Mr. Wechsler,

You shoudl bear in mind that the topic of the state of isolated ions of atoms is very disputable from the point of view of correlation to real experimental conditions.

Miss Bojidarka,

And if I bear in mind, what does that help me?

Best regards!

Good question, Sofia. Zero is already continuous spectrum. It is named in chemistry predissociation. The whole science. Bojidarka is also right. Problems are severe.

Dear Eugene,

please be more generous with details. I need these things for an article that I write. Which problems? Please describe.

By the way, are you familiar with Rydberg states? I would like to know that there is somebody from whom I can ask advice.

My PhD thesis was devoted to the continuous nuclear spectrum (resonanses). There were publications in Nuclear Physics in 1979 and 1980 years.

Rydberg states is the special case because of the 1/r potential. Here are the plenty of discrete states near zero energy (your original question). Unfortunately, I was not engaged in solving concrete problems in this field and hardly can help You.

P.S. Continuous spectrum is the severe problem in quantum mechanics as the whole.

Dear Sofia,

Thank to ask me on this question but I never use the uncertainty principle. Let me say “one has to lean on the principles at the end they collapse”, A fortiori in physics.

Best regards,

Xavier

The lowest energy state is not achieved by adding photons. It is achieved at low temperature and low density. The state you describe has unfilled orbitals in the lower electron shell. The convention of assigning negative energy to bound electrons does not change the physical situation, it only obscures the fundamentals.

Your question can be answered by revealing your zero energy state to be greater than zero as shown by the temperature that supports it. Then the Heisenberg problem disappears. You are using a relative energy scale which leads to the wrong answer.

@ Sofia - You write: "If the atom gets more than E0, the electron leaves the atom." Actually, if the atoms gets that energy, the electron has left the atom because to get to Eo it has to get to an infinite distance from the atom - or effectively infinite distance. It means there is no longer an influence from the nucleus. The uncertainty in the energy actually can be infinite because it can take any positive value as far as the nucleus is concerned. An important point of the Uncertainty Principle is the uncertainty has to be with reference to something else because there is no absolute reference point or reference energy. In the case of ionization, it is only the Coulomb interaction with that nucleus that is relevant.

Having said that, I do not find the Uncertainty Principle to of much help regarding stationary states. The lifetime of an excited state has, as far as I can tell, nothing to do with the Uncertainty Principle and more to do with the external influence and the fact htat other state wave functions occupy the same space. As an example of this, microwave phase tethering of specific excited states of lithium (N = 71, from memory is suitable for a convenient microwave) can hold the excited state indefinitely.

@ Ian,

" . . . if the atoms gets that energy, the electron has left the atom because to get to Eo it has to get to an infinite distance from the atom - or effectively infinite distance."

Ian, the infinite distance of which you speak, is on the microscopic scale, not macroscopic. And the atom gets only the energy E0, s.t. the electron can't fly into the space because it has no kinetic energy; it remains attached to the atom at a (microscopically) big distance from the nucleus.

Here is the problem with the uncertainty principle. As we know, highly excited states have a very short life-time (s.t. the electron should fall immediately to a lower state with emission of a photon). Given that the life-time of the zero energy state is so short, the uncertainty in energy of this state should be very big. But we didn't irradiate the atom with photons of totally undefined energy, but of energy E0 (with a very small uncertainty in energy).

Sofia D. Wechsler - What the Uncertainty Principle does is define the line width - the variation in photon energies that will permit an electron to change states. Since a photon itself has infinite lifetime (unless it gets absorbed) it has a precise energy. As you approach the ionization energy, the lines pack closer, and then reach a continuum at the ionization energy. It is that continuum that I was referring to, and it has infinite line width because any photon that will give the electron net positive energy will ionize it.

Actually, the lifetime of the zero state is also infinite because by definition the electron no longer feels a net attraction from the nucleus. By analogy with gravitational mechanics, it has reached its escape velocity. Since the electric fields also go to infinity, the escape velocity concept is a better way of looking at it. The electron most certainly does not have zero velocity, and is not sitting outside the atom stationary - as an aside, that concept violates the Uncertainty Principle because it has precise position and precise velocity, but that is irrelevant as the same thing happens in classical mechanics. You cannot throw an object into space and have it stationary because the fields go to infinity

@ Ian,

The uncertainty principle does not regard the photon lifetime, but the imprecision about when the photon was emitted. If the wave-packet is very long, the energy of the photon is very well defined because the imprecision in the emission time of the photon is ver big.

"As you approach the ionization energy, the lines pack closer, and then reach a continuum at the ionization energy."

I understood NOTHING. If the photon energy distribution is narrower, that means the uncertainty in energy is narrower.

". . . the lifetime of the zero state is also infinite because by definition the electron no longer feels a net attraction from the nucleus."

