I Schrödinger Potential Fields with no Energy Quantisation?

The scattering case, i.e. a situation where the particle's energy is higher than the maximum value of the potential ##V(\mathbf{r})##. A particular example is the continuous solution of the hydrogen atom:

• http://journals.aps.org/pr/abstract/10.1103/PhysRev.28.936
• http://www.ncbi.nlm.nih.gov/pmc/articles/PMC522475/

 

We have observed this, starting many years ago:

"... in the exceptional case of the star AC +70 d 8247, surface gravity is about 3 million times the Earth's ... the hydrogen lines would be broadened to such an extent that they would flow together into a fairly uniform continuum of absorption. This might explain why there are white dwarfs with purely continuous spectra. According to Kuiper, AC +70 d 8247 has such continuous spectrum."

From "White Dwarfs", Otto Struve, Sky and Telescope, December 1953

 

Probably some people out there have already observed this in collision experiments, but I don't know for sure. Anyway, continuum states are unbounded state. Given that an atom in reality is not floating alone in the universe, I imagine it must be hard to maintain a stable quasi-isolated hydrogen atom with positive energy without being quickly ionized.

The continuous spectrum in the case of a star's emission sounds more like due to the usual line broadening mechanism, which is further due to a collective motion of an ensemble of atoms.

 

blue_leaf77, you're right. Struve says it's the "broadening of spectral lines when an element is influenced by an electric field" - due to other ionized atoms and electrons in the vicinity. Not the same as the continuous-spectrum case you're referring to, but seems close enough to greswd's request for "any other potential field functions whereby the energy of the particle is not quantized" to be worth mentioning.

 

Ah I see, so you intended to answer the original question of the thread. I thought you were responding to grewsd's last question in post #3 where the discussion is narrowing down to the issue of an isolated H atom's continuum states.

 

It's the ionization, the liberation of an electron from an atom following the addition of energy exceeding its binding energy.

 

You are drifting away from the current discussion. In post #5, I brought up ionization to present a possibility of the reason why the continuum state of hydrogen atom have been difficult to observe, I myself am not sure if there have been an observation out there that's why I also expressed my uncertainty in the same post. It has nothing to do with the spectrum of discharge lamp.

Continuum states, by the way, are essentially not a bound state anymore and in fact they don't correspond to physically realizable state because they do not go to zero at infinities (probably I should also have added this beforehand to the possible reason of the difficulty of observing continuum states).

 

I don't think so, the reason why continuum states are not realizable is due to the non-vanishing wavefunction at infinities, and this is a consequence of satisfying the Schroedinger equation. Nevertheless, along with the discrete, bound states, continuum states can serve as the basis function of any realizable wavefunction.

 

When you say "in this state", do you mean one of the continuum state? Haven't I said that this state is not a realizable state.

 

Because it's not normalizable. Pretty much like the wavefunction for a free particle.

 

In which way, if it's alone in the universe? Moreover about the "freeness" of that electron, its wavefunction in reality does not exactly equal to that of the theoretical free particle's wavefunction.

 

They can lose energy and drop down to a lower level by interaction with vacuum field (spontaneous emission), stimulated emission, and non-radiative energy transfer.

 

I'm a little confused about whether the original poster, greswd, considers his/her question answered? I think the answer is about the depth of the potential well. Depending on the shape of the well, we have the following possibilities for energy eigenstates:

1. There are only continuum states. This is true if the well is too "shallow" to bind anything.
   
2. There are energy bands of continuum states, and other energy ranges where there are discrete states.
3. There are only discrete energy eigenstates.

The first situation holds for "shallow" potentials (or repulsive potentials). For example, [itex]V(x) = -A e^{-\lambda x}[/itex]. If [itex]A[/itex] or [itex]\lambda[/itex] is too small, I don't think that there will be any bound states.

The second situation holds for deep wells that don't become shallower with distance (for example, the Coulomb potential). [itex]V(x) = -A/r[/itex].

The third situation holds for deep wells that remain deep at large distances (for example, the harmonic oscillator potential). [itex]V(x) = -\frac{1}{2} k x^2[/itex]

 

How many times do I have to repeat, such a state is not normalizable.