Fourier Transform - Dirac Delta

Question

How can I compute this Fourier transform?

$$\int \frac{d^3q}{(2\pi)^3}(\vec q \cdot \vec a)(\vec q \cdot \vec b)\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }\,$$

My idea was to write it as

$$-( \vec a \cdot \vec \nabla)(\vec b \cdot \vec \nabla)\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }$$

and use the fact that $\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }=\frac{1}{4\pi r}$, together with $\Delta \frac{1}{r} = -4\,\pi \,\delta(\vec r)$ but is is not exactly the same.

Answer 1

So the other answers are correct only for $x≠0$. In the sense of distributions, however, there is indeed a Dirac delta appearing in the result.

The first steps in your approach are correct. both functions $\frac{1}{q^2}$ and $\frac{(q·a)\,(q\cdot b)}{q^2}$ are indeed locally integrable functions and bounded outside a compact set, so are tempered distributions: one can take their Fourier transform in the sense of distributions (remark however that these functions are not in $L^1$ or in $L^2$, so one cannot take the Fourier transform in the usual sense).

In the sense of distributions (writing with abuse of notation the Fourier transform as an integral) $$ \frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q = -(a\cdot\nabla)(b\cdot\nabla)\, \frac{1}{4π|x|}. $$ Taking a first derivative of $1/|x|$, one obtains $-x/|x|^3$ (even in the sense of distributions) which is still a locally integrable function. However, taking two derivatives leads to a non locally integrable function, and so one has to be careful as you suspect. A formula that generalizes the formula of the Laplacian $-\Delta \frac{1}{4\pi|x|} = \delta_0$ is the Formula of the Hessian (where $\nabla^2 = \nabla\nabla$) $$ \nabla^2\left(\frac{1}{4\pi|x|}\right) = \frac{1}{4π} \,\mathrm{pv.}\,\frac{3x\otimes x - |x|^2 \,\mathrm{Id}}{|x|^5} - \frac{1}{3}\, \delta_0 \,\mathrm{Id} $$ where pv. denotes the principal value, see e.g. p.55 here. (In particular, if you sum the coordinates $(j,j)$ of this matrix, you obtain the formula for the Laplacian). Now remark that $(a\cdot\nabla)(b\cdot\nabla) = (a\otimes b):\nabla^2$ (if you prefer coordinates, $\sum_{ij} a_i\partial_i\, b_j\partial_j = \sum_{ij} a_ib_j\,\partial_i\partial_j$) and so $$ \begin{align} \frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q &= -(a\otimes b):\nabla^2\left(\frac{1}{4\pi|x|}\right). \\ &= \frac{1}{4π} \,\mathrm{pv.}\,\frac{|x|^2 (a·b) - 3\,(a\cdot x)(b\cdot x)}{|x|^5} + \frac{a\cdot b}{3}\, \delta_0 \end{align} $$

The physicist usually do not write the principal value, and so in shorter notations (with $r=|x|$) $$ \boxed{\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q = \frac{1}{4πr^3} \left(a·b - 3\,(a\cdot \tfrac{x}{r})(b\cdot \tfrac{x}{r})\right) + \frac{a\cdot b}{3}\, \delta_0} $$

Answer 2

I would go with spherical polar coordinates. Perhaps with $\vec{r}=(r,0,0)$, then the dot products can be evaluated using trigonometry functions of $\phi$ and $\theta$. It would eventually look like a Dirac delta modulated by the dot products among $\vec{a}$, $\vec{b}$ and $\vec{r}$.

Edit: It is indeed more straightforward with your original idea.

$-(a\cdot\nabla)(b\cdot\nabla) \Delta^{-1} \delta(\vec{r})=-(a\cdot\nabla)(b\cdot\nabla) \frac{1}{4\pi r}=b\cdot\nabla(a\cdot \frac{\vec{r}}{r^3})=\frac{1}{4\pi r^3}[\vec{a}\cdot\vec{b}-3(\vec{a}\cdot\frac{\vec{r}}{r}) (\vec{b}\cdot\frac{\vec{r}}{r})].$

So it's the mutual energy between two dipoles a and b.

