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Free Particle

  1. Jun 27, 2008 #1
    Problem
    A free particle of mass [tex]m[/tex] moving in one dimension is known to be in the initial state

    [tex]\psi(x, 0) = \sin(k_0 x)[/tex]

    a) What is [tex]\psi(x, t)[/tex]?
    b) What value of [tex]p[/tex] will measurement yield at the time [tex]t[/tex], and with what probabilities will these values occur?
    c) Suppose that [tex]p[/tex] is measured at [tex]t=3 s[/tex] and the value [tex]\hbar k_0[/tex] is found. What is [tex]\psi(x, t)[/tex] at [tex]t > 3 s[/tex]?

    Attempt at Solutions
    Well one question I have is this: how is this a valid state function for a free particle if it is non-square integrable? Generally, for any free particle, doesn't the wavefunction have to be square-integrable?
     
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  3. Jun 27, 2008 #2

    Dick

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    Exactly the opposite. In general, free particles having wavefunctions like exp(ikx) are NOT square integrable. Since they aren't localized.
     
  4. Jun 28, 2008 #3
    Okay, so part a) would then just be

    [tex]\psi(x,t) = e^{-i \omega_0 t} \sin k_0 x[/tex]​

    since [tex]\psi(x,0)[/tex] is an eigenfunction of the momentum operator. Now, when I try to evaluate part b), I get

    [tex]b(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \sin k_0 x e^{-ikx} dx[/tex]

    where [tex]\int_{a}^b |b(k)|^2 dk[/tex] represents the probability of a particle having a wavenumber between [tex]a[/tex] and [tex]b[/tex]. How do I actually evaluate the above integral? Also, the question asks for the probability of certain momenta... how is this allowed if momenta aren't discrete but are, instead, continuous?
     
  5. Jun 28, 2008 #4

    Dick

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    You are making this much harder than it actually is. sin(k0*x) is a linear combination of exactly two momentum eigenfunctions exp(ikx). Which two? Don't think 'integral', think 'deMoivre'.
     
  6. Jun 28, 2008 #5
    Aah, I see it now... sin(k0*x) = 1/2 exp(i*k0*x) - 1/2 exp(-i*k0*x). But we need to normalize this... so we finally have

    [tex]\psi(x, t) = e^{-i \omega_0 t} \cdot \left(\frac{\sqrt{2}}{2} \phi_{k_0} - \frac{\sqrt{2}}{2} \phi_{-k_0}\right)[/tex]​

    Thus, the particle has a 50% chance of having momenta [tex]\pm \hbar k_0[/tex]. This solves part b). As for part c), the wavefunction collapses onto

    [tex]\psi(x, 0) = e^{-i \omega_0 t} \cdot \frac{1}{\sqrt{2\pi}} e^{i k_0 x}[/tex]​
     
  7. Jun 28, 2008 #6

    Dick

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    Actually, it's sin(kx)=(exp(ikx)-exp(-ikx))/(2i). And you don't need to do any particular normalization. The only thing thats important is that the magnitude of the two coefficients are equal.
     
  8. Jun 28, 2008 #7
    Heh, yes that's what I meant :P Thanks a bunch :smile:
     
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