Fourier Transform and Dirac Delta Function

Question

A Fourier transform of function R into function Q is defined as: $$Q(\underline{k}) = \int_{}^{}R(\underline{x}) e^{-i\underline{k}·\underline{x}} \mathrm{d}\underline{x}.$$ where I've underlined $\underline{x}$ and $\underline{k}$ to denote that they are a set of vectors.

Suppose that $R(\underline{x})$ is only a function of coordinate differences, for example, $R(\underline{x})=$($x$₁-$x$₂)($x$₃-$x$₁), where $\underline{x}$=($x$₁,$x$₂,$x$₃).

Why must the Fourier transform $Q(\underline{k})$ contain a Dirac Delta function?

Note: The fact that the Fourier transform must contain a Dirac Delta function was mentioned on the bottom of page 25 of the following set of notes by my Physics professor, http://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/qftMT2012.pdf

Answer 1

A somewhat physical explanation is this:
Dependence on coordinate differences only implies translation invariance, which by Noether's theorem implies conservation of momentum, which in turn is represented as $\delta(\sum_k p_k)$ in $p$-space.

A mathematical example in 1 dimension: $$\begin{align} \mathcal{F}\{f(x_1-x_2)\}(p_1,p_2) &= \iint f(x_1-x_2) \, e^{-i(p_1 x_1+p_2 x_2)} \, dx_1 \, dx_2 \\ &= \{ \text{ variable change: } s = x_1 - x_2,\ t = x_2 \} \\ &= \iint f(s) \, e^{-i(p_1(s+t)+p_2t)} \, ds \, dt \\ &= \int \left( \int f(s) \, e^{-ip_1s} \, ds \right) e^{-i(p_1+p_2)t} \, dt \\ &= \int \mathcal{F}\{f\}(p_1) e^{-i(p_1+p_2)t} \, dt \\ &= \mathcal{F}\{f\}(p_1) \int e^{-i(p_1+p_2)t} \, dt \\ &= \mathcal{F}\{f\}(p_1) \, 2\pi \, \delta(p_1+p_2) \end{align}$$

Answer 2

This is not totally rigorous, but for Physics it should be enough :-)

In general, if $Q(\underline{k}) $ is the transform of $R(\underline{x})$, then the transform of $R(\underline{x} + a \underline{u})$ where $ \underline{u}=(1,1,1)$ and $a$ is some scalar, is $$Q'(\underline{k})=\exp(i 2 \pi a \, \underline{u} \cdot \underline{k}) Q(\underline{k}) = \exp(i 2 \pi a ( k_1+k_2 +k_3)) \, Q(\underline{k}) \tag{1}$$

But if $R(\underline{x})$ is only a function of coordinate differences, then $R(\underline{x} + a \underline{u})=R(x)$, and $Q'(\underline{k})=Q(\underline{k})$ for all $a$.

Then, from $(1)$, $$Q(\underline{k})\ne 0 \implies \exp(i 2 \pi a ( k_1+k_2 +k_3))=1 \hskip{5mm} (\forall a) \implies k_1+k_2 +k_3 = 0$$

That is, $Q(\underline{k})$ must be zero everywere except (at most) a measure zero set (a plane in $\mathbb{R}^3$). Then either $Q(\underline{k})$ is identically zero (trivial) or it's "degenerate" (has Dirac deltas).