# Does electron physically move in an interband transition?

How does electron move say from ground state energy level to first excited state? Is there any actual displacement in terms of motion?Is there a way to logically think about this by the help of creation and annihilation operators?

I know this is a quantum mechanical point of view but is there some way we can think in a classical sense?

• The short answer is that the shape of the wavefunction changes, with the higher energy state being typically "more wiggly". In principle the deformation of the shape of the wavefunction during the excitation is a continuous process, although calculating the shape of the wavefunction during such a transition can be non-trivial, and measuring it is even harder. Commented Aug 18, 2021 at 16:54
• Electrons do move, but not to go between the conduction band and the valence band. Commented Aug 18, 2021 at 16:55
• @BySymmetry okay but how will you explain the position of electrons when the transition takes place? Commented Aug 18, 2021 at 17:20
• @Gandalf73 his point is that all we ever have is a 3-dimensional shape indicating where the electron is most likely to be found. They never "sit" in one location, nor do they follow anything like a classical orbit path. Commented Aug 18, 2021 at 17:51
• Is it possible that the question arises from a simple misunderstanding of the typical graphical depiction of such transitions? E.g. here. It should be stressed that the locations on the diagram do not correspond do locations in space; they are different states. the x-axis is momentum, the y-axis energy. Sure, if momentum is changed the location may change (or stop changing, for that matter!). But that's not the point of the diagram. Commented Aug 19, 2021 at 8:53

The bands are the many-to-one relation between energy and momentum for electrons in a crystal. This relation is not position dependent. (I am not talking about band warping near crystal borders.) To go from one band to another just means that an electron's energy, and possibly momentum, changes. For a so-called vertical transition only the energy changes. There is no movement in real space, only a move in ($$E$$, $$\vec k$$) space.

For an atomic transition in general the spatial distribution changes upon excitation. In this sense the electron moves.

I am answering your question where you ask about the electron transitioning from ground state to excited state. The electron (and the atom as a whole) is a quantum mechanical entity and its position around the nucleus is best described by quantum mechanics.

Over time, $$c_{1}(t)$$ smoothly decreases from one to zero, while $$c_{2}(t)$$ smoothly increases from zero to one. So everything happens continuously, and there are no jumps. (Meanwhile, the expected number of photons in the electromagnetic field also smoothly increases from zero to one, via continuous superpositions of zero-photon and one-photon states.)

Do electrons really perform instantaneous quantum leaps?

The state of the electron (and the atom as a whole) evolves from the ground state to the exited state smoothly. The probability distribution of the electron will likewise smoothly change over time.

It is very hard to imagine this in a classical way, and to describe it with classical motion.

For the sake of argument, let's imagine that you move to a new house next door. The GPS map (let's say we record your location every day a few times and then look at it later) showing your location (based on your GPS signals) will first show you to be located mostly (with the highest probability) at (in the vicinity of) the old house. Then, the map will start showing your location sometimes at (in the vicinity) of the new house (you go there, check it out). Then, your location on the map will be more and more at the new house, and less and less at the old house (as you move over). After a while, your location on the map will be mostly (with the highest probability) at the new house. You could still show up at the old house of course (if you forgot something), but you have transitioned to the new location.

• In your analogy, you walked continuously from one house to another, occupying the space in between. So then the electron does the same?
– Evan
Commented Aug 18, 2021 at 21:38
• You are saying that it's a continuous transition from one state to the other. But can we somehow think of by help creation and annihilation operators. Like while transition a electron(when energy is supplied to it) is annihilation at the ground state and created at 1st excited state.? Commented Aug 18, 2021 at 21:47
• @Evan that would be true for a classical particle, but not for an electron, which is a quantum object. Classical trajectories are hard to explain for electrons, please see here physics.stackexchange.com/questions/649238/… Commented Aug 19, 2021 at 0:33
• @Evan The analogy is not about continuously walking once from one house to the other. It's about the process of you relocating your address to the other house over the course of a few weeks. "(let's say we record your location every day a few times and then look at it later)" The point of the analogy is that you can't continuously observe an electron, you can only take snapshots of it at specific instants of time every once in a while. If you do a "still-motion" montage of those snapshots, the smooth transition from mostly one state to mostly another will emerge. Commented Aug 19, 2021 at 16:15

Energy levels are eigenstates of the Hamiltonian, so their probability density remains constant in time. But a superposition of two (non-degenerate) Hamiltonian eigenstates has time-dependent relative phase, which makes the density change with period $$\propto E_1-E_2$$. Speaking classically, this means something like an oscillatory motion.

