Group velocity of Quantum state

[Relativeee]

Homework Statement

Ok hello, The problem is there is a system in the state characterized by

[tex]\Psi[/tex](x,t) = [tex]\int[/tex] dk e^(ikx)e^(-iwt)f(k), which is the time dependent representation of a wave packet, where individual waves travel with velocity
vp = w/k and f(k) = (2*[tex]\alpha[/tex]/[tex]\pi[/tex])^(1/4)e^(-[tex]\alpha[/tex]*k^2). The problem is to find the group velocity(associated with the state) along with the probability for measuring the particle at time t=t0>0 at position x=x0>0 and also probability per unit length at same position. Also what exactly is the meaning of these results?

Homework Equations

The Attempt at a Solution

Ok so the first part, the idea in the book was if f(k) has a peak at k = k0 and w is a function of k then there can be an expansion of w around the point k = k0
w = w(k) = w0 + (dw/dk)(k-k0) = w0 + vg(k-k0), where vg = (dw/dk) at w = w0
Ok now to show that vg is group velocity, we can note that since dominant contributions of [tex]\Psi[/tex](x,t) come from region where k = k0 the above expansion can be kept at first order in (k-k0) and so from the integral we can write kx - wt = kx - [w0 +vg(k-k0)]t
and therefore the integral can be written as
[tex]\Psi[/tex](x,t) =e^[i(k0vg-w0)t] [tex]\int[/tex] dk e^[ik(x-vgt)]f(k)

[tex]\Psi[/tex](x,t) =e^[i(k0vg-w0)t][tex]\Psi[/tex](x-vgt,0) so we note that the packet travels as a whole velocity vg that maintains the same probability density and therefore we can call vg the group velocity.
So that was i was thinking they were asking for the first part. For finding the probability at x = x0 > 0 and t = t0> 0 is it enough to say that the probability is
|[tex]\Psi[/tex](x0-vgt0,0)|^2? I am not quite sure on the probability per unit length or The exact meaning of these results. If anyone could give some insight that would be great thanks alot

 

[someguy61]

Anyone has a solution?