Do you know the "Reaction theory" of Feshbach (of which the decay is a particular case)? What makes the energy of the bound levels to have a width, and, by the uncertainty principle, a finite life-time before de-excitation, is the coupling with the evironment. The atom is not an isolated system, but an open system, coupled with the electromagnetic field of the vacuum. This is the way we write the Hamiltonian of the decay in QED.

There is a law that the higher is a level, the greater is the energy width, s.t. the shorter the life-time, but I don't recall the proof. Maybe you recall the proof.

With best regards!

@ Sofia, to the best of my knowledge, nobody has recorded single photon spectra. What you see is the result of many photons, and the reason why the is a line width is because photons can come from slight variations in the energy of the state. I do not believe anyone has ever observed frequency variation in a single photon, but I could be mistaken. Yes, there is coupling with the environment, and that is actually why the excited states are not stable. The phase tethering experiment I mentioned above works by smoothing that out, as I understand. Details:

Maeda, H., Norum, D. U. L., Gallagher, T. F. 2005. Microwave manipulation of an atomic electron in a classical orbit. Science 307:1757-60.

@ Ian,

There is no such thing as the spectrum of a single photon. A quantum particle can be measured only once. If it is not absorbed, any subsequent measurement produces the same result as the first one - the famous collapse of the wave-function.

The energy of a bound state of an atom is not infinitely precise, it has a width, as predicted by the Feshbach's theory, because the coupling with the environment. (Feshbach's theory is for nuclear reactions but in what concerns the decay of atom levels there is principial analogy with the nuclear decay.)

Now, this is elementary material. Let's not waste time reviewing it. My question is the zero-energy level vs. the uncertainty principle: how can a state of well-defined energy (=0) be also short-lived.

With best regards!

@ Sofia,

I think we are at cross purposes. My point was, a state is measured in a spectrum as the sum of a number of photons. My other point is you do not really have a state of E = 0 exactly except by sheer accident, and it is not a stationary state - it is metastable at best. So it is not a well-defined state, other than in theory.

Sofia,

time-energy uncertainty relation has different sense than impuls-coordinate because time is not quantum mechanical operator. This was well explained by Mandelstamm (published after his death in UFN in 1944 by I.E. Tamm).

@ Ian,

To ensure a highly clean energy E0 of the photon beam, we have to make the beam very long. The longer the beam, the longer is the uncertainty at which time, if at all, did the atom absorb the photon.

Once the photon was absorbed a couple of excitation channels open: the atom may rise not to the level E = 0, but to a lower excited level. In thts case a, photon will exit the interaction, carying the rest of energy. Or, the atom may absorb the photon in entirety, in which case the electron rises to the level E = 0.

The sheer accident is the absorption of the photon by the atom when the photon wave-packet is VERY long. Also a sheer accident is the atom chosing the excitation level E = 0. But iff the photon was absorbed in entirety, i.e. the electron rose to the level E = 0, the uncertainty principle begins to ask the question.

The level E = 0 can't be metastable. Metastability occurs as a result of a particular configuration of the electron cloud. I didn't speak of such a case.

As I understand it, a photon must be absorbed or emitted in its entirety. Another photon may be re-emitted, e.g. in cases of fluorescence, but you can't absorb part of a photon.

@ Ian,

Didn't you hear of inellstic scattering?

@ Sofia,

All the scattering experiments I have carried out did not alter the frequency, other than when fluorescence was present. You have to be careful that the photon was not absorbed and re-emitted, as when that happens, there can be a frequency change

@ Dear Ian,

The experiments done in the physics community are not limited to your experiments. Inellastic scatteriang is text-book material. Two particles meet with certain individual kinetic energies, internal structures, and linear momenta, and leave with modified internal structures, energies, and linear momenta. Everything according to the laws of conservation.

Ian my dear, I won't discuss elementary material, I am too much busy.

I am aware that you try to help and I am grateful.

Probably the answer to my question is simple, but I don't see it. Did you ever have the feeling that some truth looks at you laughing and saying "I am here, how can't you see me?". This is my feeling.

You know, the Nature is simple, and sometimes we don't understand it exactly because we look for complicated explanations. Maybe this is the case.

With most kind regards,

Sofia

Dear Sofia D. Wechsler .

Ian has a very valid point. You can NOT absorb partial photon in any material. Working principle of fluorescence and phosphorescence are the live example. Better you read these topic very well before passing any judgment. I don't think you know even Franck Codon principle, that heavily uses in Display industries for lighting .

Gokaran,

I repeat that I won't discuss elementary topics as in ellastic scattering. The theory of scattering can be found in any text-book. I am sorry that you didn 't read my reply to Ian. I am also not willing to repeat replies.

