infinite line.

This next video does a wonderful job of deriving the Fourier transform from this physical starting point.

To Understand the Fourier Transform, Start From Quantum Mechanics

The video 'To Understand the Fourier Transform, Start From Quantum Mechanics' by Physics with Elliot will show us how the Fourier transform naturally emerges when we generalize the problem of a 'particle on a circle' to a 'particle on a line'.

Watch the segment from 17:55 to 22:09. This part shows how the discrete sum of the Fourier series becomes a continuous integral (the Fourier Transform) as the radius of the circle goes to infinity, making the allowed momentum values continuous.

This leads us to the formal definition of the Fourier transform and its inverse.

The Formal Definition

Based on the intuition from the videos, we can now state the formal definitions. We will work with a function of position, $f (x)$ , and its transform, $f ~ (k)$ , which is a function of wavenumber $k$ . Note that $k$ is directly proportional to momentum in quantum mechanics via the de Broglie relation, $p = ℏ k$ .

The convention we will use throughout this course is common in physics:

Fourier Transform:

f ~ (k) = \int_{- \infty}^{\infty} f (x) e^{- ik x} d x

Inverse Fourier Transform:

f (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} f ~ (k) e^{ik x} d k

We say that $f (x)$ and $f ~ (k)$ form a Fourier pair. The function $f (x)$ lives in "position space," and its transform $f ~ (k)$ lives in "wavenumber space" or "momentum space." These two functions are two sides of the same coin; they contain the exact same information, just represented in a different basis.

For a rigorous walkthrough of this definition and its origin from Fourier series, you can refer to the following document.

FOURIER ANALYSIS: LECTURE 6

The handout 'FOURIER ANALYSIS: LECTURE 6' from the University of Edinburgh provides a concise derivation of these formulas.

Please read Section 3.1, 'Fourier transforms as a limit of Fourier series', and the brief notes on 'k space and momentum space'. This formalizes the argument you saw in the last video and confirms the definitions we will be using.

The Fourier Transform of a Gaussian

Now, let's tackle the main event: calculating the Fourier transform of a Gaussian function. This is a crucial result for your goal of understanding Gaussian wave packets. We want to compute the transform of $f (x) = N e^{- a x^{2}}$ , where $a$ is a positive constant and $N$ is a normalization factor.

The integral we need to solve is:

f ~ (k) = \int_{- \infty}^{\infty} (N e^{- a x^{2}}) e^{- ik x} d x = N \int_{- \infty}^{\infty} e^{- a x^{2} - ik x} d x

Look at the exponent: $- a x^{2} - ik x$ . This is a general quadratic form, just like we handled in the last lesson! The only new feature is that one of the coefficients is imaginary. The "completing the square" method works perfectly.

Let's work through the derivation, following the steps outlined in the handout.

FOURIER ANALYSIS: LECTURE 6

We'll now calculate the Fourier Transform of a Gaussian. The derivation is a direct application of the 'completing the square' technique you learned in the last lesson.

Read Section 3.2.2, 'The Gaussian'. Follow the steps under 'Deriving the FT'. Pay close attention to how they complete the square in the exponent and use the standard Gaussian integral result \int e^{-u^2}du = \sqrt{\pi} to arrive at the final answer.

Let's summarize the derivation from the resource.

Set up the integral: $f ~ (k) = N \int_{- \infty}^{\infty} exp (- a x^{2} - ik x) d x$
Complete the square in the exponent: $- a x^{2} - ik x = - a (x^{2} + \frac{ik}{a} x) = - a (x + \frac{ik}{2 a})^{2} - a (\frac{ik}{2 a})^{2} = - a (x + \frac{ik}{2 a})^{2} - \frac{k ^{2}}{4 a}$
Substitute back into the integral: $f ~ (k) = N \int_{- \infty}^{\infty} exp (- a (x + \frac{ik}{2 a})^{2} - \frac{k ^{2}}{4 a}) d x = N e^{- k^{2} /4 a} \int_{- \infty}^{\infty} exp (- a (x + \frac{ik}{2 a})^{2}) d x$
Change variables: Let $u = x + \frac{ik}{2 a}$ , so $d u = d x$ . The limits of integration remain $- \infty$ to $\infty$ . (This step involves a contour integration argument in the complex plane, but for our purposes, we can accept that the shift by a constant imaginary number does not change the result of the definite integral from $- \infty$ to $\infty$ ). $f ~ (k) = N e^{- k^{2} /4 a} \int_{- \infty}^{\infty} e^{- a u^{2}} d u$
Use the standard result from our last lesson, $\int_{- \infty}^{\infty} e^{- a u^{2}} d u = \frac{π}{a}$ : $f ~ (k) = N \frac{π}{a} e^{- k^{2} /4 a}$

This is a remarkable and beautiful result: The Fourier transform of a Gaussian function is another Gaussian function!

