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There are many questions on this site relating to the spreading of Gaussian wave packets and the time dependence of the uncertainty in position ($\Delta x$). For example, see Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? or Does Heisenberg's uncertainty under time evolution always grow? (and the links questions therein). However, I am struggling to follow the arguments presented in the thread above.

Specifically, if I measure a free particle described by a Gaussian wave packet with uncertainty $\Delta x$ at some time, does the positional uncertainty increase or decrease (or neither)? I am not sure I appreciate the physical significance of time-reversed solutions $\Psi^*(x,-t)$ - surely we either observe increasing positional uncertainty or not? If not, why do people even consider the dispersion of Gaussian wave packets?

Edit 1: From the comments and discussion in the answer here and Propagating a Gaussian wavepacket backwards in time I can accept that wave packets do not always spread and states can be prepared that narrow or spread with time. My refined confusion therefore is, what implicit initial conditions do quantum textbooks/lectures use when they describe Gaussian wave packet spreading in time?

For example, see the line from this lecture:

According to Eq. (112), the width of our wave packet grows as time progresses. Indeed, it follows from Eqs. (79) and (105) that the characteristic time for a wave packet of original width ${\mit\Delta} x$ to double in spatial extent is ...

The questions linked do not this issue (or at least not explicitly enough for me to work out).

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  • $\begingroup$ It depends on the potential. You cannot say how a gaussian wavepacket will behave unless you also specify the Hamiltonian that determines the time evolution of the wavepacket. In a purely harmonic potential, a gaussian wavepacket will keep its shape, in any other potential you will see a complex time dependency and the wavepacket may as well quickly take on a shape that is no longer gaussian at all. Or is your question only about a free particle with a constant zero potential ? $\endgroup$
    – Hans Wurst
    Mar 2 at 14:01
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    $\begingroup$ @HansWurst: the second paragraph mentions the particle being free. $\endgroup$
    – Javier
    Mar 2 at 14:05
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    $\begingroup$ Does this answer your question? Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? I know that you have already cited this question as a similar question that doesn't answer your question but I don't see how that's the case -- thus, voting to mark as a duplicate. $\endgroup$
    – ACat
    Mar 3 at 20:48
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    $\begingroup$ The answer is actually very simple. Textbooks assume the wavefunction is real at $t = 0$. That implies that it will immediately start spreading for $t > 0$. (A real wavefunction is also time-reversal invariant, so spreading also happens for $t < 0$.) $\endgroup$
    – knzhou
    Mar 3 at 21:44
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    $\begingroup$ If you want a wavefunction that starts shrinking at $t = 0$, you need to have the complex phase spiraling forward in the back, and backwards in the front (because momentum corresponds to phase rotation rate). You can achieve that with a formal Gaussian with complex variance. There's nothing wrong with such states, they're just very slightly more complicated than real-valued Gaussians, so textbooks sometimes don't mention them. $\endgroup$
    – knzhou
    Mar 3 at 21:44

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The width of a Gaussian wavepacket for a free particle evolves as (see Eq 112 of [1]) \begin{equation} \sigma^2(t) = (\Delta x)^2 + \frac{\alpha^2 t^2}{4 (\Delta x)^2} \end{equation} where $\alpha = d^2 \omega(k_0) / dk^2$ and $\Delta x$ is the initial width of the wavepacket.

This function first decreases with time, then hits a minimum at $t=0$ (when $\sigma^2 = (\Delta x)^2$), and then increases. The width always increases with time as $t \rightarrow \infty$, which is why people say the width of a free particle's wave packet tends to increase.

The width evolves in a way that is time reversal invariant, since it is the same if we send $t\rightarrow - t$. As an aside, time reversal invariance of the Schrodinger equation doesn't actually imply every solution is invariant -- it just implies that if $\psi(x, t)$ is a solution, then so is $\psi^\star(x, -t)$.

To conclude, the fact that $\sigma^2(t)$ is invariant under $t\rightarrow -t$ is completely consistent with time-reversal invariance, even though the wavepacket expands for $t>0$.

[1] https://farside.ph.utexas.edu/teaching/qmech/lectures/node26.html

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  • $\begingroup$ Given a time-invariant Hamiltonian, the implication is that $\psi^\ast(x,-t)$ will also be a solution if $\psi(x,t)$ is a solution -- not that $\psi(x,-t)$ will also be a solution. $\endgroup$
    – ACat
    Mar 3 at 21:46
  • $\begingroup$ @DvijD.C. D'oh! Fixed. Thank you! $\endgroup$
    – Andrew
    Mar 3 at 21:46
  • $\begingroup$ Why is t=0 special? $\endgroup$ Mar 4 at 3:28
  • $\begingroup$ @flippiefanus It's the time when the wavepacket has its minimum width. Of course you can shift this time, but the minimum was chosen to be $t=0$ in the lecture notes I linked to for simplicity. $\endgroup$
    – Andrew
    Mar 4 at 4:23
  • $\begingroup$ Is the minimum width the minimum uncertainty state (at $t=0$)? So could I conclude that a Gaussian wave packet will always spread from its minimum uncertainty state? But one could be prepared such that it wasn't a minimum uncertainty state (not sure how?!) and in that case it would shrink first? $\endgroup$
    – user246795
    Mar 4 at 9:48

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