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Does a radio receiver "collapse" a wave function when listening to a radio broadcast generated via a transmitting antenna?

Background:

There has been much discussion on this forum (here) about the nature of radio waves, photons and quantum mechanics. I have reviewed many of these but haven't found an answer. (I especially liked this post and the part describing Willis Lamb's "anti-photon" article.) I apologize if all this has already been covered and I missed it in my search.

In the classic double-slit experiment, "individual photons" are directed at a double slit, and the wave equation interacts with itself such that the distribution of strikes on the far wall show the pattern of a wave interference. This has been explained to me to mean that the wave equation is in all the places at once, but once a measurement has been made (i.e. the photon interacts with the far wall), then it collapses to one particular location. (As an aside, I personally prefer the De Broglie–Bohm theory as considered by the YouTuber Veritasium (here), though it likely is not correct or complete.)

I understand that radio waves and light are the same thing, but just with differing energy. But it seems that it is more convenient to think about radio signals as waves, and many illustrations of radio antennas show expanding waves, like ripples on the surface of a pond after a rock is thrown (e.g. here). This seems consistent with the idea that when energy pertubates the electromagnetic (EM) field, that this "ripple" of excitation propagates outward. And a radio receiver can pick up this disturbance, and convert it back to sound waves for the listener. And we know that radio waves can travel far around the world reaching potentially millions of listeners, or even out into space.

My confusion is how to consider all this from a classic (Copenhagen) interpretation of quantum mechanics. Would we say that the wave equation collapses when a particular radio receiver picks up a signal? And that each "photon" of the RF energy only goes to one location? That there are trillions of photons going in all directions, each wave function collapsing at a particular listener? This doesn't make sense to me. I imagine a radio wave from a NASA radio transmitter propagating far out to two different space probe on opposite sides of our solar system, but then "collapsing" to just one of the probes like the EPR paradox of entangled particles.

It seems like this would be easily testable: Generate a very controlled radio wave with limited energy. Then see if many receivers pick up their small portion of that wave energy. Or does one receiver get all of the energy from one photon's amount of quantized energy?

The more I think about this, the more that the concept of a photon (as opposed to a wave) seems only to apply in particular circumstances. The rant by Willis Lamb quoted in the article above rings increasingly true.

Go figure, the wave-particle duality is not easily understood.

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    $\begingroup$ Welcome to PSE! This is a great question and I think it comes from the limits of regular quantum mechanics, where the realm of quantum field theory is needed to have a better intuition/understanding of what is likely to happen. $\endgroup$ Commented Nov 20, 2022 at 15:11
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    $\begingroup$ @JeanbaptisteRoux Quantum field theory, as applied to radio, isn't even wrong. $\endgroup$
    – John Doty
    Commented Nov 20, 2022 at 15:16
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    $\begingroup$ Radio is extremely accurately modeled as classical electromagnetic radiation. No quantum effects need to be considered. $\endgroup$
    – John Doty
    Commented Nov 20, 2022 at 15:18
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    $\begingroup$ Does this answer your question? Quantum description of radio antenna $\endgroup$
    – Roger V.
    Commented Nov 20, 2022 at 20:25
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    $\begingroup$ @JeanBaptisteRoux I am aware of QFT but have not studied it in detail. My gross understanding is that everything (matter or energy) is considered to be oscillations of various quantum fields. I am thinking about photons / radio waves as fluctuations of the EM field. Are these different concepts? $\endgroup$
    – kdtop
    Commented Nov 20, 2022 at 21:54

4 Answers 4

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There is no conceptual difference between the setups of a light wave and a photodectector and a radio wave and a receiving antenna. These situations are exactly analogous, they just live in different frequency ranges of the electromagnetic spectrum.

The notion of single photons becomes only relevant when the intensity of the electromagnetic wave is low enough that you conceivably only get one photon per whatever time intervals your detector can resolve and when your detector is able to detect a single photon in the first place. Ordinary radio antennas - much like common photo sensors - are not sensitive enough for this.

But if you have such a low-intensity, high-sensitivity setup, then yes, the radio detectors will see single photon blips, randomly distributed among them in accordance with the classical intensity, not a continuous low-intensity signal, just like the double slit with single photons sees single-photon blips on the detector screen instead of a continuous pattern.

But again, any practical radio antenna does not operate in the sensitivity range where this would be relevant. A classical electromagnetic wave is a coherent state with an indeterminate but high number of photons, and it is perfectly appropriate to model even far-away antennas in this classical regime. This is again exactly like with visible light - even though the stars in the night sky are very far away, we still don't see their light arriving as single photons - those that are so far away we just don't see at all because our eyes do not reliably resolve single photons under ordinary circumstances.

