Several questions have been posted on Physics SE regarding the relationship between photons and electromagnetic waves, and several good answers have been given. Some of those questions are listed below, but I didn't find any that requested a mathematically explicit analysis of what happens — in terms of photons — when an oscillating current generates an electromagnetic wave with a macroscopic wavelength, such as a radio wave.

I'm attempting to fill that gap by posting this question-and-answer.

I haven't found an equally explicit / equally narrated analysis anywhere else, but less-explicit / less-narrated references include:

  • Itzykson and Zuber, Quantum Field Theory, section 4-1: "Quantized electromagnetic field interacting with a classical source";

  • Cohen-Tannoudji, Dupont-Roc, and Grynberg, Atom-Photon Interactions, exercise 17: "Equivalence between a quantum field in a coherent state and an external field", and also exercise 9.

Some related posts, from newest to oldest:

Does a photon require an EM field to exist?

How to interpret a wavepacket in quantum field theory: is it one particle or a superposition of many?

Can the equation $E = h\nu$ be used not only for light, but for radio waves? (Since they are all part of the electromagnetic spectrum)

Electromagnetic waves and photons

Photons of a radio wave

What exactly is meant by the wavelength of a photon?

Why call it a particle and not a wave pulse?

Is double slit interference due to EM/de Broglie waves? And how does this relate to quantum mechanical waves?

What is the physical nature of electromagnetic waves?

Relation between Wave equation of light and photon wave function?

Sequence of E and B field in radio waves and in single photons

Photon Quantum Field proportional to Electromagnetic Field?

Light Waves and Light Photons gedanken Experiment

Do photons occupy space?

How is the classical EM field modeled in quantum mechanics?

Are coherent states of light 'classical' or 'quantum'?

Amplitude of an electromagnetic wave containing a single photon

Radio waves within an atom

Radio waves and frequency of photon

Reconciling refraction with particle theory and wave theory

Properties of the photon: Electric and Magnetic field components


In QED, the electromagnetic (EM) field and the charged matter are both quantum entities. This answer uses a semiclassical model instead, with a quantum field coupled to a prescribed classical current. This is an exactly solvable model inspired by QED. As a further simplification, the quantum field will be a scalar field instead of the EM field. By analogy, the quanta of this scalar field will be called "photons".

In the context of the free (non-interacting) quantum EM field, the word "photon" is typically used to mean a quantum of energy, and that's how I'm using the word here. The current will be active only during a finite time interval, and I'll only apply the word "photon" at the times when the current is not active, so that the meaning of "quantum of energy" is unambiguous.

To help limit the length of this post, familiarity with introductory QFT is assumed. The notation will be similar to that used in chapter 2 of Peskin and Schroeder's An Introduction to Quantum Field Theory.

The model and its exact solution

The Heisenberg picture will be used, so the state-vector is independent of time, but its physical significance still changes in time because the observables do. The equation of motion in the Heisenberg picture is \begin{equation} \partial_\mu\partial^\mu\phi(t,\mathbf{x}) = J(t,\mathbf{x}) \tag{1} \end{equation} where $\phi$ is the quantum field and where $J$ is a prescribed function that will be called the "current" by analogy with the EM case. The equal-time commutation relation for the quantum scalar field is \begin{equation} \big[\phi(t,\mathbf{x}),\,\dot\phi(t,\mathbf{y})\big]=i\delta^3(\mathbf{x}-\mathbf{y}). \tag{2} \end{equation} The quantum field $\phi(t,\mathbf{x})$ is the local observable corresponding to field-amplitude measurements.

Equations (1)-(2) can be solved exactly. The solution is \begin{equation} \phi(t,\mathbf{x}) = \phi_0(t,\mathbf{x})+\phi_J(t,\mathbf{x}) \tag{3} \end{equation} where:

  • $\phi_J$ is a real-valued solution to (1), which commutes with everything;

  • $\phi_0$ is an operator-valued solution to the $J=0$ version of (1) that satisfies the commutation relation (2).

