Sub-photon Radio wave experiment - Your predictions

[Theoretical]

Anyone dare to predict the results? This experiment is still being built. So I don't know the results, yet. I make no claims what the experimental results will be.

 

Definition:
EDP - exponentially decaying pulse.

Goal of experiment 1 - one transmitter, *one* receiver:
1) To emit sub-photons.
2) To possibly know when the EDP (exponentially decaying pulse) is absorbed by the receiving antenna.
3) To absorb sub-photons from one receiving antenna.

Goal of experiment 2 - one transmitter, *two* receivers:
1) To emit sub-photons.
2) To possibly know when the EDP (exponentially decaying pulse) is absorbed by both receiving antennas.
3) To have two separated receiving antennas simultaneously absorb some of the energy in the emitted sub-photon. The software will either multiply both signals together or subtract both signals to guarantee that both antennas received a EDP at the same time every time.


First radio wave experiment:

The following video shows an animation of experiment 1 along with notes and descriptions:
https://youtu.be/oZf9bHqfbu0

If this forum supports animaged gifs, here's an animation of experiment 1. You might have to click on the image.

Experiment consist of one transmitter and one receiver. Both are physically separated by roughly one wavelength, but facing each other. The transmitting antenna is a simple dipole antenna (a straight wire or tube, center fed). The receiving antenna is also a simple dipole.

Every so often the transmitting antenna will transmit an EDP (exponentially decaying pulse). The center frequency of the EDP will be ~ 49MHz.

Example of an EDP:

Important: The total transmitted energy for each entire EDP will be significantly less than one h*f, where h is planck's constant, and f is the center frequency. The oscilloscope will display a spectrum and highlight the center frequency, in addition to displaying the time domain signal.

I'll try to make the *delay* between each EDP as long as possible.



Second radio wave experiment:

Similar to the first experiment, except there will be two receivers. Both receivers will be separated from each other by at least one wavelength, and at least one wavelength away from the transmitter. The second experiment will accomplish the same as experiment 1 with the addition of knowing if both receiving antennas can absorb a sub-photon *simultaneously*.


Again, I'm not saying what the results will be, and actually at this point I'm not sure.

By all means please post your prediction for experiment 2 as well. It'll be interesting.

BTW, as mentioned in the video, the purpose of the IR transmitter & detector is to signal the oscilloscope that an EDP is about to be emitted. This is fed to the oscilloscopes trigger, which causes the oscilloscope to collect a certain amount of data. The data is then fed to the PC, where my software processes the data.

[Sensei]

Important: The total transmitted energy for each entire EDP will be significantly less than one h*f, where h is planck's constant, and f is the center frequency.

Unlikely.

Do you have any idea how much energy has photon at 49 MHz?

 

[math]6.62607004*10^{-34} * 49*10^6=3.2467743196*10^{-26} J[/math]

 

[math]4.135667*10^{-15} * 49*10^6=2.02647722512406*10^{-7} eV[/math]

 

Green photon f.e. 532 nm has 2.33 eV. Nearly 12 millions more energy than photon at 49 MHz.

 

We are flood with photons with higher energy. They are absorbed, emitted, absorbed, emitted, all the time. And heating environment.

Piece of metal used for antenna, will absorb and emit tremendous more energy (black body emission at room temperature) than such single photon..

 

Power equation is:

P=I*U

if you multiply it by t both sides, there is equation for energy (energy lost by electron flowing through conductor, later emitted as f.e. MW,IR,or visible spectrum as photons to cool down resistor/conductor):

P*t=I*t*U

E=I*t*U

 

But I*t=Q is charge, so

E=Q*U

and charge is quantized to e=1.602176565*10^-19

so you have

E=e*U

 

Substitute

h*f=e*U

and solve for your photon @ 49 MHz,

U will be 0.2 uV (micro volt)

 

Try this: take voltmeter, set to the smallest voltage possible to get, connect to some metal piece, and observe. It's jumping back and forth, randomly.

 

ps. I am not trying to discourage you, if you really want to perform experiments. You should perform them, to learn and gain knowledge.

