Electron "clouds" in an atom

Question

I have a very naïve understanding of atomic structure and quantum mechanics. But from what I have heard about the double slit experiment with an electron, we can say that if we shoot photons on the electron to detect its path, the pattern goes away as if the electron starts behaving like a particle.

But also every textbook writes that the electrons around the nucleus are not localised but present in the form of clouds (probability clouds).

But isn't the electron constantly interacting with the nucleus (i.e. exchanging photons)? So why wouldn't it behave like a particle and be localised all the time?

Answer 1

I like this question, because it combines 3 intro-level quantum analogies that are fine for chemists, but are insufficient for physicists.

Let's start with the orbital: it is not a probability cloud. This gives the impression of a point electron bee-bopping around the atom, and if you were to look where it is, you would just find it where it is, with a probability distribution.

That is problematic. 1) because you can't look where it is. It's a measurement that can't be done. It's a thought experiment. 2) because a probability cloud is classical, and hence: wrong.

The electron is described by a wave function which is the complex square root of a probability cloud--whatever that means. I recommend using it as a tool to do calculations, and then maybe think about what it means later.

(On a side note, it also gives the impression that the nucleus is a rock solid point, with the electron doing all the quantum stuff--no, we solve the hydrogen atom in reduced mass coordinates: the proton is also in a matching wave function)

Then we have the part about the electron exchanging virtual photons with the nucleus. That is a quantum field theory thing, so hold on.

The probability cloud also gives the impression that the electron is moving. An important part of eigenstates of the Schrödinger equation is that they are stationary states: they never change, other than an arbitrary unobservable global phase factor that rotates at $E/\hbar$. Now there is a probability current (that is not velocity), that rotates around the nucleus in $L\ne 0$ states, but it is also stationary.

Now the virtual photons: the quantum mechanical solution to atomic orbitals is an approximation in a static background electric field. When you "turn the field on", e.g., make it a dynamical variable, several things happen:

The eigenstates are no longer exact. There is an interaction hamiltonian that links states, and hence they can transition. They are no longer stationary states, and the energy is no longer exact. This is good: it allows atoms to change.

So now we no longer have an exact analytical description of an atom, since the system includes the EM field out to infinity. It's too much.

Nevertheless, you can still break it into approximate pieces and do calculations. This involves perturbation theory, and that is described in powers of $\alpha$ (maybe squared), and these terms can be written with diagrams of virtual particles. This does not make virtual particles real. They are just pieces of an approximation to a quantum field linking states.

They are useful, when used properly, and cause all kinds of confusion when taken seriously, esp. in a non-relativistic fashion where they "borrow energy" from the vacuum (in relativistic QFT, they don't do that).

Finally: the double slit (YDSE). The amplitude to go from initial ($i$) to final state ($f$) is the sum (or path integral) of all possible paths linking $i\rightarrow f$, and in the 2 path approximation to the YDSE, you add the amplitudes for both paths and get interference.

If you observed the electron going through one slit, then that is the path it takes, so you can't add any other amplitudes. It doesn't matter how you measure it: photon, entangled partner, whatever. If you know it, there is no 2-slit interference pattern.

Now when it goes through one slit, you still don't know where it went through the slit, which has a finite width. In that case, you have to integrate over all positions in the slit, et voila: that gives you the Fourier transform of the slit, and that is a diffraction pattern.

So the takeaway is: don't take classical description of QM too literally, don't take virtual particles too literally, and know what approximations you're description of a system is making (and you have to make them, or the problem cannot be solved).

Answer 2

Short answer: Because the electron is entangled with the nucleus.

Longer answer: For simplicity, consider the decay of a pion particle. The pion has spin zero. If you don't know what spin is, don't worry. The important thing is that each particle can carry a certain number we call spin and that the total spin is always conserved. The pion decays into two particles (call them particle A and particle B), each of which have a spin with magnitude $1/2$. Since the pion has a spin of zero and spin is conserved, the spins of the other two particles have to add up to zero. So either particle A has spin $1/2$ and particle B has spin$-1/2$, or particle A has spin $-1/2$ and particle B has spin $1/2$.

