Does Each Component of a Vector Have an Independent Fourier Transform?

[redtree]

TL;DR

Is the Fourier transform of a function whose argument is a vector equivalent to performing a Fourier transform on the function of each vector component separately?

Given $f(\vec{x})$, where the Fourier transform $\mathcal{F}(f(\vec{x}))= \hat{f}(\vec{k})$.
Given $\vec{x}=[x_1,x_2,x_3]$ and $\vec{k}=[k_1,k_2,k_3]$, is the following true?

\begin{equation}

\begin{split}

\mathcal{F}(f(x_1))&= \hat{f}(k_1)

\\

\mathcal{F}(f(x_2))&= \hat{f}(k_2)

\\

\mathcal{F}(f(x_3))&= \hat{f}(k_3)

\end{split}

\end{equation}

such that

\begin{equation}

\begin{split}

f(\vec{k})&= [\hat{f}(k_1), \hat{f}(k_1), \hat{f}(k_1)]

\end{split}

\end{equation}

 

[S.G. Janssens]

Your notation is not entirely clear to me. (Also, I think there are typos in (2)?)

In any case, I don't think what you want will work. Let's say that $f$ maps vectors to scalars. Then the Fourier transform $\hat{f}$ is still scalar-valued, but your vector of partial Fourier transforms is vector-valued.

 

[redtree]

You are correct; the equation should be the following:

\begin{equation}

\begin{split}

\hat{f}(\vec{k})&= [\hat{f}(k_1), \hat{f}(k_2), \hat{f}(k_3)]

\end{split}

\end{equation}

The Fourier transform maps vectors to vectors; otherwise one could not transform back from the Fourier conjugate space to the original vector space with the inverse Fourier transform. Information would have been lost in the mapping of a vector to a scalar.

That the Fourier conjugate of a vector is also a vector is seen in the variance relationship between Fourier conjugates, such as for example between position $\vec{x}$ and momentum $\hbar \vec{k}$, where both are 3-vectors.

 

[jasonRF]

In your example $f$ is a scalar-valued function of 3 variables. It's Fourier transform is therefore a scalar-valued function of 3 variables. The forward and inverse transforms are then
$\begin{eqnarray*} \hat{f}(\vec{k}) & = & \int_{-\infty}^\infty f(\vec{x})\, e^{-i \vec{k}\cdot\vec{x}} \, d^3x\\ & = & \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty f(x_1,x_2,x_3)\, e^{-i (k_1 x_1 + k_2 x_2 + k_3 x_3)} \, dx_1 dx_2 dx_3 \\ f(\vec{x}) & = & \frac{1}{(2\pi)^3} \int_{-\infty}^\infty \hat{f}(\vec{k})\, e^{i \vec{k}\cdot\vec{x}} \, d^3k \\ & = & \frac{1}{(2\pi)^3} \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \hat{f}(k_1,k_2,k_3)\, e^{i (k_1 x_1 + k_2 x_2 + k_3 x_3)} \, dk_1 dk_2 dk_3 \end{eqnarray*}$

jason

 

[redtree]

Got it. My mistake.