But you also have to be realistic.

Edited January 14, 2016 by Sensei

[Theoretical]

 

Unlikely.

Do you have any idea how much energy has photon at 49 MHz?

 

[math]6.62607004*10^{-34} * 49*10^6=3.2467743196*10^{-26} J[/math]

 

[math]4.135667*10^{-15} * 49*10^6=2.02647722512406*10^{-7} eV[/math]

 

Green photon f.e. 532 nm has 2.33 eV. Nearly 12 millions more energy than photon at 49 MHz.

 

We are flood with photons with higher energy. They are absorbed, emitted, absorbed, emitted, all the time. And heating environment.

Piece of metal used for antenna, will absorb and emit tremendous more energy (black body emission at room temperature) than such single photon..

 

Power equation is:

P=I*U

if you multiply it by t both sides, there is equation for energy (energy lost by electron flowing through conductor, later emitted as f.e. MW,IR,or visible spectrum as photons to cool down resistor/conductor):

P*t=I*t*U

E=I*t*U

 

But I*t=Q is charge, so

E=Q*U

and charge is quantized to e=1.602176565*10^-19

so you have

E=e*U

 

Substitute

h*f=e*U

and solve for your photon @ 49 MHz,

U will be 0.2 uV (micro volt)

 

Try this: take voltmeter, set to the smallest voltage possible to get, connect to some metal piece, and observe. It's jumping back and forth, randomly.

 

ps. I am not trying to discourage you, if you really want to perform experiments. You should perform them, to learn and gain knowledge.

But you also have to be realistic.

 

I get that a lot. Actually I've done a lot of single photon radio wave experiments, albeit different. With sample averaging it's pretty easy to detect sub-nano volts on the digital scope. What most people fail to understand is that blackbody radiation / noise is incoherent, while the signal the oscilloscope receives is coherent. Therefore you can narrow down the frequency bandwidth to unbelievable values that cuts out the noise. That's why four times as many samples of a coherent signal doubles the signal to noise ratio.

 

So I've already done the math. A 49MHz photon is 3.2e-26 joules. Lets see what classical physics predicts using a real example. The antennas radiation resistance according to the NEC2 engine is 0.28 ohms and at 5.0 meters away from the transmitting antenna field is 0.83 V/m per amp-peek. When we enter the antenna into LTspice and set the voltage source so that the transmitting antenna produces 3.2e-26 joules per pulse, the antenna peek current is 1.8nA, produced by a 35nV source pulse (easy task, easy circuit). So given past experiments I can gladly say that a signal of 1.8nA is extremely easy to detect given my setup. This was confirmed hundreds of times with correct predictions. So for example if the antenna is set to radiate a strong signal, and we then decrease the antenna current to say by a 100,000 times, the oscilloscope software shows a drop in signal by 100,000 with a very clear signal spike in the spectrum. The aforementioned example is just an example. I could very well change the details such as frequency, distances, etc.

 

So that's what classical physics predicts. And I ask everyone here, what does QM predict according to your understanding of physics? No need to give specific values. Just a simple experiment result such as "The oscilloscope will not detect the signal."

[swansont]

 

I get that a lot. Actually I've done a lot of single photon radio wave experiments, albeit different. With sample averaging it's pretty easy to detect sub-nano volts on the digital scope. What most people fail to understand is that blackbody radiation / noise is incoherent, while the signal the oscilloscope receives is coherent. Therefore you can narrow down the frequency bandwidth to unbelievable values that cuts out the noise. That's why four times as many samples of a coherent signal doubles the signal to noise ratio.

What is this "unbelievable" value?

