21  The Fourier transform

The Fourier transform of a square-integrable function is a very useful tool for the study of quantum chemistry. It phrases in mathematical terms the idea that a function can be decomposed into plane wave components with varying weights and frequencies.

21.1 Fourier transform for functions over \(\mathbb{R}^n\)

The Fourier transform is an integral transformation, taking a function into a new function. It is therefore an operator between function spaces.

The transform is most easily defined for \(L^1\) functions, but can be extended to a much larger class of functions, e.g., \(L^2\) functions.

Definition 21.1 (Fourier transform on \(L^1\)) We define the Fourier transform of \(u \in L^1(\mathbb{R}^n)\) as the linear operator \(\mathcal{F} \in L(L^1(\mathbb{R}^n), L^\infty(\mathbb{R}^n))\), with an auxiliary notation \(\mathcal{F}u = \hat{u} \in L^\infty(\mathbb{R}^n)\), and given by the formula

\[ \hat{u}(\mathbf{k}) = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} e^{-\mathrm{i}\mathbf{k}\cdot\mathbf{x}} u(\mathbf{x})\; \mathrm{d}\mathbf{x}. \tag{21.1}\]

We define the inverse Fourier transform of \(u \in L^1(\mathbb{R}^n)\) as the function \(\mathcal{F}^{-1}u = \check{u} \in L^\infty(\mathbb{R}^n)\) given by the formula

\[ \check{u}(\mathbf{k}) = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} e^{\mathrm{i}\mathbf{k}\cdot\mathbf{x}} u(\mathbf{x})\; \mathrm{d}\mathbf{x}. \tag{21.2}\]

As defined, the inverse transform is actually only defined as a transform on \(L^1(\mathbb{R}^n)\), not \(L^\infty(\mathbb{R}^n)\). Therefore it is not actually the inverse. However, it can be extended to the actual inverse on \(L^\infty(\mathbb{R}^n)\).

To see that the definitions of \(\hat{u}\) and \(\check{u}\) as integrals make sense, consider a function \(u \in L^1(\mathbb{R}^n)\), and the integral in Equation 21.1. This integral exists for all \(\mathbf{k}\), since \[|\hat{u}(\mathbf{k})| \leq \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} | u(\mathbf{x})| \; \mathrm{d}\mathbf{x} = \frac{1}{(2\pi)^{n/2}}\|u\|_1.\] Hence, \(\hat{u} \in L^\infty(\mathbb{R}^3)\). The same argument applies to Equation 21.2.

Can we define the Fourier integral for any \(u \in L^\infty(\mathbb{R}^n)\)? No, just pick the constant function \(u(\mathbf{x}) = 1\). The function \(\exp(-i\mathbf{k}\cdot\mathbf{x})\) is measurable but not integrable.

We can, however, extend the definition of the Fourier integral to \(L^2(\mathbb{R}^n)\). The classical way to do this, is via Plancherel’s Theorem. The result is a unitary operator \(\mathcal{F}\) on \(L^2(\mathbb{R}^n)\), and in that case \(\mathcal{F}^{-1}\) is actually given by Equation 21.2:

Theorem 21.1 (Fourier Transform on \(L^2\)) For \(u \in L^2(\mathbb{R}^n)\), the Fourier integral Equation 21.1 is almost-everywhere defined, and \(\hat{u} \in L^2(\mathbb{R}^n)\). The Fourier transform is unitary, i.e., \(\|u\|_2 = \|\hat{u}\|_2\). Moreover, we have the following properties: Let \(u,v\in L^2(\mathbb{R}^n)\). Then,

  1. \(\left\langle u,v\right\rangle_2 = \left\langle\hat{u},\hat{v}\right\rangle_2 = \left\langle\check{u},\check{v}\right\rangle\) unitarity

  2. \(\widehat{\partial^\alpha u} = (\mathrm{i}\mathbf{k})^\alpha \hat{u}\) partial derivatives

