Is the electron considered to be at rest within an atom?

Question

According to older models of the atom, the electron moved around the nucleus at great speeds. If this were so, would this not mean under the laws of special relativity that it would cause significant accumulation of its mass? I am assuming under current theories that the electron is considered to be at rest WRT to the nucleus and therefore exempt from any such effects.

Answer 1

Let's consider Hydrogen for concreteness. Then, when the Hydrogen atom is in its ground state, the electron's probability is mostly bound within a distance $a_0$ of the nucleus of the Hydrogen atom, where \begin{equation} a_0 = \frac{\hbar}{\alpha m_e c} = 5.3 \times 10^{-11}\ {\rm m} \end{equation} where $\hbar$ is Planck's constant, $c$ is the speed of light, $\alpha$ is the fine structure constant, and $m_e$ is the mass of the electron. The length scale $a_0$ is the Bohr radius.

Because the electron is localized to a region of order $\Delta x \approx a_0$, because of the uncertainty principle $\Delta p \Delta x >\hbar/2$, there must be a corresponding uncertainty in momentum \begin{equation} \Delta p > \frac{\hbar}{2a_0} \end{equation} In fact, in a more careful calculation of the variance of the electron's momentum in Hydrogen's ground state, it turns out that the electron approximately saturates this bound.

From this information, we can estimate the "velocity" of the electron (I put scare quotes around velocity for reasons we'll get to in a minute) \begin{equation} v \approx \frac{\Delta p}{m_e} \approx \frac{\hbar}{2 a_0 m_e} = 1.1 \times 10^6\ {\rm m\ s^{-1}} = 3.6\times 10^{-3} c \end{equation} So in fact, the electron is not moving relativistically in a Hydrogen atom. There are some relativistic effects that are visible in the fine structure of Hydrogen, but these are small effects.

Now, let's get back to the scare quotes. It's important to recognize that a quantum particle does not have a trajectory $x(t)$ like we have in classical mechanics, so the concept of velocity as the time derivative of a trajectory simply doesn't exist in quantum mechanics. Being able to define both the particle's position and velocity precisely violates the Heisenberg uncertainty principle.

However, we can define the (magnitude of the) "velocity" in analogy to classical mechanics as "a measurement of the (magnitude of the) particle's momentum, divided by its mass." In this sense, the electron is not at rest, since a typical measurement of velocity would give a value of order $3.6\times 10^{-3} c$, rather than $0$. In fact, the uncertainty principle tells us that since we have localized the position of the electron within the Hydrogen atom, there must be some spread in momenta (or velocities) that include a non-zero value. The only way to have a quantum particle perfectly at rest -- meaning that a measurement of its momentum would always yield zero -- would be to have a completely de-localized particle. (Actually, that's the only way to have any definite value of momentum, not just zero).

Answer 2

The reason the planetary model of the atom would not work was that in classical electrodynamics the rotating electrons would radiate away their kinetic energy and fall on the nucleus , so no atoms could exist. Bohr in his successful model that explained the Balmer series of the hydrogen spectrum, imposed by hand angular momentum quantization . (second page).

This model is outdated by the theory of quantum mechanics which solved the hydrogen atom using its postulates. In quantum mechanics there are no orbits for the electrons to have to decide on their velocity . There are orbitals, given by the probability of the electron to be in a specific (x,y,z,t), see here for experiment..

Thus your question

Is the electron considered to be at rest within an atom?

"At rest" has no meaning within the successful theory of quantum mechanics. The relativistic effects will be inherent in the wavefunction solution for the particular atom, and will affect the orbitals (probabilities) accordingly.

Answer 3

Electrons do indeed move in atoms at great speeds (thousands of meters per second, as you can see from this answer). However, it does not mean they "accumulate mass", since fast-moving objects don't gain mass in special relativity. See this question. The point is that if you say "the object gains mass", then some Newtonian formulae retain their familiar form. However, this leads to confusing paradoxes,* so most modern authors have stopped teaching this.

Relativistic effects will still be present, of course, which shows up experimentally as fine structure.

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*For example, say there's a weighing machine, and if its reading exceeds a certain value, a bomb connected to the weighing machine goes off. We put the weighing machine on a spacecraft moving at 0.99c. To you as the lab-frame observer, does the bomb go off or not? (The answer is "no", even though one might naively think it is yes, if using relativistic mass.)