 

So I've already done the math. A 49MHz photon is 3.2e-26 joules. Lets see what classical physics predicts using a real example. The antennas radiation resistance according to the NEC2 engine is 0.28 ohms and at 5.0 meters away from the transmitting antenna field is 0.83 V/m per amp-peek. When we enter the antenna into LTspice and set the voltage source so that the transmitting antenna produces 3.2e-26 joules per pulse, the antenna peek current is 1.8nA, produced by a 35nV source pulse (easy task, easy circuit). So given past experiments I can gladly say that a signal of 1.8nA is extremely easy to detect given my setup. This was confirmed hundreds of times with correct predictions. So for example if the antenna is set to radiate a strong signal, and we then decrease the antenna current to say by a 100,000 times, the oscilloscope software shows a drop in signal by 100,000 with a very clear signal spike in the spectrum. The aforementioned example is just an example. I could very well change the details such as frequency, distances, etc.

 

So that's what classical physics predicts. And I ask everyone here, what does QM predict according to your understanding of physics? No need to give specific values. Just a simple experiment result such as "The oscilloscope will not detect the signal."

.

 

I don't understand your math here. How does 1.8nA and 35nV give you 10^-26 joules? What is the 1/e duration of the pulse? What's the time separation between the pulses?

[Sensei]

So I've already done the math. A 49MHz photon is 3.2e-26 joules. Lets see what classical physics predicts using a real example. The antennas radiation resistance according to the NEC2 engine is 0.28 ohms and at 5.0 meters away from the transmitting antenna field is 0.83 V/m per amp-peek. When we enter the antenna into LTspice and set the voltage source so that the transmitting antenna produces 3.2e-26 joules per pulse, the antenna peek current is 1.8nA, produced by a 35nV source pulse (easy task, easy circuit). So given past experiments I can gladly say that a signal of 1.8nA is extremely easy to detect given my setup. This was confirmed hundreds of times with correct predictions. So for example if the antenna is set to radiate a strong signal, and we then decrease the antenna current to say by a 100,000 times, the oscilloscope software shows a drop in signal by 100,000 with a very clear signal spike in the spectrum. The aforementioned example is just an example. I could very well change the details such as frequency, distances, etc.

What oscilloscope model do you have?

 

Is it digital or analog? "Oscilloscope software" suggest it's digital model with USB.

 

What is its vertical resolution? How many bits? Rate?

 

Which oscilloscope model have nano volt resolution?

Edited January 14, 2016 by Sensei

[swansont]

.83 V/m is 1.8 mW/m^2, and that's per amp. You say you have 1.8 nA, so we have 3.24 x 10^-12 W/m^2 at the receiving antenna. (or 10^-16 W/cm^2)

[Theoretical]

What is this "unbelievable" value?

Depends on sample time. One hour (3600 seconds) of sampling is 1/3600 Hz bandwidth given that it's a coherent signal.

 

 

 

I don't understand your math here. How does 1.8nA and 35nV give you 10^-26 joules? What is the 1/e duration of the pulse? What's the time separation between the pulses?

It's an exponentially decaying pulse. As stated 1.8nA is the peek. The 35nV is the DC voltage across the LCR antenna, which is switched on and off. This produces a square wave, which causes the LCR circuit to oscillate. The only way I can show you how it comes to that joule amount is to show you a screenshot of the LTspice power plot, but I'm on mobile right now.

 

 

 

 

What oscilloscope model do you have?

 

Is it digital or analog? "Oscilloscope software" suggest it's digital model with USB.

 

What is its vertical resolution? How many bits? Rate?

 

Which oscilloscope model have nano volt resolution?

Since I require custom software that I wrote to analyze the signal I use a Hantek DSO-2090 PC USB 9bit with 100MB/s sample rate and ~24KB data. So each trigger takes ~ 24 thousand samples. From that you get a spectrum. Take the average of just one thousand spectrums and you'll be there.

 

Like I said. It works. Hundreds of times I've measured sub-nano volts using this method. As stated in my previous post to you it's due to taking a lot of samples.l and taking the average.

[swansont]

Depends on sample time. One hour (3600 seconds) of sampling is 1/3600 Hz bandwidth given that it's a coherent signal.

You imply here that your antenna will reject a signal that's at 49.000000001 MHz. I don't believe that. You also haven't said what the pulse repetition rate is, so that's another reason I don't believe this. If you have one pulse per hour, what happens?