From now on, suppose the current is non-zero only within the finite time interval $0<t<T$: \begin{equation} J(t,\mathbf{x})=0 \hskip1cm \text{ except for }0<t<T \tag{4} \end{equation} and choose \begin{equation} \phi_J(t,\mathbf{x})=0 \hskip1cm \text{ for }t\leq 0. \tag{5} \end{equation} These conditions are all satisfied by \begin{equation} \phi_0(t,\mathbf{x}) =\int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, \frac{e^{-i\omega t}a_0(\mathbf{p})+e^{i\omega t}a^\dagger(-\mathbf{p})}{ \sqrt{2\omega}} \tag{6} \end{equation} and \begin{equation} \phi_J(t,\mathbf{x}) = \int ds\ \theta(t-s) \int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, i\,\frac{e^{-i\omega (t-s)}-e^{i\omega (t-s)}}{2\omega}\, \tilde J(s,\mathbf{p}) \tag{7} \end{equation} with \begin{equation} \omega\equiv \sqrt{\mathbf{p}^2} \hskip2cm \tilde J(s,\mathbf{p}) \equiv \int d^3x\ e^{-i\mathbf{p}\cdot \mathbf{x}}\,J(s,\mathbf{x}), \tag{8} \end{equation} and where the operators $a_0(\mathbf{p})$ and their adjoints satisfy \begin{equation} \big[a_0(\mathbf{p}),\,a_0^\dagger(\mathbf{p}')\big]=(2\pi)^3\delta^3(\mathbf{p}-\mathbf{p}'). \tag{9} \end{equation} The operators $a_0$ and $a_0^\dagger$ are just a basic set of operators in terms of which everything else in the operator algebra may be expressed. Define a state-vector $|0\rangle$ by the conditions \begin{equation} a_0(\mathbf{p})\,|0\rangle = 0 \hskip2cm \langle 0|0\rangle = 1 \tag{10} \end{equation} for all $\mathbf{p}$, and suppose that the state of the system is the one represented by $|0\rangle$. The Heisenberg picture is being used here, so the state-vector has no time-dependence, but the physical state that it represents still changes in time because the observables do.

The rest of this answer addresses the interpretation of the state-vector (10) both for $t<0$ and for $t>T$, first in terms of photons and then as it relates to radio waves.

The interpretation in terms of photons

Equation (5) says that for $t<0$ we have the familiar free scalar field, and then we recognize the state defined by (10) as the vacuum state — the state of lowest energy, with no photons. This, of course, was the motive for choosing the state (10).

The question is what happens at $t > T$ in the aftermath of the temporary current $J$. For these times, equation (4) says that the factor $\theta(t-s)$ may be omitted in equation (7), because it is already enforced by the current itself. Therefore, for these late times, the solution (3) may be written \begin{equation} \phi(t,\mathbf{x}) =\int\frac{d^3p}{(2\pi)^3}\ e^{i\mathbf{p}\cdot \mathbf{x}}\, \frac{e^{-i\omega t}a(\mathbf{p})+e^{i\omega t}a^\dagger(-\mathbf{p})}{ \sqrt{2\omega}} \tag{11} \end{equation} with \begin{equation} a(\mathbf{p}) \equiv a_0(\mathbf{p}) + a_J(\mathbf{p}) \hskip2cm a_J(\mathbf{p}) \equiv \frac{i}{\sqrt{2\omega}}\int ds\ e^{i\omega s}\tilde J(s,\mathbf{p}). \tag{12} \end{equation} The complex-valued function $a_J$ encodes the effect of the current.

Before we can interpret the state (10) in terms of photons at times $t>T$, we need to determine which operators represent photon creation/annihilation operators at these times. The Hamiltonian associated with the equation of motion (1) is \begin{equation} H(t) = \int d^3x\ \left(\frac{\dot\phi^2(t,\mathbf{x}) +(\nabla\phi(t,\mathbf{x}))^2}{2}-\phi(t,\mathbf{x}) J(t,\mathbf{x})\right). \tag{13} \end{equation} Equations (9) and (12) imply \begin{equation} \big[a(\mathbf{p}),\,a^\dagger(\mathbf{p}')\big] =(2\pi)^3\delta^3(\mathbf{p}-\mathbf{p}'). \tag{14} \end{equation} Any any time for which $J=0$, equations (11) and (13)-(14) imply \begin{align} H(t)= \int\frac{d^3p}{(2\pi)^3}\ \omega\,a^\dagger(\mathbf{p})a(\mathbf{p}) +h(t) \tag{15} \end{align} where $h(t)$ is a real-valued function that doesn't affect this analysis. Whenever $J=0$, these equations all have the same form as they do in the free-field case (where $J$ is zero for all times). Based on this, we can interpret $a(\mathbf{p})$ and its adjoint as the operators that annihiliate and create (respectively) a photon with the indicated momentum at times $t>T$. The justification for this interpretation is identical to the corresponding justification for $a_0$ at times $t<0$.