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    $\begingroup$ Comments are not for extended discussion; this conversation about characteristics of real-world radio receivers has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Nov 21, 2022 at 12:24
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    $\begingroup$ This answer hinges on the statement that a radio receiver and a photodetector are "exactly analogous". That's not the case. Radio receivers can and do detect the phase angle. This is the working principle of things like radio interferometry where the data from two or more telescopes are combined to improve angular resolution by many orders of magnitude. The entire "trick" is, that the telescopes need to timestamp their signals with synchronized atomic clocks, so that phase is recorded consistently between the telescopes. $\endgroup$ Commented Nov 22, 2022 at 17:11
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Classically: light is waves In this sense the meaning of the two-slit experiment for electrons (which classically are particles) and photons is different. For electrons the appearance of the interference pattern demonstrates their wave-like behavior, which is unexpected classically. On the other hand, diffraction pattern of electromagnetic wave on double slits is not surprising - it had been known and studied well before the advent of quantum mechanics. What is surprising for photons is that the diffraction pattern vanishes, when we try to detect through which slit the photons pass, which proves their particle-like nature.

Photodetectors and radio receivers do not measure the same thing Photodetectors measure photon number - absorption of each photon corresponds to an event in a photodetector: an electron jumping from valence to conduction band in a photo-diode, a chemical change in a single molecule of a photographic film and likewise. In other words, the Hamiltonian describing the measurement is $H_{detector} =\lambda a^\dagger a$, where $a, a^\dagger$ are the photon annihilation and creation operators, whereas $a^\dagger a$ is the photon number operator.

Radio receiver measures something similar to photon phase (keeping in mind that the operator of photon phase is poorly defined in quantum mechanics) - the radio wave excites oscillations in the LC circuit. That is the measurement operator is more like $H_{receiver}=\lambda a^\dagger + \lambda^* a$.

Wave function collapse means projecting the wave function on a basis state of the Hamiltonian of the measurement device. The wave function of radio waves is not a photon number eigenstate, but a coherent state, in which the variance of the photon number is as big as the average number of photons. Measurement of coherent states by the operator $H_{receiver}$ is equivalent to shifting them. In other words, the photon coherent state collapses on a shifted coherent state. This might look very different from how one usually thinks of the wave function collapse, but it is wave function collapse in the very real sense.

Are photons absorbed by radio receiver?
We could go a step further and model the receiver as a quantum device (rather that treat it as an external observer, which is always implied to be classical). We could then describe the receiver interacting with photons by a Hamiltonian similar to Jaynes-Cummings model:
$$ H=\hbar\omega_0 a^\dagger a + \Delta b^\dagger b + \lambda(a^\dagger+ a)(b^\dagger + b). $$ If we were describing a photodetector, $b,b^\dagger$ would be the operators of a two-level system, and we would be aiming at calculating the occupancy of the excited state, $b^\dagger b$, whereas in case of radio receive these are bosonic operators describing the excitations in the LC circuit, and we are interested in the voltage or current given by $V, I \propto b \pm b^\dagger$.

The Hamiltonian above is in fact a Hamiltonian of two coupled oscillators - we could diagonalize it in terms of joint photon-circuit modes (polaritons), using polaron transformation. Absorption can then be described as projecting the coherent states of photons on these polariton states - the collapse becoming to look more real.

Finally, the detection of oscillating current/voltage is usually done using a non-linear element (like a diode), which essentially means measuring $(b^\dagger \pm b)^2\sim b^\dagger b$. That is, in the end the detection of radio waves is indeed reduced to a number operator (like in a photodetector), but well after the electromagnetic wave itself has "collapsed".

The circuit of a simple radio receiver:
enter image description here

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    $\begingroup$ Thank you for this detailed and well-contructed replay. I appreciate the effort you put into this. Unfortunately, it is a bit above my level of understanding. I will let what you have written percolate in my brain and see if I can gain some understanding. :-) $\endgroup$
    – kdtop
    Commented Nov 21, 2022 at 17:46
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    $\begingroup$ Are the mentioned photo detectors practical for counting photons? I thought it was only photomultipliers. $\endgroup$ Commented Nov 21, 2022 at 22:17
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    $\begingroup$ @PeterMortensen nist.gov/pml/quantum-networks-nist/… $\endgroup$
    – John Doty
    Commented Nov 21, 2022 at 23:15
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    $\begingroup$ @JohnDoty nowadays it is technically possible to create such quantum radio - I wouldn't be surprised, if somebody tried it. Anyhow, the question it to explain radio in quantum terms, even though classical theory works well. $\endgroup$
    – Roger V.
    Commented Nov 22, 2022 at 6:07
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    $\begingroup$ @PeterMortensen what differs photomultiplier from a photodiode or photographic film is the high sensitivity and the low error, but they all count photons: one photon - one absorption even, one electron added to the current (or one chemical change in photofilm) $\endgroup$
    – Roger V.
    Commented Nov 22, 2022 at 6:12
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In the transmitting antenna many many electrons are accelerated and many many individual photons are created. IN the receiving antenna a portion of the many many photons are absorbed by some of the electrons in the antenna .... and for these photons you can say their wave function collapsed. But all the remaining photons continue to propagate ... you can say the radio wave is made up of many many small waves.

In your satellite experiment signals would reach both .... if a directional antenna was designed then each satellite could be contacted individually.

