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Electron in a One-Dimensional Box

  1. Jun 24, 2008 #1
    An electron in a one-dimensional box with walls at [tex]x=(0,a)[/tex] is in the quantum state

    [tex]\psi(x) = A, 0<x<a/2[/tex]
    [tex]\psi(x) = -A, a/2 < x < a[/tex]

    Obtain an expression for the normalization constant, [tex]A[/tex]. What is the lowest energy of the electron that will be measured in this state?


    Well, we know that

    [tex]\int_{0}^A |\psi(x)|^2 dx = 1 \iff A^2 \cdot a = 1 \iff A = \sqrt{\frac{1}{a}}[/tex]

    so this is our normalization constant, [tex]A[/tex]. Now, to find the lowest energy of the electrong that will be measured in this state, we have:

    [tex]-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi[/tex]

    but the left hand side evaluates to 0 since [tex]\psi[/tex] is independent of [tex]x[/tex]. So this is the lowest (and only observed) energy in this state... is this right?
  2. jcsd
  3. Jun 24, 2008 #2
    Well, [tex]\psi[/tex] is independent of [tex]x[/tex] everywhere except around [tex]x = a / 2[/tex]. Around there things could get a little weird. (In fact, as this problem is stated, [tex]\psi[/tex] changes infinitely quickly from one side of [tex]a / 2[/tex] to the other, so around there things could get very weird.)

    Even without doing any calculations, though, you should know that since the electron is stuck in the box, it must have nonzero energy. If nothing else, the box gives it [tex]\Delta x < \infty[/tex], so it must have [tex]\Delta p > 0[/tex] by Heisenberg.

    If you've already studied the infinite potential well (one-dimensional box), you might also know that the energy eigenfunctions (or stationary states) are all [tex]sin[/tex] functions (since one of the walls is at [tex]x = 0[/tex]). Now, the given wavefunction [tex]\psi[/tex] is not such a [tex]sin[/tex] function, so to get the measurable energies we have to rewrite it as a sum of the energy eigenfunctions -- that is, as a Fourier series of [tex]sin[/tex] functions.

    Hopefully that's enough to get you started.
  4. Jun 25, 2008 #3
    Well, its a step function so if you compute the derivative you get a delta function...and then a derivative of a delta function and so on.

    By the way, if you solve for the wavefunction of a particle in a 1D box, you get sin or cosine functions. Any state the particle is in can be written as a linear combination of these eigenfunctions. This is a rather strange state...and interesting too :smile:
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