Now that we know which operators create and annihilate photons at $t>T$, we can interpret the state $|0\rangle$ at these times. Equations (10) and (12) imply \begin{equation} a(\mathbf{p})\,|0\rangle = a_J(\mathbf{p})\,|0\rangle, \tag{16} \end{equation} which is the defining equation of a multi-mode coherent state. The state $|0\rangle$ was chosen because it represents the vacuum state for $t<0$, but equation (16) says that it is no longer the vacuum state with respect to observables at $t>T$. The vacuum state at $t>T$ is represented instead by the state-vector $|T\rangle$ that satisfies \begin{equation} a(\mathbf{p})\,|T\rangle = 0. \tag{17} \end{equation} Equation (14) implies that the relationship between the coherent state (16) and the vacuum state (17) is \begin{equation} |0\rangle \propto \exp\big(A^\dagger \big)\,|T\rangle =|T\rangle + A^\dagger|T\rangle + \frac{1}{2!}(A^\dagger)^2|T\rangle + \frac{1}{3!}(A^\dagger)^3|T\rangle +\cdots \tag{18} \end{equation} with \begin{equation} A^\dagger \equiv \int\frac{d^3p}{(2\pi)^3}\ a_J(\mathbf{p}) a^\dagger(\mathbf{p}). \tag{19} \end{equation} In words, the state at times $t>T$ is a special superposition of different numbers of identical photons, all with this same profile described by the complex-valued function $a_J(\mathbf{p})$.

The interpretation as a radio wave

At any time $t$, equations (3)-(10) imply \begin{equation} \langle 0|\phi(t,\mathbf{x})|0\rangle=\phi_J(t,\mathbf{x}) \tag{20} \end{equation} and \begin{equation} \langle 0|\phi(t,\mathbf{x})\phi(t,\mathbf{y})|0\rangle - \langle 0|\phi(t,\mathbf{x})|0\rangle\, \langle 0|\phi(t,\mathbf{y})|0\rangle = \langle 0|\phi_0(t,\mathbf{x})\phi_0(t,\mathbf{y})|0\rangle. \tag{21} \end{equation} Equation (20) says that the expectation value of the quantum field behaves just like a classical wave generated by the current $J(t,\mathbf{x})$. Equation (21) says that the fluctuations in the outcomes of field-amplitude measurements are just as small as they would be in the vacuum. If the current $J$ is large enough, so that the expectation value (20) is large enough, then the square root of (21) will be negligible compared to (20). In this case, we have a classical wave for all practical purposes. By choosing the oscillation frequency of the current, we can make it a radio wave.

Altogether, this shows that if we start with the vacuum at time $t<0$ and turn on a current during the interval $0<t<T$, then the state at times $t>T$ is a coherent state of photons, and the same state can also be interpreted as an effectively classical wave.

Classical superposition versus quantum superposition

Note that a classical superposition of two such effectively-classical waves is obtained by adding the corresponding single-photon profiles in the exponent of equation (18), like this: \begin{equation} \exp\big(A_1^\dagger + A_2^\dagger\big)\,|T\rangle. \tag{22} \end{equation} This follows from the fact that such a superposition is produced by a classical current of the form $J=J_1+J_2$, where $J_1$ and $J_2$ may be localized in different regions of space (for example). In contrast, a quantum superposition of two effectively-classical waves has the form \begin{equation} \exp\big(A_1^\dagger\big)\,|T\rangle + \exp\big(A_2^\dagger\big)\,|T\rangle. \tag{23} \end{equation} In this state, equation (21) doesn't hold; fluctuations in field-amplitude measurement-outcomes are typically just as large as the expectation value, so a quantum superposition of two effectively-classical waves not like a classical wave at all.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.