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So, I'm following the MIT OCW lectures on 8.04 quantum mechanics by Prof. Allan Adams. I have the expression for the probability distribution of a gaussian wavepacket for a free particle situation. No initial momentum is imparted. This is a non-relativistic treatment.

$$\mathbb{P}(x,t) = \frac{1}{\sqrt{\Pi}} \frac{1}{\sqrt{a^{2}+(\frac{\hbar}{2ma})^2t^{2}}} e^{\frac{-x^{2}}{2(a^{2}+(\frac{\hbar}{2ma})^2t^{2})}}$$

Say at time $t=5$, I calculate the gaussian form. If I then ask what would the gaussian have looked like at time $t=-5$, the answer would be the same, because of the quadratic factors of $t$. Basically, as you decrease $t$ from $t=5$, the gaussian gets tighter till $t=0$, and then disperses again for negative times.

If someone at time $t=-5$ had indeed actually prepared a gaussian wavepacket of the form we found above at time $t=5$, and propagated the system forward in time, that person would have to update $a$ in the expression for $\mathbb{P}(x,t)$ and would find that the gaussian disperses instead of getting tighter.

There seems to be a contradiction here. How do I reconcile the two scenarios?

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    $\begingroup$ Take a look at the actual wave function for a Gaussian wave packet. It doesn't have a second power of time in there. You are propagating the wave function, not the probability distribution. $\endgroup$
    – CuriousOne
    May 13, 2016 at 7:42
  • $\begingroup$ I calculated this using the wavefunction and arrived at a consistent result. I think the logic above was wrong because $\mathbb{P}(x,t)$ assumes you start with a gaussian wavefunction of width $a$. But in my case, I had to start with a gaussian wavefunction of width $\sqrt{a^2-i\frac{5\hbar}{m}}$, which changes the expression for probability distribution to $\mathbb{P}(x,t-5)$ $\endgroup$
    – Matrix23
    May 13, 2016 at 12:58
  • $\begingroup$ I hear a lot about the dispersion of the gaussian. Doesn't the above analysis show that the probability density of the gaussian wavepacket particle "can" get narrower - before eventually starting to disperse? $\endgroup$
    – Matrix23
    May 13, 2016 at 14:08
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/54534/2451 and links therein. $\endgroup$
    – Qmechanic
    May 13, 2016 at 16:28

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Yes, the gaussian wavepacket can get narrower as the time passes indeed. It's a matter of phases. You know that a gaussian wavepacket is the superposition of plane waves, each one having a precise wavevector. So it really depends on how you "prepare" this superposition, i.e. on how you set the phase of each chromatic component. If at $t=0$ all the plane waves have the same phase, the wavepacket will be maximally narrow and will get wider and wider both in the past and in the future, simply because the various chromatic components will go more and more out-of-phase (recall that the angular velocity depends on $k^2$, so it's different for every chromatic component). On the other hand, if, at $t=0$, the various chromatic components have their relative phases arranged in a particular way, then, as the time passes, the phase difference will get smaller and smaller. When all of them will have the same phase you will have maximum localization, and, after that, broadening i.e. increasing delocalization.

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  • $\begingroup$ You seem to have a crucial point here. However, I can't understand it. If the math is simple, could you please provide an explanation as to how this occurs mathematically in the equations. Alternatively, if there's an online or text resource that treats this aspect, could you please let me know. $\endgroup$
    – Matrix23
    May 14, 2016 at 5:35

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