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Gaussian Distribution

  1. Jun 23, 2008 #1
    Problem
    Let us define a wave function [tex]\phi = A \exp \(\frac{-(x-x_0)^2}{4a^2}\) \exp \(\frac{i p_0 x}{\hbar}\) \exp(-i \omega_0 t)[/tex]. Show that [tex](\Delta x)^2 = a^2[/tex]. Also, calculate the uncertainty [tex]\delta p[/tex] for a particle in the given state.

    Attempt at a solution
    I honestly have no idea as to how to proceed... could someone give me a hint without giving away the answer?
     
  2. jcsd
  3. Jun 23, 2008 #2

    nicksauce

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    Well what exactly is giving you problems?
    Do you know how to normalize the wavefunction?
    Do you know in theory how to calculate the expectation values, and uncertainties?
    Or is all that fine and the mathematics is giving you problems?
     
  4. Jun 24, 2008 #3
    Okay, well I know how to normalize the function... we just have

    [tex]\int_{-\infty}^{\infty} A^2 e^{\frac{-(x-x_0)^2}{2a^2}} = 1[/tex].

    So if we let [tex]X = (x-x_0)/a[/tex], then we get

    [tex]A^2 \cdot a \cdot \int_{-\infty}^{\infty} e^{-X^2/2} dX = 1[/tex],

    and this just yields

    [tex]A^2 \cdot a \cdot \sqrt{2\pi} = 1 \iff A = \frac{1}{a^{1/2} (2\pi)^{1/4}}[/tex].

    So the wavefunction is just

    [tex]\phi = \frac{1}{a^{1/2} (2\pi)^{1/4}} \cdot \exp \(\frac{-(x-x_0)^2}{4a^2}\) \exp \(\frac{i p_0 x}{\hbar}\) \exp(-i \omega_0 t)[/tex]

    From here, I tried to set [tex](\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2[/tex], but nothing happened from there...
     
  5. Jun 24, 2008 #4

    nicksauce

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    What did you find for <x> and <x^2>? It's not clear what you mean by nothing happened from there.
     
  6. Jun 24, 2008 #5
    Well, I tried to calculate <x^2>, but it ended up that I got an integral of the form

    [tex]\int_{-\infty}^{\infty} x^2 e^{x^2} dx[/tex]

    However, <x> was pretty easy to find... it's just [tex]x_0[/tex]. I'm just wondering... maybe I set up the integral the wrong way... to find <x^2>, is it just

    [tex]\int_{-\infty}^{\infty} \phi^* \hat{x}^2 \phi dx = \int_{-\infty}^{\infty} \phi^* x^2 \phi dx[/tex]?
     
  7. Jun 24, 2008 #6

    nicksauce

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    Yes that's what it should be. To do an integral of the form
    [tex]\int{x^2e^{-x^2}}dx[/tex], here is a trick.
    Let [tex]I(a) = \int_{-\infty}^{\infty}e^{-ax^2}dx[/tex]. Then [tex]I'(a)=\int_{-\infty}^{\infty}-x^2e^{-ax^2}dx [/tex]. You should already know I(a), so there shouldn't be a problem.
     
  8. Jun 24, 2008 #7
    Wow that's clever! Thanks very much =)
     
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