Answer 4

The motion of an object is an idea, which reaches deeply in classical physics. By describing an objects by its position and momentum we were able to predict the position of the planets, as well as do ballistic calculations. However, this idea of motion is not part of the basic building blocks of quantum mechanics. In fact, in quantum mechanics it is not meaningful to ask, what an object does, if we are not observing it. Quantum mechanics does not tell us what the object does, but only tells us the probability of observing a given measurement result. Therefore, the wave function is all we have as a description of the electron, which is part of an hydrogen atom -- there is no mechanical model behind quantum mechanics, which tells us why the electron behaves this way.

Think of the two-slit experiment: There is no apparatus, which allows us to tell through which slit the object (=light, electron, ...) went without destroying the interference pattern on the screen. If we which to maintain the interference pattern, we are unable to determine what the object does between its emission and our observation on the screen. This idea can be transferred to your question: If we were able to continuously observe the electron inside the hydrogen atom, we would alter its wave function. Therefore, we should expect that the obtained result differs from the one we obtain, if we are not continuously observing the electron. Does this make sense to you?

Answer 5

To answer your question I'm going to have to describe some aspects of how quantum theory describes systems.

In classical physics the position of a particle is described by a position vector $(x(t),y(t),z(t))$ and this vector is also the result of measuring the particle's position. Similarly, the momentum of a particle is described by a vector $(p_x(t),p_y(t),p_z(t))$ that gives the result of measuring the particle's momentum.

In quantum theory the situation is very different. The position of a quantum particle is described by a set of matrices $(\hat{x}(t),\hat{y}(t),\hat{z}(t))$ called observables and the possible measurement results are eigenvalues of these matrices. The momentum is also described by a set of observables $(\hat{p}_x(t),\hat{p}_y(t)(t),\hat{p}_z(t))$ whose eigenvalues are the possible measurement results. In general the outcomes of experiments on quantum systems depend on what is happening to all of the possible states of the system and you can't describe this by just describing the quantity you're interested in using a single number. See section 2 of this paper for an example:

https://arxiv.org/abs/math/9911150

Quantum theory predicts the expectation values of observables. For each possible outcome of a measurement $a$ quantum theory predicts a probability $p_a$ and the expectation value is $\sum ap_a$.

When information is copied out of a quantum system interference is suppressed and this effect is called decoherence:

https://arxiv.org/abs/1911.06282

Decoherence selects a set of possible states for a system. In everyday life the systems you see around you have information copied out of them a lot faster than they change so interference is very effectively suppressed and on the scales you can see they can be described by classical physics to a good approximation. Electrons bound to an atom are interacting a lot more weakly with their environment and they do undergo interference so it is not a good approximation to look on them as having a single position or momentum. When a system interacts weakly with its environment, decoherence tends to select energy eigenstates:

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

Those energy eigenstates don't change over time and so the expectation values of the observables don't change. However, the possible measurement results of an electron might sometimes include momenta that correspond to the electron moving at relativistic speeds. There are aspects of chemistry that are explained by electrons having relativistic momenta such as the colour of gold:

https://www.researchgate.net/profile/Pekka-Pyykkoe-2/publication/221690646_Relativistic_Effects_in_Chemistry_More_Common_Than_You_Thought/links/54f6bb550cf2ca5efefefda6/Relativistic-Effects-in-Chemistry-More-Common-Than-You-Thought.pdf

Answer 6

Several answers given are quite to the point but the answer can be formulated in more precise terms in my opinion.

There are two questions here. Is the electron, say in the hydrogen ground state, at rest according to the Schrödinger equation (Q1)? Are there any relativistic effects in an atom (Q2)?

Q1. The Schrödinger equation tells us that the electron is associated with a kinetic energy distribution $E_k( r ) = \psi {p^2 \over 2m} \psi$, so it is not at rest. Note that $\psi$ is real and only depends on $r$. The distribution of ${\bf p}$ is zero, because at any point it is equally likely that ${\bf p}$ is found to be parallel or antiparallel to $\hat z$. It must be aligned with $\hat z$ because $l^2=0$.

Q2. The kinetic energy of the electron on average is 13.61 eV. Its average potential energy is -27.2 eV. This is only 2.7 $\times$ 10$^{-5}$ times its rest energy. This still corresponds to formidable speeds but the relativistic effects are very small and neglected in the Schrödinger equation.

Note that for heavier atoms relativistic effects may become significant and a relativistic equation, such as the Dirac equation, is needed.

Answer 7

No. Electrons are not considered to be at rest within an atom. In quantum mechanics, electrons occupy atomic orbitals, which are probability distributions rather than fixed paths. These orbitals represent regions where an electron is likely to be found. This means that electrons are in constant motion within the atom and do not have a definite, stationary position.