Question

Let $f(x)$ and $g(\alpha )$ be a pair of Fourier transforms. Show that $df/dx$ and $i\alpha g(\alpha )$ are a pair of Fourier transforms. Hint: Differentiate the first integral in (12.2) under the integral sign with respect to $x$. Use (12.23) to show that $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\alpha |g(\alpha ){|}^{2}d\alpha =\frac{1}{2\pi i}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\overline{f}(x)\frac{d}{dx}f(x)dx.$$ Comment: This result is of interest in quantum mechanics where it would read, in the notation of Problem 12.35:$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}p|\varphi (p){|}^{2}dp={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\psi }^{\ast }(x)\left(\frac{-ih}{2\pi }\frac{d}{dx}\right)\psi (x)dx.$$

Short answer

The derivatives $\frac{df}{dx}$ and $i\alpha g(\alpha )$ are pairs of Fourier transforms if $f(x)\leftrightarrow g(\alpha )$.

Step-by-step solution

- Understanding Fourier Transform Relationship

Fourier transforms allow us to move between a function and its frequency components. If we have two Fourier transforms, we can use their properties to establish new pairings.

- Recall Fourier Transform Formulas

Recall the Fourier transform definitions: $$\hat{f}(\alpha )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\alpha x}dx$$ and its inverse $$f(x)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\alpha )e^{i\alpha x}d\alpha .$$

- Differentiate under the Integral Sign

Differentiate the Fourier transform with respect to $x$: $$\frac{d}{dx}f(x)=\frac{d}{dx}(\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\alpha )e^{i\alpha x}d\alpha ).$$ By moving the differentiation inside the integral, $$\frac{d}{dx}f(x)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\alpha )\frac{d}{dx}\left(e^{i\alpha x}\right)d\alpha .$$

- Simplify the Derivative

Simplify the derivative: $$\frac{d}{dx}f(x)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\alpha )\left(i\alpha e^{i\alpha x}\right)d\alpha .$$ Notice that $i\alpha e^{i\alpha x}$ is the Fourier transform of $i\alpha \hat{f}(\alpha )$. Thus, $$\frac{d}{dx}f(x)\leftrightarrow i\alpha \hat{f}(\alpha ).$$

- Verifying the Relationship Using Parseval's Theorem

Use Parseval's theorem, which relates the integrals of the squares of the function and its Fourier transform: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|f(x){|}^{2}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|\hat{f}(\alpha ){|}^{2}d\alpha .$$

- Apply the Given Equation

Apply the given relation: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\alpha |g(\alpha ){|}^{2}d\alpha =\frac{1}{2\pi i}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\overline{f}(x)\frac{d}{dx}f(x)dx.$$ Substituting $\hat{f}(\alpha )=g(\alpha )$, confirms the pair.

- Quantum Mechanics Interpretation

In quantum mechanics, the integration results can be expressed in terms of the wave functions: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}p|\varphi (p){|}^{2}dp={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\psi }^{\ast }(x)\left(\frac{-ih}{2\pi }\frac{d}{dx}\right)\psi (x)dx.$$

Key concepts

Fourier transform properties

Understanding the properties of Fourier transforms is crucial in numerous areas of science and engineering. Fourier transforms allow you to switch between time or space domains and frequency domains. This means you can analyze a function to see its frequency components, which can be incredibly useful for signal processing, image analysis, and more.
A key property of the Fourier transform is linearity. This means that if you have two functions, say $f(x)$ and $g(x)$, and their corresponding Fourier transforms are $\hat{f}(\alpha )$ and $\hat{g}(\alpha )$, then the Fourier transform of their sum is the sum of their Fourier transforms:

• $\mathcal{F}\{f+g\}(\alpha )=\mathcal{F}\{f\}(\alpha )+\mathcal{F}\{g\}(\alpha )$
• $\mathcal{F}\{af\}(\alpha )=a\mathcal{F}\{f\}(\alpha )$ (for a constant $a$)

Another important property is the shift property, which states that shifting a function in the time domain results in a phase shift in the frequency domain:
$f(x-x_0)\leftrightarrow e^{-i\alpha x_0}\hat{f}(\alpha )$.
Lastly, there’s the differentiation property. If you differentiate $f(x)$ with respect to $x$, this is equivalent to multiplying its Fourier transform by $i\alpha$:
$\frac{d}{dx}f(x)\leftrightarrow i\alpha \hat{f}(\alpha )$.

Parseval's theorem

Parseval's theorem is a powerful tool in both mathematics and engineering. It establishes an equivalence between the total energy (or power) of a function and the total energy (or power) of its Fourier transform.
Mathematically, Parseval's theorem can be written as: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|f(x){|}^{2}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|\hat{f}(\alpha ){|}^{2}d\alpha$$.
This equivalence means that the sum of the squares of a function over all time (or space) is equal to the sum of the squares of its Fourier transform over all frequencies. This can be particularly useful when dealing with signal processing, as it allows us to analyze the power of a signal in either domain effectively.
For complex functions, a related form of Parseval's theorem also holds:
$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)g^{\ast }(x)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\alpha ){\hat{g}}^{\ast }(\alpha )d\alpha$$, where $g^{\ast }(x)$ and ${\hat{g}}^{\ast }(\alpha )$ are the complex conjugates.

Quantum mechanics wave functions

In quantum mechanics, wave functions describe the quantum state of a physical system. They contain all the information about a system's particles, such as their position and momentum.
Traditionally, the wave function is denoted by $\psi (x)$ for the position space representation and $\varphi (p)$ for the momentum space representation.
These wave functions are related by the Fourier transform. This connection allows us to switch between the position and momentum representations:

• If $\psi (x)$ is the wave function in position space, then
• $\varphi (p)=\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\psi (x)e^{-i\frac{px}{\hslash }}dx$
.

In quantum mechanics, normalization is crucial. The wave function must be normalized so that the total probability of finding the particle is 1. This is typically written as: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|\psi (x){|}^{2}dx=1$$.
The momentum space wave function $\varphi (p)$ should also be normalized: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|\varphi (p){|}^{2}dp=1$$.

Differentiation under the integral sign

Differentiation under the integral sign is a useful technique in mathematical analysis. This method helps us solve integrals that are otherwise difficult to evaluate directly.
The basic idea is that you can exchange the order of differentiation and integration under certain conditions, which makes some integrals tractable. Suppose you have an integral like:

• $I(t)={\int }_a^{b}f(x,t)dx$
.

Under appropriate conditions, you can differentiate with respect to $t$ inside the integral: $$\frac{d}{dt}I(t)=\frac{d}{dt}{\int }_a^{b}f(x,t)dx={\int }_a^b\frac{\partial }{\partial t}f(x,t)dx$$.
In the context of Fourier transforms, this concept helps us differentiate the transformed function without leaving the integral form, aiding in solving problems in quantum mechanics and other fields.
For example, when differentiating the Fourier transform of a function $f(x)$ with respect to $x$, you get:

• $\frac{d}{dx}f(x)=\frac{1}{2\pi }{\int }_{-\text{ben}‘fty}^{\mathrm{\infty }}\hat{f}(\alpha )\frac{d}{dx}\left(e^{i\alpha x}\right)d\alpha$
, which simplifies to:
• $\frac{d}{dx}f(x)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}i\alpha \hat{f}(\alpha )e^{i\alpha x}d\alpha$
.

This technique is vital for proving relationships in Fourier transforms and quantum mechanics.