Physically, I like the conclusión or hypothesis: "It seems that a very fast transition occurs..." The trouble is that it would hardly be observable. 

   There is not much more that I can add.

Hello Daniel!

First of all, the spectrum at high pressure is not a sharp line because phenomena as "pressure broadening" and "Doppler broadening".

You mentioned the pressure broadening. It's cause stands in he interruption of an event of emission by an atom, by a collision with another atom. The shortened emission time produces an uncertainty in energy, ∆E ≥ ħ/∆t, where ∆t is the shortened duration of the emission. Of course that one single atom won't display pressure broadeing, because it does not collide. See explanation at:

https://en.wikipedia.org/wiki/Pressure_broadening#Pressure_broadening

As to the Doppler broadening, if the atom is at rest you won't see this broadening.

However, the phenomenon is more general, and maybe you'd want to know. From all the states of the atom, only the ground state is stable. All the excited states are unstable states, usually called "resonant states". The energy distribution of these states is not an represented by infinitely sharp line, but by a Breight-Wigner distribution, essentially a Lorentzian curve. The width of the Lorentzian, ∆E, and the time of life of the excited state, ∆t, are also connected by the uncertainty principle.

A few more words: the instability of the excited states is explained by the so-called "coupling with the continuum". The atom is considered as interacting with the vacuum, which displays an infinite and continuous spectrum of energies of the electromagnetic field. Thus, in the quantum optics, the Hamiltonian of an excited atom is written as:

Htotal = Hatom + He.m. + Hcoupling

Hatom has a discrete spectrum of sharp energies, but He.m., the Hamiltonian of the e.m. field, has a continuous spectrum. In all, instead of displaying an infinitely sharp energy, the excited states display a Breit-Wigner distribution of energies.

Thanks very much Simon. I agree.

It is not even clear what it means to observe such photon. But at least the quantum evolution equation should describe some transition process. However Schrödinger time dependent equation does not show any energy transition because 1.- It ignores the photon and 2.- It actually is energy conservative. This means that there is no energy carrier, and that the energy of the system remains constant. Once more, thanks.

Sophia, your comments are very informative and I am grateful about them.

On first sight you indicate what seems the correct approach. Unfortunately I do not know how to write any useful Hamiltonian He.m. or Hcoupling. A very illuminating procedure should be the calculation of their eigenvalues and eigenfunctions, together with those of Htotal.  And then, from these results --this is the physically most relevant part-- to obtain the unitary evolution that shows the atom going from a given stationary state to another one, in such manner that the energies e( \psi_t ) of the intermediate states appear as continuous functions of time. For radiative transitions one would expect that this energy function starts from a value equal, or very near, to -\lambda_n and decrease (continuously) until the value -\lambda_m is reached.

Simultaneously to the decreasing values of the energy of the hydrogen system, the quantum unitary equations should also show a photon evolving in time. Beginning with zero or very small energy, this initially minuscule photon continuously develops into a full fledged photon of energy \lambda_m-\lambda_n.

Again, I am most grateful to you, Sophia, and please let me know of any available reference where such important quantum results as decreasing energy of the hydrogen system, and photon continuous "birth" --here barely sketched by me-- are worked out from any unitary quantum evolution equation.

With most cordial regards for both,

Daniel Crespin

Dear Daniel,

First of all my name is Sofia - without "ph".

Yes, there is plenty of material. What I told you I learnt both from quantum optics, and from the nuclear theory of decay. Note that de-excitation of an atom by emission of a photon is in principle analog with the decay of an excited nuclid by emission of a gamma particle.

Now it's late in my country, but tomorrow I'll give you references. I happen to have worked in the past on this topic.

Just, please note that there is a polemic among scientists whether the decay process (be it of an atom or of a nucleus) is a unitary process or a NON-unitary one. I am sure that you heard of the Gamow decay. Gamow used complex eigenvalues for energy. The real part of the energy represents the central value of the Breit-Wigner distribution, while the imaginary part gives the width of the Lorentzian.

Well, let's continue tomorrow. Best regards!

Hello dear Sofia

I have little idea of the use of energy observables that are operators having eigenvalues equal to non-real complex numbers. It seems a trick to avoid unitary evolution without making the crucial statement "linear unitary evolution is wrong".

Vaguely I recall the disquieting concept of "imaginary times" in relation with Feynman or Kac integrals. But it is better for me to stay clear from sophisticated formulations and limit myself to the very basic and fundamental ideas and models. If one keeps trowing objects to the heap it will just grow, will never go away, and whatever is hidden underneath may remain unknown forever.

Very recently I wrote an answer to why were complex numbers introduced in QM. It could be relevant for the present question as well, and is attached below as ComplexNumbresinQM2.pdf.

Introducing complex numbers as eigenvalues for the energy observable and keeping Schrödinger type of unitary evolution equation implies introducing imaginary times.

I thank you Sofia for your kind interest and give a warm welcome to your opinions and help.

With most cordial regards

Daniel Crespin

Dear Daniel,

If this topic is important for you, then, there is some material to read. I indicate it below, and you can pick from it according to how deeply you decide to go into the topic.

