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Wave packet expansion

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    The text states:
    "Let us consider a wave packet whose Fourier inverse ##\phi (\vec{k})## is appreciably different from zero only in a limited range ##\Delta \vec{k}## near the mean wave vector ##\hbar \vec{\bar{k}} ##. In coordinate space, the wave packet ##\psi(\vec{r}, t)## must move approximately like a classical free particle with mean momentum ##\hbar \vec{\bar{k}}##. To see this behavior we expand ##\omega (\vec{k})## about ##\vec{\bar{k}}##:

    ## \omega (\vec{k}) = \omega(\vec{\bar{k}}) + (\vec{k} - \vec{\bar{k}}) \cdot (\vec{\nabla_k} \omega)_{k= \vec{\bar{k}}} + ...##

    ##=\bar{\omega} + (\vec{k} - \vec{\bar{k}})\cdot \vec{\nabla_{\bar{k}}} \bar{\omega} +... ## with obvious abbreviations"

    I'm afraid not much of the above is obvious to me ... have searched all over but no progress ... please help me understand -

    What expansion are they using and where do they get it from?
    What are the abbreviations mentioned?
     
  2. jcsd
  3. Feb 6, 2016 #2

    George Jones

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    A multivariable Taylor expansion has been used.

    ##\omega \left( \bar{k} \right)## has been abbreviated to ##\bar{\omega}##.
     
  4. Feb 7, 2016 #3
    Thanks George, a couple of things still puzzle me:
    1. what is the 2nd variable? Is it ##\vec{\bar{k}}## .... how can they expand about a variable (##\vec{\bar{k}}##) if it is?
    2. the text talks earlier about the width/spread of the wave packet as ##k_x - \bar{k}##, shouldn't that be ##|k_x - \bar{k}|## to keep both sides positive? (its a length)
     
  5. Feb 7, 2016 #4

    George Jones

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    No.

    If you post the Taylor expansion of a function of three variables in the form that appears in a calculus text, then we can discuss how the form in the original post is exactly the same.
     
  6. Feb 7, 2016 #5
    Thanks, I'm assuming we are talking about 2 variables, then the expansion is:

    ## f(x,y,z) = f(a,b,c) + (x-a)f_x(a,b,c) + (y-b)f_y(a,b,c) + (z-c)f_z(a,b,c) + \frac{1}{2!} ## [quadratic and higher terms - which are ignored bu the text]
     
    Last edited: Feb 7, 2016
  7. Feb 7, 2016 #6

    George Jones

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    No, three variables are needed. Don't worry about retyping the expansion.

    One more thing: Are you familiar with vector calculus and the gradient, divergence, and curl operations?
     
  8. Feb 7, 2016 #7
    Thanks George, Edited original, and am familiar with those operators
     
  9. Feb 7, 2016 #8

    George Jones

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    The three variables ##x##, ##y##, and ##z## can be thought of as comprising one vector, i.e., ##\vec{r} = \left( x,y,z \right)## and ##\vec{r_0} = \left( a,b,c \right)##. Then ##f\left( x,y,z \right)## and ##f\left( a,b,c \right)## become ##f \left(\vec{r}\right)## and ##f \left(\vec{r_0}\right)##.

    Also, ##\vec{r} - \vec{r_0} = \left( x - a,y-b,z-c \right)## and the gradient of ##f## evaluated (gradient first, then evaluation) at ##\vec{r_0} = \left( a,b,c \right)## is the vector

    $$\left(\nabla f\right)_{\vec{r_0}} = \left( f_x \left( a,b,c \right), f_y \left( a,b,c \right), f_z \left( a,b,c \right) \right).$$

    Consequently,

    $$\left( \vec{r} - \vec{r_0} \right) \cdot \left(\nabla f\right)_{\vec{r_0}} = \left(x - a\right) f_x \left( a,b,c \right) + \left(y - b\right) f_y \left( a,b,c \right) +\left(z - c\right) f_z \left( a,b,c \right), $$

    and, in new mod attire, the Taylor expansion for ##f## that you posted becomes

    $$f \left(\vec{r}\right) = f \left(\vec{r_0}\right) + \left( \vec{r} - \vec{r_0} \right) \cdot \left(\nabla f\right)_{\vec{r_0}} + ...$$

    If anything in the above is unclear, let me know.
     
