Stargazing Phase of radio waves received by a radio telescope dish

[jordankonisky]

I understand that a radio telescope can be tuned to receive radio waves generated by neutral hydrogen atoms present in galactic gas, for example, within the spirals of the milky way. I think I understand that the incoming radio waves will be a mixture of red- and blue-shifted photons depending on the whether the photon sources are moving towards or away from the dish. But what about the phases of the collection of photons? Since the sources may differ in distance and time of photon release, is the signal a collection of photons in different phases? And, if so, how does the telescope take this into account in processing the data? Or perhaps the phase does not matter at all. Am I thinking about this in the right way?

 

[Charles Link]

This is something that I don't know that the people who are experts in the Quantum Theory will be in agreement on, but I believe, even in a laser, the phases of the individual photons, (assuming you can assign a phase to an individual photon), would be completely random. If the individual photons were in phase, this would necessarily violate energy conservation, because the energy (irradiance (watts/m^2)) of the electromagnetic wave is proportional to the square of the electric field amplitude, while the energy must also be proportional to the number of photons $N$. If the photons are in phase, we have a real dilemma, because $E_{total}=NE_o$ making the energy proportional to $N^2$. Alternatively If the phases are random, the phasor diagram becomes a 2-D random walk, with $E_{total} \approx \sqrt{N} E_o$, making the energy proportional to $N$ as it needs to be. $\\$ Hopefully this post doesn't get "pinged" for being a personal theory, but it is the best answer I have for what otherwise is a very big dilemma.

 

[Tom.G]

...radio telescope...

...photon sources...

Huh?

 

[Charles Link]

Huh?

I thought the OP is asking a reasonable question. The radio telescope is basically a large antenna that looks in one particular direction, and radio waves will get picked up as a sinusoidal voltage on an oscilloscope. You can expect fluctuations in amplitude and phase if you have an electronic filter/amplifier system that looks at a particular frequency. $\\$ The amplitude and phase of the radio waves is a result of r-f (radio frequency) photons that are getting picked up by the receiver at any given time. There will be fluctuations in photon counts.

 

[kimbyd]

I understand that a radio telescope can be tuned to receive radio waves generated by neutral hydrogen atoms present in galactic gas, for example, within the spirals of the milky way. I think I understand that the incoming radio waves will be a mixture of red- and blue-shifted photons depending on the whether the photon sources are moving towards or away from the dish. But what about the phases of the collection of photons? Since the sources may differ in distance and time of photon release, is the signal a collection of photons in different phases? And, if so, how does the telescope take this into account in processing the data? Or perhaps the phase does not matter at all. Am I thinking about this in the right way?

The phase of the photons is only of interest when combining the signals from multiple radio telescopes. With multiple detectors, the relative phase between the different detectors can be used for interferometry.

The absolute phase is irrelevant in all contexts that I'm aware of. But the relative phase can be of tremendous interest.

 

[vela]

I understand that a radio telescope can be tuned to receive radio waves generated by neutral hydrogen atoms present in galactic gas, for example, within the spirals of the milky way. I think I understand that the incoming radio waves will be a mixture of red- and blue-shifted photons depending on the whether the photon sources are moving towards or away from the dish. But what about the phases of the collection of photons? Since the sources may differ in distance and time of photon release, is the signal a collection of photons in different phases? And, if so, how does the telescope take this into account in processing the data? Or perhaps the phase does not matter at all. Am I thinking about this in the right way?

The telescope doesn't detect individual photons. It detects the combined effect of the massive number of photons incident on the telescope, which we call a radio wave. The fact that the photon phases are random just means the light coming from the sources is incoherent.

 

[f95toli]

The telescope doesn't detect individual photons. It detects the combined effect of the massive number of photons incident on the telescope.

Yes, or at least this would be true for the hydrogen hyperfine (1420 MHz)
However, the next generation of radio telescopes will have detectors that ARE capable of detecting single photons, albeit at a higher frequency (say above 100 GHz or so, which is still radio astronomy). The detectors for these are currently under development (kinetic inductance detectors or improved transition edge sensors) and at least in the lab they are able to detect individual photons as "event"; similar to what has been possible with optical photons for a long time.

Hence, the question from the OP sort-of makes sense.

That said, the "phase" of a single photon is not a well-defined concept meaning in this context it is probably easier to just use classical theory; the full QM theory would work but would just agree with the classical result for e.g. coherent radiation.

 

[Charles Link]

Yes, or at least this would be true for the hydrogen hyperfine (1420 MHz)
However, the next generation of radio telescopes will have detectors that ARE capable of detecting single photons, albeit at a higher frequency (say above 100 GHz or so, which is still radio astronomy). The detectors for these are currently under development (kinetic inductance detectors or improved transition edge sensors) and at least in the lab they are able to detect individual photons as "event"; similar to what has been possible with optical photons for a long time.

