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Neutrons in a One-Dimensional Box

  1. Jun 25, 2008 #1
    Problem
    One thousand neutrons are in a one-dimensional box, with walls at x = 0, x = a. At t = 0, the state of each particle is

    [tex]\psi(x, 0) = Ax(x-a)[/tex]

    a) Normalize [tex]\psi[/tex] and find the value of the constant [tex]A[/tex].
    b) How many particles are in the interval [tex](0, a/2)[/tex] at [tex]t=0[/tex]?
    c)How many particles have energy [tex]E_5[/tex] at [tex]t=0[/tex]?
    d)What is [tex]\langle E \rangle[/tex] at [tex]t=0[/tex]?

    Solutions
    a) This is quite straightforward... just set [tex]\int_0^a \psi^2 dx = 1[/tex] and solve for [tex]A[/tex], which yields [tex]A = \sqrt{30/a^5}[/tex].
    b)Just find [tex]\int_{0}^{a/2} \psi(x, 0)[/tex], which is 0.5. Now, 0.5 * 1000 = 500 neutrons.
    c) This is the part where I have problems... could someone help out? Exactly what eigenstates am I supposed to break [tex]\psi[/tex] up into? Is it a one-dimensional particle in a box scenario? If so, should I express the function [tex]\psi[/tex] in terms of

    [tex]\sqrt{\frac{2}{a}} \sin \frac{n \pi x}{a}[/tex]

    ? I already tried to express [tex]\psi[/tex] in terms of the momentum eigenstates,

    [tex]\phi_n = \frac{1}{\sqrt{2\pi}} e^{ikx}[/tex]

    This didn't work out too well (without TI-89)... how are we supposed to find the number of particles in [tex]E_5[/tex] exactly? I know that the energy levels of the neutrons are discrete, but how are we supposed to model this without an integral? If it's an infinite summation, could someone tell me how to do this?
    d) This part is pretty simple... we have to just do

    [tex]\int_{0}^{a} \psi^* \hat{E}\psi dx = \int_{0}^{a} Ax(x-a) \cdot -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} dx = \int_{0}^{a} Ax(x-a) \cdot 2A dx = -\frac{30}{a^5} \cdot \frac{\hbar^2}{2m} \cdot -\frac{a^3}{3} = \frac{5\hbar^2}{ma^2}[/tex]

    I'm just having a bit of an issue with part c... could someone help out? :smile:
     
  2. jcsd
  3. Jun 25, 2008 #2
    I would go with trying to do a Fourier expansion of the position wave function with sine terms mostly because your box extends from 0 to a, (ie has odd symmetry). (Like you had mention doing before trying to transform the problem into momentum space.)

    Although changing to momentum space should also work, although you'll have to think about it a little more carefully.

    --------------------
    theUndergrad

    http://www.theUndergraduateJournal.com/
     
  4. Jun 25, 2008 #3
    Hmm... yes I think I got it. It's actually a discrete Fourier series, which turns out to be MUCH easier than the momentum way.
     
  5. Jun 25, 2008 #4
    Hmm... does this work for part C:

    Let

    [tex] \psi = \sum_{n=1}^\infty b_n \phi_n[/tex]

    where [tex] \phi_n [/tex] denote the eigenstates of the particle in a box scenario:

    [tex]\phi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi x}{a}[/tex]

    Now,

    [tex] \psi = \sum_{n=1}^\infty b_n \phi_n[/tex]
    [tex] |\psi\rangle = \sum_{n=1}^\infty b_n | \phi_n \rangle[/tex]

    if we change notation. From here,

    [tex] \langle \phi_j|\psi\rangle = \sum_{n=1}^\infty b_n \langle \phi_j | \phi_n \rangle[/tex]

    [tex] \iff \langle \phi_j|\psi\rangle = \sum_{n=1}^\infty b_n \delta (j - n) = b_j \iff b_n = \langle \phi_n | \psi \rangle[/tex]

    Now, here's the thing... in our case, we have that the right hand side of this last equation, [tex]b_n = \langle \phi_n | \psi \rangle[/tex], must be expanded discretely, ie

    [tex]\langle \phi_n | \psi \rangle = \sum_{x = 1}^\infty \phi_n^* \psi (x)[/tex]

    but this turns out to be very discouraging... you can't really evaluate it that well. To approximate it, can I go ahead and evaluate

    [tex]\int_{1}^\infty \phi_n^* \psi (x) dx [/tex]

    ?
     
  6. Jun 25, 2008 #5
    Hmm... does this work for part C:

    Let

    [tex] \psi = \sum_{n=1}^\infty b_n \phi_n[/tex]

    where [tex] \phi_n [/tex] denote the eigenstates of the particle in a box scenario:

    [tex]\phi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi x}{a}[/tex]

    Now,

    [tex] \psi = \sum_{n=1}^\infty b_n \phi_n[/tex]
    [tex] |\psi\rangle = \sum_{n=1}^\infty b_n | \phi_n \rangle[/tex]

    if we change notation. From here,

    [tex] \langle \phi_j|\psi\rangle = \sum_{n=1}^\infty b_n \langle \phi_j | \phi_n \rangle[/tex]

    [tex] \iff \langle \phi_j|\psi\rangle = \sum_{n=1}^\infty b_n \delta (j - n) = b_j \iff b_n = \langle \phi_n | \psi \rangle[/tex]

    Now, here's the thing... in our case, we have that the right hand side of this last equation, [tex]b_n = \langle \phi_n | \psi \rangle[/tex], must be expanded discretely, ie

    [tex]\langle \phi_n | \psi \rangle = \sum_{x = 1}^\infty \phi_n (x)^* \psi (x)[/tex]

    but this turns out to be very discouraging... you can't really evaluate it that well. To approximate it, can I go ahead and evaluate

    [tex]\int_{1}^\infty \phi_n (x)^* \psi (x) dx [/tex]

    ?
     
  7. Jun 26, 2008 #6
    You could try this:

    [tex]b_n = \int^{a}_{0} \phi_n (x)^* \psi (x,0) dx [/tex]

    solve for [tex]b_n[/tex] , then you can find probability that particle is in E5 .
     
    Last edited: Jun 26, 2008
  8. Jun 26, 2008 #7
    Oooh, I completely forgot that the wavefunction was 0 everywhere else heheh...
     
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