You obviously don't know which spin particle A is going to have until you measure it. And once you measure it, you automatically know the spin of particle B. Nothing strange about that.

The strange thing is this: Before you measure particle A's spin, the particle itself doesn't know which spin it has. It is in a superposition of spin $1/2$ and spin $-1/2$, and so is particle B. So the whole system is in a superposition of the state (spin$_A$ = 1/2, spin$_B$ = -1/2) and the state (spin$_A$ =-1/2, spin$_B$ = 1/2)

This phenomenon is called entanglement. The point that when you have two entangled particles, you cannot treat them as individual systems.

Let's apply this to a hydrogen atom. For simplicity, we will do this in the framework of quantum mechanics, not quantum field theory.

First, think about the hydrogen atom classically. The electron orbits the nucleus, but the nucleus is not really stationary. The nucleus and the electron both orbit around the system's center of mass. So classically, the position of the nucleus will depend on the position of the electron.

When we quantize the system, this is still true, except we know think of the state of the electron as a superposition of all possible classical states. But each classical state involves the electron being in a certain position and the nucleus being in a certain position. Thus, you have to describe the whole system as a single wavefunction.

And it only becomes localized when we measure it.

Some people claim that when you measure the position of the electron, you also get entangled with the system. Meaning, there is now a version of you which saw the electron at position 1, and another version of you which saw the electron at position 2. This is the core idea behind the many-worlds interpretation of quantum mechanics.

Answer 3

An isolated quantum system described by a wavefunction governed by an equation of motion, such as the Schrodinger equation: $$i\hbar {\frac {\partial }{\partial t}}\Psi (\mathbf{r},t)=\left[-{\frac {\hbar ^{2}}{2m}}\nabla^2+V(\mathbf{r},t)\right]\Psi (\mathbf{r},t)$$ If you drop the time dependence in both the potential and the wavefunction you get a time independent equation. If you solve that equation for the Coulomb potential and the resulting wavefunction can be displayed to look like the clouds you see in many illustrations.

Now if you measure an electron during an interference experiment, then information about the electron spreads to other systems often called the environment, including whatever detector you use to measure the photon, and this produces an effect called decoherence that suppresses the interference:

https://arxiv.org/abs/quant-ph/0105127

https://arxiv.org/abs/1111.2189

Decoherence doesn't make quantum systems into particles it just suppresses interference on macroscopic scales for systems that interact heavily with the environment, like the objects you see around you.

How does this fit in with electron clouds? If an atom is interacting relatively weakly with its environment, then the evolution of the atom is dominated by its self Hamiltonian that describes the interaction between the proton and the electron. In these circumstances, decoherence tends to lead to the system being in an energy eigenstate, which doesn't change over time:

https://arxiv.org/abs/quant-ph/9811026

So treating the electron using the strategy outlined at the start of my answer can shed some light on features of real atoms. You can improve the approximation by using relativistic equations and introducing corrections for multiple electrons and so on.

Answer 4

Your question has received many answers, some of which contain part of the information blurred by additional and unnecessary concepts (virtual particles, photons, decoherence, entanglement, perturbation theory, path integrals, quantum field theory, and so on). Therefore, I'll try to use only the necessary concepts.

Particles with mass can be described in the position representation of ordinary Quantum Mechanics (QM) by a wavefunction $\Psi$, i.e., generally, by a complex wavefunction of the space coordinates of the particles and time. The wave function, if normalized, provides a probability density function in the space of particle coordinates through its modulus squared. The probability density is interpreted as a weight assigned to each small region around some values of the spatial coordinates. From such a weight, we can extract the probability that a measurement of positions performed with high precision on many equally prepared systems will find particles around those values of the coordinates. Ideally, each measurement can be performed with arbitrary precision. This is not entirely correct. There are reasons it is impossible to measure positions with a precision larger than the Compton length of a particle, but for an electron, such a quantity is much smaller than the typical orbital size.

