Photons, light and electricity

Question

Light is ultimately composed of photons. Photons are also force carriers of the electrical force. When an electric motor is turning it is photons which are turning it. What is the relation between photons that make up light and the ones that make up the electric force? Are they just photons in a higher energy state? I am aware this question has been asked before-- and has been answered as being at a different energy level. I would like a more detailed mathematical answer.

Answer 1

Trying to explain all electromagnetic interactions in terms of photons is like trying to explain all atmospheric phenomena in terms of tornadoes.

The foundation for our current understanding of electromagnetic interactions is not expressed in terms of photons at all. It is expressed in terms of quantum fields. Photons are just one of many phenomena that can occur, like tornadoes are just one of many phenomena that can occur in the atmosphere. The electric force between two electrons is not naturally expressed in terms of photons, just like other atmospheric phenomena (calm weather, clouds, thunderstorms) are not naturally expressed in terms of tornadoes.

There is one practical issue, though. We humans are not mathematically omniscient: we don't automatically know the correct answer to every well-posed mathematical question. Instead, we must usually resort to awkward and complicated computational procedures in order to extract approximate solutions. In the case of quantum field theory, one of those awkward-and-complicated approximation schemes — the one that dominates almost all of the literature — is a scheme that uses mathematical devices called "virtual particles" (including "virtual photons") to transform otherwise intractable equations into something that we mortal humans can handle with a humble pencil. It is an intellectually painful compromise, just like expressing all atmospheric phenomena in terms of "virtual tornadoes" would be an intellectually painful compromise. The intellectual pain is partly alleviated by using a pretty diagrammatic notation ("Feynman diagrams") for the various terms in that awkward-and-complicated approximation scheme, but that's like swallowing a pill to suppress the pain of a broken arm. It makes the experience more tolerable, but it doesn't address the root cause of the pain.

So, instead of trying to describe the force between electrons in terms of photons, I'll outline an alternative approach, one that starts with the foundation of quantum electrodynamics itself, in which both phenomena — photons and the force between electrons — are expressed in terms of the quantum electromagnetic field. Graphically:

We still need to make an approximation if we want calculations to be manageable, but we don't need to express everything in terms of photons. Here's an outline of what we can do instead.

We can use an approximation in which the electrons are immobile. This might seem too drastic, but remember that we can express "force" in terms of the gradient of potential energy. So if we calculate the energy of a configuration of immobile electrons as a function of the distance between them, then we have implicitly determined the force between them, and we can infer that if the electrons were mobile then they would be affected by this force in the usual way. With the immobile-electron approximation, the equations governing the quantum electromagnetic field become almost as easy to handle as Maxwell's equations. In fact, they are Maxwell's equations, except that the components of the electromagnetic field are replaced by operators that act on a Hilbert space (as usual in quantum theory). The electric components of the field don't commute with the magnetic components of the field, so they satisfy an "uncertainty relation" analogous to the position/momentum operators in single-particle QM. This non-commutativity is what leads to the "photon" phenomenon. More explicitly, the commutation relations between the electric field operators $E_k(t,\mathbf{x})$ and the magnetic field operators $B_{ij}(t,\mathbf{x})$ are $$ \big[E_k(t,\mathbf{x}),\,B_{ij}(t,\mathbf{y})\big]\propto (\delta_{ki}\partial_j-\delta_{kj}\partial_i)\delta(\mathbf{x}-\mathbf{y}). \tag{1} $$ (I'm writing the magnetic field with two indices because, mathematically, the magnetic field is really a bivector field. Only in the most-relevant special case of three-dimensional space can we get away with pretending that it is a vector field. To relate the bivector formulation to the usual "vector" formulation, just replace $B_{12}\rightarrow B_3$, $B_{23}\rightarrow B_1$, $B_{31}\rightarrow B_2$.) Except for the commutation relations (1), the equations governing the quantum electromagnetic field are just Maxwell's equations, which in the absence of current (immobile electrons) are \begin{align} \sum_k\frac{\partial}{\partial x_k} E_k\propto\rho \hskip2cm \frac{\partial}{\partial x_i} E_j \propto \frac{\partial}{\partial t} B_{ij} \\ \sum_i \frac{\partial}{\partial x_i} B_{ij} \propto \frac{\partial}{\partial t}E_j \hskip2cm B_{ij}\propto \partial_i A_j-\partial_j A_i \tag{2} \end{align} where $\rho$ is the charge density. As usual, the energy in the electromagnetic field is $$ U = \int dx\ \frac{\sum_k E_k^2 + \sum_{i<j} B_{ij}^2}{2}. \tag{3} $$ Here's the point: given a function $\rho$, such as the one that describes two point charges separated by some distance $r$, we can solve equations (1)-(2) almost as easily as we can solve the classical version of Maxwell's equations. The new twist is that the solution is operator-valued, not real-valued. Using this solution in (3) gives an expression for the energy operator with the given configuration of charges. Then we can find the state-vector $|\psi\rangle$ that minimizes the value of $$ U_0\equiv \frac{\langle\psi|U|\psi\rangle}{\langle\psi|\psi\rangle}. \tag{4} $$ The minimization ensures that the state $|\psi\rangle$ does not contain any background fields or electromagnetic radiation (which may or may not be in the form of photons!), so that the energy is due only to the electrostatic interaction between the charges. This gives us the potential energy $U_0(r)$ as a function of the distance $r$ between the charges, which implicitly gives us the force.

By the way, a similar approach is sometimes used in quantum chromodynamics (QCD) to study the confining force between quarks.

I've glossed over several details, but this is the basic idea. The message is that we can understand the force between electrons directly in terms of the quantum electromagnetic field, which is the same foundation that we use to understand other electromagnetic phenomena like photons.

Answer 2

Current in an electrically conductive wire consists of mobile electrons that are being acted on by an electric field and have thereby been set into motion. Since all the mobile electrons in the wire are acted on by that field, they all individually experience that force which is causing them to move, and they do not electrostatically communicate or transmit their motion to one another on the wire via real photons, like the ones that form beams of light in space. Those photons cannot travel through electrically conductive wires.

The transmission of electrostatic (coulombic) forces on the surface of a conductor or in free space is mediated instead by the exchange of virtual photons, which cannot be directly observed in any way and are often thought of as representing a calculational convenience without a real physical existence.

This is a simplified picture. The difference between real and virtual photons is a big deal in the world of physics and I invite others here to weigh in with their perspectives.