Quantum fluctuations at radio frequencies

[f95toli]

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.

 

[Ponderer]

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?

No. Your power spectral density equation is wrong:

http://www.ecircuitcenter.com/Circuits/Noise/Noise_Analysis/res_noise.htm

http://en.m.wikipedia.org/wiki/Johnson–Nyquist_noise

 

[Ponderer]

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.

 

[f95toli]

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.

 

[sheaf]

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
(can't find it on the arxiv unfortunately).

 

[Ponderer]

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.

In electronics we have amazing control over the signal and know to high accuracy how much energy per unit of time the circuit has produced. Take for instance a circuit that produces just one sine wave, say 50MHz, through an antenna that has say 0.01 ohm of radiation resistance at 50MHz. Additionally, as stated in the original post in this thread, there is a resistor of high resistance to keep thermal noise current down. Let's say the resistance is 1Mohm. And let's set the voltage of our 50MHz source, which only produces one wavelength, is 0.13V. If you do the math you'll see this produces a bit more than the required energy for one photon. That antenna will either produce one photon or no photons. Maybe on rare occasional it will emit two photons in one wavelength since the energy is a bit above two h*f.

Over the years I've heard nothing but great news about QED. Can you please tell me exactly what the current signal will look like in the aforementioned example when repeated once ever second? And also what would the signal look like when absorbed by a similar antenna? Thanks

 

[Ponderer]

Sorry, my previous post made a few minutes ago might be asking too much of QED. If that's the case, then please detail what the electromagnetic near field will look like around the transmitting antenna. Maybe we can see this with a sensor next to the antenna.

 

[f95toli]

If you do the math you'll see this produces a bit more than the required energy for one photon. That antenna will either produce one photon or no photons. Maybe on rare occasional it will emit two photons in one wavelength since the energy is a bit above two h*f.

No it won't. If the current drive is reasonably coherent (this is usually a good approximation) it will on give you a signal where the power corresponds to an average photon number of about 1 photon per second, but it is impossible to say anything about "number of photons per second" in this case, not because we don't know but because the number of photons is not fixed and has a Poissonian distribution.

If you wanted to make a QED model of this you could start by drawing an equivalent lumped element circuit (R, C and L), write down the (classical) Lagrangian, quantize it and then write down the Hamiltonian. There are some nice review papers out there that includes some examples.



.

 

[Ponderer]

No it won't. If the current drive is reasonably coherent (this is usually a good approximation) it will on give you a signal where the power corresponds to an average photon number of about 1 photon per second, but it is impossible to say anything about "number of photons per second" in this case, not because we don't know but because the number of photons is not fixed and has a Poissonian distribution.

If you wanted to make a QED model of this you could start by drawing an equivalent lumped element circuit (R, C and L), write down the (classical) Lagrangian, quantize it and then write down the Hamiltonian. There are some nice review papers out there that includes some examples.

So you're saying it will emit one photon, but not all the time?

 

[f95toli]

Sorry, my previous post made a few minutes ago might be asking too much of QED. If that's the case, then please detail what the electromagnetic near field will look like around the transmitting antenna. We can see this with a sensor next to the antenna.

it will look like the classical field. Maxwell's equations are to some extent inherently "quantum" and it is usually only when you focus on the photon statistics (correlation functions) that you see any deviations from the classical result (which of course does not include photons). This actually works even when you are working with true single photon sources, in e.g. a cavity resonator the classical anti-node is where you are most likely to interact with a single photon.

 

[f95toli]

So you're saying it will emit one photon, but not all the time?

No, I am saying it will emit something which if you detect the signal will have e.g a Poissonian distribution. Again., you can't talk about an exact "number of photons" because the field is not in a number state.

Look up some information about photon anti-bunching on e.g. Wikipedia.

 

[Ponderer]

it will look like the classical field. Maxwell's equations are to some extent inherently "quantum" and it is usually only when you focus on the photon statistics (correlation functions) that you see any deviations from the classical result (which of course does not include photons). This actually works even when you are working with true single photon sources, in e.g. a cavity resonator the classical anti-node is where you are most likely to interact with a single photon.

In electronics we can know how much energy goes into each wave, given the noise is low enough. This circuit is only giving the antenna one photons worth of energy. So if you feel it will emit more than one photon, please tell where this free energy comes from.