 

It's an exponentially decaying pulse. As stated 1.8nA is the peek. The 35nV is the DC voltage across the LCR antenna, which is switched on and off. This produces a square wave, which causes the LCR circuit to oscillate. The only way I can show you how it comes to that joule amount is to show you a screenshot of the LTspice power plot, but I'm on mobile right now.

 

Yes, I understand it's exponentially decaying. I will ask again: What's the duration of the pulse before it gets to the 1/e value? How long are you waiting between pulses?

[Theoretical]

 

 

 

 

.83 V/m is 1.8 mW/m^2, and that's per amp. You say you have 1.8 nA, so we have 3.24 x 10^-12 W/m^2 at the receiving antenna. (or 10^-16 W/cm^2)

As stated that's the peek of the exponentially decaying pulse, not the average. That's why I wrote 0.83 V/m per amp-peek.

[swansont]

As stated that's the peek of the exponentially decaying pulse, not the average. That's why I wrote 0.83 V/m per amp-peek.

Factors of two (or similar) are inconsequential here

[Sensei]

Since I require custom software that I wrote to analyze the signal I use a Hantek DSO-2090 PC USB 9bit with 100MB/s sample rate and ~24KB data. So each trigger takes ~ 24 thousand samples. From that you get a spectrum. Take the average of just one thousand spectrums and you'll be there.

 

One of the cheapest I saw...

http://www.hantek.com/en/ProductDetail_2_44.html

It's not 100 MB/s, but 100 MS/s (mega samples per second),

and according to producer website it's 8 bits resolution.

It's supporting USB 2.0, which has max 480 Mbps (max 60 MB/s). Effective 280 Mbps (35 MB/s) according to wikipedia.

But here there is said communication speed is 12 Mbps http://www.circuitspecialists.com/usb-oscilloscope-dso-2090.html

Edited January 14, 2016 by Sensei

[Theoretical]

You imply here that your antenna will reject a signal that's at 49.000000001 MHz. I don't believe that. You also haven't said what the pulse repetition rate is, so that's another reason I don't believe this. If you have one pulse per hour, what happens?

Consider this example to understand how it works. A signal repeats thousands of times and you know when the signal starts, which triggers the scope. The part of the signal you're interested in is say flat and it's 1 nV. That signal will be there every time. Sum a million samples and you get 1mV, which comes to of course 1mV / 1000000 = 1nV average. The reason it works is because you will always know that the signal will be there. Of course there's jitter due to noise but that makes no appreciable difference. Now the noise on the other hand decreases by sqrt(1000000) on average.

 

 

 

Yes, I understand it's exponentially decaying. I will ask again: What's the duration of the pulse before it gets to the 1/e value? How long are you waiting between pulses?

I haven't calculated the EDP's 1/e.

 

 

 

 

One of the cheapest I saw...

http://www.hantek.com/en/ProductDetail_2_44.html

It's not 100 MB/s, but 100 MS/s (mega samples per second),

and according to producer website it's 8 bits resolution.

It's supporting USB 2.0, which has max 480 Mbps (max 60 MB/s). Effective 280 Mbps (35 MB/s) according to wikipedia.

But here there is said communication speed is 12 Mbps http://www.circuitspecialists.com/usb-oscilloscope-dso-2090.html

The 9th bit is polarity. It works great for me and does as advertised. Best part is they provide the source code to allow people to write their own custom software.

It's a PC scope, so manufacture doesn't have to provide all of the mechanical stuff found in non-PC scopes.

Sorry. That's right the scope is one byte from 0 to 255, 8bits. 9bits got stuck in my head from the seller advertising it as 9bits.

[Sensei]

Read this f.e.

https://blog.adafruit.com/2012/01/27/why-oscilloscope-bandwidth-matters/

You can't measure 49 MHz signal with oscilloscope that has 40 MHz bandwidth IMHO..

