How is Coulomb's Law compatible with quantum physics?

[Mayan Fung]

TL;DR

How to obtain the Coulomb potential if we treat the electron in the quantum regime

In classical physics, we treat an electron as a point charge with a Coulomb potential $V = \frac{q}{4\pi\epsilon_o r}$.

However, in quantum mechanics, we treat it as an electron cloud. In this situation, how shall we describe the Coulomb potential? Shall we treat the electron as a charge distribution with the density $\rho(\vec r) = qP(\vec r)$ where $P(\vec r)$ is the probability density at $\vec r$?

If that's the case, then when we make an observation and the wave function collapses, the electron is certain to locate at some position similar to a point charge. In order words, there is a sudden change in the Coulomb potential or the electric field. Does it imply that our measurements of an electron's position can have an effect on other charges? (Note that I am not focusing on the propagation speed of the changes. The change in the electric field can be restricted under the speed of light without affecting the above discussions)

 

[atyy]

A simple example is the helium atom, which has 2 electrons.
http://www.umich.edu/~chem461/QMChap8.pdf (Eq 3)
https://quantummechanics.ucsd.edu/ph130a/130_notes/node372.html

 

[Mayan Fung]

@atyy , I see your point. But how about when we perform a measurement so that the wavefunction collapse? The electrons will be like a delta function at a particular position

 

[atyy]

If that's the case, then when we make an observation and the wave function collapses, the electron is certain to locate at some position similar to a point charge. In order words, there is a sudden change in the Coulomb potential or the electric field. Does it imply that our measurements of an electron's position can have an effect on other charges? (Note that I am not focusing on the propagation speed of the changes. The change in the electric field can be restricted under the speed of light without affecting the above discussions)

When the wave function collapses (and the state after collapse depends on the measurement being made), the Hamiltonian (that governs the time evolution of the wave function) before and after collapse remain the same. So it depends on what one means by "effect".

 

[PeterDonis]

Summary:: How to obtain the Coulomb potential if we treat the electron in the quantum regime

when we make an observation and the wave function collapses

Whether "the wave function collapses" is a real event or just something that happens in the math depends on which interpretation of QM you adopt. But there is no need to treat it as a real event in order to make accurate predictions.

Summary:: How to obtain the Coulomb potential if we treat the electron in the quantum regime

the electron is certain to locate at some position similar to a point charge

No, it isn't. First, not all measurements are measurements of position. Second, even for a measurement of position, the measurement cannot have infinite accuracy, so the result will not be that the electron is certain to be at a single point. (More precisely, the wave function in the position representation that we assign to the electron after the measurement will not be a delta function; it will be a Gaussian with a spread that depends on how the measurement was made.)

 

[atyy]

@atyy , I see your point. But how about when we perform a measurement so that the wavefunction collapse? The electrons will be like a delta function at a particular position

Just apply the Rules of Quantum Mechanics. After the collapse, you can use the Schroedinger equation to predict how the wave function evolves.

 

[PeroK]

Summary:: How to obtain the Coulomb potential if we treat the electron in the quantum regime

In classical physics, we treat an electron as a point charge with a Coulomb potential $V = \frac{q}{4\pi\epsilon_o r}$.

However, in quantum mechanics, we treat it as an electron cloud. In this situation, how shall we describe the Coulomb potential? Shall we treat the electron as a charge distribution with the density $\rho(\vec r) = qP(\vec r)$ where $P(\vec r)$ is the probability density at $\vec r$?

If that's the case, then when we make an observation and the wave function collapses, the electron is certain to locate at some position similar to a point charge. In order words, there is a sudden change in the Coulomb potential or the electric field. Does it imply that our measurements of an electron's position can have an effect on other charges? (Note that I am not focusing on the propagation speed of the changes. The change in the electric field can be restricted under the speed of light without affecting the above discussions)

This is the sort of confusion that arises when you mix concepts from two theories. The theory of classical EM does not involve the electron being described by a wave function at all. It must be considered a point particle. There is no obligation, therefore, on classical EM or Coulomb's law to fit perfectly with QM.

Instead, classical EM is a good approximation to QM in many circumstances.

If you've been following the news, there has been much publicity over the anomalous magnetic dipole moment of the muon (which is a heavier version of the electron). In classical EM the magnetic dipole moment of the electron and muon is $1$. Whereas, in basic full-blown QFT it is approximately $2.0023$. In that respect, classical EM cannot be reconciled with QM.

