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 $P=-i\partial$ denote the momentum operator. We have $$ F[Pf]=p F[f]\tag1 $$ so that $F$ diagonalises $P$. Indeed, the plane-wave basis $\mathrm e^{ipx}$ satisfies $$ P\mathrm e^{ipx}=p\mathrm e^{ipx} \tag2 $$ which automatically implies $(1)$, as claimed.

From this we learn that any operation that includes $P$ 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 $$ F[\langle x|]=\langle p| $$ which means that both sides agree when they act on $|\psi\rangle$.

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 $|\psi\rangle$.

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 $P=-i\partial$ denote the momentum operator. We have $$ F[Pf]=p F[f] $$ so that $F$ diagonalises $P$. From this we learn that any operation that includes $P$ 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?

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 $P=-i\partial$ denote the momentum operator. We have $$ F[Pf]=p F[f]\tag1 $$ so that $F$ diagonalises $P$. Indeed, the plane-wave basis $\mathrm e^{ipx}$ satisfies $$ P\mathrm e^{ipx}=p\mathrm e^{ipx} \tag2 $$ which automatically implies $(1)$, as claimed. 

From this we learn that any operation that includes $P$ 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 $$ F[\langle x|]=\langle p| $$ which means that both sides agree when they act on $|\psi\rangle$.