# How do we visualise antenna reception of individua radiowave photons building up to a resonant AC current on the antenna?

I am a chemical/biological scientist by trade and wish to understand how quantum EM phenomena translates to our more recognizable classical world. In particular I want to get a mechanistic picture of what is going on when a tuned antenna is interacting with a photon of the desired frequency? I believe an individual electron on the antenna (many electrons) accepts a photon but how does the eventual process of a measurable AC current build up on the dipole (or 1/4 wavelength for example) to be fed with no reactance onto the transmission line? Is there a good text that discusses quantum EM with classical antenna theory?

"When photon meets antenna" is a great meeting ground for a quantum/classical bridge.

Unfortunately I do not have a serious maths background but will try anything suggested. I have read and listened to many of the Feynman's popular quantum discussions which only increases my thirst for a better understanding of how quantum EM translates to our more visible world.

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## 6 Answers

Here is an experimentalist's view of the question:

1) one photon hits the antenna and raises a molecular electron band to a higher energy level, and it will fall back to its lower one, with the characteristic electromagnetic transition time of the order of 10^-16sec, giving the energy to the antenna grid of molecules. One photon will just disappear.

2)a stream of photons that carry a signal means: a) that there is enough amplitude, b) there is coherence between photons: photons carry spin and thus polarization and in order to carry a signal the phases between all photons must be fixed and be coherent in time and space. Coherent means that there are fixed phases in the whole bunch. When such a bunch of photons hits an antenna the coherence will be transferred to the individual photon absorptions and de-excitations by conservation of spin, building up a corresponding electromagnetic wave on the molecular Fermi conduction level which can be detected further as a signal.

3) It is simpler for such problems to use the classical EM picture.

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Thank you Anna. – user6869 Dec 28 '11 at 7:50
1. "The antenna grid molecules" I presume means the outer atoms on the antenna. 2. "there is enough amplitude" I presume means sufficient number of coherent photons to get to a detectable current. 3. I did not know that classical EM explains the first steps in photon electron interaction. I am trying to feel my way from photon interaction towards measurable current as described by classical EM. 4. I will need to read more about Fermi conduction and try to picture how measurable AC voltage and current builds. Have you any suggestions? Thanks again Anna. – user6869 Dec 28 '11 at 8:46
As to point 3, what Anna may have meant is that there is really no reason to use quantum theory in practice because the EM waves that interact with antennas are large enough that quantum effects are completely negligible. – David Zaslavsky♦ Dec 28 '11 at 8:57
@DavidZaslavsky right David. Large enough in number of photons to act as an aggregate whose limiting behavior is the classical EM solutions. – anna v Dec 28 '11 at 12:17
@user6869 By grid I meant the crystal lattice, except an antenna is not an organized crystal but does have a structure. This is to contrast with energy levels on atoms and molecules which have much higher frequencies than the incoming  RF. The collective modes of the solid state pick up the energy. Yes, amplitude means enough photons to carry the signal. Classical EM is the macroscopic statistical manifestation of the many photon state at the quantum level. – anna v Dec 28 '11 at 12:29
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Understanding antenna physics in terms of photons is not trivial, because the quantum statistics of the photons means that they do not interact as separate distinguishable incoherent particles, at least not when they form a classical electromagnetic wave. They flow together, bunching up into a coherent slowly varying field state which carries them from their source to their destination in a way more analogous to a fluid flow.

To make an imperfect analogy, suppose you have a bucket of liquid helium and you punch a little hole at the bottom. You can model the phenomenon by He atoms randomly knocking about and finding the hole and escaping, but this model will fail to predict the flow rate or the emptying time, except in the wrong limiting case of an extremely dilute gas of atoms. The flow in the liquid He is determined by a profile of the classical Schrodinger field, which sets up a gradient for the mass flow along streamlines that escape through the hole.

The process with photons is only very roughly analogous, because the He atoms repel each other strongly, making an interacting quantum fluid, while the photons are non-interacting, making a Bose-Einstein condensate. But the Bose statistics are the same. When you have an antenna interacting with a classical EM field, the motion of the charges sets up a Poynting flow which directs the field energy into the antenna, when you superpose the reradiated field from the antenna with the incoming field from the far source. This superposition acts as a guide for the photons, sucking them into the antenna. The classical field picture applies, and the photon picture is in the large number fully coherent zero temperature limit, where it reproduces the field picture.

### Photon picture reproduces classical fields

The photon is never a nonrelativistic particle, because it is massless. The propagation of a photon is then never strictly forward in time, and there is no productive identification between a photon wavefunction and a classical electromagnetic field.

