What does it mean to say that linear momentum has dimension of inverse of length? Hello i was reading a book written by a physicist about Fourier Transform, and there he says that the momentum space in discrete Fourier transform has it's name because of the value   $ p=\frac{2\pi}{L}k $  that has dimension of linear momentum, and then he says that it is true even in quantum mechanics where the $x$ component of linear momentum is given by the operator $\frac{\partial}{\partial x}$ that has dimension of inverse of length. But is it true also in classical mechanics ? because he seems to imply so in the text and how does one equates this with the usual dimension of mass*length/time ?
 A: The momentum operator agrees with the classical units of momentum. To show this first work out the momentum operator.
$$p\psi=\hat p\psi=-i\hbar\frac{\partial}{\partial x}\psi $$
This means
$$[p][\psi]=[\hbar][\frac{\partial}{\partial x}][\psi]$$
The dimensions of $\psi$ cancels so we only need to plug in the dimensions of $\hbar$ and $\frac{\partial}{\partial x}$. The dimensions of  $\frac{\partial}{\partial x}$ are $L^{-1}$ and since the dimensions of $\hbar$ are $energy\cdot time$ it follows that $[\hbar]=ML^2T^{-1}$. So we find the following:
$$[p]=ML^2T^{-1}\cdot L^{-1}=MLT^{-1}$$
Which is the same as the classical case. If $\hbar$ were set to 1 the dimensions would be the inverse of length.
Notation: [x] means the dimensions of x, L means length, M means mass and T time.
A: This is in essence the de Broglie relation,
$$p=\frac h\lambda,$$
rephrased in a more useful form in terms of the wavenumber $k=2\pi/\lambda$ and the reduced Planck constant as
$$p=\hbar k,$$
and then reduced to
$$p=k$$
in natural units by setting $\hbar=1$.
Normally, this relationship is exclusive to quantum mechanics, and cannot be extended to classical mechanics. To know what the author meant by that we'd have to see the full reference.
