# Wave Packet problem

1. Sep 9, 2004

### tornpie

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

Consider a wave packet defined by

$\vec{A}(\vec{r},t)=\int \hat{\mathcal{A}}(\vec{k}-\vec{k_0}) \frac{e^{i(\vec{k}\cdot\vec{r}-\omega(k)t)}}{(2\pi)^{3/2}}d\vec{k}$

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

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

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?

Last edited: Sep 9, 2004
2. Sep 9, 2004

### 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

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

3. Sep 9, 2004

### 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 hints[/i] not answers... I am sorry, I shall better go sleeping before making another mystake.

4. Sep 9, 2004

### 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.

5. Sep 9, 2004

### 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.

6. Sep 9, 2004

### tornpie

Quite a problem to be on Homework #1 lol.