 

[Ponderer]

No, I am saying it will emit something which if you detect the signal will have e.g a Poissonian distribution. Again., you can't talk about an exact "number of photons" because the field is not in a number state.

Look up some information about photon anti-bunching on e.g. Wikipedia.

This doesn't make sense to me because as stated the signal is not a continuous signal. It's one wavelength long and we're only given it one h*f. Please can you be specific what to expect. I know why poissonian distribution is. So you're saying that most of the time it will emit one photon, but less often it will not or may emit two photons?

 

[f95toli]

In electronics we can know how much energy goes into each wave, given the noise is low enough. This circuit is only giving the antenna one photons worth of energy. So if you feel it will emit more than one photon, please tell where this free energy comes from.

No you can't. No unless you are using a single photon source. Every "normal" measurement of energy will only tell you the average power applied. It is quite easy -if you have the right equipment- to apply powers which -on average- correspond to something much less than one photon per second, and you will still get a signal out.

Also, you should be careful when mixing the concepts of "wave" and photons. You can't apply a signal that is "one wavelength long", even in classical physics (long in what sense? time?). If you are referring to short pulses you get into trouble because the shorter you make the pulse the wider the frequency spectra; meaning you end with a very broad distribution of photon energies.

 

[Ponderer]

No you can't. No unless you are using a single photon source. Every "normal" measurement of energy will only tell you the average power applied. It is quite easy -if you have the right equipment- to apply powers which -on average- correspond to something much less than one photon per second, and you will still get a signal out.

Also, you should be careful when mixing the concepts of "wave" and photons.

Yes you can! Nearly all of the energy in the circuit goes to the resistor, not the antenna. Since the resistor and antenna are in-series, we know what the current signal is through the antenna. If we know the antennas radiation resistance, then we know the energy for the sine pulse or wave.

 

[Ponderer]

Also, you should be careful when mixing the concepts of "wave" and photons. You can't apply a signal that is "one wavelength long", even in classical physics (long in what sense? time?). If you are referring to short pulses you get into trouble because the shorter you make the pulse the wider the frequency spectra; meaning you end with a very broad distribution of photon energies.

Yes. A single sine pulse has a bandwidth about equal to the center frequency.

 

[f95toli]

Yes. A single sine pulse has a bandwidth about equal to the center frequency.

I am not sure this is even a QM issue.

What do you mean by "sine pulse"? If you are talking about simply creating current pulses that are one period long in time, you will then end up with a very wide frequency distribution, i.e. you can no longer even talk about photons of a specific energy.

Also, you do realize that "one wavelength" does not correspond to "one photon"? The former is a classical concept and can't be applied directly to photons. Photons do not have a well defined "size" in that way, neither in time nor space.

 

[Ponderer]

I am not sure this is even a QM issue.

What do you mean by "sine pulse"? If you are talking about simply creating current pulses that are one period long in time, you will then end up with a very wide frequency distribution, i.e. you can no longer even talk about photons of a specific energy.

Also, you do realize that "one wavelength" does not correspond to "one photon"? The former is a classical concept and can't be applied directly to photons. Photons do not have a well defined "size" in that way, neither in time nor space.

A sine pulse is a dampened single sine wave. When it's about as short in duration as possible the bandwidth is equal to about that of the center frequency.

 

[Ponderer]

It's a Quantum Mechanics issue because the energy to the antenna is that of one 50MHz photon. There is a center frequency to the sine pulse., which is 50MHz

 

[Ponderer]

Here's a thought. Instead of sending one wavelength to the antenna, what if we send say 100 wavelengths, but set the energy per wavelength to 1/100th? That way the bandwidth is shorter, and according to Quantum Mechanics the probability is there that somewhere in the 100 wavelengths one photon will be emitted. This is what's wonderful about radio frequencies. We have amazing control over the current and signal. :)

 

[f95toli]

It's a Quantum Mechanics issue because the energy to the antenna is that of one 50MHz photon. There is a center frequency to the sine pulse., which is 50MHz

No, thing don't become "quantum mechanical " just because you reduce the power,. Try the following "gedanken" experiment, Imagine a setup with a 50 MHz sine-wave signal generator that is outputting a perfectly coherent signal. If you connect this to a spectrum analyzer you will see a peak at 50 MHz. Now gradually reduce the signal; as you reduce the power the amplitude of the peak will decrease and if you had a perfect analyzer you could go on doing this forever. At some point you might start to observe the effects of shot noise because the of the "discrete nature" of the incoming photons, but other than that nothing special happens; you can keep going until the average number of photons per second become arbitrarily small.

