11.3 Distributions

11.3 Distributions

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

While the use of the subspace $\mathcal{S}(\mathbf{R})\subset {\mathcal{L}}^2(\mathbf{R})$ as state space gives a well-behaved momentum operator $P$ and a formalism symmetric between positions and momenta, it still has the problem that eigenfunctions of $P$ are not in the state space. Another problem is that, unlike the case of $L^2(\mathbf{R})$ where the Riesz representation theorem provides an isomorphism between this space and its dual just like in the finite dimensional case (see 4.3), such a duality no longer holds for $S(\mathbf{R})$

For a space dual to $\mathcal{S}(\mathbf{R})$ one can take the space of linear functionals on $\mathcal{S}(\mathbf{R})$ called the Schwartz space of tempered distributions (a certain continuity condition on the functionals is needed, see for instance [89]), which is denoted by $S^{\prime }(\mathbf{R})$ . An element of this space is a linear map

$$T:f\in \mathcal{S}(\mathbf{R})\to T[f]\in \mathbf{C}$$

$\mathcal{S}(\mathbf{R})$ can be identified with a subspace of $S^{\prime }(\mathbf{R})$ , by taking $\psi \in \mathcal{S}(\mathbf{R})$ to the linear functional $T_{\psi }$ given by

$$T_{\psi }:f\in \mathcal{S}(\mathbf{R})\to T_{\psi }[f]={\int }_{-\infty }^{+\infty }\psi (q)f(q)dq\text{\hspace{1em}(11.8)}$$

Note that taking $\psi \in \mathcal{S}(\mathbf{R})$ to $T_{\psi }\in {\mathcal{S}}^{\prime }(\mathbf{R})$ is a complex linear map.

There are however elements of $S^{\prime }(\mathbf{R})$ that are not of this form, with three important examples

• The linear functional that takes a function to its Fourier transform at k:

$$f\in \mathcal{S}(\mathbf{R})\to \tilde{f}(k)$$

• The linear functional that takes a function to its value at $q:$

$$f\in \mathcal{S}(\mathbf{R})\to f(q)$$

• The linear functional that takes a function to the value of its derivative at $q:$

$$f\in \mathcal{S}(\mathbf{R})\to f^{\prime }(q)$$

We would like to think of these as “generalized functions”, corresponding to $T_{\psi }$ given by the integral in equation 11.8, for some $\psi$ which is a generalization of a function.

From the formula 11.3 for the Fourier transform we have

$$\tilde{f}(k)=T_{\frac{1}{\sqrt{2\pi }}{e}^{-ikq}}[f]$$

so the first of the above linear functionals corresponds to

$$\psi (q)=\frac{1}{\sqrt{2\pi }}{e}^{-ikq}$$

which is a function, but “generalized” in the sense that it is not in $\mathcal{S}(\mathbf{R})$ (or even in $L^2(\mathbf{R}))$ ). This is an eigenfunction for the operator $P,$ and we see that such eigenfunctions, while not in $S(\mathbf{R})$ or $L^2(\mathbf{R})$ , do have a meaning as elements of $S^{\prime }(\mathbf{R})$ .

The second linear functional described above can be written as $T_{\delta }$ with the corresponding generalized function the “$\delta$-function”, denoted by the symbol $\delta (q-q^{\prime })$ , which is taken to have the property that

$${\int }_{-\infty }^{+\infty }\delta (q-q^{\prime })f(q^{\prime })d{q}^{\prime }=f(q)$$

$\delta (q-q^{\prime })$ is manipulated in some ways like a function, although such a function does not exist. It can however be made sense of as a limit of actual functions. Consider the limit as $ϵ0$ of functions

$$g_{ϵ}=\frac{1}{\sqrt{2\pi ϵ}}{e}^{-\frac{(q-q^{\prime }{)}^2}{2ϵ}}$$

These satisfy

$${\int }_{-\infty }^{+\infty }{g}_{ϵ}(q^{\prime })d{q}^{\prime }=1$$

for all $ϵ>0$ (using equation 11.6).

Heuristically (ignoring problems of interchange of integrals that don’t make sense), the Fourier inversion formula can be written as follows

$$\begin{array}{rl}\psi (q)& =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{ikq}\tilde{\psi }(k)dk\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{ikq}\left(\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ik{q}^{\prime }}\psi (q^{\prime })d{q}^{\prime }\right)dk\\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }\left({\int }_{-\infty }^{+\infty }{e}^{ik(q-q^{\prime })}\psi (q^{\prime })dk\right)d{q}^{\prime }\\ & ={\int }_{-\infty }^{+\infty }\delta (q-q^{\prime })\psi (q^{\prime })d{q}^{\prime }\end{array}$$

Physicists interpret the above calculation as justifying the formula

$$\delta (q-q^{\prime })=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{e}^{ik(q-q^{\prime })}dk$$

and then go on to consider the eigenvectors

$$|k⟩=\frac{1}{\sqrt{2\pi }}{e}^{ikq}$$

of the momentum operator as satisfying a replacement for the Fourier series orthonormality relation (equation 11.1), with the $\delta$-function replacing the ${\delta }_{nm}$ :

$$⟨k^{\prime }|k⟩={\int }_{-\infty }^{+\infty }\overline{\left(\frac{1}{\sqrt{2\pi }}{e}^{i{k}^{\prime }q}\right)}\left(\frac{1}{\sqrt{2\pi }}{e}^{ikq}\right)dq=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{e}^{i(k-k^{\prime })q}dq=\delta (k-k^{\prime })$$

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