Quantum fluctuations at radio frequencies

In summary, the probability of a photon being emitted from a radio antenna where the energy per wavelength is below the threshold to emit photons can be calculated by starting with the equation for classical electromagnetism and then subtracting out thermal noise.
  • #1
Ponderer
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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.
 
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  • #2
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
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
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.
 
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  • #5
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
Ponderer said:
Yes, I'm concerned that Quantum Mechanics equation might be taking thermal noise into account.

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.

Ponderer said:
In the excited atom example, the sufficient energy is already there. It's excited. Although in the antenna example the energy is insufficient.

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.

Ponderer said:
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.

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.

Ponderer said:
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.

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
The_Duck said:
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.
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.
The_Duck said:
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.
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.

The_Duck said:
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.
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.
The_Duck said:
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.
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!
 
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  • #8
Ponderer said:
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.

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.)

Ponderer said:
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.

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.

Ponderer said:
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.

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.
 
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  • #9
The_Duck said:
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.)
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.
The_Duck said:
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.
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?
 
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  • #10
The_Duck said:
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.
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
Ponderer said:
The classical prediction for photons being emitted at UV due to thermal energy is *wrong.*

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.

Ponderer said:
The energy from each cycle in the antenna does not accumulate because the antenna circuit is nearly entirely resistive, not reactive.

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.
 
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  • #12
The_Duck said:
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.
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.
The_Duck said:
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.
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?
The_Duck said:
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.
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
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.
 
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  • #14
Ponderer said:
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.

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.

Ponderer said:
That's obviously wrong.

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.

Ponderer said:
See UV catastrophe. The thermal energy is still there trying to emit those photons,

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?

Ponderer said:
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?

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.
 
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  • #15
The_Duck said:
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.
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.
The_Duck said:
In QM the energy is not gradually transferred, rather there is a gradual buildup of the probability that energy has been transferred. This is worked out in QM textbooks under the name "Fermi's golden rule."
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?
The_Duck said:
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 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 frequency.

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.

Note that this sort of coherence is not present in a thermal bath, which is why quantum blackbodies radiate differently from classical ones
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?
 
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  • #16
Ponderer said:
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.

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.

Ponderer said:
Your only point now is that photons are emitted

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
The_Duck said:
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.
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.
The_Duck said:
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.
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.
 
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  • #18
Ponderer said:
And you haven't shown the math to prove this.

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
The_Duck said:
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.
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
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
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
Ponderer said:
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.
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
Derek Potter said:
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 :)

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
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
Ponderer said:
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.
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
[tex]\Phi(f) = \frac{2 R h f}{e^{\frac{hf}{kT}} - 1} \,\, {\rm V}^2/{\rm Hz}[/tex]
So I guess the power spectral density of the noise current is
[tex]\frac{\Phi(f)}{R^2} = \frac{2 h f}{R \left(e^{\frac{hf}{kT}} - 1\right)} \,\, {\rm A}^2/{\rm Hz}[/tex]
(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
[tex]\frac{\Phi(f)}{R^2} \times f^2 = \frac{2 h f^3}{R \left(e^{\frac{hf}{kT}} - 1\right)}[/tex]
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.
 
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  • #26
The_Duck said:
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
[tex]\Phi(f) = \frac{2 R h f}{e^{\frac{hf}{kT}} - 1} \,\, {\rm V}^2/{\rm Hz}[/tex]
So I guess the power spectral density of the noise current is
[tex]\frac{\Phi(f)}{R^2} = \frac{2 h f}{R \left(e^{\frac{hf}{kT}} - 1\right)} \,\, {\rm A}^2/{\rm Hz}[/tex]
(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
[tex]\frac{\Phi(f)}{R^2} \times f^2 = \frac{2 h f^3}{R \left(e^{\frac{hf}{kT}} - 1\right)}[/tex]
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.
Sorry but that's the wrong answer lol. You need to arrive at the same results that academic scientists already know as being factual, which is classical mechanics does not correctly predict blackbody radiation. Blackbody radiation at higher frequencies drops significantly because the energy to emit a photon is higher.
 
  • #27
Ponderer said:
classical mechanics does not correctly predict blackbody radiation.