Now imagine the same setup except that you introduce a pulse modulator after the generator. This "chops" the signal by turning it in and off at some rate. If you now send this into your analyzer. You will now observe lots of peaks ; both harmonics of the 50 MHz signal and other peaks at frequencies related to the modulation frequency and the rise and fall time of your modulator. Each peak corresponds to distributions of photons of different frequency reaching your analyzer. Hence, you no longer have a "pure" signal with photons of a a single frequency.
You can also reduce the power, but as long as you focus at a single peak it is exactly the same as for a continuous wave.

Here's a thought. Instead of sending one wavelength to the antenna, what if we send say 100 wavelengths, but set the energy per wavelength to 1/100th? That way the bandwidth is shorter, and according to Quantum Mechanics the probability is there that somewhere in the 100 wavelengths one photon will be emitted. This is what's wonderful about radio frequencies. We have amazing control over the current and signal. :)

Again, you can't send "wavelengths" into a circuit. That terminology does not make sense even in classical EM. A longer pulse will obviously result in a narrower frequency distribution, but why not then simply use a continuous wave?

The point is that creating single RF photons is very difficult, and it is very much an active field or research. The first experiment to demonstrate single microwave photons wa done only a few years ago; and there are several ongoing projects where people are trying to do this as efficiently as possible (I work on one such European project)

 

[Ponderer]

Honestly, one's definition of when something falls under a Quantum Mechanics is not so interesting to me. ;)

I think we get too caught up in frequency. Wikipedia says radiation resistance is caused by radiation reaction, which redirects to Abraham–Lorentz force, which in terms of an individual photon causes a spike, not sine wave. If a continuos sine wave current signal is produced in the antenna, then a sine signal is emitted. If it's a square wave given a certain bandwidth, then that's the signal that's emitted. Early this year I wrote some software that shows how spikes given the probability of producing a sine wave, say 50MHz, will produce a signal that has 50MHz in the spectrum even if on average there's only one spike every 1000 cycles. This is in reference to a continuous signal.

Anyhow I disagree with you on single radio wavelength photons. I think you're getting to caught up on trying to nail each photon down to a specific frequency. It's a packet of energy that has a spectrum given a starting and ending time domain, which changes depending on the FFT input time range even if it there's no trailing or leading signal. At radio frequencies it's very easy to limit the time duration of that packet of energy. :)

BTW I've sent you and Duck a private conversation message, but haven't heard from either one of you. Did you receive it?

 

[Ponderer]

For several months I've been designing a single radio wavelength photon experiment. Hopefully this thread can remain open until I publish the results.

If this thread closes, then you can periodically check my profile for a new thread. Or you could follow me, although I'm not sure if that would send you alerts if I start a new thread.

Soon we'll know for certain the results of this experiment. :)

 

[Ponderer]

I made a major improvement to the experiment that will allow us to tell if The_Duck and f95toli are correct. They're saying that even if thermal noise is low and not an issue, and if the antenna current is set considerably below the one photon per wavelength threshold, that photons will be emitted but just not every wavelength. So for example if the antenna current is set to 1/10th the level that produces one photon per wavelength, they are saying one photon is emitted once every ten wavelengths on average with a poisson distribution.

I disagree with them. Here's an example that clearly explains why I disagree. Imagine a stiff button. If you tap the button, then you can tap it a billion times till you're blue in the face and it's not going to press the button. Yes, normally I believe the photon would be emitted every so often in a classical way because thermal energy brings the antenna current beyond the threshold. But as started from the start of my experiment, there are resistors of high resistance that bring the thermal noise current significantly below the threshold. And hence the title for this thread. Will quantum fluctuations somehow cause the photon to emit even if thermal noise is too low?

Anyhow, hopefully soon we'll see the results of the experiment. Maybe within a week if all goes well.

 

[jtbell]

This thread is now closed, as per PF policy on (not) discussing unpublished research. When you publish your results in a professional peer-reviewed journal, feel free to use the "Report" link at the bottom of this post to ask the Mentors to re-open this thread if you want.