Hence, the question from the OP sort-of makes sense.

That said, the "phase" of a single photon is not a well-defined concept meaning in this context it is probably easier to just use classical theory; the full QM theory would work but would just agree with the classical result for e.g. coherent radiation.

Very interesting. For the OP @jordankonisky : When the classical r-f electromagnetic wave is viewed on an oscilloscope as a sinusoidal signal, it may be the result of thousands or millions of photons or more contributing to the sinusoidal r-f signal. When a single photon is detected, (at least this is how it is done in the U-V (ultraviolet)), it causes a disturbance on the detector and hundreds of thousands of electrons may be collected as a result. There is no observation of any sinusoidal electromagnetic field for this latter case.

 

[jordankonisky]

Thanks all. I think I am starting to get it.

Based on responses, allow me to restate my question. Consider the receiving parabolic dish of a radio telescope and the collection radio waves of same wavelength and phase that arrive perpendicular to the vortex line. No matter where each of these waves lands on the dish, each will then travel the same distance from its landing position on the dish to the telescope focus point. Thus, the collective beam will arrive at the focus point both in phase and in focus. On the other hand, members of a collection of radio waves of same wavelength and phase that arrive at an angle to the vortex line will not arrive on the dish at the same time, and, therefore, not all will arrive at the focus point simultaneously.

Suppose, a radio telescope is tuned to detect radio waves derived from neural hydrogen atoms at some region of the sky. My question is “do all the radio waves from a specific area of the sky arrive at the surface of the dish in phase?” It would seem to me that the flood of radio waves comprises of a collection of radio waves of same wavelength, but in all possible phases such that the dish sees radio waves of same wavelength, but in all phases, all of which reach the focus point.

 

[Charles Link]

To the OP @jordankonisky : For this particular problem of all the photons(scratch that=all of the wavefront) arriving in phase at the collection point, my fascination as well as the part that puzzled me was the energy conservation. It seemed to violate energy conservation because the various portions of the wave would constructively interfere as they were brought to focus=all having equal path distance. It is perhaps a somewhat routine optics calculation in diffraction theory of imaging, that solves the dilemma=the focused spot size, (and it does have a finite diffraction-limited spot size), decreases in area as the area of the parabolic dish is increased. (It actually took me a quite a while to figure this one out, after I had an advanced Optics course with diffraction theory). (And if the wavefront does not have constant phase, energy would still be conserved in the focusing=the spot size would simply be somewhat larger than the diffraction-limited spot size, and the focused intensity (watts/m^2) somewhat lower). $\\$ See also the Insights article that I authored about a year ago: https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/ $\\$ And, yes, I think the wavefront can be treated classically as a plane wave with constant phase across the whole plane as it arrives at the telescope dish. Alternatively, as @kimbyd has pointed out, it is possible to do interferometric measurements, with multiple telescopes, where the intensity and/or phase across the whole plane is found to not be constant. (I need to review the Hanbury Brown -Twiss experiment, which I think was the first experiment in stellar interferometry).$\\$ Hopefully this helps answer your question, and doesn't get too far off topic.

 

[anorlunda]

I believe this thread is relevant
https://www.physicsforums.com/threa...retations-of-electromagnetic-radiation.903551
especially this post
https://www.physicsforums.com/threa...lectromagnetic-radiation.903551/#post-5689306

 

[marcusl]

You have gotten some good responses from vela, f95toli and kimbyd. I take issue with this, however:

I believe, even in a laser, the phases of the individual photons, (assuming you can assign a phase to an individual photon), would be completely random. If the individual photons were in phase, this would necessarily violate energy conservation, because the energy (irradiance (watts/m^2)) of the electromagnetic wave is proportional to the square of the electric field amplitude, while the energy must also be proportional to the number of photons $N$. If the photons are in phase, we have a real dilemma, because $E_{total}=NE_o$ making the energy proportional to $N^2$. Alternatively If the phases are random, the phasor diagram becomes a 2-D random walk, with $E_{total} \approx \sqrt{N} E_o$, making the energy proportional to $N$ as it needs to be. $\\$ Hopefully this post doesn't get "pinged" for being a personal theory, but it is the best answer I have for what otherwise is a very big dilemma.

Stimulated emission preserves phase, so the EM field builds up coherently. This is why lasers are such a "powerful" source (in numerous senses).

 

[Charles Link]

You have gotten some good responses from vela, f95toli and kimbyd. I take issue with this, however:

Stimulated emission preserves phase, so the EM field builds up coherently. This is why lasers are such a "powerful" source (in numerous senses).

