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Nonsinusodial wave function problem

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    I am having a problem with an example problem in my physics book. The example goes like so:

    a.)Show that
    [tex]\psi(x) = Ax + B[/tex]
    [tex] A, B, constant [/tex]
    is a solution of the Schrodinger equation for an E = 0 energy level of a particle in a box. b.) what constraints do the boundary conditions at x = 0 and x = L place on the constants A and B?

    2. Relevant equations
    [tex]\frac{-\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}= E\psi(x)[/tex]

    3. The attempt at a solution
    Part a i understand completely -- i just take the second derivative of the wave function and find its eigenvalue to be 0, which corresponds to the energy. However, for part b, I am not quite understanding how the book applies the boundary conditions. They claim the following:

    "applying the boundary condition:
    [tex] x = 0[/tex]
    [tex]\psi(0) = A = 0[/tex]
    [tex] A = 0, and \psi(x) = Bx[/tex].
    Then applying the boundary condition:
    [tex] x = L[/tex]
    gives [tex] \psi(L) = BL = 0[/tex]
    so B must be zero. How are they finding this for the IVP? If i plug zero into the wave function i get B = 0, not A = 0. I must be missing something. Could someone please tell me what I'm not seeing? Thank you!
  2. jcsd
  3. Apr 18, 2008 #2


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    Agreed. If psi is zero at x = 0, then there no constant term (the linear wavefunction starts at the origin). ==> B = 0.

    The other boundary condition would give you AL = 0. ==> A = 0.

    So, the wavefunction is identically zero. Which means that there is no chance of a particle being in the box. So what was the point of this problem exactly???

    Edit: By this, what I mean is that, isn't this the TRIVIAL solution to the eigenvalue equation? If there is no particle in the box, the energy of the system is zero (inside the box).

    Edit 2: In fact, I thought that a zero eigenfunction was not allowed as a valid solution to the eigenvalue equation, because it would have infinitely many possible eigenvalues.
    Last edited: Apr 18, 2008
  4. Apr 18, 2008 #3
    The problem was intented to be trivial i think to show that you cant have a valid wave function of this form since you would get zero probability of finding your particle anywhere. But, im glad that you also get AL = 0 and B = 0 instead of A = 0 and BL = 0 (they lead to the same thing but them getting it in that form is somewhat bothersome for me because I feel like i'm missing somehing). Thank you for the help!
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