But in a space-time picture with classical current and charge sources, there is an identification of the probability amplitude of finding the 4 dimensionally propagating photon at a certain point with the vector potential set up by the sources. This identification is four dimensional, meaning the photon can zig-zag in time, and the amplitude is only for quantum propagation along the world-line of the photon, which is not directly observable, since we only see superpositions of all incoming proper times. This is the Schwinger-Feynman picture of relativistic particles, which applies to all quantum field theories.

The Lagrangian is

$$S=\int {1\over 4} F^2 + J\cdot A$$

and the path integral in Feynman gauge gives vacuum persistence amplitude (the quantum partition function) in the presence of J

$$Z[J] = \int e^{iS} \propto e^{\int J^{\mu}(x) G_{\mu\nu}(x-y) J^{\nu}(y) dx dy}$$

Where G(x-y) is the photon propagator in Feynman gauge, which, in x-space is

$$G_{\mu\nu}(x) = {1\over 2\pi^2} {ig_{\mu\nu}\over x^2}$$

Up to an $i\epsilon$ prescription along the light-cone which resolves the singularity of photon propagation along $x^2=0$ (this formula is often written with the delta-function singularities separated out, leaving a principal value for the part which is $1/x^2$, but I don't like this convention too much because both parts come from the same expression, which is just the 4d solution to Laplace's equation) The Z[J] functional tells you what all the particle propagation properties are, since it describes how a J source, which produces an A particle (photon) then reabsorbs the photon at a different location.

The actual photon propagator can only be seen to be a particle propagation in the relativistic picture in a full 4-dimensional form. In Euclidean space. Ignoring the $g_{\mu\nu}$ polarization factor (which is somewhat nontrivial, because the time component has the wrong sign, but irrelevant for the discussion here, which is about the propagation)

$$G(k) = {1\over k^2+i\epsilon} = \int_0^\infty d\tau e^{-\tau (k^2+i\epsilon)}$$

This is Schwinger's proper time representation of the Feynman propagator, central to the modern point of view. The function G(k) has an immediate probability intepretation as a probabilistic superposition over all intermediate proper times of a spreading Gaussian (a shrinking Gaussian in k space which is equal to 1 at the origin is a spreading Gaussian in x-space with a unit integral, a spreading probability distribution). This spreading Gaussian probability process is a random walk of a point particle, and it equivalently describes the Euclidean propagator in a point-particle picture.

The analytic continuation to real time can be gotten by analytically continuing $G(x)$, which is standard, and also by analytically continuing $\tau$, which is less commonly presented (but still in the literature, usually in introductory string theory texts as a warm up for the string). The result of continuing in $\tau$ produces a $\tau$ quantum propagation which makes a freely four-dimensionally propagating point-particle with quantum amplitudes to get from one point to another which, when summed over all intermediate times, reproduces the Feynman propagator of the free field. This is best understood as abstractly as possible, from the equivalence of stochastic processes in imaginary time to quantum amplitudes, and this connection is quickly reviewed here: Correct application of Laplacian Operator ( the question is involved and intimidating looking, but irrelevant for this discussion, I am just using the relation of quantum mechanics in real time to stochastic evolution in imaginary time, which is the general principle explained there )

This back-and-forth between particle picture and field picture is well known since Schwinger's era, but it is not often presented nowadays, perhaps because the picture is so acausal, involving sums over four-dimensional paths for particles which zig-zag in time.

### Particle view of antenna

In the case of an antenna, the classical solution A(J) in the Feynman gauge gives an alternate expression for the path integral:

$$Z[J] = e^{i\int A[J]\cdot J}$$

In other words, the entire photon partition function is determined by knowing the classical field in response to the source J. This determines both the amplitude for photons to go from source to source (during their 4-d acausal propagation), and all the correlation functions of the field (by infinitesimally varying J at different points).

Since everything is determined by the classical field, you might as well solve the classical equations to find the behavior of the field in response to J. This is because the photon field is free. The manipulations here, although formally trivial, are the content of the equivalence between the modern photon and the classical field.

### Antenna emission/absorption

Now consider an actual antenna responding to a far away source. In the classical picture, in order to know that energy is flowing into the antenna and not out, you need to know that the current distribution is produced in response to the field (in a causal field picture). The energy flowing out of or into the antenna is determined by the interaction Lagrangian, once you have dynamics for the degrees of freedom of the antenna:

$$L_i = \int J(x) A(x)$$

The interaction Lagrangian is the covariant generalization of $\int \rho(x) \phi(x)$ for the electrostatic source terms. It cannot be written in terms of E,B fields, only the vector potential is a local Lagragian variable.