The Reciprocal Relationship and a Glimpse of Uncertainty

Take a closer look at our result. Let's define the width of the initial Gaussian $f (x) = e^{- x^{2} / (2 σ_{x}^{2})}$ by the parameter $σ_{x}$ . In this notation, our constant $a = 1/ (2 σ_{x}^{2})$ .

Substituting this into our FT result, $f ~ (k) \propto e^{- k^{2} /4 a}$ , we get:

f ~ (k) \propto exp (- \frac{k ^{2}}{4 ( 1/2 σ _{x}^{2} )}) = exp (- \frac{k ^{2} σ _{x}^{2}}{2})

This is a Gaussian in $k$ -space with a width $σ_{k}$ defined by $σ_{k}^{2} = 1/ σ_{x}^{2}$ , or $σ_{k} = 1/ σ_{x}$ .

This reveals a profound inverse relationship:

A narrow Gaussian in position space ( $σ_{x}$ is small) transforms into a wide Gaussian in momentum space ( $σ_{k}$ is large).
A wide Gaussian in position space ( $σ_{x}$ is large) transforms into a narrow Gaussian in momentum space ( $σ_{k}$ is small).

This is the heart of the Heisenberg Uncertainty Principle. If a particle's position is known very precisely (narrow $ψ (x)$ ), its momentum is very uncertain (wide $ψ ~ (k)$ ), and vice-versa.

Test your understanding!

Consider the Gaussian function $f (x) = e^{- 3 x^{2}}$ . Without worrying about the normalization constant, what is the functional form of its Fourier transform, $f ~ (k)$ ? Which is wider, $f (x)$ or $f ~ (k)$ ?

Show answer

Here, $a = 3$ . Using our formula $f ~ (k) \propto e^{- k^{2} /4 a}$ , the Fourier transform is:

f ~ (k) \propto e^{- k^{2} /12}

The original function $f (x) = e^{- 3 x^{2}}$ falls off faster than the transform $f ~ (k) = e^{- k^{2} /12}$ . Therefore, $f (x)$ is the narrower function, and its Fourier transform $f ~ (k)$ is wider.

Another Example: The Top-Hat Function

To solidify your understanding, let's compute the transform of another simple function: the "top-hat" or "box" function. It is defined as:

Π (x) = {10 if ∣ x ∣ \leq a if ∣ x ∣ > a

The derivation is a straightforward integral.

FOURIER ANALYSIS: LECTURE 6

Let's work through one more example: the Fourier transform of a top-hat function. This is another classic example that clearly illustrates the properties of the transform.

Please read Section 3.2.1, 'The top-hat'. The derivation is shorter and simpler than the Gaussian case. The result is the 'sinc' function, which you should also learn to recognize.

As the resource shows, the calculation is:

Π ~ (k) = \int_{- a}^{a} 1 \cdot e^{- ik x} d x = [\frac{e ^{- ik x}}{- ik}]_{- a}^{a} = \frac{e ^{- ik a} - e ^{ik a}}{- ik} = \frac{e ^{ik a} - e ^{- ik a}}{ik}

Using Euler's formula, $e^{i θ} - e^{- i θ} = 2 i sin (θ)$ , we get:

Π ~ (k) = \frac{2 i sin ( k a )}{ik} = 2 a \frac{sin ( k a )}{k a} = 2 a sinc (k a)

where $sinc (z) = \frac{s i n ( z )}{z}$ . This oscillating, decaying function is the Fourier transform of a sharp, localized pulse. Again, notice that as the width of the box $2 a$ gets smaller, the central lobe of the sinc function (between $k = - π / a$ and $k = π / a$ ) gets wider.

Conclusion

In this lesson, we have demystified the Fourier transform, a pivotal concept connecting the position and momentum descriptions in quantum mechanics.

Key Takeaways:

The Fourier transform decomposes a function into its constituent frequencies (or wavenumbers).
It is formally defined by the integral $f ~ (k) = \int f (x) e^{- ik x} d x$ , with a corresponding inverse transform to return to the original function.
The Fourier transform of a Gaussian is another Gaussian. We proved this by applying the "completing the square" method from our previous lesson to an integral with a complex exponent.
There is a fundamental reciprocal relationship between the width of a function and its Fourier transform. This is the mathematical root of the Heisenberg Uncertainty Principle.

You are now equipped with three fundamental mathematical tools: complex exponentials, Gaussian integrals, and Fourier transforms. The next lesson in our syllabus is to solve second-order linear homogeneous ordinary differential equations with constant coefficients. This may sound like a pure math topic, but it is the final prerequisite skill you'll need before we can write down and solve the most important equation in quantum mechanics: the Schrödinger equation.

Recommended resources

Introduction

What is the Fourier Transform? A Visual Intuition

From Fourier Series to the Fourier Transform

The Formal Definition

The Fourier Transform of a Gaussian

The Reciprocal Relationship and a Glimpse of Uncertainty

Another Example: The Top-Hat Function

Conclusion