# Bound States

1. Feb 10, 2005

### RPI_Quantum

To my understanding, when a particle is in a bound state, it is "stuck" because its total energy is less than the surrounding potential.

I am confused on how to prove a particular potential has no bound states. For example, in one problem, I am asked to show that there is no bound state in a spherical finite well, if the potential inside the well is less than some quantity. I guess I would like to know what a bound state "looks" like mathematically. In a simple case like the harmonic oscillator, I can see what the bound state is, but I do not know how to show that they are bound states. This is what I need help with for the case of a finite spherical well.

2. Feb 10, 2005

### ZapperZ

Staff Emeritus
Let's try a specific example. Look at the finite square well case with a depth of V_0. When you solve for the eigen energies for this case, you get a series of values that depends on 1/L, where L is the width of the well. It means that if you make the well narrower (L getting smaller), the magnitude of the ground state energy is now larger, i.e. the energy for n=1 state will be greater. At some point, the potential well is so narrow that E(n=1) = V_0, and the well can no longer have a state entirely within itself. So you will no longer have a bound state.

So the condition for a bound state depends not only on the depth of the potential well, but also on its physical size. To know rigorously if a potential well can have a bound state, you need to find out what is the energy of the lowest energy state it can have. If it is less than V_0, then you have a bound state. If not, you have no bound state.

Zz.

3. Feb 11, 2005

### reilly

This is basic: you will find bound states discussed in any QM text book. Read, and you will be able to answer your question.
Regards,
Reilly Atkinson

4. Feb 21, 2005

### broegger

In Griffiths' Introduction To QM he states that in the case of the finite square well "there is always one bound state, no matter how "weak" the well becomes". Am I missing something here?

5. Feb 21, 2005

### ZapperZ

Staff Emeritus
I don't have access to Griffiths right now, so I can't double check. But let's try running this scenario where the well is so "weak", it is essentially V=0. I do not see how one can still get a bound state out of that one.

On the other hand, if you look at the rigorous solution to this problem, you will get, as one of the solution to the energy eigenvalues, a transcendental equation of the form of $$k/\kappa = -tan(ka)$$ for the even solution. One could get a solution for k=0, pi/2 no matter how weak the potential is. However, one must also realize that the exponential decaying "tails" of psi penetrate deeper and deeper into the classically forbidden region as $$\kappa$$ gets smaller and smaller until a point where the sinusoidal property of psi at the center of the well becomes negligible (the wavefunction decays away exponentially from a centre cusp). You get a solution whose stability depends very much on the "flatness" of V.

Zz.

6. Jun 21, 2010

### sirenayka

Bound state implies the "classically expected" state, so bound state energy should be smaller than V(r = infinity). For definition, Shankar says "a particle's wave function should go to zero as x -> infinity in bound state"