Does coulomb field contain photons?

As 1 THz and 1 kHz fields differ only in frequency, how does 0 Hz field differ from them?


No, it does not (admittedly, arguably). The Coulomb fields of charges set the part of the electric field that is irrotational (has no curl). The photons are excitations in the part of the electromagnetic field (electric field and vector potential) that has no divergence.

Consider the quantum harmonic oscillator (simple harmonic oscillator = SHO). It has discrete energy levels because the Hamiltonian, \begin{align} H = \frac{p^2}{2m} + \frac{1}{2} k x^2, \end{align} has both the momentum, $p$, and the position, $x$, in it. When the Hamiltonian has just one or the other ($m=\infty$ or $k=0$), its spectrum is continuous.

The 'particles' in quantum field theory (QFT) are ultimately the result of exactly that sort of discretization of the possible values of the Hamiltonian. One way to write the Hamiltonian for the electromagnetic field is $$H = \int \left[\frac{\epsilon_0}{2}\mathbf{E}_{\mathrm{div}}^2 + \frac{\epsilon_0}{2}\mathbf{E}_{\mathrm{sol}}^2 + \frac{1}{2\mu_0} \left(\nabla\times \mathbf{A}_{\mathrm{sol}}\right)^2\right] \operatorname{d}^3x.$$ In this Hamiltonian $\mathbf{A}_{\mathrm{sol}}$ plays the role of a coordinate (like $x$ in the SHO) and $\mathbf{E} = \mathbf{E}_{\mathrm{div}} + \mathbf{E}_{\mathrm{sol}}$ plays the role of momentum ($p$ in the SHO). Notice how $\mathbf{A}_{\mathrm{div}}$ doesn't contribute? That means that the $\mathbf{E}_{\mathrm{div}}$ part of the field doesn't support particle-like excitations because that part of the energy can have any energy level. Photons live in the parts of the Hamiltonian that have the $\mathrm{sol}$, for solenoidal, subscript.

Now, I never said anything about the scalar potential, $\phi$. Unlike $\mathbf{A}_{\mathrm{div}}$ which lacks a potential term but has a kinetic energy term, $\phi$ appears in the Hamiltonian/Lagrangian but it's time derivative does not, so it's a different form of weird like where something has a potential energy part but no kinetic energy part. Worse, it's actually tied up with $\mathbf{A}_{\mathrm{div}}$, but understanding all of that isn't necessary to understand why the part of the field that depends on them doesn't have photons in it.

Now, what did I mean by "arguably"? Well, you'll often hear about "virtual photons", and Feynman explicitly says that the Coulomb potential arises from the action of such virtual photons. Feynman was no dummy, so he's not outright wrong, he's just committing an abuse of terminology that has, admittedly, become standard.

My understanding of what a virtual photon, or really any virtual particle, is is intimately tied into the Fourier transformation of the field and satisfying constraints (especially boundary conditions). Specifically, they're a mathematical tool that allows you to satisfy constraints in real space after changing to Fourier space. If you dig in to quantum electrodynamics, for example, you'll find that the Maxwell's equation equivalent to Coulomb's law, $\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$, is treated as a constraint equation that the fields have to obey rather than a result of dynamics. Thus, you need virtual photons to satisfy it.

Consider an example: a string with both ends fixed a distance $L$ apart and with a fixed tension, $T$. Now, tug on the string at some point, $x'$, with a force, $F$. In this analogy, the position of the string plays the role of the field and the force applied plays the role of an electron ($F$ is the charge). We assume that the string obeys the inhomogeneous wave equation (for simplicity) $$\mu \frac{\partial^2 y}{\partial t^2} - T \frac{\partial^2 y}{\partial x^2} = F \delta(x-x'), \tag1$$ where $\mu$ is the mass per unit length of the string and $\delta(x-x')$ is the Dirac delta function. In this case, it is possible to show that the shape of the string, $y(x)$, is a tent function given by \begin{equation} y(x) = \left\{\begin{array}{ll} \frac{F}{T} \left(1 - \frac{x'}{L}\right) x & x \le x' \\ \frac{F}{T} \left(\frac{x'}{L}\right) (L-x) & x \ge x'. \end{array}\right. \tag2 \end{equation}