Which text book telling you that you can absorb partial photon from a single photon?.If any author telling you this CRAP and have written a book then better to SUE him/her. Neither he/her nor you have any understanding to display materials and it functioning.

In any elastic or inelastic scattering, first material absorb complete single photon. Material goes to higher energy states. If photon energy is greater than the band-gap energy then particle will excite to higher energy (greater than the conduction band minima) states.Higher energy states has finite slope ( look up any material band structure, and see how slope vary with wave-vector k) and then particle will falls back on the conduction band minima within femto-second, assuming there is no inter-system crossing, if there is inter-system crossing then emission of photon will take place from triplet state, which is forbidden and thus system will take longer time for emitting the photon. This time could be mili-second, or even minutes. This process is called phosphorescence.), after that particle goes back to ground states after emitting single photon (exactly equal to the band-gap energy). This photon is very DIFFERENT from the absorbed photon (whose energy is greater than the band-gap energy). This process is called Fluorescence in display technology. This whole process come under IN-ELASTIC scattering . Here NO, Partial photon absorption from single photon has taken place at any time-step.

It is absolutely inconceivable, that some people are so confused with basic fundamentals of spectroscopy that they think that any system can absorb partial photon from single photon in any process. Not only this, they think that they can use multiple photons in term of continuous wave-train and minimized uncertainty in either energy or in time scale.

Gokaran,

The question is what happens with the state E=0 vs. the uncertainty principle.

My atom does not make part of a solid body. I don't have problems with bands, gaps or phosphorescence. Who wishes to open other problems may open another thread.

I am terribly busy and absolutely not interested in a chat about everything.

Dear Sofia,

When we talk about energy, we almost always mean the expectation value of the energy. The expectation value can be very well defined, yet the uncertainty (e.g. as measured by the standard deviation) can be extremely large. If you consider a uniform distribution over wide range of energies, the average is right in the middle, but the uncertainty is enormous.

In the particular case of atomic energy levels, I usually think about it the other way around: the higher the energy level the shorter the lifetime (typically) so delta-t is small, and thus the measured range of energies is larger -- but the mean energy is as you expect. For the E=0 energy you have the same physics (well, if it's slightly bound anyway) so the mean energy is zero but the lifetime is short so it would be quite a broad spectrum. It isn't clear to me why you think it should be zero lifetime/infinitely broad, but perhaps I have overlooked something.

Hope that helps,

Phil Hasnip

@ Philip James Hasnip

You see, IF the uncertainty of the energy level E = 0 is very large, the same should be the uncertainty in the energy of the irradiating beam of photons. In that case, for example, the atom might be ionized and the ejected electron may have a high kinetic energy.

But this is not what happens. The irradiating beam has a tight energy spectrum. For achieving this, we make the photon wave-packet very long. Therefore, you won't find an emitted electron with high kinetic energy, the probability for such a thing would be practically zero.

The measured lifetimes of Rydberg atoms (large principal quantum numbers) increases as n increases. They do not decrease as you assert. There is no inconsistency.

For large values of n, the core electrons do not perfectly screen the nuclear charge, so the energy levels do not scale like 1/n^2 but like 1/(n+d)^2, where d is a phenomenological term called the 'quantum defect.' As a result, the singular behavior that you contemplated does not occur.

@ Mark - I disagree with the screening defect - basically the screening is fairly good, although there is some residual defect. What happens is the wave functions are not simple solutions of the Schrödinger equation, but rather superpositions of waves, one of which has no radial nodes in the ground state. See Miller, I. J. 1987. The quantization of the screening constant. Aust. J. Phys. 40 : 329-346. , My more detailed explanation is in my ebook "Guidance Waves", and the value of this is shown in the ability to calculate covalent bond energies as in my ebook on covalent bonds.

Ian Miller: I agree that the naive single-particle wave functions are no longer eigenfunctions of the Hamiltonian; the eigenfunctions are indeed superpositions. My real point, though, is that the lifetime of states is not a monotonically decreasing function of n as suggested by Sofia D. Wechsler in her original post. There is no singular behavior that arises as n approaches infinity. Her confusion arises, I believe, from application of a model that does not reflect experimental observations.

Additionally, in a later response, she suggests that the bandwidth of the light source is somehow connected to the linewidth of the quantum state. That is also incorrect. For a real light source with finite bandwidth and a real detector with finite resolution, as one increases the photon energy to E0 and beyond, one will see ejected electrons before reaching E0 due to the fact that finite temperature atoms will see the photons blue-shifted or red-shifted depending upon their direction of motion. This is not a demonstration of energy non-conservation. As one surpasses E0, there will be more ejected electrons. There is no singular behavior anywhere.

@ Mark. I agree with what you are saying. I just thought I would add in this little additional feature that does not contradict the basic points you are raising.