# Finding range of bound/non bound state energies of 1D finite

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1. Dec 1, 2016

### MxwllsPersuasns

1. The problem statement, all variables and given/known data
I'm currently working on a homework set for my intermediate QM class and for some reason I keep drawing a blank as to what to do on the first problem. I'm given three potentials, V(x), the first is of the form {A+Bexp(-Cx^2)}, the others I'll leave out. I'm asked to draw the potential (easy enough) and also then to state the range (if any) of the bound-state energies (discrete eigenvalues of the Hamiltonian) and also state the range (if any) of the non-bound state energies (continuous range of eigenvalues of the Hamiltonian).

2. Relevant equations

3. The attempt at a solution

I think I understand that a bound state is one which (simply put) decays to zero at or before infinity, in other words the energy of the particle in the 1D well is less than 0 but greater than the negative potential energy.

So for my example V(x) above I notice that the max for the potential is achieved at (0, A+B) and that my potential decays to A quickly then remains there until +(inf) and as x trends towards -(inf) V(x) trends towards +(inf). So then would the whole of the plane be a bound state for this potential? Also how would I determine these discrete eigenvalues (bound state energies) and continuous eigenvalues (non bound state energies)? Any help is GREATLY appreciated. Thanks for taking the time to read through my post :)

2. Dec 1, 2016

### PeroK

A bound state is one where the particle remains in a finite region. A plane cannot be a bound state.

3. Dec 1, 2016

### MxwllsPersuasns

Okay yes perhaps a poor choice of words on my part. The way I saw it was that from (around) the origin to +(inf) the potential is constant at the value A and from the origin to -(inf) the potential rapidly increases towards +(inf). Knowing this it seems, to me, that from the origin to -(inf) would definitely be a bound state due to the rapid approach to infinity and to the right of the origin would also be a bound state, so long as the kinetic energy of the particle does not surpass a value of A. Is this analysis correct? Also then what is meant by find the range of the (non-)bound state energies? I'm not sure how exactly I approach that question.

4. Dec 2, 2016

### PeroK

This makes no sense to me. Bound and scattering states are identified by their energy levels and the behaviour of the particle. I'm assuming that $A, B, C$ are positve constants. If so, you have a steep "hill" at the origin.

That said, your description of the potential suggests you might not have sketched the function correctly. The potential is a function of $x^2$, hence symmetrical about the origin.

If you have a bound state, which finite region would the particle remain in?

When you posted this first, I asked you to consider the classical case. Would there be any bound states for a classical particle?