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This is a Homework problem so please feel free to not answer and just give pointers.

A localized wavepacket is given as:

$$\Phi(r,t=0 ) = \frac { e^{-\large\frac {r^2}{2s^2}} e^{\large\frac{i\pi x}{4a}} } {(2\pi^{\frac{3}{4}}s^\frac{3}{2})}$$

Find the group velocity (given by: $v_g = \frac{d\omega}{dk}$ (I can't see how this will help))

Also, given a time-dependent homogeneous electric field in z-direction $E_z(t) = E_0sin(\omega t)$, give the center of mass movement $r_0(t)$ of the wavepacket for $t>0$, neglecting all the scattering events.

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  • $\begingroup$ what is $s$s supposed to denote? $\endgroup$
    – Danu
    Mar 10, 2014 at 17:26
  • $\begingroup$ s is a positive real number, the square of the width of the wavepacket. $\endgroup$
    – Hasan
    Mar 10, 2014 at 17:30

1 Answer 1

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If you don't want the answer you could look in the wiki Group Velocity. In particular:

One derivation of the formula for group velocity is as follows.

Consider a wave packet as a function of position $x$ and time $t$: $α(x,t)$. Let $A(k)$ be its Fourier transform at time $t=0$:

$\alpha(x,0)= \int_{-\infty}^\infty dk \, A(k) e^{ikx}$, By the superposition principle, the wavepacket at any time t is:

$\alpha(x,t)= \int_{-\infty}^\infty dk \, A(k) e^{i(kx-\omega t)}$, where $\omega$ is implicitly a function of $k$. We assume that the wave packet $\alpha$ is almost monochromatic, so that $A(k)$ is nonzero only in the vicinity of a central wavenumber $k_0$. Then, linearization gives:

$\omega(k) \approx \omega_0 + (k-k_0)\omega'_0$ where $\omega_0=\omega(k_0)$ and $\omega'_0=\frac{\partial \omega(k)}{\partial k} |_{k=k_0}$. Then, after some algebra,

$\alpha(x,t)= e^{it(\omega'_0 k_0-\omega_0)}\int_{-\infty}^\infty dk \, A(k) e^{ik(x-\omega'_0 t)}$. The factor in front of the integral has absolute value 1. Therefore,

$|\alpha(x,t)| = |\alpha(x-\omega'_0 t, 0)|$ i.e. the envelope of the wavepacket travels at velocity $\omega'_0=(d\omega/dk)_{k=k_0}$. This explains the group velocity formula.

The trick is to know what to do with the fourier transform I think.

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  • $\begingroup$ I'm still lost. $\endgroup$
    – Hasan
    Mar 10, 2014 at 22:26

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