 

"But what does the 50MHz or 100MHz really mean? If I purchase a 50MHz scope can I accurately capture and measure 50MHz worth of data? The answer (like everything else in engineering) is: it depends. You should be able to measure frequency up to and even beyond the maximum rated value, so if determining frequency is all that matters (checking how accurate the output of an oscillator is, the pixel clock on an LCD controller, etc.) you can safely go up to the maximum. Where things become more fuzzy is amplitude (the upper and lower voltage values measured by the scope)."

 

 

[Theoretical]

Read this f.e.

https://blog.adafruit.com/2012/01/27/why-oscilloscope-bandwidth-matters/

You can't measure 49 MHz signal with oscilloscope that has 40 MHz bandwidth IMHO..

 

"But what does the 50MHz or 100MHz really mean? If I purchase a 50MHz scope can I accurately capture and measure 50MHz worth of data? The answer (like everything else in engineering) is: it depends. You should be able to measure frequency up to and even beyond the maximum rated value, so if determining frequency is all that matters (checking how accurate the output of an oscillator is, the pixel clock on an LCD controller, etc.) you can safely go up to the maximum. Where things become more fuzzy is amplitude (the upper and lower voltage values measured by the scope)."

 

 

Sure you can because the sample rate is 100MS/s. The drop isn't that much at 49MHz. But it doesn't matter because the measurements are relative.

Also it can be calibrated by the software if you know the scopes characteristics.

I use a Panasonic VP-8174A signal generator to calibrate my 2090 oscilloscope.

[swansont]

Consider this example to understand how it works. A signal repeats thousands of times and you know when the signal starts, which triggers the scope. The part of the signal you're interested in is say flat and it's 1 nV. That signal will be there every time. Sum a million samples and you get 1mV, which comes to of course 1mV / 1000000 = 1nV average. The reason it works is because you will always know that the signal will be there. Of course there's jitter due to noise but that makes no appreciable difference. Now the noise on the other hand decreases by sqrt(1000000) on average.

Let me put it this way: how does your system differentiate between signals that are a little larger or smaller than your chosen tone? What is the resolution, and how are you determining that? How pure of a tone is your signal?

 

What if your noise isn't white, so it doesn't average down?

[Sensei]

You ignored my link. And at the end of article was link to PDF, where is showed comparison between 100 MHz, 500 MHz, 1 GHz, 2 GHz bandwidths, which are all working on the same signal @ 100 MHz.

http://cp.literature.agilent.com/litweb/pdf/5989-5733EN.pdf

Pages 6-7

 

You're completely not interested in frequency of signal, since 49 MHz, will still be 49 MHz. You are interested only in amplitudes.. So bandwidth does matter..

Edited January 14, 2016 by Sensei

[Theoretical]

Let me put it this way: how does your system differentiate between signals that are a little larger or smaller than your chosen tone? What is the resolution, and how are you determining that? How pure of a tone is your signal?

 

What if your noise isn't white, so it doesn't average down?

I think I know what you're asking. If the scope (no external amp) has a 40uV resolution, and the signal oscillates +/- 1uV, you're wondering how the scope can detect that. Very simple. Noise There's no way the 1uV signal won't show up in the spectrum. I've spent years taking analyzing spectrums on this scope. There's never been a time when the signal suddenly vanished lol. Is that what you're referring to? This scope is noisy. There's a small 50MHz sine wave signal. That's why I amplify the signal with an amp circuit to go above the scopes digital noise.

 

The only non-white noise that's coherent to the sampling is the scope digital noise. Remember it doesn't matter how wobbly the signal is. In fact the delay between EDPs can be random because the IR will always let the scope know when the next EDP will occur.

 

 

 

 

You ignored my link. And at the end of article was link to PDF, where is showed comparison between 100 MHz, 500 MHz, 1 GHz, 2 GHz bandwidths, which are all working on the same signal @ 100 MHz.

http://cp.literature.agilent.com/litweb/pdf/5989-5733EN.pdf

Pages 6-7

 

You're completely not interested in frequency of signal, since 49 MHz, will still be 49 MHz. You are interested only in amplitudes.. So bandwidth does matter..