Right; the reason I got the correct spectrum is that my expression for ##\Phi(f)## takes into account quantum effects.

Ponderer said:
Blackbody radiation at higher frequencies drops significantly because the energy to emit aphoton is higher.

Yes, and I think the derivation I gave above gives a deeper understanding of this. As Derek Potter mentioned, at high frequencies the energy levels of the current oscillations in the antenna are quantized just like the energies of electromagnetic waves are quantized. To create a photon with frequency f, thermal fluctuations must first excite an electric current with frequency f. But this current is hard to excite for the reason you give: the quantization of energy means that the smallest possible energy of a current oscillation with frequency f is hf. So for kT << hf this will only happen very rarely. This is why the quantum form for ##\Phi(f)## that I gave above is different from the classical form ##2RkT##. This step, creating an oscillating current at frequency f, is the hard part. Once you have some charge oscillating at frequency f, which must have energy at least hf, there's no particular trouble to create a photon at frequency f. *Given that you already have some current* at frequency f, photons are created at the same rate as they would be classically.

Said another way, QM implies that there is a minimum energy hf for oscillations at a frequency f. Thus these oscillations will be hard to produce at high f, which is why blackbody radiation falls off for high f. But it's not particularly hard to transfer energy from one type of oscillation (electric current) to another type oscillation (electromagnetic waves) at the same frequency.
 
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  • #28
That's why the blackbody radiation equation answered my question, since the antenna example is the same. The energy is quantized. It takes a minimum energy to emit the photon. In my example if thermal energy is insufficient, and the current source is insufficient, the radiation drops significantly below the classical prediction, and quantum fluctuations are insignificant as evident in the UV catastrophe. Agreed?
 
  • #29
BTW if you study the blackbody radiation and UV catastrophe page at Wikipedia it says the problem in classical mechanics is because the energy is quantized.
 
  • #30
I claim that *if you know the current*, you can use classical mechanics to predict the rate of photon emission. (I demonstrated this above for blackbody radiation). Is that compatible with what you're saying?
 
  • #31
I don't think the question is complete. There is not enough information.

You need to specify how the photons are created. If you just apply an AC current to antenna you won't create single photons regardless of how low the energy is; what you will end up is radiation that is roughly coherent or possible thermal. In either case you often experimentally have a situation where the average photon number in the antenna is much smaller than one but the antenna is still emitting radiation. T|his is a very common situation experimentally when working with quantum circuitry (qubits etc).
It is possible to create single RF (well, microwave) photons and there is no reason why you couldn't emit them via an antenna, but the details of the "conversion" between the applied energy (which could come from a current) and the creating of a photon matters.
Creating single photon (Fock) states is actually very difficult and just applying an AC current to an antenna won't work.
 
  • #33
f95toli said:
I don't think the question is complete. There is not enough information.

You need to specify how the photons are created. If you just apply an AC current to antenna you won't create single photons regardless of how low the energy is; what you will end up is radiation that is roughly coherent or possible thermal. In either case you often experimentally have a situation where the average photon number in the antenna is much smaller than one but the antenna is still emitting radiation. T|his is a very common situation experimentally when working with quantum circuitry (qubits etc).
It is possible to create single RF (well, microwave) photons and there is no reason why you couldn't emit them via an antenna, but the details of the "conversion" between the applied energy (which could come from a current) and the creating of a photon matters.
Creating single photon (Fock) states is actually very difficult and just applying an AC current to an antenna won't work.
If you're working with qubits, then I'd agree. Not that I know much about them, mind you. In terms of an antenna that has exceptionally high resistance as I've described, I disagree with you. But that's my opinion, because it seems apparent to me the detailed nature of the "photon" is yet to be understood. I have my theories, which I'll keep to myself for now. :) No sense in starting another debate or heavy discussion over it now. We'll just have to wait and see when the experiments come in.
 
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  • #34
We have a very good understand of what photons are (QED), it is just that it is very difficult to explain using normal language you just need some pretty sophisticated math.