The laser has basically a single phase, but it cannot be the result of $N$ photons all having the same phase. It may be completely incorrect to assign a sinusoidal electric field and phase to a single photon, but if such a model is used, the in-phase superposition of $N$ photons results in a violation of the conservation of energy. The energy density in the electromagnetic wave is proportional to $E^2$, and that would make the energy of the $N$ photons proportional to $N^2$.

 

[marcusl]

You are mixing classical and quantum descriptions. Stick to one. In the quantum world, stimulatied emission preserves quantum numbers of incoming and outgoing photons. Classically, the phase and frequency of the emitted radiation matches that of the stimulating wave. I think you are misapplying energy conservation as well. The atoms in the excited state supply the additional energy.

 

[Charles Link]

You are mixing classical and quantum descriptions. Stick to one. In the quantum world, stimulatied emission preserves quantum numbers of incoming and outgoing photons. Classically, the phase and frequency of the emitted radiation matches that of the stimulating wave. I think you are misapplying energy conservation as well. The atoms in the excited state supply the additional energy.

I welcome other people's inputs on this. I have put considerable time and effort into the previously mentioned energy calculations. I actually prefer the classical picture, but it is important, especially when working with photodiodes (in the IR and visible) to have consistent photon models as well.

 

[jartsa]

I welcome other people's inputs on this. I have put considerable time and effort into the previously mentioned energy calculations. I actually prefer the classical picture, but it is important, especially when working with photodiodes (in the IR and visible) to have consistent photon models as well.

If we pretend that every hydrogen atom sends a tiny classical radio-wave at random times, then if 10 of those identical waves hit a radio-telescope at the same time and in-phase, the absorbed energy is only ten times one waves energy, right?

 

[Charles Link]

If we pretend that every hydrogen atom sends a tiny classical radio-wave at random times, then if 10 of those identical waves hit a radio-telescope at the same time and in-phase, the absorbed energy is only ten times one waves energy, right?

Yes. $\\$ Edit: Classically, one can work with sources that have larger amplitudes such as 10 r-f antennas, each (by themselves) radiating equal powers in a spherically symmetric pattern. By proper array design, there may be locations (solid angles) where the signals (sinusoidal electric fields) are all in phase with each other, so that the energy density\power per unit solid angle is 100x higher than that of a single source. In any case the total power radiated over the $4 \pi$ steradians remains 10x that of a single radiator, and that means there are necessarily solid angles where some destructive interference/cancellation occurs in the signals. The energy and number of photons is conserved, but by adjusting the geometry/relative position of the sources, the available energy/number of photons can be steered into specific directions rather than getting a spherically symmetric pattern.

 

[sophiecentaur]

The telescope doesn't detect individual photons. It detects the combined effect of the massive number of photons incident on the telescope, which we call a radio wave.

furthermore, you cannot talk in terms of different photons hitting different parts of the receiving dish. The energy in each photon is spread all over the dish (all of space, actually) and the wave interferes with itself to produce a probability density of where it will be detected) After enough photons have arrived, there will be a peak in received signal power at the feed from a source on the boresight .

I think I understand that the incoming radio waves will be a mixture of red- and blue-shifted photons depending on the whether the photon sources are moving towards or away from the dish.

I have skimmed through this thread an I haven't yet found the term "bandwidth". However you choose to model the source, it will have a bandwidth and the spectrum of the resulting wave that arrives at your telescope will have a peak and a gaussian distribution about that peak. At one-photon-at-a time rate, the same spectral shape would build up. If you want to discuss photons that are created by stimulated emission, the situation is different because the coherence is greater.

If we pretend that every hydrogen atom sends a tiny classical radio-wave at random times,

There is a problem with trying to think of this as a wave and a particle phenomenon at the same time. For instance, what would be the pulse shape of this burst of waves that you want each photon to consist of? If they are all identical in nature then that would imply that the little pulsed oscillators that generate them must have identical 'circuit characteristics'. That's nonsensical and it's been discussed endlessly in threads about ' how big is a photon?', how long does a photon last for'? or 'what is the length of a photon?".
A classical model with a set of random pulsed oscillators is OK and the result will give you a centre frequency that's noise modulated. But it's too far to go to involve photons.

 

[jartsa]

Yes. $\\$ Edit: Classically, one can work with sources that have larger amplitudes such as 10 r-f antennas, each (by themselves) radiating equal powers in a spherically symmetric pattern. By proper array design, there may be locations (solid angles) where the signals (sinusoidal electric fields) are all in phase with each other, so that the energy density\power per unit solid angle is 100x higher than that of a single source. In any case the total power radiated over the $4 \pi$ steradians remains 10x that of a single radiator, and that means there are necessarily solid angles where some destructive interference/cancellation occurs in the signals. The energy and number of photons is conserved, but by adjusting the geometry/relative position of the sources, the available energy/number of photons can be steered into specific directions rather than getting a spherically symmetric pattern.

