Wave Packet problem

[tornpie]

I was wondering if anyone can give me some assistance on a homework problem. Here it is,

Consider a wave packet defined by


\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}


where


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

is a function that is peaked at \vec{k}=\vec{k_0}.

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


\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}
,

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

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

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

(d) Suppose we had \omega(k)=bk^2, where b 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?

[humanino]

(a) First, Taylor expand \omega(k) as \omega(k) = \omega(k_0) + \vec{\nabla}_k\omega(k)|_{k_0} \cdot (\vec{k}-\vec{k_0}) + \cdots

Then insert this Taylor expansion, shift dummy variable \vec{k}=\vec{k'}+\vec{k_0} and use the all-important expression of the Dirac peak as its Fourier transform :

\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}

=\int \hat{\mathcal{A}}(\vec{k'})
\frac{e^{i\left[(\vec{k'}+\vec{k_0})\cdot\vec{r}-[\omega(k_0) + \vec{k'} \cdot \vec{\nabla}_k\omega(k)|_{k_0} ]t\right]}}{(2\pi)^{3/2}}d\vec{k'}+\cdots

=e^{i(\vec{k_0}\cdot\vec{r}-\omega(k_0)t)}
\int \hat{\mathcal{A}}(\vec{k'})
\frac{e^{i\left[\vec{k'}\cdot (\vec{r}- \vec{\nabla}_k\omega(k)|_{k_0}t)\right]}}{(2\pi)^{3/2}}d\vec{k'}+\cdots

=e^{i(\vec{k_0}\cdot\vec{r}-\omega(k_0)t)}
\int \hat{\mathcal{A}}(\vec{k'})
\frac{e^{i\left[\vec{k'}\cdot (\vec{r}-\vec{v}_gt)\right]}}{(2\pi)^{3/2}}d\vec{k'}+\cdots

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

(b) I need to lower the level of rigor in order to save my time. Damn latex :wink:


\vec{A}(\vec{r},t)=
e^{i(k r-\omega t)}\mathcal{A}(r-v_gt)+\cdots
The wave packet will not spread if
\vec{A}(r,t)=\vec{A}(r+\delta r,t + \delta t) where \delta r = v_g \delta t. It follows that the argument of \mathcal{A} is automatically unchanged. So you only need to ensure the invariance of the exponanetial's argument :
k r-\omega t = k (r+\delta r)-\omega (t + \delta t) from which k\delta r -\omega\delta t =0 and hence \frac{\delta r}{\delta t}=\frac{\omega}{k}=v_g

(c)For light in vacuum, v_g=c so \omega = k c
From the invariance of \mathcal{A}, you only need to work with the exponential (again) when you deal with differential equations.
You can readily see that the operator (adjust in case you use a different metric signature) \square = \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial r^2}=\frac{\omega^2}{c^2}-k^2=0

(d) With this other dispersion relation, the velocities become unequal. v_p=\omega/k is always valid, but v_g = \frac{d\omega}{d k}=b k.
The new differential equation can be expected to be
\frac{1}{b^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^4}{\partial r^4}=0 and the wavepacket will spread.

[humanino]

This is not a great do for me today. I abandonned my fight against the craniale-size/intelligence lobby, and I forgot the basic rule in homework help : provide only [b]hints[/i] not answers... I am sorry, I shall better go sleeping before making another mystake.

[tornpie]

Thanks a million. Don't worry about ruining it for me. I will learn each step. I need to learn this packet stuff in a hurry for the future homeworks and tests.

I gave it a pretty fair shot, and I was close to getting it.

[humanino]

You're welcome. It took me a little while, but it was worth for me too. Except that, i am not absolutely certain for the last question, especially the differential equation.

[tornpie]

Quite a problem to be on Homework #1 lol.