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Wave Packet problem

  1. Sep 9, 2004 #1
    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})


    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


    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?
    Last edited: Sep 9, 2004
  2. jcsd
  3. Sep 9, 2004 #2
    (a) First, Taylor expand [tex]\omega(k)[/tex] as [tex]\omega(k) = \omega(k_0) + \vec{\nabla}_k\omega(k)|_{k_0} \cdot (\vec{k}-\vec{k_0}) + \cdots[/tex]

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

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

    [tex]=\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[/tex]

    \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[/tex]

    \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[/tex]


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

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

    (c)For light in vacuum, [tex]v_g=c[/tex] so [tex]\omega = k c[/tex]
    From the invariance of [tex]\mathcal{A}[/tex], 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) [tex]\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[/tex]

    (d) With this other dispersion relation, the velocities become unequal. [tex]v_p=\omega/k[/tex] is always valid, but [tex]v_g = \frac{d\omega}{d k}=b k[/tex].
    The new differential equation can be expected to be
    [tex]\frac{1}{b^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^4}{\partial r^4}=0[/tex] and the wavepacket will spread.
  4. Sep 9, 2004 #3
    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.
  5. Sep 9, 2004 #4
    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.
  6. Sep 9, 2004 #5
    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.
  7. Sep 9, 2004 #6
    Quite a problem to be on Homework #1 lol.
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