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Quantum well, time evolution of probabilities

  1. Aug 6, 2016 #1
    this thread might be a bit longer, but I have already calculated everything and I am quiet sure that it is right :)
    Just need someone who confirms me or not :)

    1. The problem statement, all variables and given/known data

    Time evolution of probabilities. An electron inside a quantum well of length L is at time t = 0 in the state
    [tex] \psi(x,0) =\frac{1}{ \sqrt{L} } ( \sin( \frac{\pi x}{ L} ) + \sin( \frac{2 \pi x}{L} )) [/tex]

    Outside the well [tex] \psi(x,0) = 0. [/tex] The state [tex] \psi(x,0) [/tex] is thus a superposition of two stationary states[tex] \psi_{1}(x) \ and \ \psi_{2}(x) [/tex] with the energies [tex] E_{1} = \frac { \pi^2\hbar^2 }{2mL^2} \ and \ E_{2} = 4 E_{1} [/tex]

    Calculate the porbability that the electron is in the interval (0, L/2) at the time [tex] t = \pi \hbar / E_{1} [/tex]

    3. The attempt at a solution

    So, that is the homework, now I show you how i get a result:

    [tex] \psi(x,t) = \psi(x,0) e^{-iHt/\hbar} [ \psi_1(x,0) + \psi_2(x,0) ] e^{-iHt/\hbar} =\psi_1(x,0)e^{-iE_1 t/\hbar } + \psi_2(x,0)e^{-iE_2 t/\hbar } [/tex]

    [tex] \Rightarrow | \psi(x,t)|^2 = |\psi_1(x,0)e^{-iE_1 t/\hbar } + \psi_2(x,0)e^{-iE_2 t/\hbar }|^2 [/tex]
    [tex] = | \psi_1(x,0)|^2 + |\psi_2(x,0)|^2 + \psi_1(x,0)^* \psi_2(x,0) e^{i(E_1 - E_2) t / \hbar } + \psi_2(x,0)^* \psi_1(x,0) e^{i ( E_2 -E_1) t / \hbar} [/tex]
    Now we put this into the integral:
    \int_0^b \! | \psi(x,t)|^2 \, d x [/tex] where b = L/2 and we geht the result
    [tex] 0,5 + \frac{2}{3\pi} e^{i(E_1 E_2)t/\hbar} + \frac{2}{3\pi} e^{-i(E_1 E_2)t/\hbar} [/tex]
    Now I insert: [tex] t = \pi\hbar/E_1 = \pi \hbar 2mL^2 / \pi^2 \hbar^2 = 2mL^2/(\pi\hbar) [/tex]
    [tex] E_1 = \pi^2\hbar^2 / (2mL^2) [/tex]
    [tex] E_2 = 4E_1 = 2\pi^2 \hbar^2 / (mL^2) [/tex]

    And after doing that, we find out that the probability that the electron in in the interval (0,L/2) for the specific time is 0,076.

    Thank you :)
    Last edited: Aug 6, 2016
  2. jcsd
  3. Aug 6, 2016 #2


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    Is the wavefunction at ##t=0## what you are already given? It has an incorrect unit, for 1D system like this the unit of the wavefunction should be ##m^{-0.5}##.
  4. Aug 6, 2016 #3
    Yes it is already given. I forgot the root over the "L". Sorry for that, i corrected it.
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