# Quantum fluctuations at radio frequencies

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1. Apr 20, 2015

### Ponderer

Hi. I'd like to learn how to calculate the probability of a photon being emitted from a radio antenna where the energy per wavelength is below the threshold to emit photons.

Let's assume the electrical thermal noise is insignificant. The antenna temperature could be sufficient low or the antenna could have sufficient resistance from resistors such that the thermal noise current is insignificant, Inoise = sqrt(4*k*T*B/R). I'm talking about placing actual resistors in the antenna, not the antennas radiation resistance. Higher resistance from resistors decreases the antennas thermal noise current.

Let's say the antenna's AC current is set at a level such that it would produce one photon per wavelength on average. Now we decrease the power by half, which is half the required photon energy per wavelength. If we assume no appreciable thermal noise or quantum fluctuations, then the antenna would not emit photons. However, if quantum fluctuations can somehow occasionally push it beyond the minimum threshold to emit a photon, then how can I calculate this probability?

Thanks for any help. I appreciate it.

2. Apr 20, 2015

### The_Duck

I think the number of photons emitted in a given time interval is a Poisson distribution, and the average emitted power is the same emitted power you would calculate in classical electromagnetism. There's a derivation of this in the simpler case of a scalar field theory in Peskin & Schroeder section 2.4 and problem 4.1.

3. Apr 20, 2015

### Ponderer

I think that might be true for radio antennas at normal power levels. What if the energy required to emit a photon per wavelength is not coming from the transmitting source and the thermal noise current is insignificant?

4. Apr 20, 2015

### The_Duck

To make sure I understand, you are worrying that the answer I gave above may be wrong in the case that the power emitted averages less than one photon per cycle of the AC current? I think the answer I gave is true in that case also. It comes from a proper quantum mechanical calculation (though the current is modeled as classical and any back-reaction of the radiation on the current is ignored). If the emitted power is much less than one photon per cycle, that just means that on average it will be many cycles between emitted photons.

A somewhat similar situation happens in spontaneous emission from atoms. Excited electrons orbit the nucleus quite quickly and it takes them many orbits, on average, before they manage to emit a photon and drop to a lower energy level. You can get a decent semi-classical estimate of the excited state lifetime by modeling the electron as a classical point charge moving in a circle, using the classical formula for radiated power, and calculating how long it will take until the total radiated energy is equal to one photon of the transition frequency.

Last edited: Apr 20, 2015
5. Apr 20, 2015

### Ponderer

Yes, I'm concerned that Quantum Mechanics equation might be taking thermal noise into account. In the excited atom example, the sufficient energy is already there. It's excited. Although in the antenna example the energy is insufficient. Blackbody radiation is a better example. Back in the classical mechanics days they ran into what is called the UV catastrophe. Quantum Mechanics correctly predicted that blackbody radiation would drop at a certain frequency because of a lack of energy to emit the photon.
I would like to know more about the Quantum Mechanics calculation you're talking about. What is the equation and how was it derived. I'm wondering if they took into account thermal energy. In my antenna example thermal noise current is insignificant.

6. Apr 21, 2015

### The_Duck

Nope, the calculation I'm talking about starts with "suppose there is a certain time-dependent classical electromagnetic current J(t)" and then calculates the probability of emitting N photons. Thermal noise doesn't enter into it unless you insert that into J(t) by hand.

In the antenna example, I'm assuming that the source that is driving the AC current is perfectly capable of providing the energy to create a photon. Or if the power source is so weak that the back-reaction from a single photon will halt the current, I don't know what will happen.

Sure, in the case of blackbody radiation quantum effects do reduce the radiated power at high frequencies compared to the classical prediction. An antenna isn't a blackbody, and I don't think that happens here.

I'm not sure how much the details will help you unless you know some quantum field theory. If you do you can look up the reference I gave. The calculation goes like this. We start off with the electromagnetic field in a state of zero photons. Then we turn on a specified time-dependent electric current, say in an antenna, for some amount of time. Afterward we look at the quantum state of the electromagnetic field. We find that with probability P1 it's in a state with 1 photon, with probability P2 it's in a state with 2 photons, etc. If we average over all these possibilities to get the expected number of photons and thus the average total radiated energy, we get the same answer as if we had done a classical calculation of the radiated energy from the same current. Thermal effects appear nowhere in the calculation; as I said if you wanted to include them I guess you would put some random jitter into the source current by hand.

