I used to think $\lvert\langle p\lvert\psi\rangle\rvert^2$ had the meaning of some likelihood of the particle's momentum being $p$ (within some tolerance interval $\Delta p$). Now I'm just confused.
I'll use momentum $p$, wavenumber $k$, and velocity $v$ interchangeably here (i.e., I'll use the convention $m=1$, $\hbar=1$). I begin with the momentum distribution at the time $t=0$. Further, let's limit ourselves to wavefunctions that are initially real, and also describe a stationary state in some potential well configuration.
I want to compare the evolution of the above state, with the evolution of the state of a free particle. As soon as the clock moves forward just an instant beyond $t=0$, these two different systems evolve completely differently.
I believe a theory that allows the probability distribution of velocities to depend on $\frac{\partial \psi}{\partial t}$ at time $t=0$ would make more sense than a theory by which the above distribution depends only on the initial form of the wave-function $\psi_0$.
A) For exemplification of the evolution of a free particle consider the spreading of a wave-packet with the initial state $\psi_0(x)= \exp(-\lvert x\rvert)$,
$$\psi(x,t)\propto\int \mathrm{d}k \; \frac{\exp(i k x-i \omega_k t)}{k^2+1}, \tag{1}$$ where $\omega_k=k^2/2$.
B) However, if the same initial state would describe the particle in the well, the particle is in a bound, stationary state, and $\lvert\psi(x,t)\rvert^2$ just remains constant over time. Nothing similar to the evolution $\text {(1)}$.
Discussion : My point was to pick an initial state (real, no current) and show that there is a great amount of current within a very short time, i.e. it seems to me that the difference between $|\psi_A (x,t)|^2$ and $|\psi_B (x,t)|^2$ happens just about instantaneously.
In the state $\text {(1)}$ we have no current of probability, $j(x,t)=0$. (The current is given by the density of probability times the Bohmian velocity $j(x,t)=\lvert\psi\rvert^2 v_{\text{avg}}$, where $v_{\text{avg}}=\frac{\hbar}{m}\frac{\partial}{\partial x}\,\arg(\psi)$ is a quantity central in Bohmian interpretations of QM.) But the zero-current doesn't mean there isn't any movement, simply the rate of particles moving to the bright counter-balances the rate of leftward going particles. And I have a hard time believing the distribution of velocities in the cases A and B is exactly equal.
(There are similar well problems, one being the distribution $\psi_0=\exp(-x^2)$ and compared with a particle in a ground state harmonic oscillator well. This would be a better one to use if we want to talk about the velocity distribution and the kinetic energy. And, since there are higher bound state energy states, about the classical limit.)
C) An additional thought: in the classical limit (large number of nodes within some space $\Delta x$), one should be able to talk about a velocity distribution near a position $x$. It should be something like $0.5\delta(v-v_c)+0.5\delta(v+v_c)$ where $v_c=\sqrt{2T/m}$. How does this distribution compare to any meaning we can gather from the positionless distribution $\lvert\langle p\lvert\psi\rangle\rvert^2$? Thinking classically I believe that a factor $\delta t$ times the difference in acceleration between the two systems should explain the difference in evolution.
Would there be any way to obtain a position dependent velocity distribution, or even a quantity similar to this, that would agree in the classical limit?