Well, can't we tell that same story, with some small modifications, about ten photons emitted by ten hydrogen atoms?By the way, I meant to show that the same 'problem' exists with classical radio-waves too. (It's an interesting problem)

 

[sophiecentaur]

about ten photons emitted by ten hydrogen atoms?

Each photon will have its own probability function if it's spontaneous emission (omnidirectional pattern). If there's stimulated emission then the radiation pattern / probability function will have a maximum in the direction of the progressive phase of the atoms.

 

[kimbyd]

furthermore, you cannot talk in terms of different photons hitting different parts of the receiving dish. The energy in each photon is spread all over the dish (all of space, actually) and the wave interferes with itself to produce a probability density of where it will be detected) After enough photons have arrived, there will be a peak in received signal power at the feed from a source on the boresight .

This is an important point, and to me it stresses why it's more useful to think of this situation classically.

All of the behavior relevant to radio telescopes is completely described by the classical description of electromagnetic radiation as a wave. The quantum behavior only enters into it when your radiation source is so dim, and your detector so sensitive, that you're detecting individual photons. This typically isn't possible at radio wavelengths (I believe the lowest energy waves for which this is possible are in the millimeter wave range, as bolometers are capable of it down to about 100GHz or so). Even with bolometers, however, it's typically not useful to detect photons individually so they typically aren't designed for that. Thus ignoring the quantum nature of the radiation is still usually appropriate even there.

In the mean time, the quantum behavior will always, over time, average out to the classical behavior. The correct situation, then, is to think of the incoming radiation detected by the telescope as an incoming wavefront with some level of anisotropy in terms of phase. You can generally expect that the anisotropy of the incoming wavefront will tend to be on the order of or larger than the wavelength of the radiation as it comes in. For example, if your incoming wave has a wavelength of 1m, you should expect significant changes in phase only over transverse distances comparable to or larger than 1m. Much of the time, the incoming wave might have far less anisotropy, so that you might have very small changes in phase even over distances of many meters. The amount of anisotropy will depend upon the source.

 

[jartsa]

I welcome other people's inputs on this. I have put considerable time and effort into the previously mentioned energy calculations. I actually prefer the classical picture, but it is important, especially when working with photodiodes (in the IR and visible) to have consistent photon models as well.

I have a question: There is some energy in the E-field of a classical radio-wave pulse that is in its way from A to B.

What is the velocity of that energy?

 

[Charles Link]

I have a question: There is some energy in the E-field of a classical radio-wave pulse that is in its way from A to B.

What is the velocity of that energy?

The speed of light. The Poynting vector in classical E&M theory is used to describe this energy flow. $\\$ In Optics, they often use a simplified type of units, where intensity (basically some kind of watts/m^2, i.e. it will be proportional to the Poynting vector) $I=nE^2$ where $n$ is the index of refraction, and $E$ is the electric field amplitude. These units simplify computations of interference effects, such as a plane wave source with multiple layers of dielectrics= thin films, etc. (e.g. Given $E_o=E_{incident}$ and $I_o=I_{incident}$, oftentimes the item of interest is $I_{transmitted}$ or $I_{reflected}$. The chosen units greatly simplify the calculations).

 

[alantheastronomer]

I understand that a radio telescope can be tuned to receive radio waves generated by neutral hydrogen atoms present in galactic gas, for example, within the spirals of the milky way. I think I understand that the incoming radio waves will be a mixture of red- and blue-shifted photons depending on the whether the photon sources are moving towards or away from the dish. But what about the phases of the collection of photons? Since the sources may differ in distance and time of photon release, is the signal a collection of photons in different phases? And, if so, how does the telescope take this into account in processing the data? Or perhaps the phase does not matter at all. Am I thinking about this in the right way?

The radio radiation from a mass of neutral hydrogen is basically incoherent, in much the same way as radiation from a light bulb is incoherent. The signal IS a collection of photons in different phases. The radio receiver doesn't deconstruct the photons into their individual phases, it just collects the resultant radio wave. The radio receiver is tuned to a rather narrow band of wavelength. In order to observe radiation that's been doppler shifted due to galactic rotation, the receiver has to be re-tuned to follow the doppler shifted wavelengths.

 

[Charles Link]

I would like to have an additional response to comments around post 12 and 13: In a vector model that considers the electric field of a single photon, energy conservation as well as a maintaining of order in the phase of the original beam would be the case if the photon that enters the beam as a result of stimulated emission always enters with a phase of $90^o$ relative to the main beam. This idea might be useful to have a workable model, although admittedly, it is rather speculative.

 

[sophiecentaur]

The signal IS a collection of photons in different phases.

