Atom excitation: what changes, amplitude or frequency?

Question

When an atom is excited by a photon and there is an electron transition from ground to excited state, from energy level 1 (E1) to energy level 2 (E2), I understand that the energy of the exciting photon must be equal to E2 – E1 = hf (h being Planck constant and f being the photon’s frequency). But my doubt is whether:

a) the electron was vibrating at frequency f and it experiences an increase in amplitude, i.e. it vibrates farther away from the nucleus or

b) the electron was vibrating at say f1 and its frequency is raised to f2 because the exciting photon has frequency f2 – f1.

I have read the two explanations.

The one I certainly prefer is a), because it is more in line with the classical analogy of a standing wave, where 1st incoming energy can only interact with a mode vibrating at same frequency as the stimulus and 2nd vibration modes are orthogonal, so incoming energy cannot propitiate a change of mode, only a change of amplitude within the mode in question…

Thus the peculiarity of quantum realm would be just that there is a minimum unit of light stimulus which is related to its frequency and defined as hf, which we call a photon. Incidentally, I would say that this would not be a great peculiarity because I can imagine that if I want to excite a string fastened at both ends, to make it vibrate at one of its natural frequencies or tones (f), I must do it with a stimulus whose intensity would be calculated in view of the nature of the challenge: as product of f (which is after all what you want the string to do, i.e. vibrate at this frequency but more intensely) times some factor playing the role of h and related (I gather) to the characteristics of the string…

But please don’t pay much attention to the latter ramblings…, the question is only: is a) or b) the right explanation about what changes in an electron transition: amplitude or frequency?

Edit: I've realized that there may be some ambiguity in the question. A likely scenario is that the atom contains no oscillation at a given frequency and after the incidence of the photon it acquires such frequency. I would say that this falls under category a) (change in amplitude), because formerly amplitude at such frequency was zero and it becomes whatever, but someone might interprete that this is a change of frequency (one frequency that was acceptable, but was latent, becomes actual). Anyhow, let us call this case c) for convenience.

I would also like to clarify what case b) means in my question: it means that a mode of oscillation existed at f1 and (because of a photon oscillating at f2 - f1) such mode f1 disappears and gives way to a mode f2.

Answer 1

The right answer is b).

"Vibrating" is not the word that I would use for a stationary state, but those states have phase factors $e^{-iE_n t/\hbar}$.

Superpositions of eigenstates with different parity have a charge density distribution that is oscillating with a frequency that is given by the energy difference. Here an animation of a particle in a box in a superposition of the ground state and the first excited state. So it will radiate or absorb radiation from an oscillating electric field with that frequency.

Answer 2

It is now over one hundred years that the narrative has become entrenched: namely, that the photon drives the electron into a higher orbital, and subsequently the electron drops back to the ground state, emitting a photon.

THERE IS NO EXPERIMENT WHICH CONFIRMS THIS NARRATIVE. No one has ever shot a photon at an isolated hydrogen atom. No one has ever seen the atom jump into the excited state. No one has ever seen the atom drop back into the ground state. And no one has ever seen or measured the new photon leaving.

Here is what quantum mechanics tells us. If you use perturbation theory to solve the Schroedinger equation for the hydrogen atom in the presence of an electric field, you find that the ground state has a dipole moment, most easily approximated as the superposition of the 1s state and the 2p state. If you then make the electric field oscillate, the ground state will follow the oscillation.

There will be a time delay, and in general the oscillation of the dipole moment will be very slight. But at certain frequencies the oscillation can become quite strong. In particular, the hydrogen atom oscillates quite strongly at the difference frequency of 1s and 2p states.

So if the whole system is put in a box…the hydrogen atom and the oscillating electric field…the effective ground state is the oscillating hydrogen atom sitting in the oscillating electric field. For small fields, if you double the field, the hydrogen atom oscillates twice as hard. If you halve the field, the oscillation drops by half.

THERE IS NO MINIMUM FIELD AND THERE IS NO MINIMUM OSCILLATION. If you drive the atom with a very weak oscillating field for a very short period of time, the atom oscillates a little bit and then quiets down again. Just the same as if you drive a receiving antenna for a short period of time with a radio wave. And just like a receiving antenna, during the time when the atom is oscillating, it also functions as a transmitting antenna, giving off radiation in the classical donut pattern familiar to student of electromagnetic theory.

There is no difference between the hydrogen atom and the radio antenna. It oscillates when it is driven by an oscillating field, and during that time it emits energy. There is no miniumum “quanta” of energy required to initiate the oscillation, and no minimum “quanta” of energy which is scattered.

There is absolutely no experimental way to distinguish the standard narrative (with its photons and quantum leaps) from the simple (and frankly obvious) mechanism which I have explained here