7. Apr 21, 2015

### Ponderer

Fine. Let's just assume they did not take thermal energy into consideration. I'm certain they did not intend the equation to work at sub-photon energy levels. If it doesn't take thermal energy into account, then it's definitely not intended for sub-photon energies. Again, look at UV catastrophe. Things change when the energy level is low enough.

Not in the setup I mentioned. The resistance is too high, which is exactly what I would do to bring it below the photon per wavelength level. Even if the source voltage is brought to it's highest level it could not.

Actually it is. Radiation resistance from antennas is entirely related to blackbody radiation. An antenna that has 10 ohms and 10uAmps rms thermal noise will radiate 1nW of blackbody radiation. And electrical resistance from resistors cause thermal noise, which also has a Gaussian distribution, produces noise current, which is a cause of blackbody radiation being emitted from the object. Blackbody radiation is caused by charges oscillating back and forth that emit photons. The blackbody radiation begins to decrease at a rapid rate toward UV, and by the time it reaches upper UV it's nearly zero. The only reason it doesn't turn off like a switch at a specific frequency is because blackbody radiation has a Gaussian distribution. So blackbody radiation is a perfect example for the antenna example. :) In my antenna example I mentioned that the resistance is high enough such that noise current would not cause photons to emit blackbody radiation through the antenna at the desired frequency.

Just out of curiosity, does it specify what it means by turning on the current. If it's DC, then yes it will definitely always emit photons initially because if we look at the spectrum we'll see it covers a wide range of frequencies that will be low enough to emit a photon, E = h*f. Lower frequencies require lower energy. If it's AC, then it will always emit at least one photon initially because the signal increases from zero to some value.

But anyhow, I appreciate your replies. I gave you a like. Replying to you made me answer my own question lol. Blackbody radiation is exactly what I was looking for. It is perfectly clear in the math along with a world of clear detailed text on blackbody radiation that it is caused by the thermal energy having insufficient energy to emit the photon. So it's true that there are no quantum fluctuations that will somehow push the thermal energy above the threshold to emit photons to match the sub-photon level. Example, if thermal energy is 1/100th one photon per wavelength, then quantum fluctuations won't bump it up to cause it to emit one photon every 100 wavelengths. This truth is clearly seen in blackbody radiation curves. We can see the blackbody radiation curve, especially when looked at linearly and not logarithmically. It dips rather fast. Like I was saying, in my antenna example this is exactly the problem I was looking for. :)

BTW, I'm not suggesting such an antenna experiment I mentioned would emit photons as predicted classically. If it did, then Houston we have a problem lol!

Last edited: Apr 21, 2015
8. Apr 21, 2015

### The_Duck

It's a general calculation that works fine for all frequencies. (I'm not sure what "sub-photon energies" means here? All you specify in this calculation is a source current.)

OK, so I guess you are imagining an antenna at finite temperature with basically no power source. I agree that this will radiate like a blackbody. Since I thought you wanted to exclude blackbody radiation, I have been imagining an antenna at zero temperature with a decent power source. This should not emit any blackbody radiation but should still emit in the manner I described.

Or a less ambiguous example of the sort of thing I've been picturing: suppose you have a point charge on the end of a stick. If you wave the stick back and forth at some frequency the charge will emit radio waves at that frequency. Thermal noise is not present here; if we are worried about it we can specify that the stick is at 0 K. If you reduce the amplitude of the oscillation the amplitude of the emitted waves will go down. Eventually you can reach the point where a photon will only occasionally be emitted. But no matter how small the amplitude, the time-average emitted power in these photons should be equal to the time-average emitted power you would get classically.

You specify how the current turns on when you give the current J(t) as a function of time. You can avoid the effects you are worrying about by turning on the AC current arbitrarily gradually, so that there is essentially only one frequency component, and leaving it on for a very very long time, so that any transient effects from the turn-on will be comparatively small compared to the energy emitted over the full run-time of the antenna.

Last edited: Apr 21, 2015
9. Apr 21, 2015

### Ponderer

For all frequencies, sure, why not. For all levels of energy for a given frequency, no. :) What is meant by "sub-photon energies" is the amount of energy for a given situation (I'm referring to a specific frequency) that is below the energy for one photon.

I don't think we're on the same page. The antenna example produces AC current through the antenna. Blackbody radiation has thermal energy, which produces thermal noise *AC current*. The antenna example mentions setting the AC current such that it's enough to produce a photon per wavelength, but then lowering the current so that the energy per wavelength is below one photon per wavelength. Blackbody radiation is the same. See UV catastrophe. See how it's the same example. Please note that when I say the antenna AC current is decreased, this is not saying the energy per wavelength decreases at the same rate. Classical mechanics fails at predicting Blackbody radiation, just as it *should* fail at predicting my antenna example. Do you agree?