That is an only assertion and involves an invalid mix of classical and quantum ideas. All you can say about the photons involved is that an individual photon can react with a detector of some sort and the detector can be 'tuned' to a particular frequency (with whatever bandwidth it's designed with).
Physics does not deal with what things 'really are', only with how they behave under certain circumstances.
A spectrum analyser with sufficient resolution can show the spectrum of microwave radiation resulting from a particular transition. That's all.
QM is relevant each end of the transmission path but trying to treat the path that way merely involves the same Maths as with the classical calculation.

I would like to have an additional response to comments around post 12 and 13: In a vector model that considers the electric field of a single photon, energy conservation as well as a maintaining of order in the phase of the original beam would be the case if the photon that enters the beam as a result of stimulated emission always enters with a phase of $90^o$ relative to the main beam. This idea might be useful to have a workable model, although admittedly, it is rather speculative.

I find it difficult to get my head around this, despite your convincing argument. The energy conservation requirement seems to support the idea of 90° phase difference but, inside a laser, there are photons traveling up and down the cavity, stimulating emission from excited atoms in random places on the way. What, then would be the net effect on the resultant phase of the emerging beam if every stimulated emission introduces a quadrature component? Where would the coherence come from? In fact, what would be the result of such a large number of randomly phased photons. A similar thing occurs in traveling wave amplifiers (TWTs) where the energy from a beam of electrons, couples with an EM wave (slowed down by a helical guide). The 90° phase must happen there too (?) yet the Power in the EM wave grows along the tube.

 

[Charles Link]

That is an only assertion and involves an invalid mix of classical and quantum ideas. All you can say about the photons involved is that an individual photon can react with a detector of some sort and the detector can be 'tuned' to a particular frequency (with whatever bandwidth it's designed with).
Physics does not deal with what things 'really are', only with how they behave under certain circumstances.
A spectrum analyser with sufficient resolution can show the spectrum of microwave radiation resulting from a particular transition. That's all.
QM is relevant each end of the transmission path but trying to treat the path that way merely involves the same Maths as with the classical calculation.

I find it difficult to get my head around this, despite your convincing argument. The energy conservation requirement seems to support the idea of 90° phase difference but, inside a laser, there are photons traveling up and down the cavity, stimulating emission from excited atoms in random places on the way. What, then would be the net effect on the resultant phase of the emerging beam if every stimulated emission introduces a quadrature component? Where would the coherence come from? In fact, what would be the result of such a large number of randomly phased photons. A similar thing occurs in traveling wave amplifiers (TWTs) where the energy from a beam of electrons, couples with an EM wave (slowed down by a helical guide). The 90° phase must happen there too (?) yet the Power in the EM wave grows along the tube.

It really puzzled me for quite a number of years as well. I could not make any sense out of $N$ photons in phase that had an electromagnetic wave energy density that was proportional to the square of the wave amplitude. If anyone has something that resolves this dilemma with using either of the two possibilities that I mentioned=either random phases or a $90^o$ phase difference, I would very much like to hear it. These two possibilities didn't occur to me until quite a number of years of wondering just what might be going on here. I studied mostly visible and IR (infrared) early on, but later also worked with r-f (radio frequency), and I never saw any textbook that presented any type of model to solve this dilemma. The books on quantum theory had the wave function getting a $\sqrt{N}$ factor on the vector potential $A$ for an electromagnetic wave that contained $N$ photons, (see e.g. J.J. Sakurai's book Advanced Quantum Mechanics Chapter 2), but there was nothing ever presented in a classical type model that provided any explanation.

 

[sophiecentaur]

I just realized that, with two coherent sources, the resulting spatial radiation pattern can be narrower, so the power flux can actually be four times that of a single source. It works for two radio antennae with the right spacing and phases.

 

[Charles Link]

I just realized that, with two coherent sources, the resulting spatial radiation pattern can be narrower, so the power flux can actually be four times that of a single source. It works for two radio antennae with the right spacing and phases.

Yes. The separate coherent sources in an array, as well as the diffraction pattern from a diffraction grating conserve energy. A completely classical picture with energy density proportional to the square of the electric field amplitude works fine for both of these cases. $\\$ Also for the diffraction-limited focused spot in focusing with a parabolic mirror. The spot size gets smaller as the area of the mirror and beam is increased. For these cases, there is complete energy balance using the classical picture and diffraction theory. $\\$ The same is the case for the Fabry-Perot interference with plane waves. Energy is conserved when the Fresnel coefficients are employed, and once again, the energy density is proportional to the square if the electric field amplitude. $\\$ For a couple of these cases, it actually also took me quite a few years of working in the optics field before I knew how the energy was conserved, but I ultimately figured out each case, with a completely classical textbook type explanation. $\\$ Meanwhile, multiple photons in the same mode, as that which occurs in a laser, does not balance energy with a classical picture of the photons being in phase with each other.

