# Bound states

1. Oct 24, 2007

### silimay

I have a question about bound states as they relate to a question on my homework...

From what I can see, bound states in quantum mechanics are associated with energies that are discrete, not continuous. I don't really understand why...

In my homework problem we are given a set of potential curves, and one question is for which curves can you definitely exclude the existence of a bound quantum state. I don't understand how to do this...I can draw wavefunctions for the potential curves given some energy, but I don't know how to tell if the energies are discrete rather than continuous.

2. Oct 24, 2007

### Avodyne

The first question is whether the potential could produce a classical bound state. If there are no classical bound states, then there are no quantum ones either. Usually, if there is a classical bound state, there is a quantum bound state, but there are some exceptions; physically, the zero-point energy of a putative bound state might be enough to render it unbound. (If I recall correctly, and I may not, this doesn't happen in one dimension, but may in higher dimensions.)

So, you can definitively exclude a quantum bound state if there are no classical bound states. A classical bound state requires a region of low potential energy bounded on all sides by a region of higher potential energy, so that the particle just bounces around inside.

3. Oct 24, 2007

### blechman

Not quite. Bound states are states for which the energy eigenvalues are negative. You can have a non-bound state with discrete energy eigenvalues or a continuum of bound state energies. That all depends on the details of your boundary conditions. But the proper definition of a bound state is negative energy eigenvalue (so therefore the particle cannot escape).

Hope that helps without doing the HW for you ;-).

4. Oct 24, 2007

### Count Iblis

Well, if a particle is in a bound state then that means that it is practically confined within some volume. The probability that it is outside that volume is small and the larger you take the volume the smaller that probabilty becomes.

Now, if you have a continuous spectrum, then the above picture cannot be true. Given some eigenstate with energy E, there must exist another eigenstate with energy E + delta E and you can take delta E arbitrary small. But the eigenfunctions corresponding to E and E + dE will look more and more similar as dE becomes smaller and smaller. Yet these eigenfunctions have to be orthogonal to each other, which means that the integral of the product must be zero. But if the wavefunctions only differ appreaciably from zero inside a finite volume and within that volme they are practically he same, this is impossible.

Last edited: Oct 24, 2007
5. Oct 24, 2007

### blechman

This may be true, but I find it a little confusing. There are bound states that are very weak (deuteron, for example) where it is actually more probable that the particles are "outside" the bound-state radius. Also Cooper-pairs in BCS superconductors, where the electrons form bound states that can be the size of the entire sample! Perhaps I'm wrong in thinking of Cooper pairs as true bound states, however...

This might be true in one dimension, but I can think of a possible flaw in higher dimensions: it is no longer the case that two states of energy E and E+dE must become the same state as dE goes to zero. In 1D I agree that this must happen, but in higher dimensions, you can have degeneracies. Thoughts?

I don't pretend to be an expert, but I was thinking of electron conduction bands in crystals (or more generally, a periodic potential), where you can have a discrete set of bands, but each band is a continuum of energy states. Do you call these "bound states" (electrons are confined to the crystal) or "free states" (electrons are moving from one atom to another)?

6. Oct 24, 2007

### Count Iblis

Blechman, I agree that it is a bit more complicated due to possible degeneracies. I'm not an expert about these subtle issues either. But basically, if we say that a particle is bound if its wavefunction falls of expontially or faster, then the number of eigenstates below some energy E is bounded. You can put the whole system in a big box without changing anything significantly. But for free particles inside the volume the number of states follows from the density of states formula which is proportional to the volume:
V d^3k/(2 pi)^3

The total number of bound states below some energy E must be less than what yo can derive from the free density of states formula for suficiently large V.

About the electrons in the crystal, I guess that's just a matter of definition. The electrons are confined to the crystal and the volume of the crystal then determines the density of states.

Last edited: Oct 24, 2007
7. Oct 24, 2007

### blechman

I'm afraid I cannot agree with your definition of bound state: what about a gaussian wavepacket? such a state is normalizable, so its wavefunction vanishes at infinity (and so does the probability, obviously) and yet it is the canonical free-state! I really believe that the definition of bound state is that

$$E-V_{\rm max}<0$$

where $V_{\rm max}$ is the maximum value of the potential. Your condition of the wavefunction vanishing is nothing more than the normalization condition required by all wavefunctions, bound or free.

8. Oct 24, 2007

### Count Iblis

Yes, you are right. The reverse is true. If the energy of the eigenstate is negative then it will fall off exponentially. Of course, the (heuristic) argument that there can only be a finite number of such states below some energy E is still holds because there can onlyt be a finite number of orthogonal functions that all fall of faster than some given exponential Exp(-r/r0).