A resonant transition between two energy levels happens through gradual increase of the amplitude of the target level and corresponding decrease of the amplitude of initial level. After the target level has amplitude $$1$$ and the initial level's amplitude goes to $$0$$, if no decoherence happens, this system will transition back to the initial state and then repeat the process again in Rabi cycles.

So, the transition from the ground state will begin by a small oscillation of the peaks of the probability density. Then this oscillation becomes larger, and after some time the probability density peaks from the target state appear. These peaks grow, while those of the ground state diminish. After some time the oscillation stops, so that we get the probability density of the target state. Then the reverse transition happens, unless something inhibits it.

Classically this looks a bit counterintuitive: the most active "motion" (change of probability density in time) happens during the transition, not when the system completely gets to the target, more energetic, state. And the initial and final states look "motionless", because they are stationary.

• @Rusian I will read more about the Rabi Cycle, but in way we can term with help of creation and annihilation operators? Commented Aug 19, 2021 at 5:33

In classical electromagnetism, an electron in an orbit due to the Coulomb force around a proton ( the hydrogen atom), would be continually in motion, because of the angular acceleration it would radiate its energy and slowly the radius of the orbit would would become smaller, until the electron fell on the proton. No hydrogen atom, no atoms in general. As experimentally atoms and molecules exist, it became necessary to abandon classical electromagnetism at the micro level. Bohr by imposing ad hoc quantization of the angular momentum, manage to reproduce the spectrum of Hydrogen, creating stable orbits at a given radius.

Thus in the Bohr model, the electron moves from one level to another and the energy/momentum lost or gained is taken or given by an appropriate energy photon.

The Bohr model was superseded by the Schrodinger mode for the Hydrogen atom, which gave the same prediction for the spectra, but has a completely different mathematical basis. There is though correlation between the hypothetical orbits of the Bohr model and the orbitals of the Schrodinger solution.

The orbitals are the loci where the electron with the given (l,n,m) has a finite probability to be found in a measurement, but the average should conform with the Bohr radii. See this for example.

When discussing quantum mechanical states , the concept of motion is not relevant. Look at the hydrogen orbitals"

Fig.6 Representation of the hydrogen 1s orbital

When a photon is absorbed, it is one point in this plot for the electron location, it can be anyplace within the orbital, and occupy any similar single point in the orbital it ends up . It is only averages that could have any meaning.

• What I think the op really wants to know is how to describe the position of electron when it is in transit from one shell to another. Commented Aug 18, 2021 at 18:37
• @MRKhan quantum mechanics is about probabilities. The only mathematical description we have. So we can calculate the probability of transition to an orbital, but individual electrons cannot be mathematically described, and as mathematics is what we have to "see" inside a quantum system.... The only 'visual" tool is the Bohr model. There one can "see" the electron dropping by a photon going away. Commented Aug 19, 2021 at 3:36

Disclaimer: I'm not a physicist, just an interested layman.

According to the YouTube video "What Happens During a Quantum Jump", experiments indicate that the transition is a continuous process.

My intuitive explanation is that, as an electromagnetic wave propagates through an atom, the stationary quantised electric wave of the atom (the "electron") interacts with (a quant of) the electromagnetic wave (a "photon") and enters a transitional, superimposed state in which you cannot really talk about electrons and photons as separate entities. Both the electron and the photon (the waves!) are smeared over the space occupied by the atom, and probably beyond. Based on the wavelength of the visible light absorbed by atoms and the speed of light, my guess is that this transitional state lasts about $$10^{-14} s$$ (at least one, probably several full periods of the electromagnetic oscillation). Eventually, the system (electron, nucleus, photon) settles down in a different energy configuration, with a higher potential energy for the electron.

The best "classical" analogy I can think of is as of a droplet of water (an "electron") sitting on a surface ("energy level"). When hit by thermal energy (a "photon"), the droplet evaporates, remains in the evaporated state for a while, and eventually condensates on a higher surface (higher "energy level"). In the evaporated state you cannot really say that "the droplet" is moving in space, because it doesn’t exist. But, yes, the molecules of water (the probability density of the electron) move around. However, I admit that the analogy is imperfect, for many reasons.

• I usually follow PBS, but I haven't seen this video before, so like he said it's open for interpretations and new development. Thanks btw. Personally, I also think it should be a continuous spectrum as random discrete theory is kind of hard to accept. Commented Aug 19, 2021 at 13:32