Bound states in 1D potentials

  • #1

Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
410
0
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.
 
  • #3
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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.
 
  • #4
410
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The variational method does not require the wavefunction be a solution of the schrodinger equation.
 
  • #5
pam
455
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.!
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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