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Negative energy in the Schrödinger equation

  1. Dec 9, 2012 #1
    When my book deals with bound states and scattering states it puts:

    E<0 bound state, E>0 scattering state. What reference for the potential have been used for these?
     
  2. jcsd
  3. Dec 9, 2012 #2

    K^2

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    Potential energy at infinity.
     
  4. Dec 9, 2012 #3
    but can you just play around with sign in front of the energy just like that? I thought you only obtained solutions if E>Vmin?
     
  5. Dec 9, 2012 #4

    K^2

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    Your Vmin is also negative for scattering/bound-state problems.
     
  6. Dec 9, 2012 #5
    I see. Makes sense. But sinse you can pick any reference for your potential what if you then picked one such that the energy switches sign. That seems to alter everything since you would then get a sinusoidal wave function instead of an exponential. What am I missing out on?
     
  7. Dec 9, 2012 #6
    And also. It seems my book treats the problem where you have a delta function potential at the origin both with negative and positive energy. But how would it be possible to have negative energy with a potential that goes to zero at infinity?
     
  8. Dec 9, 2012 #7

    K^2

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    Why don't you try it? Take a potential that you know how to solve. Replace V -> V + constant. Solve again, and see what happens to the eigen values.

    Delta function potentials always bothered me too, so I'm afraid I can't help you visualize it. Just trust the math.
     
  9. Dec 9, 2012 #8
    Problems with a delta-like potential may be solved by applying a Fourier transform.

    A delta-like potential [itex]V(x) = \alpha \, \delta(x - x_0)[/itex] has a Fourier transform:
    [tex]
    \tilde{V}(k) = \alpha \, e^{-i k \, x_0}
    [/tex]

    The Schroedinger equation in momentum space is:
    [tex]
    \frac{\hbar^2 \, k^2}{2 \, m} a(k) + \int_{-\infty}^{\infty} \frac{dk'}{2 \pi} \, \tilde{V}(k - k') a(k') = E \, a(k)
    [/tex]
    where [itex]a(k) = \int_{-\infty}^{\infty} \psi(x) e^{-i k x} \, dx[/itex] is the wave function Fourier transform. The inverse Fourier transform is:
    [tex]
    \psi(x) = \int_{-\infty}^{\infty} \frac{dk}{2 \pi} a(k) e^{i k \, x}
    [/tex]

    Substituting the Fourier transform of the potential in our Schrodinger equation, and using the definition of an inverse Fourier transform of the wave function, we get:
    [tex]
    \left(E - \frac{\hbar^2 \, k^2}{2 \, m} \right) \, a(k) = \alpha \, \psi(x_0) \, e^{-i k \, x_0}
    [/tex]
    Substituting this into the definition for [itex]\psi(x_0)[/itex], we get the self-consistency condition:
    [tex]
    \psi(x_0) = \alpha \, \psi(x_0) \, \int_{-\infty}^{\infty} \frac{dk}{2\pi} \frac{1}{E - \frac{\hbar^2 k^2}{2 \, m}}
    [/tex]
    If we assume that [itex]\psi(x_0) \neq 0[/itex] (otherwise [itex]a(k) \equiv 0, \forall k[/itex], which is a trivial solution), and we introduces the energy parameter [itex]\epsilon[/itex] (with a dimension [k]2):
    [tex]
    \epsilon \equiv \frac{2 m \, E}{\hbar^2}
    [/tex]
    [tex]
    1 = -\frac{2 m \, \alpha}{\hbar^2} \, \int_{-\infty}^{\infty} \frac{dk}{2\pi} \frac{1}{k^2 - \epsilon}
    [/tex]
    This is an implicit equation for [itex]\epsilon[/itex]. Can you find its solutions?
     
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