Last edited: Apr 21, 2015
10. Apr 21, 2015

### Ponderer

I don't understand how you could believe that when the UV catastrophe clearly disagrees with your statement. The classical prediction for photons being emitted at UV due to thermal energy is *wrong.* The math is simple. E = h*f is the energy to emit a photon at frequency f. If the energy is not there, then please explain where the energy comes from. The energy from each cycle in the antenna does not accumulate because the antenna circuit is nearly entirely resistive, not reactive.

11. Apr 21, 2015

### The_Duck

Right. In the case I described of waving around a point charge on a stick, the energy isn't coming from thermal fluctuations--as I said we can take the stick to be at 0 K. The energy is coming from whatever is waving the stick back and forth.

The thing that accumulates each cycle is the probability amplitude to have created a photon. If you perturb a quantum system in a periodic way, the probability of producing a transition (in this case, creating a photon) accumulates over time, even if the amplitude of the perturbation is very small.

The situation with a charge on a stick really is pretty similar to the case of an excited atom. If whatever is waving the stick around is capable of providing energy hf, then just like in the atom the needed energy is there; it's just a question of how long it takes before a photon is likely to have been created. Just like in the atom, the energy will be taken from the power source in one chunk of size hf.

Last edited: Apr 21, 2015
12. Apr 21, 2015

### Ponderer

You're saying the *energy* in you're stick example decreases.
As I pointed out, I'm talking about a decrease in AC current, not power or energy per wavelength. That seems the confusion here. Just because the current decreases doesn't mean the energy per wavelength decreases at the same rate-- impedance can change.

That's obviously wrong. See UV catastrophe. The thermal energy is still there trying to emit those photons, but it's no where as linear as classical mechanics predicts, and actually drops exponentially. I would ask you again, how does the energy that is coming from the AC source somehow magically and 100% efficiently store that energy and sum it up in phase every other cycle or every 100th cycle or one billionth cycle?

No that's not like my antenna example. Again, I said the AC current decreases. If it can't emit photons, then the radiation resistance would go away because the emission of photons is what causes radiation resistance.

13. Apr 21, 2015

### Ponderer

BTW I apologize because in my first post I said the antenna *power* decreases, but later on in this discussion I said the *current* decreases. The later is what I meant. Maybe that's the confusion here. Impedance would not be a constant if the emitted photons per unit of time decreased at a faster rate than the I^2 * R, where R is total impedance including the resistors, because we must remember an antenna has impedance as well, plus the resistors as mentioned in the example.

Last edited: Apr 21, 2015
14. Apr 21, 2015

### The_Duck

I'm with you here, I think, though I don't quite know what you mean by "power or energy per wavelength." In my stick example, the decrease in amplitude of the oscillation is just a decrease is just amplitude of the AC current.

This really is how QM works! Very small periodic perturbations can add up coherently over time to eventually produce a large probability of a transition.

I've completely eliminated thermal energy from my example by taking the temperature to be 0 K.

The QM prediction for blackbody radiation is a result of how energy is distributed in a system in thermal equilibrium at finite temperature. It doesn't have anything to say about how energy transfer works in non-thermal processes. I'm sure you agree that if I take a high-power antenna and cool it down to 0 K, it will still emit plenty of power, even though the blackbody radiation formula gives 0 watts of backbody radiation at 0 K?

In QM the energy is not gradually transferred, rather there is a gradual buildup of the probability that energy hf has been transferred. This is worked out in QM textbooks under the name "Fermi's golden rule."

Consider the following classical scenario. Suppose you have a resonator with very little energy loss. Say a guitar string with negligible friction that can keep vibrating essentially forever. Then from a nearby speaker you play a very low-amplitude sound that is in resonance with the guitar string's natural frequency. Over time the guitar string will absorb energy from the sound and eventually it will be vibrating at very high amplitude.

In QM the wave function of the electromagnetic field is like the guitar string. Perturbing it with a periodic perturbation in resonance with the frequency of a photon will gradually build up the amplitude of the wave function, which is related to the probability of having created a photon. Note the following difference between the quantum and classical situation: in the QM case, there is NOT a gradual transfer of energy. Rather there is a gradual buildup of the probability that one unit of energy has been transferred.

In both situations it is crucial that the perturbation is in resonance with the natural frequency of the thing being excited (the guitar string, or the photon wave function). That is what allows the coherent transfer of energy, in the classical case, or probability, in the quantum case.

