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Discrete energy level of schrodinger's equation

  1. Feb 28, 2009 #1
    Is it true that the discrete energy level of schrodinger's equation is due to boundary condition?
    What is the definition of boundary condition?
  2. jcsd
  3. Feb 28, 2009 #2
    A simple example is provided by a (classical) guitar string. The guitarist controls the boundary conditions by fingering the strings in various positions. The notes and harmonics for a given fingering will be a discrete set.
  4. Feb 28, 2009 #3


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    For the hydrogen atom, the boundary condition is that [itex]\Psi \rightarrow 0[/itex] as [itex]r \rightarrow \infty[/itex]. This gives discrete solutions for the radial part of [itex]\Psi[/itex], with correspondingly discrete energies.

    For the one-dimensional "particle in a box" that most all undergraduates learn as their first example of solving the Schrödinger equation, the boundary condition is that [itex]\Psi = 0[/itex] at the "walls" of the box. This likewise leads to discrete solutions with discrete energies.
  5. Feb 28, 2009 #4
    Boundary conditions for a second order differential equation are two numerical conditions for the general solution containing two arbitrary constants. These conditions "pick up" (select) the physically right solution. For a string oscillations they signify that the string is fixed at its ends (does not move). So any perturbation of the string is reflected from the ends making survive only discrete (proper) frequencies due to constructive/destructive interference.

  6. Feb 28, 2009 #5
    Is the discrete solutions due to quantum confinement or boundary condition?
    They are not the same, right?
  7. Feb 28, 2009 #6


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    what is "quantum confinement"?
  8. Feb 28, 2009 #7
  9. Feb 28, 2009 #8


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    It is due to the boundary conditions. If you have large systems, the discrete solutions are however so close to each other that they are often treated as a quasicontinuum. For example the phonon dispersion of a bulk solid is usually drawn as a continuous curve of energy versus k although the allowed values are discrete. There are so many possible states close to each other, that this seems sensible. Strong confinement now leads to a large difference in energy between the energy levels (see the particle in a box problem), so that the discreteness becomes easily noticeable.
  10. Mar 1, 2009 #9


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