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Math Help - Problem #14: Schrodinger Eigenvalue problem

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    Problem #14: Schrodinger Eigenvalue problem

    Given the Schrodinger equation:
    -\frac{\hbar ^2}{2m} \frac{d^2 \psi}{dx^2} + V(x) \psi (x) = E \psi (x)

    and V(x) = -a \delta (0)
    find the energy eigenvalue, E, for a bound state in terms of a > 0.

    To make an appropriate energy eigenfunction we require that \psi (x) is continuous on the real line, \lim_{x \to \pm \infty} \psi (x) = 0 and \lim_{x \to \pm \infty} \frac{d \psi }{dx} = 0. Also, since the potential contains a delta function, the first derivative of the wavefunction will not be continuous.

    -Dan
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    Forum Admin topsquark's Avatar
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    Re: Problem #14: Schrodinger Eigenvalue problem

    I forgot to mention....for a bound state, E < 0.

    -Dan
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    Forum Admin topsquark's Avatar
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    Re: Problem #14: Schrodinger Eigenvalue problem

    No takers? I was hoping for a good old fashioned Math derivation using Green's functions or something. Well, here's how a Physicist does it!


    Start with x \neq 0. Schrodinger's equation becomes:
    -\frac{\hbar ^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi

    This has a general solution of
    \psi (x) = C e^{kx} + D e^{-kx} where k^2 = -\frac{2mE}{\hbar ^2}
    where C and D are constants to fit the boundary conditions.

    Note: We are looking for a bound state for the system, so E < 0 and thus k^2> 0.

    Since we've got the delta function at x = 0 it makes sense to provide two solutions: one for x < 0 and one for 0 < x. To keep the solutions from blowing up at infinity we choose:
    \psi (x) = \begin{cases} Ae^{kx} & x < 0 \\ Be^{-kx} & 0 < x \end{cases}

    Continuity at x = 0 requires that B = A. So the wavefunction and its first derivative are:
    \psi (x) = \begin{cases} Ae^{kx} & x < 0 \\ Be^{-kx} & 0 < x \end{cases}

    \frac{d \psi }{dx} = \begin{cases} \phantom{-} Ake^{kx} &  x < 0 \\ -Ake^{-kx} & 0 < x \end{cases}
    where k is as defined above. Note: The wavefunction has a discontinuity in its first derivative at x = 0. This kind of behavior is expected for potentials that contain any type of discontinuity in V(x). In reality we require the first derivative be continuous as well and requiring continuity in the first derivative can lead to some "non-trivial" solutions. Let's keep it simple.

    At this point we typically normalize this wavefunction, but there is no need to in this case, so I'm going to skip it. If you really want to you can normalize it and you find that A = \sqrt{k}.

    So far so good. We can easily find E as a function of k, but we want it in terms of a...we can fix this problem by considering the wave equation at x = 0.
    -\frac{\hbar ^2}{2m} \frac{d^2 \psi }{dx} - a \delta (0) \psi = E \psi

    Integrate this equation over the 1 dimensional ball with radius \epsilon centered on the origin, then take the limit as \epsilon \to 0:
    \lim_{\epsilon \to 0} -\frac{\hbar ^2}{2m} \int_{-\epsilon}^{\epsilon} \frac{d^2 \psi }{dx^2}~dx - \lim_{\epsilon \to 0} \int_{-\epsilon}^{\epsilon}a~\delta (0) ~\psi~dx = \lim_{\epsilon \to 0} \int_{-\epsilon}^{\epsilon}E \psi ~dx

    Skipping forward a bit we find that
    \lim_{\epsilon \to 0}-\frac{\hbar ^2}{2m} \left . \frac{d \psi}{dx} \right | _{-\epsilon}^{\epsilon} - a \psi (0) = 0

    Plugging in the wavefunction we obtain:
    \frac{\hbar ^2 k A}{m} - aA = 0

    Or
    \frac{\hbar ^2 k}{m} = a \implies E = -\frac{ma^2}{2 \hbar ^2}

    -Dan
    Last edited by topsquark; July 22nd 2014 at 12:58 PM.
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