 

[sophiecentaur]

Meanwhile, multiple photons in the same mode, as that which occurs in a laser, does not balance energy with a classical picture of the photons being in phase with each other.

The sources of the multiple photons (or the sources and the passing photons) are not in the same location so (even if we think it's totally valid to discuss the phases) there will never be enhancement in all directions.

 

[Charles Link]

The sources of the multiple photons (or the sources and the passing photons) are not in the same location so (even if we think it's totally valid to discuss the phases) there will never be enhancement in all directions.

If they are multiple point sources creating a diffraction pattern that is very much a narrow pattern in one direction, then the antenna array theory applies. The model I am considering is that of the plane wave modes in a cavity, where the plane waves fill the cavity uniformly. This is the type of modes that get counted in a rectangular cavity in statistical physics, where the Planck function is computed. Using the idea of many point sources as in an antenna array, the idea is workable to say the sources are "in phase". Exactly what a photon is supposed to be is the subject of the quantum theorists. In any case, if the "photons" are uniform plane waves making up a cavity mode, a dilemma does result if it is assumed the photons are all in phase.

 

[sophiecentaur]

If they are multiple point sources creating a diffraction pattern that is very much a narrow pattern in one direction, then the antenna array theory applies. The model I am considering is that of the plane wave modes in a cavity, where the plane waves fill the cavity uniformly. This is the type of modes that get counted in a rectangular cavity in statistical physics, where the Planck function is computed.

If we can assume that there is, in fact an answer to this, could it not be due to the approximation that assumes a perfect plane wave in the cavity when, in fact, an extra source of EM (from just one atom) has been introduced into that cavity? We are in this zone of trying to reconcile QM and classical in this case when that is well known for being problematical.

 

[Charles Link]

If we can assume that there is, in fact an answer to this, could it not be due to the approximation that assumes a perfect plane wave in the cavity when, in fact, an extra source of EM (from just one atom) has been introduced into that cavity? We are in this zone of trying to reconcile QM and classical in this case when that is well known for being problematical.

It is at this point, that I am satisfied to have a model that is reasonably consistent. It may be completely impractical and/or incorrect to consider the phase of a single photon, but I do like the classical approach to E&M. The result of the energy density being proportional to the square of the electric field amplitude is an important one, and the diffraction (and interference) theory does give results that are consistent with this. $\\$ Is the individual photon from a stimulated emission in phase with the main beam in a laser? If it comes from a point source, it very well could be, especially if it is viewed as one of the sources in an antenna array. My approach of a random phase was more for considering the case of superimposed plane waves. $\\$ As you stated in the previous post, it is here that it may be impossible to completely reconcile QM with a classical picture.

 

[sophiecentaur]

it may be impossible to completely reconcile QM with a classical picture.

It is clearly a very dodgy thing to try to characterise a photon in classical terms. There is a great temptation to treat an emitting atom as some sort of transmitting antenna but the nature of a conventional transmitter will affect the detailed nature of the transmitted signal. But all photons are identical (apart from in their Energy / Frequency. Hence the analogy or equivalence is dead as soon as you start.
Ah,the perils of trying to say what things 'really are'.

 

[Charles Link]

It is clearly a very dodgy thing to try to characterise a photon in classical terms. There is a great temptation to treat an emitting atom as some sort of transmitting antenna but the nature of a conventional transmitter will affect the detailed nature of the transmitted signal. But all photons are identical (apart from in their Energy / Frequency. Hence the analogy or equivalence is dead as soon as you start.
Ah,the perils of trying to say what things 'really are'.

In coming up with an idea such as the phases of its component parts perhaps being random, e.g. in a plane wave of large amplitude, I was trying to make it completely compatible with the classical picture that is well established that the energy density is proportional to the square of the electric field amplitude. It remains somewhat elusive just what is the nature of the component parts when it is taken down to the level of a single photon. $\\$ The antenna model that says that the photon resulting from stimulated emission should be in phase with the original signal also has considerable merit, and the narrowed pattern that results from many sources in phase is justification for that. $\\$ At this point, I am glad we gave the problem a thorough discussion, but I think it is fair statement, that we were unable to completely correlate the classical picture with the quantum description.

 

[sophiecentaur]

The antenna model that says that the photon resulting from stimulated emission should be in phase with the original signal also has considerable merit, and the narrowed pattern that results from many sources in phase is justification for that.

Also, of course, there is no specified place for this antenna-like mechanism to be radiating from. Irrespective of the phase (referenced to what?), there is the position for the source. It is easy to place an antenna , fed with any phase you like, at a place where the added wave is in phase with the passing signal. (Waves are time and position dependent) So the idea of the necessity for the added photon to be in 'quadrature' because of some sort of resonance (or whatever) disappears.

