I took quantum where I learnt the usage of momentum space. I then understood why momentum space could be so useful in computation. However, what I still did not understood was the mean to define a momentum grid/space.

i.e. I have this equation. for iteration. where $k^2$ in the middle was the step where, through the Fourier transformation on the right side, calculate the iteration in momentum space.

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In the 1D example, the textbook defined $dx$(separation) in position space and define the position space by $x=(-n,..,n)\cdot dx$.(with unit in meter.)

However, it defined the momentum space through $dkx=\pi/(n\cdot dx)$ and obtain the momentum space $kx=(-n,..,n)\cdot dkx$.

I looked through my quantum textbooks, physics.stachexchange and google, but I could not find the reason why:

  1. Why was the coefficient in $dkx$ was $\pi$? My guess was that it might because the entire $kx$ axis would be $2\pi$, but why not it be $\sqrt{2\pi}$ as we usually seen from the quantum mechanics? Further in 2D case, would it still be $2\pi$?

  2. How come $dkx$ was obtained by dividing $dx$? Notice momentum had units $kg\cdot m\cdot s^{-1}$. If it was dividing, won't the unit be $m^{-1}$ and no longer match?

  3. In general, how to obtain the momentum space when the position space was known in a numerical fashion? I.e. the position space was intuitive. But how do we acknowledge the "size" and definition of momentum space?

(I knew this looked kind of computational, but the other exchange sits won't understand the topic.)

  • $\begingroup$ Which textbook? $\endgroup$
    – Qmechanic
    Apr 7, 2018 at 19:03
  • $\begingroup$ @Qmechanic A primer on quantum fluids by Carlo Barenghi and Nick Parker, Appendix A 2. $\endgroup$
    – J C
    Apr 7, 2018 at 20:23

1 Answer 1


There were several notices:

  1. Momentum was bounded. In most of the cases, i.e. this one, was the position space/grid was bounded, it eventually implied that the momentum was bounded(Assume there was no singularities).

  2. Momentum was quantized. In the same sense that the position space/grid was bounded, the corresponding (allowed) momentum space/grid was also quantized and only allowed to take discrete values.

  3. Since the momentum space/grid was quantized, the function of Fourier transformation was actually discrete Fourier transformation.(In the recent decades, the DFT was usually replaced by FFT fast furier transformation.

  4. Although FFT was faster, it provided similiar accuracy as DFT, and in some cases, even a better resolution.

  5. When talking about the momentum space/grid or momentum transformation in computation, we almost never actually did a momentum transformation. Instead, the option was a "wave number space" or "wave number transformation", k, same as what was written in Griffith's textbook. The reason was that

    a. the wave number transformation was much simpler to deal with than the momentum transformation. i.e. $e^{ikx}$ was either than $e^{i\frac{p}{\hbar}x}$.

    b. It's easier for computer to deal with. Keep in mind that although we wrote $\hbar$ a lot $6.62607004 × 10-34 m^2\cdot kg \cdot s^{-1}/(2\pi)$ was a very small number in decimal place. Vise versa, $1/\hbar$ was a comparable large number in decimal place. By taking $\hbar$ in to the calculation of momentum transformation, there was a risk in running out of memory, low accuracy, and low computational efficiency. (And by doing FFT such effect could be maximized by magnitude of power of at your grid size (100^2,1000^2) e.t.c.)

  6. Notice $k=2\pi/\lambda =p/\hbar=\sqrt{2mT}/\hbar$(from wiki). A usual approach for the discrete wave number space was by iterate $k$ from $-\pi$ to $\pi$ with $\Delta k=2\pi/L=2\pi/(2n\cdot dx)$.(In the sense that the smallest changes in position space was $\lambda=dx$ and the smallest change in wave number space/grid was $k=2\pi/L$) (This part I wasn't very clear why the boundary was from $-pi$ to pi )


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