I would answer the above comment here (I cannot make comments as I'm new to the site): The Dirac delta is there at $r=0$, it's just that it is also nonzero at other places.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ In order to deal with possible issues related to integrals convergence, I'll include the $\ds{\underline{\it parameter}\,\,\, \beta}$. Its relevance is discussed at the very end. \begin{align} &\bbox[5px,#ffd]{{1 \over \pars{2\pi}^{3}} \iiint_{\mathbb{R}^{3}}\pars{\vec{q}\cdot\vec{a}} \pars{\vec{q}\cdot\vec{b}} {\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2 + \beta}\,}\,\,\dd^{3}\vec{q}} \\[5mm] = &\ {1 \over \pars{2\pi}^{3}}\,\vec{a}\cdot\bracks{% \iiint_{\mathbb{R}^{3}}\vec{q}\,\vec{q}\, {\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2}}{\dd^{3}\vec{q} \over q^{\beta}}}\cdot\vec{b} \\[5mm] = &\ {1 \over \pars{2\pi}^{3}}\,\vec{a}\cdot\bracks{% \int_{0}^{\infty}\pars{% \int_{\Omega_{\vec{q}}}\hat{q}\,\hat{q}\, \expo{\ic\vec{q}\cdot\vec{r}}\,\dd\Omega_{\vec{q}}} q^{2 - \beta}\,\,\dd q}\cdot\vec{b}\label{1}\tag{1} \end{align} Lets perform the angular integration where, for simplicity, $\ds{\vec{r}}$ is chosen along the $\ds{\hat{z}}$-axis. Later on, the $\ds{\vec{r}}$-general direction can be suitable restored: \begin{align} &\int_{\Omega_{\vec{q}}}\hat{q}\,\hat{q}\, \expo{\ic\vec{q}\cdot\vec{r}}\,\dd\Omega_{\vec{q}} = \int_{0}^{2\pi}\int_{0}^{\pi} \hat{q}\,\hat{q}\expo{\ic qr\cos\pars{\theta}}\, \sin\pars{\theta}\,\dd\theta\,\dd\phi \\[5mm] = &\ \int_{0}^{\pi} \bracks{\sin^{2}\pars{\theta}\,\pi\,\hat{x}\,\hat{x} + \sin^{2}\pars{\theta}\,\pi\,\hat{y}\,\hat{y} + \cos^{2}\pars{\theta}\pars{2\pi}\hat{z}\,\hat{z}}\ \times \\ &\ \phantom{\,\,\int_{0}^{\pi}} \,\,\,\expo{\ic qr\cos\pars{\theta}}\,\,\sin\pars{\theta} \,\dd\theta \\ = &\ \pi\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}} \int_{-1}^{1}\pars{1 - \xi^{2}}\expo{\ic qr\xi}\,\dd\xi + 2\pi\,\hat{z}\,\hat{z} \int_{-1}^{1}\xi^{2}\expo{\ic qr\xi}\,\dd\xi \\ = &\ 4\pi\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}} \on{f}\pars{qr} + 4\pi\,\hat{z}\,\hat{z}\,\on{g}\pars{qr} \\[5mm] &\ \mbox{where}\quad \left\{\begin{array}{rcl} \ds{\on{f}\pars{\xi}} & \ds{\equiv} & \ds{\sin\pars{\xi} - \xi\cos\pars{\xi} \over \xi^{3}} \\[2mm] \ds{\on{g}\pars{\xi}} & \ds{\equiv} & \ds{\pars{\xi^{2} - 2}\sin\pars{\xi} + 2\xi\cos\pars{\xi} \over \xi^{3}} \end{array}\right. \end{align}

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(\ref{1}) becomes \begin{align} &\bbox[5px,#ffd]{{1 \over \pars{2\pi}^{3}} \iiint_{\mathbb{R}^{3}}\pars{\vec{q}\cdot\vec{a}} \pars{\vec{q}\cdot\vec{b}} {\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2 + \beta}}\,\dd^{3}\vec{q}} \\[5mm] = &\ {1 \over 2\pi^{2}}\,{1 \over r^{3 - \beta}\,} \,\,\vec{a}\,\cdot \\[2mm] &\ \bracks{\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}} \int_{0}^{\infty}\on{f}\pars{\xi}\,\xi^{2 - \beta}\, \,\dd\xi + \hat{z}\,\hat{z} \int_{0}^{\infty}\on{g}\pars{\xi}\,\xi^{2 - \beta}\, \,\dd\xi} \cdot\vec{b} \end{align} The integrations are given by \begin{align} &\int_{0}^{\infty}\on{f}\pars{\xi}\,\xi^{2 - \beta}\, \,\dd\xi = {\Gamma\pars{2 - \beta}\sin\pars{\pi\beta/2} \over \beta} \,,\quad\,\,\,\,\,\, 0 < \Re\pars{\beta} < 1 \\[5mm] &\ \int_{0}^{\infty}\on{g}\pars{\xi}\,\xi^{2 - \beta}\, \,\dd\xi = -\,{\Gamma\pars{3 - \beta}\sin\pars{\pi\beta/2} \over \beta} \,,\quad 1 < \Re\pars{\beta} < 3 \end{align} The first integral $\ds{\to {\pi \over 2}}$ as $\ds{\beta \to 0^{+}}$ while such a limit is not allowed for the second integral -unless we assume an analytical continuation- which we left as a discussion for the OP.