 

[Charles Link]

Also, of course, there is no specified place for this antenna-like mechanism to be radiating from. Irrespective of the phase (referenced to what?), there is the position for the source. It is easy to place an antenna , fed with any phase you like, at a place where the added wave is in phase with the passing signal. (Waves are time and position dependent) So the idea of the necessity for the added photon to be in 'quadrature' because of some sort of resonance (or whatever) disappears.

In general, you do get a narrower beam pattern if the antenna are all in phase with each other relative to some direction in the far field. The more sources that have the same phase, the narrower the pattern and the stronger the signal. The path difference to a point in the far field is computed, and if that path difference is zero or an integer number of wavelengths, you get constructive interference, and correspondingly a narrower pattern. $\\$ Such a description looks to be very accurate, but it doesn't explain how, if we were to double the electric field amplitude of each of our $N$ sources, how we get the same pattern, but with 4x the power.

 

[sophiecentaur]

Such a description looks to be very accurate, but it doesn't explain how, if we were to double the electric field amplitude of each of our NN N sources, how we get the same pattern, but with 4x the power.

Why not? 2² = 4 (Power =V²/R)

 

[Charles Link]

Why not? 2² = 4 (Power =V²/R)

Yes, it is, but if a single photon has $E_s=E_o$, and we now have 2 photons in phase from each source, making $E_s=2 E_o$, how did we get 4x the power with twice as many photons? To have consistency, for $n$ photons from each source, we need $E_s=\sqrt{n} E_o$.

 

[Charles Link]

It may also be of interest that it is possible to superimpose two macroscopic beams that are in phase with each of equal intensity using a 50-50 beamsplitter with an interferometer type geometry. In this case, with a 50-50 energy split, the Fresnel coefficients for reflection off of the beamsplitter are $\pm \frac{1}{\sqrt{2}}$. The Fresnel transmission coefficients are $\tau=\frac{+1}{\sqrt{2}}$. For this case, there is complete conservation of energy and the beams are successfully overlaid. One can envision that there may exist in nature a similar mechanism if you try to overlay one photon on top of another that are in phase with each other. In any case, the $E_s =\sqrt{N}E_o$ with a $\sqrt{N}$ factor needs to get in there somehow. $\\$ Perhaps it is jumping to conclusions to try to assign the $\sqrt{N}$ factor to a random phase. Edit: Note: Random phases for multiple photons in a single photon mode does result in a phasor diagam with a resultant electric field vector that does satisfy conservation of energy where $E_{resultant}\approx \sqrt{N} E_o$, so that intensity $I =NI_o$, with intensity $I$ proportional to $E^2$.

 

[jartsa]

It may also be of interest that it is possible to superimpose two macroscopic beams that are in phase with each of equal intensity using a 50-50 beamsplitter with an interferometer type geometry. In this case, with a 50-50 energy split, the Fresnel coefficients for reflection off of the beamsplitter are $\pm \frac{1}{\sqrt{2}}$. One can envision that there may exist in nature a similar mechanism if you try to overlay one photon on top of another that are in phase with each other. In any case, the $E_s =\sqrt{n}E_o$ with a $\sqrt{n}$ factor needs to get in there somehow. Perhaps it is jumping to conclusions to try to assign the $\sqrt{n}$ factor to a random phase.

Radio antenna sends out a spherical wave. Amplitude of the wave at some point is inversely proportional to the square of the distance to the antenna. Energy density of the wave at some point is also inversely proportional to the square of the distance to the antenna.

So I conclude that energy density of the wave at some point is proportional to amplitude of the wave at that point.

Did I make some error there?Edit: Oh, I googled that amplitude is actually inversely proportional to the distance to the antenna. Forget it then.

 

[Charles Link]

Radio antenna sends out a spherical wave. Amplitude of the wave at some point is inversely proportional to the square of the distance to the antenna. Energy density of the wave at some point is also inversely proportional to the square of the distance to the antenna.

So I conclude that energy density of the wave at some point is proportional to amplitude of the wave at that point.

Did I make some error there?

The electric field amplitude of the wave falls off as $\frac{1}{r}$. The square of the electric field amplitude, which is proportional to the intensity(power/unit area) falls off as $\frac{1}{r^2}$. The energy density is proportional to the square of the electric field amplitude.

 

[vela]

Yes, it is, but if a single photon has $E_s=E_o$, and we now have 2 photons in phase from each source, making $E_s=2 E_o$, how did we get 4x the power with twice as many photons? To have consistency, for $n$ photons from each source, we need $E_s=\sqrt{n} E_o$.

