Does increase in position uncertainty mean decrease in momentum uncertainty?

Question

Suppose a Gaussian wavepacket describes a free particle. With increasing time, the uncertainty in position increases, and the particle moves in the $x$-direction.

Does the increase in position uncertainty mean a decrease in momentum uncertainty?

It does not decrease since the uncertainty principle is satisfied without such a decrease. But I would like to see some actual "proof" or direction from which I could derive it.

Answer 1

The following is the step you can use to see it:

• Find out the propagator for the free particle i.e. $U(t)$.
• Find out it's matrix element in position basis i.e. $U(x,t;x')\equiv \langle x|U|x'\rangle$.
• Suppose a gaussian wave packet $\psi(x',0)$.
• See it's evolution using propagator i.e. $$\psi(x,t)=\int U(x,t;x') \psi(x',0)dx'$$
• Find out the propabability density. $$P(x,t)=|\psi(x,t)|^2$$
• find out $\Delta X(t)$ and see it's evolution.

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This is what you will find for gaussian wave function. The uncertainity in position will grow as time will pass. This is a reflection of the fact that any uncertainity in the initial velocity (that is to say, the momentum) will be reflected with passing time as a growing uncertainity in position.