Wave packet with increasing time

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SUMMARY

This discussion focuses on the behavior of a Gaussian wave packet over time, specifically how its wave function evolves from an initial state. The wave function is defined as ψ(x,0) = (√β/√√π)exp(-β²(x-x0)²/2)exp(ip0x/ħ), where β is a constant that indicates the wave packet is not in the ground state of a harmonic oscillator. The transformation from the initial wave function ψ(x,0)² to the time-dependent wave function ψ(x,t)² involves the replacement of Δx0 with Δx(t), which is derived from the relationship Δx0 = 1/(β√2). This indicates that the width of the wave packet changes over time.

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mcheung4
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I am trying to understand how a gaussian packet varies in time.

Suppose we have a Gaussian wave packet that is displaced form the origin by an amount x0 and given initial momentum p0. So the wave function in coordinate space is

ψ(x,0) =\frac{√β}{√√\pi}exp(-β2(x-x0)2/2)*exp(ip0x/ħ)

where β is some constant such that ψ does not correspond to the ground state of the harmonic oscillator (not an eigenstate).

Now given Δx0 = 1/(β√2), we can eliminate β and find ψ2

ψ(x,0)2 = 1/√2\pi * 1/Δx0 * exp(-(x-x0)2/2(Δx0)2)

Then

ψ(x,t)2 = 1/√2\pi * 1/Δx(t) * exp(-(x-x0)2/2(Δx(t))2)

Q1: why does Δx0 = 1/(β√2)? Isn't Δx0 a definite value so it should not have any deviation?
Q2: How does ψ(x,0)2 transform to ψ(x,t)2 and then we have Δx(t) replacing Δx0?
 
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