11.4 Linear transformations and distributions

11.4 Linear transformations and distributions

Concept links · terms present in this machine draft; source roles are unverified: vector space · linear map

The definition of distributions as linear functionals on the vector space $\mathcal{S}(\mathbf{R})$ means that for any linear transformation $A$ acting on $S(\mathbf{R})$ , we can get a linear transformation on $S^{\prime }(\mathbf{R})$ as the transpose of $A$ (see equation 4.1), which takes $T$ to

$$A^{t}T:f\in \mathcal{S}(\mathbf{R})\to (A^{t}T)[f]=T[Af]\in \mathbf{C}$$

This gives a definition of the Fourier transform on $S^{\prime }(\mathbf{R})$ as

$${\mathcal{F}}^{t}T[f]\equiv T[\mathcal{F}f]$$

and one can show that, as for $\mathcal{S}(\mathbf{R})$ and $L^2(\mathbf{R})$ , the Fourier transform provides an isomorphism of $S^{\prime }(\mathbf{R})$ with itself. Identifying functions $\psi$ with distributions

$T_{\psi },$ one has

$$\begin{array}{rl}{\mathcal{F}}^t{T}_{\psi }[f]& \equiv T_{\psi }[\mathcal{F}f]\\ & =T_{\psi }\left[\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}f(q)dq\right]\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }\psi (k)\left({\int }_{-\infty }^{+\infty }{e}^{-ikq}f(q)dq\right)dk\\ & ={\int }_{-\infty }^{+\infty }\left(\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}\psi (k)dk\right)f(q)dq\\ & =T_{\mathcal{F}\psi }[f]\end{array}$$

showing that the Fourier transform is compatible with this identification.

As an example, the Fourier transform of the distribution $\frac{1}{\sqrt{2\pi }}{e}^{ika}$ is the $\delta$-function $\delta (q-a)$ since

$$\begin{array}{rl}{\mathcal{F}}^t{T}_{\frac{1}{\sqrt{2\pi }}{e}^{ika}}[f]& =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{ika}\left(\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}f(q)dq\right)dk\\ & ={\int }_{-\infty }^{+\infty }\left(\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{e}^{-ik(q-a)}dk\right)f(q)dq\\ & ={\int }_{-\infty }^{+\infty }\delta (q-a)f(q)dq\\ & =T_{\delta (q-a)}[f]\end{array}$$

For another example of a linear transformation acting on $\mathcal{S}(\mathbf{R})$ , consider the translation action on functions $f{A}_{a}f$ , where

$$(A_{a}f)(q)=f(q-a)$$

The transpose action on distributions is

$$A_a^t{T}_{\psi (q)}=T_{\psi (q+a)}$$

since

$$A_a^t{T}_{\psi (q)}[f]=T_{\psi (q)}[f(q-a)]={\int }_{-\infty }^{+\infty }\psi (q)f(q-a)dq={\int }_{-\infty }^{+\infty }\psi (q^{\prime }+a)f(q^{\prime })d{q}^{\prime }$$

The derivative is an infinitesimal version of this, and one sees (using integration by parts), that

$$\begin{array}{rl}{\left(\frac{d}{dq}\right)}^t{T}_{\psi (q)}[f]& =T_{\psi (q)}\left[\frac{d}{dq}f\right]\\ & ={\int }_{-\infty }^{+\infty }\psi (q)\frac{d}{dq}f(q)dq\\ & ={\int }_{-\infty }^{+\infty }\left(-\frac{d}{dq}\psi (q)\right)f(q)dq\\ & =T_{-\frac{d}{dq}\psi (q)}[f]\end{array}$$

In order to have the standard derivative when one identifies functions and distributions, one defines the derivative on distributions by

$$\frac{d}{dq}T[f]=T\left[-\frac{d}{dq}f\right]$$

This allows one to define derivatives of a $\delta$-function, with for instance the first derivative ${\delta }^{\prime }(q)$ of $\delta (q)$ satisfying

$$T_{{\delta }^{\prime }(q)}[f]=-f^{\prime }(0)$$

Previous section: 11.3 Distributions · Next section: 11.5 Solutions of the Schr¨odinger equation in momentum space · 中文版本

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