# Photons, light and electricity

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.

• No, photons are not turning the electric motor. Do you understand the difference between electrical current flow in a wire and light transmission in space? Jan 1, 2019 at 8:48
• Electricity when flowing through a wire depends on the drift velocity of electrons.The instantenous transmission through the wire is the first photon nudging the second electron which in turn nudges the next electron But when we see electrical force in action like a motor turning--photons in some way must be involved as they are the force carrier particles.Light when flowing through space is like a wave and the wave function collapses when we try to measure it by hitting it with a photon.Then light behaves as if it is composed of photons. -- please enlighten my ignorance Jan 1, 2019 at 9:06

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 $$U = \int dx\ \frac{\sum_k E_k^2 + \sum_{i Here's the point: given a function $$\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.

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.

• It seems from your answer that virtual photons are involved in conduction of electricity while real photons are involved in the transmission of light through space.Am I correct?Is there any relation to energy levels?Could you illustrate your explanation with Feyman diagrams? Jan 1, 2019 at 15:57
• Most of the electrons cannot feel the E field due to Pauli's exlusion principle. They have to satisfy Fermi-Dirac statistics, and only those near the Fei surface can be affected by the applied E field. Jan 1, 2019 at 17:23
• I appreciate the difficulty in using words to describe all this .. but I will make a comment anyway: if I have two macroscopic charged bodies electrostatically attracted by exchange of many virtual photons, in what sense can I not observe them directly e.g. I can insert a charged test particle between and it will experience the electric field. Your caveats insure you against answering, but just in case you had some more insight on this - thanks. Jan 1, 2019 at 20:05
• @BruceGreetham, I recommend you search here on keyword "virtual particle", lots has been written about how they work and what they "are". Wikipedia is also useful. My take on this is that for my intents and purposes, the charged objects interact via the field- which can be detected, measured, and mapped- but at the level of subatomic particles, the most convenient tool that physicists use to account for force transmission is instead the virtual photon. Feynman's book on this topic called "QED" is a good one to read. Jan 1, 2019 at 20:23
• Casimir force has been experimentally confirmed. This is directly the consequence of the presence of virtual photons between two plates of a charged capacitor. Jan 7, 2019 at 10:43