Moreover, Coulomb's law itself breaks down for high energy interactions between electrons and/or muons. Again, therefore, classical EM cannot be fully reconciled with QM.

 

[Demystifier]

Does it imply that our measurements of an electron's position can have an effect on other charges?

Yes. More precisely, measurement of electron position is correlated with properties of other charges. Physicists don't mutually agree whether this correlation should be called "effect" because this correlation cannot be used to send message.

 

[vanhees71]

You can also ask, what's the Coulomb field of a quantized electron. For this first of all you have to put the electron in some trap so that you have a true normalizable energy-eigenstate. You can use, e.g., a 3D harmonic oscillator potential and the electron in the ground state, which is a Gaussian wave function. Then you can calculate the expectation value of the electric charge-current density. Neglecting spin the current density is 0 and the charge density is also a Gaussian, i.e., $\rho(\vec{x})=-e|\psi(\vec{x})|^2$. You can then get the electric field of this charge distribution by simple integration. You can also include spin and use the Pauli equation (non-relativistic leading-order limit for the Dirac equation). Then in addition you get a non-vanishing current density which is equivalent to the magnetic moment associated with the electron's spin (including the gyro-factor 2, when "gauging" the free electron Hamiltonian correctly via minimal coupling).

As you see, quantum theory smoothes out the artificial singularity you get from the classical-point particle point of view. One should be aware that classical point particles are real strangers in electromagnetism though the concept is quite successful until you ask for a fully consistent dynamical theory for the electromagnetic field and point charges, including radiation reaction.

It's of course a semiclassical picture, treating the electron as quantum particle and the electromagnetic field as a classical field. If you take full QED, you can solve the problem with the radiation reaction at least in a perturbative sense since QED is a Dyson-renormalizable relativistic QFT. Then you can renormalize all quantities, including the propagator of the electromagnetic field, which gives at tree level the Coulomb field for the static case $k^0=\omega=0$. Then you can add loop diagrams to the photon self-energy (aka vacuum polarization diagrams) to get the modifications of the Coulomb field due to these radiative corrections.

 

[Twigg]

Summary:: How to obtain the Coulomb potential if we treat the electron in the quantum regime

In order words, there is a sudden change in the Coulomb potential or the electric field. Does it imply that our measurements of an electron's position can have an effect on other charges?

If you have two non-interacting (distinguishable) charges (e.g., an electron and a proton that are very far away from each other at the time of the measurement), their wavefunction before measurement will be separable: $$\psi(\vec{x_1},\vec{x_2}) = \psi_1 (\vec{x_1}) \psi_2 (\vec{x_2})$$
This is because the Hamiltonian for the two non-interacting particles lacks any kind of "mixing" term and is merely a sum of two independent free-particle Hamiltonians for each charge: $$H = \frac{p_1 ^2}{2m_1} + \frac{p_2 ^2}{2m_2}$$
With the above wavefunction, a measurement of $\vec{x_1}$ only projects $\psi_1(\vec{x_1})$ and leaves $\psi_2(\vec{x_2})$ unchanged.

However, when the two charges interact, the Coulomb potential acts as a mixing term: $$H = \frac{p_1 ^2}{2m_1} + \frac{p_2 ^2}{2m_2} + \frac{q_1 q_2}{4\pi \epsilon_0 |\vec{x_1} - \vec{x_2}|}$$
And with this new "mixing" term, the Hamiltonian can no longer be written as a sum of individual particle Hamiltonians, and the wavefunction stops being separable: $$\psi(\vec{x_1},\vec{x_2}) \neq \psi_1 (\vec{x_1}) \psi_2 (\vec{x_2})$$
As a result, any projection of $\vec{x_1}$ affects the outcomes for $\vec{x_2}$ (in other words, the two charges' positions are correlated).

This is exactly what @Demystifier was saying, just little more long-winded. Also, this very much assumes an approximate, semi-classical picture as others have indicated.

P.S. If the charges are indistinguishable, then they are correlated from the get go. The helium atom papers that @atyy shared cover this nicely.

 

[Vanadium 50]

Just apply the Rules of Quantum Mechanics.

I don't think he knows the Rules of Quantum Mechanics. I don't think he can do a calculation. If he could, he'd just do it and wouldn't be asking this question.

I think he's trying to knit together a consistent picture from a bunch of popularizations. That's not going to work.