Thread: Bound States
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Feb10-05, 08:18 AM
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Quote Quote by 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.
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.