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Bound states in 1D potentials

  1. Jun 3, 2008 #1
    Hello:

    There is a well known theorem which asserts that every attractive 1D potential has at least one bound state; in addition, this theorem does not hold for the 2D or 3D cases. I've been looking for a proof in my textbooks on qm but I've been unable to find it. Can you help me out?

    Thanks!
     
  2. jcsd
  3. Jun 3, 2008 #2
    It has to do with the variational method. You can always come up with a wavefunction that is "everywhere inside the well" and "piecewise flat", (think cross-section of a muffin tin) , so that [tex]\left\langle \psi | H | \psi \right\rangle < 0[/tex]. This puts an upper-bound on the ground state of the system.

    I never really understood how the proof breaks down in higher dimensions. I think it is because in 2D and 3D, the Hamiltonian has a centrifugal barrier. Well, other than that, explicit counter-examples are known. I hope someone else fills in the gaps.
     
  4. Jun 3, 2008 #3
    I'm guessing that the bare minimum case for an attractive potential would have to be the delta function in any number of dimensions. I know what the bound state looks like in 1D; in 2d or 3d if you could solve for the delta potential and show there are no bound states, then I think that would be pretty much it.
     
  5. Jun 4, 2008 #4
    The variational method does not require the wavefunction be a solution of the schrodinger equation.
     
  6. Jun 4, 2008 #5

    pam

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    In 1D, the wave function can be finite at the origin, and can always decrease monotonically to zero at infinity. In 2D or 3D, the "equivalent 1D" wave function is [tex]u=\sqrt{r}\psi[/tex]
    for 2D or [tex]u=r\psi[/tex] for 3D. In either case, u must equal zero at the origin.
    This requires a strong enough attraction to make u turn over at some point.
     
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