Notice that such a description is the statistical interpretation of the wavefunctions, independent of the particular interpretation scheme of QM. Particles may have additional degrees of freedom beyond position, such as spin, but we can neglect them for a basic understanding of the electronic clouds around the nucleus found in textbooks.

The electronic clouds correspond to an important approximation of the many-electron wavefunctions, the so-called one-particle approximation. It corresponds to building the many-electron wavefunctions in terms of the solutions of the simplified problem where the electron-electron interaction has been neglected.

Summarizing, one-particle wavefunctions define a probability density for obtaining a value of one electron's position around a space point $\bf r$. In particular, the eigenstates of the problems correspond to stationary states, i.e., states where the probability does not change with time. The meaning of these one-particle wavefunctions implies that if we prepare many systems in the same state (same $\Psi$) and perform many independent experiments to detect the electron position, every time we obtain a point and the set of such points provides a representation of the probability density.

It is important to stress that we deal with independent localization experiments performed on equally prepared systems. The single system, after the electron position measurement will not stay in its original stationary state anymore.

Coming to the original question:

• the analogy with the double slit experiment is that it provides a visual illustration of what performing independent measurement of position on an equally prepared set of states means. It also shows that single measurements do not reveal a spatial spreading of the electron properties, but interference patterns result from individual point-like events. In this sense, the naïve duality (quantum particles behave like waves/clouds) has no real support from experiments. The dynamics of a quantum particle have a wave-like character. However, all the measurements on individual particles show well-localized quantities (in no measurement can one observe a fraction of the electronic charge).
• The usual cloud-like representation of orbitals in textbooks is a way to provide a visual insight into the region where the probability of finding an electron in a stationary state is higher/smaller. They are based on the probability density (then on $|\Psi|^2$) and represent the outcome of many independent ideal position measurements on equally prepared systems.
• The interaction with the nucleus, for hydrogenic wavefunctions, has already been considered by solving exactly the spatial Schrödinger equation. Therefore, we cannot introduce perturbation theory with virtual photons. In a way, the interaction with the nucleus localizes the electron but not over a point-like region. The only meaningful localization is the one described by the stationary wave function, i.e., the typical size of the atomic orbital. Such localization in a stationary state does not prevent the possibility that on a system prepared in such a state, one performs an experiment to determine one electron's position with the maximum precision. We have only to consider that the original stationary state will be lost forever after that measurement.

Answer 5

I think the confusion comes from how people talk heuristically about photons. I recommend you reading: What exactly is a photon?.

From the perspective you have in a usual quantum mechanic course, you don't think at all about photons. You rather think about an electron on an electrostatic potential. The axioms of quantum mechanics tell you that the electron gets localised when you try to measure its position, but you can't do that from its interaction with the nucleus.

You could in principle try to have a localised electron trying to measure its position with an external electromagnetic field (heuristically, that's somewhat analogous to what you described as "shooting photons"). Note however that quantum mechanics tell you that a localised electron would not give you a stable atom. The reason why we think of orbitals is that these are stable states (more technically, they are time independent solutions of the Schrödinger equation). If at any time you had a localised electron in an atom, it would probably start moving non-trivially and thus emitting radiation and losing energy (this was actually what motivated Bohr's and Schrödinger's models of the atom). So quantum mechanics tells you that you need non-localised electrons if you want stable atoms.

The concept of "particles exchanging photons" is usually associated with scattering processes in particle physics (i.e. when two free-traveling electrons collide, for example). The equations you need to solve for interaction processes are non-linear and hard (if not impossible) to solve analytically, so you usually solve them by some iterative or perturbative method. It is when you do this (it is called perturbation theory) that in each step/term of the computation, you get some mathematical expression which may be interpreted as particles exchanging photons, i.e. a Feynman diagram. But as you may see, the concept of "electron exchanging photons" comes from a very different perspective than that of an electron in an electrostatic potential. This picture actually is already present at classical field theory (such as Navier-Stokes for example), see here.