What is the Group Velocity of a Quantum State?

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SUMMARY

The discussion focuses on determining the group velocity of a quantum state represented by the wave packet \(\Psi(x,t) = \int dk e^{ikx} e^{-iwt} f(k)\), where \(f(k) = (2\alpha/\pi)^{1/4} e^{-\alpha k^2}\). The group velocity \(v_g\) is derived from the expansion of the angular frequency \(w\) around the peak \(k = k_0\), leading to the conclusion that \(v_g = \frac{dw}{dk}\) at \(w = w_0\). The probability of measuring the particle at position \(x_0\) and time \(t_0\) is expressed as \(|\Psi(x_0 - v_g t_0, 0)|^2\), emphasizing the significance of group velocity in maintaining probability density during wave packet propagation.

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  • Understanding of wave packets in quantum mechanics
  • Familiarity with the concept of group velocity
  • Knowledge of Fourier transforms and their applications in quantum states
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Homework Statement

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

\Psi(x,t) = \int 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*\alpha/\pi)^(1/4)e^(-\alpha*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 \Psi(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
\Psi(x,t) =e^[i(k0vg-w0)t] \int dk e^[ik(x-vgt)]f(k)

\Psi(x,t) =e^[i(k0vg-w0)t]\Psi(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
|\Psi(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 a lot
 
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