Last edited: Apr 21, 2015
15. Apr 21, 2015

### Ponderer

Indeed, and it works with blackbody radiation. I never said the photons stop emitting in blackbody radiation at UV. I said the classical prediction is wrong.

In terms of thermal energy, yes, because it has a Gaussian distribution. Again, the predicted number of emitted photons according to classical mechanics is wrong. BTW I'm not saying my antenna would stop emitting photons. I'm saying that Quantum Mechanics and classical mechanics disagree with regards to my antenna example. Agreed?

That's better described as Gaussian distribution. I would stay away from fictional frictionless guitar strings. ;) It's very simple to understand why blackbody radiation emits photons at even gamma frequencies. Because thermal energy has Gaussian distribution. We don't need all that word salad lol. But you said in another post that Quantum Mechanics shows the same results as classical mechanics for my antenna example, and I couldn't disagree more. Will my antenna emit photons? Of course, under real conditions. If there's absolutely no thermal energy, then no it would not, but that's not my real world. Your only point now is that photons are emitted, I never disagreed with that. We need to agree that Quantum Mechanics and classical mechanics get different answers in my antenna example. Classically speaking, if the AC antenna current is 1/100th the current to emit on average one photon per wavelength, then classically speaking the emitted photons will be 100^2 times less, I^2*R. But Quantum Mechanics disagrees. Can we agree on that?

Last edited: Apr 21, 2015
16. Apr 21, 2015

### The_Duck

It sounds like you're saying here that an antenna at 0 Kelvin will not radiate anything? I don't agree. An antenna at 0 K will radiate just fine according to whatever AC current you are running through it. Not all radiation is blackbody radiation.

No, I haven't backed off my original claim, which is that if you run a very small AC current through an antenna, classical mechanics gives the right time-averaged emitted power, even though the energy comes out as discrete photons.

I suppose I do need to qualify my claim to: if you run a very small AC current through an antenna *at absolute zero*, classical mechanics gives the right time-averaged emitted power, even though the energy comes out as discrete photons. I agree that the classical and quantum mechanics give different predictions for blackbody radiation. But if you eliminate blackbody radiation by cooling the antenna to absolute zero, the classical and quantum predictions agree.

17. Apr 21, 2015

### Ponderer

And you haven't shown the math to prove this. Again, where does the energy come from? It's not accumulating in the antenna. Surely you understand the difference between reactance and resistance. According to your logic, this energy would also accumulate in blackbody radiation and emit the photons as predicted classically. But it doesn't. The photon emission drops at an extremely high rate in UV. Classical is way off.

As stated my antenna example is just like the blackbody radiation example, which you just said classical mechanics is wrong. So you must agree that classical mechanics disagrees with Quantum Mechanics for all radiating bodies where the energy per wavelength falls below one photon such as my antenna example.

Last edited: Apr 21, 2015
18. Apr 21, 2015

### The_Duck

All right, I'll try to work through the math later today and post it here!

In the meantime, here's a better version of my "charge on a stick" where the power source is more obvious. Suppose you have a really, really heavy point charge moving around in a circle at some frequency. You can calculate the power radiated by this charge according to classical electromagnetism. Here the radiated energy comes out of the charge's kinetic energy: the power source is the kinetic energy of the point charge.

Now if you make the circle smaller and smaller while keeping the frequency constant the radiated power decreases. If we make the circle small enough classical electromagnetism will predict that the radiated power will be less than one photon per cycle. What's your prediction for the radiated power according to QM in this case? Do you think it will be much less than the classical prediction? Note that here it's clear that there is enough energy available to create a photon: the point charge is very heavy so it has lots of kinetic energy even though it is moving quite slowly. That is, radiation reaction isn't a big deal because the charge is so heavy.

This is the calculation I'll try to work through and post.

19. Apr 21, 2015

### Ponderer

Ugh, but you keep avoiding what I'm pointing out. Your stick example couldn't be more different because you're talking about energy. Again, of course if you pump 1/100th the *energy* to the antenna you're going to get 1/100th the amount of photons per second. Please read my previous post again.

20. Apr 21, 2015

### The_Duck

Apologies; if I'm missing the point it's not because I'm deliberately avoiding anything but because I'm a bit confused about exactly what scenario you're after.

I think the question is: as we make the AC current running through an antenna very small, how does the # of photons emitted per second change? Is that the right question?

Is it OK if I take the antenna to be at absolute zero so that I can ignore blackbody radiation?

I think the point charge moving in a circle is a nice example of an antenna with a small AC current, so I don't understand why don't you like it as an example?

