Phase of radio waves received by a radio telescope dish

[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.

 

[sophiecentaur]

When considering photons, 1 photon+1 photon=2 photons, but their electric fields don't add in a similar manner.

Yes. Energy will be conserved but where and when that energy turns up will depend on the situation.
That equation looks pretty much like a standard diffraction / interference formula with some extra jiggery pokery. I still have a problem with actually assigning phase and position to the photon and that equation seems to be doing that. (from a fuzzy inspection)

 

[Charles Link]

Yes. Energy will be conserved but where and when that energy turns up will depend on the situation.
That equation looks pretty much like a standard diffraction / interference formula with some extra jiggery pokery. I still have a problem with actually assigning phase and position to the photon and that equation seems to be doing that. (from a fuzzy inspection)

See the previous post=post 50=a part I added, from Sakurai's Advanced Quantum Mechanics book. This results from some algebra he did just before that with some photon creation and destruction operators. The uncertainty relation may be telling us that we won't be able to reach any firm conclusions on the phases of the individual photons.

 

[Charles Link]

@sophiecentaur Please see a couple of additional edited inputs I just made to post 50.

 

[Charles Link]

I would like to make a couple additional comments to this discussion which I think mathematically are quite reasonable. There seems to be two different types of "laser" type (or coherent r-f signal) patterns being considered here=one is a plane wave and the other is a narrow pattern that has a finite beam spread/divergence. For both cases, if a given signal amplitude is considered, there needs to be a correlation of the phases of the signal across/throughout the beam (i.e. the 3-D beam pattern)=whether it is considered to be of a photon nature or simply as a completely classical picture. $\\$ For both cases, (of a plane wave or a beam pattern with a finite divergence), to double the intensity of the signal, the amplitude at each location only needs to increase by a factor of $\sqrt{2}$ if the amplitudes are in phase everywhere. Somehow, a factor needs to get introduced where a factor of $N$ in energy is a factor of $\sqrt{N}$ in the resultant electric field amplitude everywhere. $\\$ One way to achieve this is by overlaying the layers and giving each layer a random phase. Another way, is simply to keep all the superimposed layers in phase, (or some arbitrary phase), with each other, and have a mathematics where $E_{total}=\sqrt{N} E_o$, where $E$ is electric field amplitude, and the usual mathematics with phasor diagrams where $E$ fields add is no longer applicable. The Q.M. description, (see also post 50), seems to use this latter approach. (edit: see the next paragraph. In this present paragraph, I was treating the photon as a fundamental sinusoidal disturbance with a given amplitude and having some phase. These disturbances would then add linearly, taking into account the phases as they are added. For random phases, the amplitude would then be proportional to $\sqrt{N}$ . This picture of a photon may or may not be realistic.)$\\$ After pondering this question further, one additional item occurred to me that perhaps resolves some or most of this: The energy density being proportional to the square of the amplitude is how waves behave in general in linear materials, so that one should not expect to lay a layer of (wave) energy in phase on a linear material and have the electric field amplitude double. This $\sqrt{N}$ factor is normal behavior for a wave in a linear material.

 

[sophiecentaur]

Somehow, a factor needs to get introduced where a factor of NN N in energy is a factor of √NN \sqrt{N} in the resultant electric field amplitude everywhere.

I think you are worrying overmuch about this. Conventional Electromagnetic theory and RF Engineering took care of this, way before QM reared its head. We mostly accept that the two follow parallel paths and that, when we care to check, we can verify this. So why not have confidence to 'let go' and follow classical antenna theory and allow it to give the answers that you want?
I am not aware of any difference in the performance of a radio dish when it is part of a comms link (coherent signals) or when its used for looking at the power coming from astronomical objects (incoherent - mostly).

 

[Charles Link]

@sophiecentaur The r-f approach is essentially the Kirchhoff diffraction theory which is also widely used and what I am most familiar with. I have found it very useful in working with photodiodes to have a photon model that is in reasonable agreement with the diffraction theory. I am satisfied, after a close inspection and review of the Q.M. ,(see post 50), with how the quantum theory actually is in close agreement to what the r-f engineers came up with long before the Q.M. was ever invented. And thank you for your inputs.

 

[davenn]

I think you are worrying overmuch about this. Conventional Electromagnetic theory and RF Engineering took care of this, way before QM reared its head. We mostly accept that the two follow parallel paths and that, when we care to check, we can verify this. So why not have confidence to 'let go' and follow classical antenna theory and allow it to give the answers that you want?

Couldn't agree more ... and in addition to over worrying/overthinking ... just making it plain overly complicated !

Classical physics was enough to answer the OP's question and unfortunately it seriously deviated into QM and the OP was lost way back in mid page 1
Cant say I blame him/her, even I got lost with the QM side of this thread

@Charles Link ... It's probably the best to remember the KISS method when answering threads

Tho it may have been a good discussion that eventuated, it didn't really answer the OP's questions in a straightforward way
as not everyone is well versed in QM

Regards
Dave