When $F=0$ the string obeys the wave equation, and has the normal modes we're used to. Those normal modes happen at discrete frequencies, and in quantum mechanics will have discrete amplitudes (this is what photons look like when you pin the field at the edges of some box - they have discrete frequencies) fixed by $E_i=n_i \hbar \omega_i$, for some integer $n_i$ at each angular frequency $\omega_i$. That $n_i$ is what we call the number of photons that have that frequency in the electromagnetism case. Importantly, those normal modes obey a dispersion relation between angular frequency and wave number, $k$, $$ \omega^2 - \frac{T}{\mu}k^2 = 0. \tag3$$ When a mode obeys that relationship we call it "on mass shell" in particle physcs.

If you do a mode expansion of equation $(2)$ you get \begin{align} y(t,x) &= \frac{F}{T}\sum_{i=1}^\infty \frac{\sin(k_i x') \sin(k_i x)}{k_i^2} \ \mathrm{for} \tag4\\ k_i & \equiv i\frac{\pi}{L}. \end{align} Note that each term in equation (4) is a separate mode, and the relationship between angular frequency ($0$ for every mode in this case) and wave number given in equation (3) is violated. Thus, we call the terms "not real" but we still need them to satisfy the boundary conditions on the derivative of $y$ at $x'$ implied by equation (1).

Admittedly, the string is not as complicated a the photon case, since the field is not divided in to separate components, one of which admits virtual photons and the other real ones, and has no gauge conditions. Even so, the essence is the same (cf. scalar field theory, which uses the same terminology as QED).

Interestingly, virtual particles aren't always enough to satisfy all of the constraints placed on a problem. When you choose a gauge in non-abelian gauge theory, you need another mathematical tool called a Faddeev-Popov ghost particle to keep the gauge constraint satisfied. The difference between ghost and virtual particles is that FP ghosts have the opposite spin-statistics from their underlying field, virtual particles have the same.

  • $\begingroup$ Thank you very much for your answer. Could you maybe give me a link or a book where I could read more about: 1) Discreteness and continuity of energy spectrum coming from having potential and kinetic parts (never heard of that and would like to know more) 2) Something more about that form of hamiltonian? Also, could I say that as photons are the building blocks of solenoidal field, electrons are blocks of coulomb field? $\endgroup$ – KabaT Jan 31 '18 at 18:17
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    $\begingroup$ @KabaT 1) What level of technical are you looking for? The best book I can think of for its breadth off the top of my head would be Weinberg, but you need considerable training in quantum mechanics and electromagnetism before you can understand it. I wouldn't say that electrons are the building blocks of the divergent field. I would say that the Coulomb electric field is the building block of the divergent field, though. $\endgroup$ – Sean E. Lake Jan 31 '18 at 18:24
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    $\begingroup$ @KabaT I'm reading that terminology I used in the comment above and I'm cringing. What I meant by 'Coulomb field' was simply $$\frac{\mathbf{x}-\mathbf{x}'}{4\pi |\mathbf{x}-\mathbf{x}'|^3} = -\nabla \frac{1}{4\pi |\mathbf{x}-\mathbf{x}'|} ,$$ and not some other fundamental field. You can say that the divergent part of the electric field is built up from adding Coulomb fields using Helmholtz decoposition. $\endgroup$ – Sean E. Lake Jan 31 '18 at 19:20
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    $\begingroup$ @KabaT The divergent part of $\mathbf{E}$ is fixed by the charge density in a relationship that is, essentially, classical because the Hamiltonian admit states of definite $\mathbf{E}_{\mathrm{div}}$, even though the charge density is, itself, quantum mechanical. Also, at your level you should be able to handle Weinberg. $\endgroup$ – Sean E. Lake Jan 31 '18 at 19:21
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    $\begingroup$ @KabaT Sounds about right. The wave equation in the Lorenz gauge is just Newton's second law with $\nabla^2 A^\mu$ as an internal force term and $\mu_0 J^\mu$ as an external force acting in the $A^\mu$ direction. $\endgroup$ – Sean E. Lake Feb 4 '18 at 19:10

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