# Schrödinger Potential Fields with no Energy Quantisation?

• I
The solution to the One-Dimensional Time-Independent Schrödinger equation for an electric potential field of constant value is an exponential function, and its energy eigenvalue can have any value, it is not quantised.

Are there any other potential field functions whereby the energy of the particle is not quantised?

Excluding the case where the potential is entirely zero, i.e. free particles.

blue_leaf77
Homework Helper
Are there any other potential field functions whereby the energy of the particle is not quantised?
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:

DrClaude
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:
Are there any other potential field functions whereby the energy of the particle is not quantised?

very interesting. I didn't know that the atom could have a continuous spectrum. Why don't we observe this?

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

blue_leaf77
Homework Helper
Why don't we observe this?
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.
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
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.

secur
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.

blue_leaf77
Homework Helper
but seems close enough to greswd's request for "any other potential field functions whereby the energy of the particle is not quantized"
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.

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.

What do you mean by ionized?

blue_leaf77
Homework Helper
What do you mean by ionized?
It's the ionization, the liberation of an electron from an atom following the addition of energy exceeding its binding energy.

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

Is ionization an issue when examining the spectra of a hydrogen discharge lamp?

blue_leaf77
Homework Helper
Is ionization an issue when examining the spectra of a hydrogen discharge lamp?
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).

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).

Ok. So maybe there is possibly something added on top of the Schrödinger equation that excludes the physical possibility of continuum states?

blue_leaf77
Homework Helper
So maybe there is possibly something added on top of the Schrödinger equation that excludes the physical possibility of 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.

I don't think so, the reason why continuum states are not realizable is due to the non-vanishing wavefunction at infinities

How does this non-vanishing prevent an electron in this state from converting its energy into a photon?

blue_leaf77
Homework Helper
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.

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.
Yes, I was asking why the non-vanishing wavefunction prevents it from being a realizable state.

blue_leaf77
Homework Helper
Because it's not normalizable. Pretty much like the wavefunction for a free particle.

Because it's not normalizable. Pretty much like the wavefunction for a free particle.
But a free particle can convert its energy into photons right?

blue_leaf77
Homework Helper
But a free particle can convert its energy into photons right?
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.

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In which way, if it's alone in the universe?

Good question. I don't even know how electrons in an atom lose energy, I just know that they do.

blue_leaf77
Homework Helper
I don't even know how electrons in an atom lose energy,
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.

They can lose energy and drop down to a lower level by interaction with vacuum field (spontaneous emission)
So electrons in continuum states can't undergo spontaneous emission? Why not? Because they're non-normalizable?

stevendaryl
Staff Emeritus
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, $V(x) = -A e^{-\lambda x}$. If $A$ or $\lambda$ 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). $V(x) = -A/r$.

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

My first question has been answered. I do wonder why we don't observe the continuum states of hydrogen though.

blue_leaf77
Homework Helper
I do wonder why we don't observe the continuum states of hydrogen though.
How many times do I have to repeat, such a state is not normalizable.

stevendaryl
Staff Emeritus
So electrons in continuum states can't undergo spontaneous emission? Why not? Because they're non-normalizable?

I wouldn't put it that way. I'm not sure what normalizability has to do with it. If you consider a free particle, with momentum $\vec{p}$, it's impossible for it to emit a photon for kinematic reasons: You have an initial momentum $\vec{p}$. After emitting a photon, you have photon momentum $\vec{K}$ and particle momentum $\vec{p'}$. For a free particle, there is just no way to choose $\vec{K}$ and $\vec{p'}$ so that both energy and momentum are conserved. If the particle is tightly bound (to a proton, for instance), then some excess momentum can be shared with the proton, and it becomes easier to satisfy the conservation laws.

dextercioby
How many times do I have to repeat, such a state is not normalizable.
I asked the question only once before, in #14, and your answer in #17 is actually a repetition that doesn't answer my question.

stevendaryl
Staff Emeritus
I asked the question only once before, in #14, and your answer in #17 is actually a repetition that doesn't answer my question.

What do you mean when you say that the continuum states are unobservable? When an electron is knocked out of its atom and becomes an unbound electron, that means that it has left the discrete states and entered into the continuum states.

 That's not exactly right. The state of the electron after being knocked out of the atom is never a pure continuum state, since those are not normalizable. But its definite can be written as an infinite superposition of continuum states, in the same way that a function $f(x)$ can be written as a superposition of plane waves: $f(x) = \int dk e^{ikx} \tilde{f}(k)$.

blue_leaf77