Fourier transform of the state? I'm trying to make sense of the following equation I saw on Wikipedia: https://en.wikipedia.org/wiki/Momentum_operator#Fourier_transform
I saw on wikipedia that
\begin{equation}
\langle\psi|\hat{p}|\psi\rangle = h \langle\psi'|\hat{x}|\psi'\rangle 
\end{equation}
where $\psi'$ is the Fourier transform of $\psi$.
What does it mean to be the Fourier transform? More explicitly, what is the meaning of "$\psi'$ is the Fourier transform of $\psi$"? Isn't the Fourier transform of a state the same state but represented in different coordinates?
 A: Usually, you would indeed consider a single state $|\psi\rangle$, expand it over different bases :
$$|\psi\rangle = \int \text dx \;\psi(x) |x\rangle=\int \text d p\;\tilde \psi(p)|p\rangle$$
and then express the fact that $\tilde\psi(p)$ is (proportional to) the Fourier transform of $\psi(x)$.
Here, what they seem to do, and which feels highly unusual to me, seems to be to take the Fourier transform $\tilde \psi(p)$ of $\psi(x)$ and then define :
$$|\psi'\rangle = \int\text dx\;\tilde\psi\left(\frac x \hbar\right)|x\rangle$$
This does satisfy the equation they write, but seems unphysical and of very little use.
A: This is a flip excuse to expose you to Condon's 1937 classic paper about Heisenberg-picture rotations in phase space, leading to the unitary equivalence of $\hat x$ and $\hat p$.
In particular, you may write the quantum oscillator hamiltonian evolution operator in nondimensionalized coordinates,
$$
F\equiv \exp \left ( \frac{i \pi} {2\hbar} \frac{\hat p^2+\hat x^2}{2} \right )\equiv e^{\frac{i\pi}{2}G}, ~~~~\leadsto \\
F \hat x F^\dagger =\hat x +\frac{i\pi}{2}[G,\hat x]/2\hbar + \left (\frac{i\pi}{2}\right ) ^2[G,[G,\hat x]]/8\hbar^2 +...\\ =\cos (\pi/2) ~ \hat x + \sin (\pi/2) ~\hat p= \hat p, \\
F \hat p F^\dagger = \cos (\pi/2) ~ \hat p - \sin (\pi/2) ~\hat x= -\hat x,
$$
the celebrated rigid rotation by a quarter cycle in phase space.
Consequently, manifestly, $F^2=P$, the parity operator, and $F^4={\mathbb 1}$, upon a full 2π rotation. This sure behaves like the F.T. operator.
This  underlies, then, the unitary equivalence between $\hat x$ and $\hat p$, leaving Heisenberg's (Born's) commutation relation invariant, which you may use as you aimed in your comment, and @SolubleFish nicely points out.  ℏ is but a unit to be absorbed into p, or, better, its square root into both x and p to completely non-dimensionalize.
But I do not wish to appear endorsing the clumsy   and pointless former WP garble. If you are interested in Fourier Analysis and its Applications, you could do worse than G B Folland's eponymous book, ISBN-13: 978-0821847909.