There are many ways to create single photons, but the situation you describe is not one of them: When people first leans about photons they often assume that this means that ALL light is made up of a "stream of photons". However, this is not the case. For most states of light there isn't a fixed number of photons, it will always fluctuate (this is why you can't get a true single photon source by simply attenuating e.g. a laser).
You can of course always talk about an average number of photons but it should be obvious that nothing unusual happens as the average photon number goes below one. Hence, in this situation you might as well use classical EM; the answer will be (nearly*) identical to what you would get with a full QM treatment *You might get some some small deviations from the classical case if you look at high order moments.
 
  • #35
f95toli said:
We have a very good understand of what photons are (QED), it is just that it is very difficult to explain using normal language you just need some pretty sophisticated math.

There are many ways to create single photons, but the situation you describe is not one of them: When people first leans about photons they often assume that this means that ALL light is made up of a "stream of photons". However, this is not the case. For most states of light there isn't a fixed number of photons, it will always fluctuate (this is why you can't get a true single photon source by simply attenuating e.g. a laser).
You can of course always talk about an average number of photons but it should be obvious that nothing unusual happens as the average photon number goes below one. Hence, in this situation you might as well use classical EM; the answer will be (nearly*) identical to what you would get with a full QM treatment*You might get some some small deviations from the classical case if you look at high order moments.
Yes, in the past I've come across this paper by Greulich, which talks about the state of light from a source in which atoms "cooperate" http://www.fli-leibniz.de/www_kog/research/physics/SPIE.pdf [Broken]
(can't find it on the arxiv unfortunately).
 
Last edited by a moderator:
<h2>1. What are quantum fluctuations at radio frequencies?</h2><p>Quantum fluctuations at radio frequencies refer to small, random fluctuations in the electromagnetic field at the quantum level. These fluctuations occur due to the uncertainty principle, which states that the position and momentum of a particle cannot be simultaneously known with absolute certainty.</p><h2>2. How do quantum fluctuations at radio frequencies affect our daily lives?</h2><p>Quantum fluctuations at radio frequencies have a minimal impact on our daily lives. However, they play a crucial role in technologies such as MRI machines and GPS systems, which rely on precise measurements of radio frequencies.</p><h2>3. Can quantum fluctuations at radio frequencies be observed?</h2><p>Yes, quantum fluctuations at radio frequencies can be observed through various experiments, such as the Casimir effect and the Lamb shift. These experiments demonstrate the effects of quantum fluctuations on the behavior of particles and electromagnetic fields.</p><h2>4. Are there any practical applications of quantum fluctuations at radio frequencies?</h2><p>Yes, there are practical applications of quantum fluctuations at radio frequencies in fields such as quantum computing and quantum communication. These technologies utilize the principles of quantum mechanics, including quantum fluctuations, to perform complex calculations and secure communication.</p><h2>5. How do scientists study quantum fluctuations at radio frequencies?</h2><p>Scientists study quantum fluctuations at radio frequencies through various experimental techniques, such as spectroscopy and interferometry. These techniques allow them to measure and analyze the behavior of electromagnetic fields and particles at the quantum level.</p>

1. What are quantum fluctuations at radio frequencies?

Quantum fluctuations at radio frequencies refer to small, random fluctuations in the electromagnetic field at the quantum level. These fluctuations occur due to the uncertainty principle, which states that the position and momentum of a particle cannot be simultaneously known with absolute certainty.

2. How do quantum fluctuations at radio frequencies affect our daily lives?

Quantum fluctuations at radio frequencies have a minimal impact on our daily lives. However, they play a crucial role in technologies such as MRI machines and GPS systems, which rely on precise measurements of radio frequencies.

3. Can quantum fluctuations at radio frequencies be observed?

Yes, quantum fluctuations at radio frequencies can be observed through various experiments, such as the Casimir effect and the Lamb shift. These experiments demonstrate the effects of quantum fluctuations on the behavior of particles and electromagnetic fields.

4. Are there any practical applications of quantum fluctuations at radio frequencies?

Yes, there are practical applications of quantum fluctuations at radio frequencies in fields such as quantum computing and quantum communication. These technologies utilize the principles of quantum mechanics, including quantum fluctuations, to perform complex calculations and secure communication.

5. How do scientists study quantum fluctuations at radio frequencies?

Scientists study quantum fluctuations at radio frequencies through various experimental techniques, such as spectroscopy and interferometry. These techniques allow them to measure and analyze the behavior of electromagnetic fields and particles at the quantum level.

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