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

[MxwllsPersuasns]

Homework Statement

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).

Homework Equations

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 :)

 

[PeroK]

I think I understand that a bound state is one which (simply put) decays to zero at or before infinity.

So then would the whole of the plane be a bound state for this potential?

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

 

[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.

Thanks for your response PeroK!

 

[PeroK]

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.

Thanks for your response 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?