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Plane wave limit of Gaussian packet

  1. Sep 7, 2004 #1
    [tex]\psi (x)= \frac{1}{(\pi\Delta^2)^\frac{1}{4}} e^(\frac{i<p>(x-<x>)}{\hbar})e^(\frac{-(x-<x>)^2}{2\Delta^2})[/tex] as [tex]\Delta\rightarrow\infty[/tex] should approach the plane wave [tex]\frac{1}{(2\pi\hbar)^\frac{1}{2}} e^(\frac{i<p>x}{\hbar}) [/tex] up to a phase factor. I guess this happens by setting the [tex]e^(\frac{-(x-<x>)^2}{2\Delta^2})[/tex] term equal to 1. However, ignoring phase factor differences, the normalization factor out front still seems to be different. One is [tex]\frac{1}{(\pi\Delta^2)^\frac{1}{4}} [/tex] while the other is [tex]\frac{1}{(2\pi\hbar)^\frac{1}{2}} [/tex]. Am I not seeing something?
  2. jcsd
  3. Sep 8, 2004 #2
    From where did you get that this ψ is a plane wave in the limit ∆ → ∞ ? This is clearly not the case. In that limit ψ(x) → 0 for all x. Moreover,

    ∫ |ψ(x)|2 dx = 1 , for all values of ∆ ,

    and thus, in particular, this is so also for the limit ∆ → ∞. A plane wave must have infinite norm, not norm equal to 1.

    Furthermore, why are you looking in position space? It is most unintuitive to do so.

    So ... let's look in momentum space. Write

    [1] |ψε> = ∫ fε(p - po) |p> dp .

    This fε(p) is chosen in such a way that

    [2] limε → 0+ fε(p - po) = δ(p - po) .

    This means that limε → 0+ε> = |po>, a plane wave with momentum po.

    In terms of a Gaussian, a suitable choice for fε(p) is

    [3] fε(p - po) = (1/ε√π) exp{-(p - po)22} .

    Now, let's look at [1] in the position space of functions:

    [4] ψε(x) ≡ <x|ψε> = ∫ fε(p - po) <x|p> dp ,

    where, of course,

    [5] <x|p> = 1/√(2πhbar) exp{ipx/hbar} .

    The RHS of equation [4] is essentially a Fourier transform of fε(p - po). The precise result of [4], in light of [3] and [5], is

    [6] ψε(x) = 1/√(2πhbar) exp{ipox/hbar} ∙ exp{-x2ε2/4hbar2} .


    limε → 0+ ψε(x) = 1/√(2πhbar) exp{ipox/hbar} ,

    as required. Also,

    ∫ |ψε(x)|2 dx = const ∙ (1/ε) ,

    so that in the limit ε → 0+, this quantity goes to ∞ , as required.

    Equation [6] is a correct rendition of the ψ(x) you originally proposed. You can now redefine ε as you wish.
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