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 Mar 31 comment Why doesn't a marble rolling on a table ever reflect back at the edge? @BenCRowell He did, but unfortunately I zoned out when he was saying it, so I missed most of it and couldn't remember the rest. That's why I asked the question here, pretty much, as I was interested in the answer, but didn't really have the chance to attend office hours. Mar 30 comment Why doesn't a marble rolling on a table ever reflect back at the edge? @BenCrowell Well, in all honesty, I cannot take the credit for making the problem picturesque. That is due to my QM professor :) Mar 30 accepted Why doesn't a marble rolling on a table ever reflect back at the edge? Mar 13 comment Why doesn't a marble rolling on a table ever reflect back at the edge? @NathanReed Why would it not be a potential step? It's a negative potential step, but the coefficient for the transmission is the same as if the particle encounters a positive one. Mar 13 asked Why doesn't a marble rolling on a table ever reflect back at the edge? Feb 22 awarded Scholar Feb 22 accepted Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? Feb 21 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? I've also done this calculation now, and it seems to work out exactly as you've shown. And, @MarkEichenlaub, that is a nice example to show the RHS term disappears, although I wonder whether it's obvious the LHS term doesn't get killed or is negative in that specific case. I've seen elsewhere that $\langle xp \rangle=\langle \frac{xp+px}{2} \rangle$, and intuitively if you have two packets as suggested, then the classical multiplication xp would be positive for both Gaussians either before or after they hit the origin, satisfying the inequality. Would that be sound logic? Feb 20 awarded Commentator Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? Thanks, I'll go over this one more time later on and do the calculation @EmilioPisanty suggested, and I'll just return here if I still have questions or find something that doesn't jive with what has been said. Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? @MarkEichenlaub I thought of that Gaussian example before, but would those two Gaussians be describing a free particle wave packet? That's what I was unsure of, and I also wasn't sure whether we can just handwave saying the uncertainty reduces when they come closer together. After all, who's to say the spread when the centers are aligned isn't even greater than what the separation was at the outset? Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? @EmilioPisanty, so it is actually between $-t_{2}$ and $-t_{1}$. I namely asked that in the first comment above, but Mark replied that it would be between $t_{1}$ and $t_{2}$. That got me a bit confused, so can we reach a consensus on this by any chance? Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? I guess what I wanted to know is whether you can show that without assuming you can make the transformation you suggested. Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? I'll try to wrap my head around this time reversal thing, but in the mean time, is there a way to show what you stated without such recourse? This question was namely posed, but never answered, in our first QM course, and I assume there has to be an answer using only those tools. And I have to say that the approach @joshphysics took is also the one I thought of first. Any way that would work? Feb 20 awarded Editor Feb 20 revised Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? added 124 characters in body Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? @MarkEichenlaub As far as that Wikipedia article goes, it itself states one may find it only in certain contexts, so why would you necessarily apply it to the Schrodinger equation then? In particular, is there a way to show what you stated using the Schrodinger equation directly without appealing to a general theoretical prediction like this? If yes, I'd really appreciate more details on this. Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? OK, I see, and thanks for the additional feedback. There are two things I don't quite understand yet, though. First, in your answer you state that $\Psi (x,-t)$ solves the Schrodinger equation, but I don't see how this is obvious. How does one show this? And secondly, I don't see how the Gaussian ceases to be a Gaussian. I mean, you basically just plug in a certain value for t for later times, and you're still left with an expression of the same form, but now different $x_{0}$ or $p_{0}$ or whatever the peak is characterized by is, aren't you? Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? Hmm, I see, but, for example, it can be proven the uncertainty in the position of a Gaussian wave packet is always increasing. So if you time reverse that wave packet and define it the way you did in terms of $\Psi (x,-t)$, then isn't that time-reversed wave packet still Gaussian, thereby leading to a contradiction? Feb 20 comment Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? Can you elaborate on that? Are you saying that if the uncertainty is increasing on, say, $t \in (t_{1}, t_{2})$, then there is a time period where the uncertainty in position is decreasing, namely $t \in (-t_{2}, -t_{1})$?