I was wondering if anyone can give me some assistance on a homework problem. Here it is,(adsbygoogle = window.adsbygoogle || []).push({});

Consider a wave packet defined by

[itex]

\begin{equation}

\vec{A}(\vec{r},t)=\int \hat{\mathcal{A}}(\vec{k}-\vec{k_0})

\frac{e^{i(\vec{k}\cdot\vec{r}-\omega(k)t)}}{(2\pi)^{3/2}}d\vec{k}

\end{equation}

[/itex]

where

[itex]

\hat{\mathcal{A}}(\vec{k}-\vec{k_0})

[/itex]

is a function that is peaked at [itex]\vec{k}=\vec{k_0}[/itex].

(a) Show that this packet can be written in the form

[itex]

\begin{equation}

\vec{A}(\vec{r},t)=e^{i(\vec{k_0}\cdot\vec{r}-\omega(k_0)t)}\mathcal{A}(\vec{r}-v_gt)+\cdots

\end{equation}

[/itex],

where [itex]\vec{v}_g=\vec{v}_{\mathrm{group}}=\vec{\nabla}_k\omega

(k)|_{k_0}[/itex] is the group velocity and [itex]\mathcal{A}(\vec{r}-\vec{v}_g t)[/itex] is a function that is peaked at [itex]\vec{r}=\vec{v}_gt[/itex] Hint: expand [itex]\omega(k)[/itex] around [itex]\vec{k}_0[/itex]

(b) Show that for a wave packet not to "spread", i.e., not change its shape from that given by [itex]\mathcal{A}(\vec{r})[/itex], it is required that [itex]\vec{v}_{\mathrm{group}}=\vec{v}_{\mathrm{phase}}[/itex]. Here [itex]\vec{v}_{\mathrm{phase}}[/itex] is the phase velocity [itex]\vec{v}_{\mathrm{phase}}\equiv\omega/k[/itex].

(c) As a consequence of the condition [itex]\vec{v}_\mathrm{phase}=\vec{v}_\mathrm{group}[/itex] show that [itex]\omega=kc[/itex] which holds for light in a vacuum. Then deduce that the wave equation [itex]\square\vec{A}=0[/itex] follows.

(d) Suppose we had [itex]\omega(k)=bk^2[/itex], where [itex]b[/itex] is some constant. Would the phase and group velocities be the same? What differential equation would you deduce? Would the wave packet maintain its shape?

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# Wave Packet problem

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