Quantum mechanical picture of "electrons orbiting the nucleus"

Question

Given the wavefunction, $\psi$ is explained as the flow of probabilities or in other words probability density over a certain region of space. In the case of electrons, say in $s$ orbital, the probability of its existence is distributed over the whole spherical region. Now, when we talk about orbits is it true that the electrons are literally orbiting around the nucleus or are they just chaotically whizzing "around" with no sense of direction unlike we'd classically understand an electron to "orbit" around a positive nucleus.

Answer 1

[...] literally orbiting around the nucleus or are they just chaotically whizzing "around" with no sense of direction [...]

Whenever I read statements like these, I often think the OP has probably jumped in at the deep end of QM, sadly without great ability to swim. The description of the electron in an atom is fairly advanced material and mathematically quite challenging.

For those reasons it is desirable to look at a much simpler quantum system and draw parallels with the hydrogenic atom.

Consider a particle of mass $m$ that is allowed to move only along the $x$-direction and its motion is confined to the region between hard and rigid walls located at $x=0$ and at $x=L$ (see figure below). Between the walls, the particle moves freely. This physical situation is called the infinite square well, described by the potential energy function $U(x)$:

$$0 \leq x \leq L \Rightarrow U(x)=0$$

and $U(x)=+\infty$ everywhere else.

1. Classical perspective (for large wells):

Since as $U=0$ inside the box, the total energy $E$ is:

$$E=K=\frac12 m\langle v \rangle^2$$

The particle bounces from left to right and back ad infinitum. Assuming the initial position ($t=0$) was known, the exact position and velocity of the particle is always known.

Note also that $E=0$ is of course also allowed (for a stationary particle)

2. Quantum mechanical perspective (subatomic wells):

If the well is sufficiently small, the above (Newtonian) description fails and Quantum Mechanics (QM) needs to used. It is a central postulate of QM that:

The wave function $\psi$ of a quantum system contains all the information of the system. The information is extracted by applying quantum operators to $\psi$.

In the case of the total energy $E$ we need to apply the Hamiltonian operator $\mathbf{H}$, so that: $$\mathbf{H}\psi=E\psi$$ More explicit and because $U=0$ inside the well: $$-\frac{\hbar}{2m}\nabla^2 \psi=-\frac{\hbar}{2m}\frac{\text{d}^2\psi(x)}{\text{d}{x^2}}=E\psi(x)\tag{1}$$ $$\psi(0)=\psi(L)=0\tag{2}$$ $(1)$ is the time-independent Schrödinger equation (TISE, here for $U=0$). Boundary conditions $(2)$ arise from the $U=+\infty$ condition at $x=0$ and $x=L$.

The solutions of $(1)$ and $(2)$ are: $$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\Big(\frac{n\pi x}{L}\Big)$$ $$E_n=\frac{n^2 \pi^2 \hbar^2}{2mL^2}$$ For $n=1,2,3,...$

The $\psi_n$ are the eigenfunctions of the TISE, the $E_n$ are its energy eigenvalues.

It's important to realise that $E=0$ is not allowed. The particle cannot be stationary in the well. This principle of non-zero energy is universal for quantum systems, including the hydrogen atom.

With regards to position of the particle we need to use the Born Rule:

$$P(x)=|\psi(x)|^2$$

where $P(x)$ is the probability density distribution.

In our case the $\psi_n$ are Real, so: $$P(x)=\psi(x)^2=\frac{2}{L}\sin^2\Big(\frac{n\pi x}{L}\Big)$$

Consider the graphical representation:

It's now clear that classical representations of the particle's 'trajectory' do not apply to quantum systems: from $n=2$ and higher, $P(x)$ shows zero points, known as $nodes$. The hydrogenic wave functions (see anna v's answer) also show nodes. The number of nodes equals $n-1$.

For a quantum system the position of the particle(s) cannot be known until measured. Only the probability density function can be calculated.

As regards the particle's velocity, we can use the quantum momentum operator $\mathbf{p}$:

$$\mathbf{p}=-i\hbar \nabla \psi$$

And of course: $p=mv$. So without even actually applying the momentum operator we can see that quantum velocity is a complex function.

Answer 2

look at these simple orbitals of the hydrogen atom, i.e. where the electrons can be found if looked for. (the calculation is for a negative Hydrogen , the 2 electron drawings.

It is seen that there is a probability of finding it, dependent on an "orbit type" location, but there is no continuity between points.

literally orbiting around the nucleus or are they just chaotically whizzing "around" with no sense of direction unlike we'd classically understand an electron to "orbit" around a positive nucleus.

It is only the probability that can be calculated, and though the probability is constrained about the nucleus in a symmetric way there is no way to calculate a classical orbit.

Answer 3

In quantum mechanics, the position of the electron is not defined until it is measured. And each time you measure the position of an electron in the atom (independently) you get a random outcome. Random in the sense that you can not predict where exactly the electron will be located before you measure it.

And if position itself can’t be defined prior to measuring it, we can’t really talk about trajectories. We have to do away with the concept of motion (trajectories) of electrons and have to talk about wavefunctions and how they evolve. Wavefunctions then are the objects that capture our entire knowledge of the state.

Answer 4

The question assumes that an electron is some kind of identifiable particle. However this has always been, and remains, one of the defining debates in quantum mechanics. Wave-particle duality comprises two mutually incompatible models for an electron, as a wave of possibilities (where $probabilty=possibility^2$) or as a particle. The standard or "Copenhagen" understanding is that the question is meaningless. Hidden-variable models assume a particle whose position is unknown but locating it statistically obeys the wave description. Others think of it as a wave which "collapses" when measured. Nobody knows the right answer, but you will sure meet some strong opinions along the way.

So provided your maths holds together it is entirely up to you whether you imagine there is a particle following a Newtonian-like orbit or whizzing around randomly or virtually everywhere at once or non-existent because the electron is currently a possibility wave or....