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) =[itex]\frac{√β}{√√\pi}[/itex]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[itex]\pi[/itex] * 1/Δx0 * exp(-(x-x0)2/2(Δx0)2)
Then
ψ(x,t)2 = 1/√2[itex]\pi[/itex] * 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?
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) =[itex]\frac{√β}{√√\pi}[/itex]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[itex]\pi[/itex] * 1/Δx0 * exp(-(x-x0)2/2(Δx0)2)
Then
ψ(x,t)2 = 1/√2[itex]\pi[/itex] * 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?