So I understand from the equations, and considering the Fourier components of a Gaussian, how the wavefunction for a free particle spreads due to dispersion. However mathematically, and from the reciprocity relation, this means the momentum is becoming more certain! Is there a physical process governing this? Or is there a way to understand why this occursvwithoit simply referring to the position wavefunction and its spreading?
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2 Answers
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As momentum is conserved for a free particle, it is not obvious to me that the momentum becomes more certain for a Gaussian wave packet. Neither is it obvious to me that an initially Gaussian wave packet will remain Gaussian.
EDIT (02/11/2018): According to http://demonstrations.wolfram.com/EvolutionOfAGaussianWavePacket/ , an initially Gaussian wave packet can remain Gaussian (at least in some cases), but the momentum uncertainty does not depend on time.
answered Feb 11, 2018 at 20:13
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$\begingroup$ Yes- turns out I made a mistake and the gaussian in momentum space remains the same width. Although I'm curious as to the implications of this.... $\endgroup$– MeepCommented Feb 11, 2018 at 22:33
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The way I understand that is to look at what kind of information the wavefunction is giving to you, it's not a very formal way of thinking, though.
If you measure the position of the particle to be at $x=x_0$ at the instant $t=0$, then, you can measure it quickly again and it will have to be close to $x_0$, but you don't know very well where it is going. That is a distribution narrow in position and broad in momentum.
If you wait more, on the other hand, you have less certainty of where the particle will be, but the second measurement will give you a better precision of the direction and speed of it's movement. So as time passes, you lose information of the position and gain information on momentum. Of course, the second measurement will narrow the distribution in position again, but I think this picture still helps to see what the wavefunction is telling us.