  10. Feb 8, 2016 #9
    Very helpful, thanks George, the chapter is now clear to me - I had been thinking only in 1-D along the x-axis
     
  11. Feb 8, 2016 #10
    afterthought question please, is my notation correct when I wrote ## \vec{\bar{k}} ## ? Or should it be ## \bar{\vec{k}} ## or something else?
     
  12. Feb 8, 2016 #11
    Apologies, spoke too soon.
    The text has a common fourier transform equation ## \psi (\vec{r},t) = \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \omega t)} d^3k ##
    They substitute ## \omega(\vec{k})=\bar{\omega} + (\vec{k} - \vec{\bar{k}})\cdot \vec{\nabla_{\bar{k}}} \bar{\omega} +... ## into the above, and get a result I can't get - ## \psi (\vec{r},t) = exp(-i \bar{\omega}t + i \bar{\vec{k}} \cdot \nabla_{\bar{k}} \bar{\omega}t) \psi(\vec{r} - \nabla_{\bar{k}}t, 0) ##

    I instead get as far as ## \frac{e^{-i \bar{\omega}t}} {(2 \pi)^\frac{3}{2}} \int \phi(\vec{k}) e^{i (\vec{k} \cdot \vec{r} - \vec{k}
    \nabla_{\bar{k}} \bar{\omega}t + \bar{k} {\nabla}_{\bar{k}} \bar{\omega}t) } d^3k ##

    Could you help me bridge that gap please?
     
  13. Feb 9, 2016 #12
    I have reviewed this and made more progress, still can't quite see all of it.
    If I extract ## exp(-i \bar{\omega}t + i \bar{\vec{k}} \cdot \nabla_{\bar{k}} \bar{\omega}t) ## I am left with something close to the original FT, except it looks like

    ## \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \vec{k}\nabla_{\bar{k}} \bar{\omega}t)} d^3k ##

    The text says this latter ## = \psi(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t, 0) ## - but the ## \vec{k} ## wasn't dotted with the ## \nabla ## term?

    Also they set t=0, but then leave the t in the ## \nabla ## term?

    Please try and explain this to me? Thanks
     
  14. Feb 12, 2016 #13

    blue_leaf77

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    So, you have
    $$
    \exp(-i \bar{\omega}t + i \bar{\vec{k}} \cdot \nabla_{\bar{k}} \bar{\omega}t)
    \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \vec{k}\cdot \nabla_{\bar{k}} \bar{\omega}t)} d^3k
    $$
    Note that ##\vec{k}## is dotted with ##\nabla_{\bar{k}} \bar{\omega}## in the integrand. The integral can further be written as
    $$
    \int \phi (\vec{k}) e^{i\vec{k}\cdot(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t)} d^3k
    $$
    At the same time you have the Fourier transform equation
    $$
    \psi (\vec{r},t) = \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \omega t)} d^3k
    $$
    If, in the above equation, you replace ##\vec{r}## with ##\vec{r} - \nabla_{\bar{k}} \bar{\omega}t## and ##t## with ##0##, you should see that
    $$
    \int \phi (\vec{k}) e^{i\vec{k}\cdot(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t)} d^3k = \psi (\vec{r} - \nabla_{\bar{k}} \bar{\omega}t,0).
    $$
     
  15. Feb 12, 2016 #14
    Thanks blue_leaf.

    1. Why can we set the t in (r, t) to 0, but leave the t in ##\Delta_{\bar{k}} \bar{\omega}t ## ?

    2. is my notation correct when I wrote ## \vec{\bar{k}} ## ? Or should it be ## \bar{\vec{k}} ##, or something else?
     
  16. Feb 12, 2016 #15

    blue_leaf77

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    You are simply comparing
    $$
    \int \phi (\vec{k}) e^{i\vec{k}\cdot(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t)} e^{-i\omega (0)}d^3k
    $$
    with
    $$
    \int \phi (\vec{k}) e^{i\vec{k} \cdot \vec{r}} e^{-i\omega t} d^3k
    $$
    I don't know if that really matters, but I personally would use the second one.
     
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