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    $\begingroup$ Thank you for this reply. You are right that these radio signals would involve many many photons, and trying to consider the behavior of just one photon is likely non-productive. Thanks again. $\endgroup$
    – kdtop
    Commented Nov 20, 2022 at 22:11
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Yes.

Classically, a source of radio waves emits electromagnetic radiation with frequencies in the radio range, spatial patterns related to the geometry of the source, and with varying amplitudes depending on the power of the source. Quantum mechanically the story is the exact same, waves are emitted with a spatial pattern related to the geometry of the emitter and frequency in the radio range. The differences are that (1) the amplitude of the waves are quantized, (2) the waves can occupy a superposition of different amplitudes, and (3) there may be quantum correlations (entanglement) between the radio electromagnetic field and the source. However these questions are largely immaterial to the question at hand.

If you have a radio wave emitted in a radial outgoing pattern and you have an array of detectors surrounding that radio wave in a spherical pattern then classically, each detector will detect a fraction of the power emitted by the radio wave related to the solid angle subtended by the detector and the exact spatial pattern of the emitter. Quantum mechanically, if the detectors are "single photon" detectors then each detector will detect a fraction the total emitted photons related to the solid angle that detector subtends and the geometric pattern of the emitted radiation. In a Copenhagen interpretation we would indeed say that, in a certain fraction of the trials (again related to the solid angle and geometry of the radiation pattern) the wavefunction collapses such that a click is detected at a given detector.

Alternatively, in an Everettian (Many Worlds) interpretation we would say that the multipartite emitter + EM field + multi-detector systems enters a large entangled superposition as the single wave with a delocalized spatial pattern interacts with an array of detectors. i.e. part of the universal wavefunction has a photon click on detector A (but not on B, C, D...) while another part has a click on detector B (but not A, C, D...) . The coefficient in the universal wavefunction expansion will be related to the geometry of the radiation and detector solid angle. But you didn't ask about an Everettian interpretation ;)

In the entire above discussion radio waves can be replaced with optical waves and all the same results will hold.

One important practical difference between optical and radio waves is that with radio waves, because we have fast electronics, we can have radio field detectors. That is we can detect the amplitude and phase of the field at any moment in time. Optical fields, however, oscillate too quickly so we are not able to (using conventional techniques) directly detect the amplitude and phase of the optical EM field. Most optical detectors are field intensity detectors. For these reasons it is more common to see single-photon optical detectors than single-photon radio wave detectors. That not to say you can't in principle make single photon radio detectors or optical field detectors. It's just that the former is not technologically interesting (actually it might be for RF astronomy applications, I'm not well-versed enough) and the latter is technologically challenging (optical homo/heterodyne detectors are kind of like optical field detectors but it's a little different). This technological difference likely explains why you see more field-oriented literature around RF than optical frequencies.

Finally a parting rant: The particle picture of photons is pretty terrible and just leads to lots of misunderstandings. Quantum Field Theory and the Standard Model are both quantum FIELD theories. Not particle theories. These fields are always spatially delocalized field modes. It is the excitation amplitude of these fields that is quantized. The picture that you get of localized particles is just plain wrong and, in my opinion, quantized excitations of delocalized field modes are different enough from localized particles that we should not use the term particle to describe them and doing so is a mistake. In short: at the level of QFT and SM I don't believe in wave-particle duality, it's just waves.

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    $\begingroup$ Electromagnetic energy incident on a photoelectric detector behaves like particles. That's a real phenomenon. That's what we base our models on. You can wave your hands and insist that it's just waves, but to capture particle phenomena that way is a prohibitively complicated approach in practice. $\endgroup$
    – John Doty
    Commented Nov 20, 2022 at 19:59
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    $\begingroup$ @JohnDoty The detectors have nonlinear interactions with the EM field. You can calculate the probability that the interaction activates the detector above a certain threshold and you’ll get the same predictions as the “particle” picture just with a non particle interpretation. $\endgroup$
    – Jagerber48
    Commented Nov 20, 2022 at 20:11
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    $\begingroup$ Really? Do you know how hard it is to do this in the accurately in the particle picture? GEANT is quite a large code. The wave picture informs certain low-level parts, but overall, to do a fully entangled wave model is beyond computation. $\endgroup$
    – John Doty
    Commented Nov 20, 2022 at 20:19
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    $\begingroup$ @Jagerber48 I loved everything about your post -- except the Many Worlds stuff which seems to me to be a non-falsifiable theory outside the realm of actual science. I agree with your rant about photons, but I think not all do. John Doty makes a good point that at the end of the day there will be just one spot that lights up on a photo detector. It seems that at that point it is acting like something localized. The points you two make after that went over my head. :-) $\endgroup$
    – kdtop
    Commented Nov 20, 2022 at 22:17
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    $\begingroup$ Follow up question: It seems that photons are shown as packets WITH DIRECTIONALITY -- i.e. a photon might be traveling in the direction of Jupiter OR towards Venus. But radio waves seem to be described as traveling BOTH ways. How does QFT handle this? $\endgroup$
    – kdtop
    Commented Nov 20, 2022 at 22:20

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