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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).

user246795
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  • 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 ? – Hans Wurst Mar 02 '22 at 14:01
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    @HansWurst: the second paragraph mentions the particle being free. – Javier Mar 02 '22 at 14:05
  • Sometimes it helps to systematically follow a good set of lectures like https://farside.ph.utexas.edu/teaching/qmech/lectures/node26.html (see eq 117) – Andrew Mar 02 '22 at 14:30
  • So you understand the spread of the stationary free Gaussian wavepacket is time-reversal symmetric, an even function of t? – Cosmas Zachos Mar 02 '22 at 14:51
  • @Andrew but why do the questions linked in the original question conclude that "nothing increases because of time-reversal symmetry" - specifically Mark's answer https://physics.stackexchange.com/a/54541/246795? – user246795 Mar 03 '22 at 19:39
  • @user246795 In Eq 112 of the notes I sent, you can see that for when $t < 0$, the wavepacket is decreasing in size, while for $t > 0$, is increasing. I think people tend to say the wavepacket increases because they have in mind starting at $t=0$; a time when the first time derivative of the wavefunction momentarily vanishes. But you are right, you can choose initial conditions, such that the Gaussian wavepacket first decreases in size to a minimum width, then increases. – Andrew Mar 03 '22 at 19:51
  • @Andrew what confuses me though is the what it means to have $t<0$ or $t>0$, what would happen physically? Surely, when a free particle is placed in space its uncertainty in position either increases or decreases? – user246795 Mar 03 '22 at 19:55
  • @user246795 The evolution will depend on the initial conditions of the wavefunction. If you had a classical particle and asked if it would go left or right, the answer is that you don't have enough information, you need to know the initial velocity. Something similar is happening here I think; it's not enough just to say "Gaussian wave packet". Essentially you can add a constant to $\phi(t)$ in the first few lines of the notes I linked to; that constant encodes the initial condition of the wavefunction. – Andrew Mar 03 '22 at 20:00
  • @Andrew in the sentence above Eq 117 in the notes you cite, it says that "the width of our wave packet grows as time progresses" but what initial conditions are they using then - this seems to go against the general claim that "nothing spreads due to time-reversal"? – user246795 Mar 03 '22 at 20:12
  • @user246795 I personally found the answer you linked to a little confusing. The function $(\tau^2 + t^2)$ (where $t$ is time and $\tau$ is a constant) is time reversal invariant, in the sense that it is invariant under $t\rightarrow - t$. If you plotted it and described the function in words, you would say it was a "U" shape that first decreases then increases. That function describes the width of a Gaussian wavepacket for a free particle. The initial condition corresponds to where on the "U" shape you start. – Andrew Mar 03 '22 at 20:38
  • However note that time reversal invariance does not mean that every solution is time-reversal invariant. It means that if $\psi(x, t)$ is a solution, then so is $\psi(x, -t)$. The Gaussian wavepacket happens to be an especially "nice" solution that is actually time reversal invariant (at least if the average momentum of the wavepacket is zero) – Andrew Mar 03 '22 at 20:39
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    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. –  Mar 03 '22 at 20:48
  • @Andrew I think the sole of my confusion is why lectures always derive Gaussian wave packets spreading with time (see Eq 117 https://farside.ph.utexas.edu/teaching/qmech/lectures/node26.html or the final statement here https://quantummechanics.ucsd.edu/ph130a/130_notes/node83.html) - they do not even consider time-reversal. I have also added an edit to the question. – user246795 Mar 03 '22 at 20:49
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    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 $t < 0$.) – knzhou Mar 03 '22 at 21:44
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    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. – knzhou Mar 03 '22 at 21:44
  • @knzhou How would one prepare such a state? It seems to me, most basic explanations always reason that if you localize a particle HUP gives uncertainty in the momentum - and that uncertainty always causes spreading? Or is this only true when the Gaussian is at a minimum uncertainty state for $t=0$? – user246795 Mar 04 '22 at 09:45

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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

Andrew
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  • 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. –  Mar 03 '22 at 21:46
  • @DvijD.C. D'oh! Fixed. Thank you! – Andrew Mar 03 '22 at 21:46
  • Why is t=0 special? – flippiefanus Mar 04 '22 at 03:28
  • @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. – Andrew Mar 04 '22 at 04:23
  • 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? – user246795 Mar 04 '22 at 09:48
  • @user246795 The next page of the lecture notes discusses that point in more detail, https://farside.ph.utexas.edu/teaching/qmech/lectures/node27.html. At $t=0$, the inequality in the uncertainty principle is saturated. For $t>0$, the uncertainty in position grows while the uncertainty in momentum stays the same, so the uncertainty principle bound is satisfied but no longer saturated. If you started evolving the system at some time $t<0$, the system would start with some uncertainty larger than the UP bound, shrink until the UP was satisfied, then grow. – Andrew Mar 04 '22 at 12:09