I think the mistake here is treating the photon like tiny classical waves. Photons apparently have a much more complicated relationship to classical electromagnetic waves.

https://en.wikipedia.org/wiki/Coherent_states

 

[sophiecentaur]

if you try to overlay one photon on top of another

This could the nub of the problem. This statement actually has no meaning. The location and extent of a photon have no meaning so two photons cannot be 'co-located'. I think it's true to say, also, that the only 'phase' that can be discussed would be when applied to the classical model of a wave.
The mutual incompatibility of QM and classical wave treatment has been discussed ad nauseam and out present conversation just serves to confirm this, I think.

 

[Charles Link]

This could the nub of the problem. This statement actually has no meaning. The location and extent of a photon have no meaning so two photons cannot be 'co-located'. I think it's true to say, also, that the only 'phase' that can be discussed would be when applied to the classical model of a wave.
The mutual incompatibility of QM and classical wave treatment has been discussed ad nauseam and out present conversation just serves to confirm this, I think.

@sophiecentaur It is likely that you have had in-depth discussions of this nature on Physics Forums and/or other places, but this is perhaps the most detailed discussion that I have had to date on this topic. It was good to hash over the topic to see what works, and what doesn't work. It seems that there indeed is some mutual incompatibility, but I found the discussion very helpful in getting a better understanding of where the limits are of the classical formalism.

 

[Spinnor]

Hopefully this post doesn't get "pinged" for being a personal theory, but it is the best answer I have for what otherwise is a very big dilemma.

I thought the same way as you and thanks to your post I now know I need to recalibrate my thinking. Fortunately the weekend is here.

 

[mfb]

Yes, it is, but if a single photon has $E_s=E_o$, and we now have 2 photons in phase from each source, making $E_s=2 E_o$, how did we get 4x the power with twice as many photons? To have consistency, for $n$ photons from each source, we need $E_s=\sqrt{n} E_o$.

Your beam gets narrower. The power at some point increases by a factor 4 but elsewhere the power goes down.

 

[Charles Link]

Your beam gets narrower. The power at some point increases by a factor 4 but elsewhere the power goes down.

A combination of post 28 and post 40 seems to summarize the solution to the whole problem. @mfb I agree with your input.

 

[sophiecentaur]

Your beam gets narrower. The power at some point increases by a factor 4 but elsewhere the power goes down.

Yes. the snag arises when people (unconsciously or not) insist on viewing photons as bullets which are also a bit like waves. Classical EM will tell you the mean Power. Any model-in-your-head of photons must allow them to follow classical theory and not the other way round. Ones personal model just has to defer to the evidence.

 

[Charles Link]

Yes. the snag arises when people (unconsciously or not) insist on viewing photons as bullets which are also a bit like waves. Classical EM will tell you the mean Power. Any model-in-your-head of photons must allow them to follow classical theory and not the other way round. Ones personal model just has to defer to the evidence.

Part of the classical theory allows the use of phasor diagrams for diffraction theory, and electric fields follow a very strict superposition principle. When considering photons, 1 photon+1 photon=2 photons, but their electric fields don't add in a similar manner. $\\$ As previously mentioned, in a quantum mechanical representation of the vector potential, Sakurai writes (equation 2.87 of his Advanced Quantum Mechanics book), $A(x,t)=\frac{1}{\sqrt{V}} \sum\limits_{k}^{} \sum\limits_{\alpha}^{} c\sqrt{\frac{\hbar}{2 \omega}}(\epsilon^{(\alpha)}e^{[i(k \cdot x- \omega t +\phi_{k, \alpha})]} \sqrt{N_{k,a}}+\sqrt{N_{k,\alpha}} \epsilon^{(\alpha)}e^{[-i (k \cdot x-\omega t+\phi_{k, \alpha})]} )$. $\\$ The $\sqrt{N}$ is simply part of the expansion, (edit) and is a very important feature that arises naturally in the Q.M description. (edit) The treatment of the superimposed planes waves with a $\sqrt{N}$ factor is built into the Q.M. formalism. Note also when there are multiple photons in a single plane wave mode, the resultant overall phase of that mode is part of the equation, but the phases of the individual photons in that mode does not enter into the calculation. $\\$ Sakurai also gives an uncertainty relation in equation 2.89: $\Delta N \, \Delta \phi \gtrsim 1$. $\\$ Edit: Note: Sakurai notes twice in footnotes on pp.29 and 39, that the above $c \sqrt{\frac{\hbar}{2 \omega}}$ will instead be $c \sqrt{\frac{2 \pi \hbar}{\omega}}$ in Gaussian unrationalized units, which uses $U=\frac{1}{8 \pi}(|E|^2+|B|^2)$ with a $\frac{1}{8 \pi}$ in place of $\frac{1}{2}$. $\\$ If you calculate the energy density using the above vector potential, you do indeed get $U=\frac{N \hbar \omega}{V}$, just as you should.