21. Apr 21, 2015

### Ponderer

It's okay. I did a poor job at explaining the example. Like I was saying my initial post was wrong. The focus is not the antenna energy, but the current. I mean, obviously due to the laws of conservation of energy if we pump 1/100th the energy to the antenna itself we get 1/100th the photons. Ignoring antenna losses of course.

But yes, it's fine to use 0K if you like. It just makes me a bit uncomfortable, but that's fine if it makes the math easier.

Can't you accept blackbody radiation for the antenna example? It's a real life example that's exactly how i would like the experiment to be. I'm sure you're aware of the dangers of just apply math to experiments. It's easy to miss something. Experiments need the appropriate interpretation. I know it would be easy to write some Quantum Mechanics equations that look pretty and shows the average emitted photons is one per 100 wavelengths. That's why we must stick to a real experiment.

So, why do I believe blackbody radiation is the best experiment for the antenna example? Because:

1) Blackbody radiation is caused by AC current, just as the antennas current source is AC current. If you prefer, we can make the antenna current source Gaussian noise. It doesn't matter because it would be analyzed with a spectrum analyzer.

2) Blackbody radiation begins to fall significantly when the thermal energy per wavelength falls below one photon per wavelength. And in the antenna example we set the current to where the antenna emits one photon per wavelength, and then drop the ***current,*** say 1/100th the current. Again, if it helps you see this in better light, we can make the antenna current source Gaussian noise.

Actually if we make the antenna current source Gaussian noise, then it is the exact same example as blackbody radiation. Antennas are a degree of freedom. Anyhow, I'm not trying simulate blackbody radiation lol. Just trying to focus on the specific aspect that photon emission drastically decreasing below the classical value when the net energy per wavelength falls below that of one photon per wavelength. Yes, using the words "net energy" better described this, yes? What do you think?

22. Apr 21, 2015

### Derek Potter

How exactly would you reduce the power to half a quantum (per whatever?) The circuitry itself must be capable of supporting oscillations of the required frequency and will be subject to quantized energy levels. A comparable situation can be set up with a beam splitter using individual visible photons. One half-beam may as well be absorbed so you then have only half a photon's worth of excitation in the other path. Of course what you then observe is 50% photon flux. The same applies to the radio antenna. The photons will be emitted at half the rate on average. Quantum fluctuations are just a (confusing) way of describing this randomness: this transform from a smooth wavefunction to a noisy pattern of discrete observations. An interesting question to illustrate the pitfalls of mixing classical and quantum ideas in a single picture :)

23. Apr 21, 2015

### Ponderer

Yes but as stated a few times my first post was in obvious error and it won't allow me to edit it. Please see my previous post. :)

24. Apr 21, 2015

### Derek Potter

Ah, well, I've arrived in this thread through some mysterious transfer performed by PF - perhaps I answered a link I was sent but in the wrong thread. So as long as you're happy about what goes on that's great.

25. Apr 21, 2015

### The_Duck

OK, sure. Let's consider an antenna with some resistance R. Wikipedia tells me that the power spectral density of the thermal voltage noise across a resistance R at a frequency f is
$$\Phi(f) = \frac{2 R h f}{e^{\frac{hf}{kT}} - 1} \,\, {\rm V}^2/{\rm Hz}$$
So I guess the power spectral density of the noise current is
$$\frac{\Phi(f)}{R^2} = \frac{2 h f}{R \left(e^{\frac{hf}{kT}} - 1\right)} \,\, {\rm A}^2/{\rm Hz}$$
(Is this true?) Classically, given an AC current $I$ at frequency $f$, the power radiated from an antenna is proportional to $I^2 f^2$. I've been claiming that, given some AC current, the classical prediction for the power radiated by that current correct and agrees with the quantum mechanical prediction. If I apply that here I predict that the radiated power per unit frequency will be proportional to
$$\frac{\Phi(f)}{R^2} \times f^2 = \frac{2 h f^3}{R \left(e^{\frac{hf}{kT}} - 1\right)}$$
And indeed this is proportional to the correct Planck spectrum!

At high frequencies the current becomes very small, but it's always true that the radiated power goes like $I^2 f^2$, even in the UV where $hf \gg kT$. The reason the blackbody spectrum falls off rapidly at high frequencies is that the *thermal noise current* falls off rapidly at high frequencies. But it's still true that the radiated power goes like $I^2 f^2$ at all frequencies, as classical electromagnetism predicts. So if you have some other situation where you know the AC current and you want to predict the emitted power, you can use the classical formulas and get the right answer no matter how small the current or how high the frequency.

Last edited: Apr 21, 2015