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Free Protons and Probability

  1. Jun 27, 2008 #1
    At t = 0, 10^5 noninteracting protons are known to be on a line segment 10 cm long. It is equally probable to find any proton at any point on this segment. How many protons remain on the segment at t = 10 s?

    Attempt at Solution
    We can create a wavefunction for this scenario:

    [tex]\psi(x, 0) = \frac{1}{10} \forall x \in (-5, 5)[/tex]

    Everywhere else, we can let [tex]\phi = 0[/tex]. Now, we need to try to find [tex]|\psi(x, 10)|^2[/tex] where [tex]x[/tex] ranges from -5 to 5. To do this, we first construct [tex]|\psi(x, t)|^2[/tex]. Since

    [tex]\psi(x, t) = \exp \left(-\frac{i\hat{H}t}{\hbar}\right) \psi(x, 0)[/tex]

    we want to have [tex]\psi(x, 0)[/tex] in terms of the eigenstates of the Hamiltonian operator so that we can have

    [tex]\psi(x, t) = \exp \left(-\frac{iE_n t}{\hbar}\right) \psi(x, 0) = \exp (-i \omega_n t)[/tex]

    However, since the eigenstates of the momentum operator are eigenstates of the Hamiltonian operator, we can find [tex]\psi(x, 0)[/tex] in terms of the eigenstates of the momentum operator. We will now do this.

    Note that

    [tex]\psi(x, 0) = \int_{-\infty}^{\infty} b(k) \phi_k dk[/tex]

    [tex]b(k) = \int_{-\infty}^{\infty} \psi(x, 0) \phi_k^* dx[/tex]

    imples that

    [tex] b(k) = \int_{-5}^{5} \frac{1}{10} \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-ikx} dx = \frac{\sin 5k}{5k\sqrt{2\pi}}[/tex]

    Now, we have

    [tex]\psi(x, 0) = \int_{-\infty}^{\infty} \frac{\sin 5k}{5k\sqrt{2\pi}} \phi_k dk[/tex]


    [tex]\psi(x, t) = \int_{-\infty}^{\infty} e^{-i \omega t}\frac{\sin 5k}{5k\sqrt{2\pi}} \cdot \frac{1}{\sqrt{2\pi}}\cdot e^{ikx} dk[/tex]

    How do I proceed to integrate from here? It turns out to be extremely ugly...
  2. jcsd
  3. Jun 28, 2008 #2
    Actually, I think I got the coefficients [tex]b(k)[/tex] incorrectly. Would the bounds actually go from - to + infinity? The protons are actually free, right? In that case, we'd have

    [tex]b(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-ikx} dx = \frac{\sqrt{2\pi}}{10} \delta (k)[/tex]

    Is this correct?
  4. Jul 3, 2008 #3


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    No, [itex]\psi(x,0)=0[/itex], outside (-5,5).
  5. Jul 3, 2008 #4
    If that were the case, then wouldn't the protons always be in the region? Or would the wavefunction eventually evolve so that it would spread out?
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