If I take a periodic wavefunction $\psi\left(\vec{r}\right)$ and then take the Fourier space dispersion of the wave function as defined below
$$
\psi(\vec{k})=\iiint_{-\infty}^{+\infty}\psi\left(\vec{r}\right)e^{-\vec{k}\cdot\vec{r}}\mathrm{d}^3\vec{r}
$$
Is there a reason for calling $\psi(\vec{k})$ the momentum space representation of the wavefunction? (I understand the fact that the vector space $\vec{k}$ gets quantized in accordance to the formulation, $\vec{k}\cdot\vec{R}=2\pi$, where in $\vec{R}$ is the lattice translation vector periodicity of $\psi(\vec{r})$ in a crystal lattice), but is there some other reason for calling it momentum space?
Best Answer
As per my username, I feel it is partially my responsibility to address this question. I said it before and I'll say it again: The Fourier Transform is not an accident. There are countless reasons it has the precise form it has.
Let $F[f]$ denote the Fourier Transform of $f$, and let $\boldsymbol P=-i\boldsymbol \partial$ denote the momentum operator. We have $$ F[\boldsymbol Pf]=\boldsymbol p F[f]\tag1 $$ so that $F$ diagonalises $\boldsymbol P$. Indeed, the plane-wave basis $\mathrm e^{i\boldsymbol p\cdot \boldsymbol x}$ satisfies $$ \boldsymbol P\,\mathrm e^{i\boldsymbol p\cdot \boldsymbol x}=\boldsymbol p\,\mathrm e^{i\boldsymbol p\cdot \boldsymbol x} \tag2 $$ which automatically implies $(1)$, as claimed.
From this we learn that any operation that includes $\boldsymbol P$ becomes trivial if we work with $F[f]$ instead of with $f$ -- if we work in Fourier space. Thus, Fourier space is known as momentum space. Convenient, right?
In a nutshell, $F[\psi]$ is to momentum what $\psi$ is to position. This is a direct consequence of the (formal) fact that $$ F[\langle \boldsymbol x|]=\langle \boldsymbol p|\tag3 $$ which means that both sides agree when they act on $|\psi\rangle$.