The interaction Lagrangian is both according to the classical field produced by the source, and it also has a direct interpretation as photon absorption/emission, from the Schwinger proper time formulation of the Feynman propagator. So the photon picture and the classical picture are equivalent for these types of problems.

The coincidence of classical absorption and emission and photon absorption emission can be extended to single photons interacting with atoms, which leads some people to speculate that photons are not necessary. This is only true if you integrate out the photon field consistently, giving a nonlocal action to matter. If you keep a local action, the photons are still required to represent intermediate field states. The coincidence of classical and quantum behavior is a special mathematical property restricted tf Gaussian path integrals, discovered by Feynman, who uses the semiclassical approach to derive the rules of QED in his 1950s book "Quantum electrodynamics". This does not imply that photons are not physical, since you could integrate out electrons the same way.

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You cannot understand how a radio antenna works by counting the number of photons that strike the copper wire. This number is many orders of magnitude too small to account for the actual power absorbed by an antenna. An antenna would not work if it depended on physically intercepting photons. I explain all this in my blog post "The Crystal Radio".

In fact it is much more useful to analyze atoms in terms of antenna theory than to analyze antennas in terms of atomic theory. I elaborate on this question in my follow-up blog article, "How Atoms are Tiny Antennas".

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Any antenna (and a receiving system) is at some temperature $T$ that determines its proper noise of generated signals. One more photon may be too few to get a distinguishable signal in such conditions. You need a certain coherent flux of photons exceeding the system threshold to be noticeable for sure.

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Let us put an antenna in a box. Then we send a photon into the box. Now the antenna in the box has absorbed a photon, and has not absorbed a photon, and has absorbed a photon and emitted a photon. When a conscious observer observes the antenna, the antenna collapses into one of the three states.

An alternative answer: The wave function of an antenna that has not absorbed a photon evolves smoothly towards a wave function of an antenna that has absorbed a photon, when the wave function of a photon passes the antenna. The amplitude of the photon wave function decreases at the point where it meets the antenna. When observed these wave funcctions jump abruply into some state.

Let us consider an antenna made of ideal conductor and a continuous EM-wave. Classically this antenna Thomson scatters the EM-wave. Quantum mechanically radio photons are absorbed into the antenna and emitted after a time proportional to the frequency.

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You need to be a little cautious about taking ideas like "the photon" too literally.

In physics all our theories are models, that is they are approximations to the real world (whatever "real" means!). Treating light as a stream of photons is a model that works well in some circumstances, like the photo-electric effect, but isn't a useful description in other circumstances. Treating light as a wave is another model and it too works well in some circumstances, like the double slit experiment, but it isn't a good description all the time either.

Anyhow, the point of all this is that you wouldn't try and explain the double slit experiment by considering single photons and you wouldn't try and explain radio wave reception by an aerial by considering single photons. You could do, and Anna's description is about the best you can do if you insist on a photon description, but generally you make your life as a physicist a lot easier if the choose the model most appropriate to the system you're looking at.

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-1: "The photon" is as real as a brick. It is not a model of anything, it's just the way things are. – Ron Maimon Dec 29 '11 at 18:50
I'm afraid I completely disagree. The photon is a concept invented by scientists as a partial description of the electromagnetic field. Of course even QED is only a partial description of the electromagnetic field, and likewise the Standard Model and (probably) String Theory. How real any of these are is debatable, though the question is probably best left to the philosophers. – John Rennie Dec 30 '11 at 13:04
You can see photons scintillate a screen, you can hear them click a photomultiplier, they are as real as an electron. This debate is well over, and there is no modeling going on. The fact that QED is a partial description is no more apropos than the fact that Newton's laws only partially describe a brick. Photons are as real as bricks. You see them (literally) and touch them. – Ron Maimon Dec 30 '11 at 13:14
If you set up a double slit experiment with a CCD as the screen then treating light as photons is an excellent description of what happens when they interact with the CCD. However it's a poor description of what happens when a single photon passes through the slits. I suspect we're not really arguing since I'm not denying that photons can be an excellent description of what happens. I’m just saying that it isn’t a useful way to treat some case, e.g. radio waves interacting with an antenna. – John Rennie Dec 31 '11 at 10:07
Ok, I might have been touchy on this, because there is another user who denies photons, and claims that electromagnetic effects can be described semiclassically (unquantized EM field interacting with quantum atoms). I think the reason people get the idea that photons don't work in the field limit is because the photons are never nonrelativistic. But if you replace photon with "metal atoms" and EM field with "Bose Einstein condensate", the map between the two is the same, regarding double slit interference. Would you say that rubidium atoms are just a model, or are they real? – Ron Maimon Dec 31 '11 at 11:28
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