I recommend paragraph 19-5 Spontaneous emission in the book "Quantum Mechanics" of Leslie E. Ballentine, 1990, Prentice Hall Advanced Reference Series, A Division of Simon & Schuster, Englewood Cliffs, New Jersey 07632. It is a very readable book.

Additional articles I mention below, but they refer to nuclear emission. As I said, the principle is the same, a Hamiltonian comprising three parts. Note that the first two articles below, of Friedrichs and Feshbach, have the historical importance of explaining the decay as resulting from the coupling between the sharp energy levels of the atom and the continuum of energies of the scattering states of the vacuum. Personally, I suggest to begin the paragraph in Ballentine's book, and then, Rotter's article "A non-Hermitian operator . . .".

1) K. O. Friedrichs, “On the perturbation of continuous spectra”, Commun. Pure Appl. Math. 1, 361, (1948).

2) H. Feshbach, “Unified theory of nuclear reactions”, Annals of. Physics (N.Y.) 5, 357 (1958); 19, 287 (1962).

3) I. Rotter, “A non-Hermitian Hamilton operator and the physics of open quantum systems”, J. Phys. A: Math. Theor. 42, 153001, (2009).

4) I. Rotter, “Unified description of resonance and decay phenomena”, quant-ph/0710.3661v1.

5) N. Hatano, K. Sasada, H. Nakamura, and T. Petrosky “Some properties of the resonant state in quantum mechanics and its computation”, Progress of Theoretical Physics, 119, No. 2, 187, (Feb. 2008).

Now, to your comments: of course, an observable is real and complex energies are a trick. It just turns out that the trick works well. But, at the base of the trick stands a certain superposition of perturbations, which widens the initial level (and also displaces it). In all, the dependence on time of the wave-function can be approximated by 

Ψ = A exp(-iEt/ħ) = A exp[(-iE0 - Γ/2)t/ħ]

E0 is the central energy of the Lorentzian, and Γ gives the half-time, i.e. the time τ after which the probabilty of finding the atom still in excited state is 1/2 from the probability at an earlier time t1.

P(t1) = |A|2 exp(-Γt1/ħ); P(t1+ τ ) = |A|2 exp[-Γ(t1 + τ )/ħ]; 

Imposing the condition P(t1+ τ)/P(t1) = 1/2, one finds  exp(-Γτ/ħ) = 1/2, therefore

τ = (ln 2)ħ/Γ

Good luck!

I do not believe in the levels of H atom other than those predicted by Schrodinger's equation. So this question has to be modified as:

What is the physics of atom-photon interaction between the absorption and emission of the photon by the atom?

The atom is in the superposition stat,e i.e. it is neither fully in the ground state nor in the excited state, which can be treated as a new state. But the atom is NOT coupled  to the vacuum while it is in the superposition state. Hence, it cannot emit a photon from this superposition state or its probability is very low.

That's why the decay from the superposition state cannot be observed. However, there are methods to manipulate the quantum states of the atom on the Bloch sphere by playing with the field intensity (pi pulse, pi/2 pulse etc.)

Dear Muhammad Anwar,

==> "The atom is in the superposition stat,e i.e. it is neither fully in the ground state nor in the excited state, which can be treated as a new state."

The issue is that the excited state is not a sharp energy state. I repeat, the uncertainty principle ΔE Δt ≥ ħ, tells us that if ΔE = 0, we get Δt = infinite, i.e. no spontaneous de-excitation. An excited atom would remain in the excited state forever.

Dear Daniel,

I can say a little on this topic only from theoretical point of view. Today we still can not exactly solve the nonstationary problem of the Schroedinger equation. The Weisskopf-Wigner theory and its decsendants of exponential decay of the excited level are only phenomenological ones. However, there exists exactly solvable model of two level atom in a cavity called the Jaynes-Cummings model. Its solution is

|Ψ(t)>= Ci(t)|i> + Cf(t)|f>, where

Ci (t) = cos(λt√n+1) and  Cf (t) = -isin(λt√n+1), f - means final state, i - initial state. The formulae are taken from C.C.Gerry, P.L.Knight - Introductory Quantum Optics,CUP 2005. In this model energy of the atom exists at every point of time t.

As for an electron or positron not bound to an atom there exist two exact solutions of nonstationary problem. The  first was given by D.M.Volkov in 1935 for an electron in the field of an electromagnetic plane wave. The second was given by me for an electron in the homogeneous electric field - present in Researchgate. They both are solutions of the Dirac equation.

Janusz

It only means that atom exists (its wave function is defined) at all instant of time ( since t is continuous variable) but its energy is restricted to the discrete eigenvalues.

Sofia..

I apologize for my answer regarding the 'biphoton' and Fock state etc. We will discuss it sometimes later.

Regards

Anwar

Hello to all

Thanks Sofia for the references, and Muhammad and Janusz for the comments. Next week I should have enough time to sit down and elaborate some answers. 

Meanwhile I suggest to give a look at the natural model described at

https://www.researchgate.net/publication/308419467_Quantum_Wave_Collapse_is_Unsolvable

Cordially

Daniel Crespin

Article Quantum Wave Collapse is Unsolvable

Dear Daniel, "Quantum  wave collapse is unsolvable" because even relativistic Dirac's equation (to say nothing of Schrodinger's) is approximation. Strictly speaking, we are not clever enough to see detailes of transition between levels of atom.