Of course BW matters, but like I said this scope doesn't change all that much from 40 to 50 MHz. And I calibrate it with a good signal generator. But it's not like I need high precision, like 99.999% of one photon. My minimum requirement is a half of one photon.

[swansont]

I think I know what you're asking. If the scope (no external amp) has a 40uV resolution, and the signal oscillates +/- 1uV, you're wondering how the scope can detect that. Very simple. Noise There's no way the 1uV signal won't show up in the spectrum. I've spent years taking analyzing spectrums on this scope. There's never been a time when the signal suddenly vanished lol. Is that what you're referring to? This scope is noisy. There's a small 50MHz sine wave signal. That's why I amplify the signal with an amp circuit to go above the scopes digital noise.

 

The only non-white noise that's coherent to the sampling is the scope digital noise. Remember it doesn't matter how wobbly the signal is. In fact the delay between EDPs can be random because the IR will always let the scope know when the next EDP will occur.

 

 

Well, no. What I'm asking is similar to what Sensei is talking about. The resolution in the time domain. There's a point were you can't distinguish between frequencies. IOW, your measurement is integrating all of the signal in some frequency range.

 

But I just realized there is a more pressing issue. You're integrating your signal. How will you distinguish between e.g. a half a photon's worth of energy detected every measurement cycle and one photon being detected every other cycle?

[John Cuthber]

OK, let's start with this- since it's accepted by all parties " A 49MHz photon is 3.2e-26 joules."

And I hope nobody will get upset when I convert it to units I'm more "comfortable with

1.997e-7 ev

Call it 0.2 µeV

Now, at room temperature the typical thermal energy is about 0.025 eV

So thermal energies will be something like 125000 times bigger than the energy you are looking for.

So, as is (I accept, very loosely) equivalent, you are only going to me able to see the RF photons if you chill the system down to about 300/25000 kelvin (I'm assuming room temp is about 300K

 

How do you chill the equipment down to 0.0024K in order to stop the signal being swamped by the noise from everything- including you, your mythical oscilloscope and thermal emission from the antenna?

(Obviously, if you want a good s/n ratio, need to get things colder still,but we are already down at about a thousandth of the temperature of deep space and the CMBR- that figures because it's of the order of GHz and you are talking about signals that are MHz)

 

However, lets run on with this flight of fantasy.

Imagine that some time as our distant descendants head towards the heat death of the universe they are still alive and sentient (but running very slowly, in order not to trash the laws of physics).

 

Imagine they actually do the experiment in their cold, quiet, dying world.

 

is there any reason at all to expect that the result will differ from the sort of outcome we (in our 300K world) get with single photon experiment using 1.3 eV visible photons?

 

Unless someone can show why there would be, I'm not buying this thread.

 

Incidentally, thermal noise on 50 ohms at 20C with a 49MHz b/w is about 6µV

Good luck measuring reliably below that.

I'm not saying it's impossible- Just that you probably only think you are doing it right

Edited January 14, 2016 by John Cuthber

[Theoretical]

Well, no. What I'm asking is similar to what Sensei is talking about. The resolution in the time domain. There's a point were you can't distinguish between frequencies. IOW, your measurement is integrating all of the signal in some frequency range.

The signal frequency is stable due to being triggered by the IR signal. Previous experiments have shown that quadrupling the total samples doubles the signal to noise ratio. This is linear so far up to at least several hours of sampling. I think the most I've sampled so far is around a half of one day.

 

 

But I just realized there is a more pressing issue. You're integrating your signal. How will you distinguish between e.g. a half a photon's worth of energy detected every measurement cycle and one photon being detected every other cycle?

Experiment 1 can't tell. Experiment 2 can tell. I'm still undecided if exp.2 is necessary because I'm not convinced that QM will allow for the transmitting antenna to emit one hf when only a half hf worth of energy was put into the antenna. I think each EDP is a done deal. How would the universe know when I'm going to emit the next EDP? Maybe I won't even emit another one. This gets into theory, I think, and that's not what I want to talk about in this thread. Although I welcome anyone who wants to post their prediction and theory or interpretation of the experiment.

