Why is Fourier space called as momentum space?

Question

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?

Answer 1

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 $|\psi\rangle$.

Answer 2

but is there some other reason for calling it momentum space?

The canonical commutation relation for the position and momentum operators is (in one dimension)

$$[X, P]|\psi\rangle = (XP - PX)|\psi\rangle = i\hbar|\psi\rangle$$

On the position basis, this is

$$[x,P_x]\psi(x) = (xP_x - P_xx)\psi(x) = i\hbar\psi(x)$$

and it follows that a position basis representation of the momentum operator is

$$P_x = -i\hbar\partial_x$$

---

A wavefunction with definite momentum $p$ is then

$$\psi_p(x) = e^{\frac{i}{\hbar}px} = \langle x|p\rangle$$

such that

$$P_x\psi_p(x) = -i\hbar\partial_x\,e^{\frac{i}{\hbar}px} = pe^{\frac{i}{\hbar}px} = p\psi_p(x)$$

Now, if we write

$$k = \frac{p}{\hbar}$$

then

$$\psi_p(x) = e^{ikx}$$

---

The momentum space wavefunction $\psi(p)$ is just the ket $|\psi\rangle$ projected onto the momentum basis:

$$\psi(p) = \langle p | \psi\rangle = \int\mathrm{d}x\, \langle p | x\rangle\langle x | \psi\rangle$$

where I have used the completeness relation

$$1 = \int\mathrm{d}x\,|x\rangle\langle x | $$ but

$$\psi(x) = \langle x | \psi\rangle$$

and

$$e^{-\frac{i}{\hbar}px} = \langle p| x\rangle$$

thus

$$\psi(p) = \int\mathrm{d}x\,\psi(x)\,e^{-\frac{i}{\hbar}px}$$

or, finally

$$\psi(k) = \int\mathrm{d}x\,\psi(x)\,e^{-ikx} $$

Answer 3

Yet another name justification:

In both crystal and de-Broglie formalism, momentum is defined as $p=\hbar k$. This creates a mapping between a pure wave with wavenumber k and a point in momentum space. If a wavefunction is a superposition of, say, three waves, its momentum representation will consist of three points.

Fourier Transform performs the same mapping. The transform "scans" all possible k and furnishes zero all over, except for the pure wave constituents of the waveform.