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For a free particle, the wave function in the position basis spreads out over time. Based on my understanding, it means the wave function in the momentum basis becomes more and more localized over time because of the uncertainty principle (or by performing Fourier Transform). But I also learned that the probability of measuring the momentum doesn't change at all with time. Isn't this contradictory to my first understanding? How does the probability stay constant when the wave function (and so the probability density function's shape) in the momentum basis is becoming more localized over time?

For example, in the graph below, as time passes, the momentum probability distribution goes from B to A and becomes more localized. The probability density of measuring the particle's momentum between p1 and p2 obviously is different, right?

enter image description here

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  • $\begingroup$ For a free particle, $P(t)=P(0)$ for the momentum operator in the Heisenberg picture implies obviously $\Delta_\psi P(t)=\Delta_\psi P(0)$. The uncertainty relation $\Delta_\psi X(t) \Delta_\psi P(t) \ge \hbar/2$ is, of course, always satisfied. See physics.stackexchange.com/q/804052 for more details. $\endgroup$
    – Hyperon
    Commented Oct 24 at 20:05

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The uncertainty principle does not guarantee that the product of uncertainties stays the same. It merely guarantees that the product of uncertainties must be greater than $\hbar/2$. So it is perfectly fine for the uncertainty in position to grow over over time while the uncertainty in momentum remains fixed, provided that the uncertainty principle holds at $t=0$. This is indeed what occurs for the wave function describing a free-particle system.

The plot that you show is not what happens under free-particle time evolution. The momentum probability distribution (square of the momentum-space wave function) does not change with time under free-particle time evolution, because the time evolution of the momentum-space wave function $\tilde{\psi}(p)$ of a free-particle is just $$ \tilde{\Psi}(p,t) = e^{-i(p^2/2m)t/\hbar}\tilde{\psi}(p)\,, $$ (where $\tilde{\Psi}(p,0) = \tilde{\psi}(p)$).

It's just not true that if you increase the uncertainty in one variable (by changing the wave function in some way), then the uncertainty in the other decreases. Equivalently, it's just not true that if you change a function to another having a larger width, then the Fourier transforms of the function has a smaller width.

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  • $\begingroup$ Thanks. Can you please clarify why a more spread-out function doesn't necessarily mean it's Fourier Transform is more concentrated? $\endgroup$
    – snxmx
    Commented Oct 24 at 22:38
  • $\begingroup$ @snxmx I'll see if I can come up with an intuitive reason for this, but it has to do with interpreting the Fourier transform as a sum over plane waves, and thinking about what happens when the coefficients on those plane waves are allowed to be complex, so that you can get interference between different momenta. In this case, it's the $e^{-i(p^2/2m)t/\hbar}$ phase factor introduces interference between the different $p$-components of the wave function that forces the position space wave function to spread out while the momentum space wave function doesn't. I'll think on this. $\endgroup$
    – march
    Commented Oct 24 at 23:55
  • $\begingroup$ @snxmx because its frequency content hasn't changed. E.g in diffraction, in the transverse direction $\Delta x$ is set at the aperture, but expands as you propagate down-range, while the spectral content of the light is the same. $\endgroup$
    – JEB
    Commented Oct 25 at 0:19

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