The main goal of experiment 1 is to see what signal, if any, the transmitting antenna radiates if each EDP is at most a half of one photon worth of energy.

The main goal of experiment 2 is to see if *two* receiving antennas detect the *same* signal on *both* receiving antennas *every time*.

 

 

 

 

OK, let's start with this- since it's accepted by all parties " A 49MHz photon is 3.2e-26 joules."

And I hope nobody will get upset when I convert it to units I'm more "comfortable with

1.997e-7 ev

Call it 0.2 µeV

Now, at room temperature the typical thermal energy is about 0.025 eV

So thermal energies will be something like 125000 times bigger than the energy you are looking for.

So, as is (I accept, very loosely) equivalent, you are only going to me able to see the RF photons if you chill the system down to about 300/25000 kelvin (I'm assuming room temp is about 300K

 

How do you chill the equipment down to 0.0024K in order to stop the signal being swamped by the noise from everything- including you, your mythical oscilloscope and thermal emission from the antenna?

(Obviously, if you want a good s/n ratio, need to get things colder still,but we are already down at about a thousandth of the temperature of deep space and the CMBR- that figures because it's of the order of GHz and you are talking about signals that are MHz)

 

However, lets run on with this flight of fantasy.

Imagine that some time as our distant descendants head towards the heat death of the universe they are still alive and sentient (but running very slowly, in order not to trash the laws of physics).

 

Imagine they actually do the experiment in their cold, quiet, dying world.

 

is there any reason at all to expect that the result will differ from the sort of outcome we (in our 300K world) get with single photon experiment using 1.3 eV visible photons?

 

Unless someone can show why there would be, I'm not buying this thread.

 

Incidentally, thermal noise on 50 ohms at 20C with a 49MHz b/w is about 6µV

Good luck measuring reliably below that.

I'm not saying it's impossible- Just that you probably only think you are doing it right

 

I believe that was already addressed:

 

I get that a lot. Actually I've done a lot of single photon radio wave experiments, albeit different. With sample averaging it's pretty easy to detect sub-nano volts on the digital scope. What most people fail to understand is that blackbody radiation / noise is incoherent, while the signal the oscilloscope receives is coherent. Therefore you can narrow down the frequency bandwidth to unbelievable values that cuts out the noise. That's why four times as many samples of a coherent signal doubles the signal to noise ratio.

 

So I've already done the math. A 49MHz photon is 3.2e-26 joules. Lets see what classical physics predicts using a real example. The antennas radiation resistance according to the NEC2 engine is 0.28 ohms and at 5.0 meters away from the transmitting antenna field is 0.83 V/m per amp-peek. When we enter the antenna into LTspice and set the voltage source so that the transmitting antenna produces 3.2e-26 joules per pulse, the antenna peek current is 1.8nA, produced by a 35nV source pulse (easy task, easy circuit). So given past experiments I can gladly say that a signal of 1.8nA is extremely easy to detect given my setup. This was confirmed hundreds of times with correct predictions. So for example if the antenna is set to radiate a strong signal, and we then decrease the antenna current to say by a 100,000 times, the oscilloscope software shows a drop in signal by 100,000 with a very clear signal spike in the spectrum. The aforementioned example is just an example. I could very well change the details such as frequency, distances, etc.

 

So that's what classical physics predicts. And I ask everyone here, what does QM predict according to your understanding of physics? No need to give specific values. Just a simple experiment result such as "The oscilloscope will not detect the signal."

[John Cuthber]

 

I believe that was already addressed:

 

I don't.

Does anyone else?

[Theoretical]

I don't.

Does anyone else?

I've placed a known voltage source on the scope that was shunted with known resistors that according to simple well known equations would produce a nano volt. In every test the oscilloscope software showed the correct predicted voltage. And the signal was far above noise. Hundreds of tests have shown that quadrupling the total samples doubles the SNR, as expected. My setup has shown this to be linear to at least several hours of sampling time.

 

If anyone has any reasonable tests they would like to prove their claims, then by all means post it.