  3. \(\widehat{u * v} = (2\pi)^{n/2} \hat{u}\hat{v}\) convolutions

  4. \(u = (\hat{u})^\vee = (\check{u})^\wedge\) (almost everywhere) inverses

Here, \(\alpha\) is a multindex \(\alpha = (\alpha_1,\cdots,\alpha_k)\), and \[\partial^\alpha = \frac{\partial ^{\alpha_1}}{\partial x_1^{\alpha_1}}\frac{\partial ^{\alpha_2}}{\partial x_2^{\alpha_2}}\cdots\frac{\partial ^{\alpha_n}}{\partial \alpha_n^{\alpha_n}}, \quad \mathbf{k}^\alpha = k_1^{\alpha_1}\cdots k_n^{\alpha_n}\]

Example 21.1 (Gaussian) Let \(A = A^T\) be an \(n\times n\) real invertible matrix, and consider the funtion \(u \in L^2(\mathbb{R}^n)\) given by \[u(\mathbf{x}) = \exp(-\mathbf{x}^T A \mathbf{x}/2)\] Then \[\hat{u}(\mathbf{k}) = \exp(-\mathbf{k}^T A^{-1} \mathbf{k}/2)\]

21.2 Fourier series and periodic Fourier transform

Another version of the Fourier transform is for square-integrable functions defined on the “unit torus” \(\mathbb{T}^n\). The 1D unit torus \(\mathbb{T}\) is the interval \([0,1[\) with periodic boundary conditions, i.e., the points \(0\) and \(1\) are identified. (This identification means a certain modification of open sets in \([0,1[\). Can you describe it?)

Thus we consider functions in \(L^2(\mathbb{T}^n)\).

We begin with the case \(n=1\). The Fourier transform of \(u \in L^2(\mathbb{T})\) is defined by \[\hat{u}_k = \int_{\mathbb{T}} e^{-2\pi\mathrm{i}k x} u(x) \; \mathrm{d}x.\] It is readily verifiable that this is equivalent to computing the basis expansion coefficients of the orthonormal set of vectors \[\phi_k (x) = e^{2\pi\mathrm{i}k x}, \quad k \in \mathbb{Z}.\] (That this is indeed a basis must be shown.) The map \(u \mapsto \hat{u}\) is an isometric isomorphism of \(L^2(\mathbb{T})\) and \(\ell_2(\mathbb{Z})\). The inverse Fourier transform is the Fourier series \[\check{c}(x) = \sum_{k\in\mathbb{Z}} e^{2\pi \mathrm{i}k x} c_k.\] It is a fact that \((\hat{u})^\vee = u\) almost everywhere, and that \(\check{c}^\wedge = c\).

The \(n\)-dimensional generalization of the Fourier transform is \[\hat{u}_{\mathbf{k}} = \int_{\mathbb{T}^n} e^{-2\pi \mathrm{i}\mathbf{k}\cdot\mathbf{x}} u(\mathbf{x}) \; \mathrm{d}\mathbf{x},\] and the Fourier series is \[\check{c}(\mathbf{x}) = \sum_{\mathbf{k}\in \mathbb{Z}^n} e^{2\pi \mathrm{i}\mathbf{k}\cdot\mathbf{x}} c_{\mathbf{k}} \; \mathrm{d}\mathbf{x}.\]

The transform \(u \mapsto c = \hat{u}\) is an isometric isomorphism between \(L^2(\mathbb{T}^n)\) and \(\ell_2(\mathbb{Z}^n)\).

Theorem 21.2 (Fourier series)  

  1. \(\left\langle u,v\right\rangle = \left\langle\hat{u},\hat{v}\right\rangle = \sum_{\mathbf{k}\in\mathbb{Z}^n} \hat{u}_{\mathbf{k}}\hat{v}_{\mathbf{k}}\) unitarity

  2. \(\widehat{\partial^\alpha u} = (2\pi \mathrm{i})^{|\alpha|} \mathbf{k}^\alpha \hat{u}_{\mathbf{k}}\) partial derivatives

  1. \(u = (\hat{u})^\vee\) (almost everyhwere), and \(c = (\check{c})^\wedge\) inverses