[Enthalpy]

I like this experiment! A few thoughts at it:

 

At 49MHz, the photon energy is 3.2e-26J. At 4K, kT is 1.6e-20J. Do you have any trick for the amplifier, the attenuator and the transmitting antenna's losses not to radiate their thermal noise? I know no way to distinguish the radiated thermal noise from the signal. Or can you chill the experiment to <8µK? You need to chill the whole experiment chamber since the receiving antennas pick noise from the surroundings.

 

A capacitive attenuator instead of a resistive one would make very little noise. Or a weak coupler provided you can remove all passive loads that would introduce noise. Then you could have superconductive antennas that make very little thermal noise. I expect a dirty trick with the bandwidth here, becoming very narrow and introducing on the pulse duration a physical limit disguised as a technological one. Dielectric antennas radiate too and have smaller losses hence produce less noise.

 

Why are your antennas so small? <1ohm radiation resistance is a source of worries. Approaching a half-wave would ease the experiment - OK, I understand you seek near-field coupling.

 

Detecting 3.2e-26J received once isn't reasonable but you'll use many pulses. If the receiver noise temperature is 4K=1.6e-20J (is it?), then 2.5e11 pulses let the signal just appear at 1 sigma; a pulse repetition of 1MHz takes 3 days. 3 sigma bring you to 26 days and 5 sigma to 72 days. Nice.

 

Why use such a low frequency? At 49GHz instead, the experiment needs 8mK, much easier than 8µK. Amplifying is still feasible (or 26GHz, more common and easier), a correlation by analog means too, and you can downconvert the frequency to an intermediate frequency limited only by the pulse duration.

 

One tiny diode laser worked at EPFL whose resonator was an LC circuit of proper size for visible or near-IR wavelength instead of the usual cavity. Metals are quite lossy at such frequency but it did resonate. Patterning the transmitting and both receiving antennas on the semiconductor would be easy. Why shouldn't you go to these frequencies? There the photon energy exceeds the ambiant temperature.

 

From the analogy with optics, where coupling by evanescent waves exists too, and these are near-field coupling too or resemble a lot, I'd dare to make a prediction or rather a bet... I expect the signal to be quantized, and one single photon to couple to one receiving antenna OR the other

BUT

I haven't understood what guarantees in the setup that one single photon is transmitted. Let's forget thermal noise at the transmitter: just with a generator and an attenuator, you get a pulse amplitude smaller than one photon. This does not guarantee that one single photon is transmitted. There can be more, this is just a statistics. The hard part is that, as you supposedly want to see whether both receivers get a signal or not, the probability to emit two photons is exactly the amount you want to exclude from the correlation.

SO

what you need is a source that guarantees that only one photon is transmitted, and the small pulse power does not do that. The lone photon must result from a lone de-excited electron, not from an attenuator. This is commonly done in optics by using one fluorescent atom, for instance nitrogen in diamond. In case radiowaves are better, something similar can probably be done. Ammonia maser, hydrogen maser, adapted rubidium clock, all with a single molecule...? With the proper cavity they must deexcite more quickly. A big quantum well with one single dopant atom?

[Theoretical]

Thanks for the nice reply. I already discovered the experiment results.

 

I'd have to write a lengthy paper or better yet an instructional video. I can't expect people to spend much time understanding the experiment.

 

I know it's an unreasonable request but if one wants to see how it's incredibly simple to see the photon over noise at such radio frequencies then you'll have to spend a good amount of time reading my initial posts, studying every sentence.

 

Unfortunately I'm no longer at liberty to discuss the experiment. Although after some thought I will say that one issue mainstream is having has to do with the nature of electromagnetic forces longitudinally (dominated above radio frequencies, photon momentum), and transverse (dominated at radio frequencies). There mainstream will discover something special.

Edited April 6, 2016 by Theoretical

[Enthalpy]

Nice to see you! I feared you had gone since mid-January.

 

Will you put the paper on arXiv please?

 

I hope you'll treat the question of unlucky multiple photon emission in the paper.