[Eternal Student (OP)]

Hi again.
     

...what made you want to ask there?

As I mentioned, my question follows from something I heard in a lecture course and I'm still reluctant to accept.   However, I'm open minded and have to give the material some serious consideration.   So I'll start by presenting some of that material.   (Just to be clear then:  Not all of this is my opinion,  I'm attempting to present their ideas.  I've got some stuff under a spoiler that mentions some of the concerns I would have and I've tried to emphasize  *their main claims*  wherever I really wouldn't want anyone to think it was my claim).   Even though you ( @yor_on ) asked for this, feel free to skip it when you're half way through and bored.

Can we observe light waves?

    Superficially you will probably think that you can, of course you can,  there are diffraction gratings and all sorts of things about the way light behaves that only makes sense if you treat it as a wave.   (That's what I thought)

    However, the lecture took an opposing view from the outset.   All detectors, for example a detector you have in a lab,  just detect energy.  A simple detector is really just detecting a heat rise or energy deposit when photons come into it and are absorbed.   Some of the more advanced detectors can just about detect a single photon but they are still only detecting an energy deposit and not the oscillation of any wave.   Much the same is true for other ways of detecting light such as the human eye.   A few individual photons must strike a rod or cone and stuff happens... electrons can be excited, chemical changes could take place etc..  Indeed any atom absorbing light is the textbook example of a quantum effect, a photon must have the right energy.  It doesn't matter how many photons you throw at it or how quickly they are thrown one after another, if no photon has the right energy then the electron does not get excited.   The key is that the word "energy" is always there,  atoms absorbing light, rods and cones in your eye reacting to light or a sophisticated detector in a lab  -  they all just react to or detect energy.  We can observe light, that's for certain, however we don't seem to observe the wave that is supposed to be in a light wave. 

   Let's refer back to some of the other replies from people on this thread.   We're probably all in agreement that light carries energy, however that falls short of establishing that we are detecting the wave aspect of the thing.

It is obvious that light carries energy*........  (and  @alancalverd   went on to describe experiments where light transfers energy to something and then argued that this could only be due to an electric field)...

Worth having a deep think about Alan’s suggestion. Light doesn’t have mass, so that’s not the source of any energy transfer....

Idea 1:     There is a loss of wave characteristics due to time averaging.

   A typical plane wave solution to the wave equation can be written as

    =  E₀ . Cos ( k.x - ωt )

      

[Eqns 1]

and  are unit vectors orthogonal to each other and to the direction of propagation.  k = wave vector.   x = position vector.   ω = angular frequency.   t = time.
   The oscillatory or wave characteristic is clearly provided by the Cosine term.

Now the energy carried in an electromagnetic wave can be determined by using the Poynting vector:

[Eqn 2]

and the intensity, I of a wave is ultimately given by     where  < >_t  denotes an average over time.   We're only interested in an average over time because (*it is claimed that*) no detector can ever find an instantaneous intensity from a light wave.  Instead, they only detect photons and we must then consider the device to be collecting the energy over a minimum period of time that allows the absorption of one complete photon.

Hence, 
         where we have used [Eqns 1] to substitute for ExB.

   The basic claim is that T must be one full time period of oscillation.  (I'm not happy with that but that's the basic idea presented in the lecture).   Anyway, take T as one full time period (T = 2π/ω) and then we can perform the integral easily enough with the substitution  θ =  ωt   to obtain:

          

[Eqn 3]

   Where we have used the relationship c² = 1/(μ₀ε₀).   The final expression for intensity should be familiar since it holds when we allow T (the time over which we average)  to approach infinity and so it's just the usual expression for the (average) intensity of an e-m wave.   The key point here is that we have only averaged over one time period, the claim is that this is the minimum amount of time for a detector to absorb a photon.   Looking at the RHS of [Eqn 3]  we can clearly see that there is no time or space dependence suggesting a wave.  There is no  Cos ωt   term for example.   The information about the wave or oscillation is simply lost.  The detector responds to the energy delivered by the photon and it really doesn't care about the oscillation in the E field.   We could have the e-m wave oscillate in some other way (e.g. a square wave instead of cosine wave, or even just hold constant) and it wouldn't matter provided the energy delivered was the same. 
   I'm not going to repeat all of the lecture so we just need to speed through and consider the possibility that the oscillation of an E field (and B field) doesn't even need to be there.  It could just be a convenient model for the energy carried by a photon.   There is certainly no information about the wave in what the detector reacts to - which is only the intensity over 1 time period.
   

Spoiler: show

  I'll break off from what the lecture was presenting to just point out one minor issue:   I don't see why the time average really did have to cover one full period of oscillation and hence become an integral of Cos² θ   over  a range of 2π.  For example, a photon does not have to be one full wavelength in actual physical length.  I was willing to let that go... there is some sort of time average required and that process of integration over time will tend to remove the time-dependant term from the final expression.

Idea 2:     We only ever observe something related to the square of the modulus of the wave, |E|²  and not  E  directly.
   Consider a generic plane wave solution to the wave equation which we can write in the following form:
E = E₀ e^iΦ          where all the oscillation is due to the complex exponent iΦ.   The time dependence resides entirely in the phase term Φ  (e.g.  Φ = kx-ωt).
   Then  |E|²  =  |E₀|²  . | e^iΦ |²  =  |E₀|²   and the phase term is again just lost, there is no term showing an oscillation.

Spoiler: show

  I really wasn't convinced and I'm probably not doing the lecture justice in my presentation here.  Most plane waves in the real world aren't of the form  e^iΦ when you're talking about an E field or B field.   That's just one convenient simple solution.  Specifically, it's just a convenient mathematical expression that isn't disallowed by the wave equation.  However, it has a real component and an imaginary component.   If you want to eliminate the imaginary component then you must do something like add a complex conjugate solution of the form  e ^{- iΦ}, which of course you can do because the wave equation is linear.  Having real valued functions is quite important because we don't really measure an E field of strength  1+2i  anywhere in the real world.   This matters because, even if you can only measure something proportional to the square of the modulus of E field   when you take the solution e^iΦ and also it's complex conjugate  then we have something like this:    E = E₀ (e^iΦ + e^-iΦ)   =  E₀ .2Cos Φ    and hence    |E|² = 4 . E₀²  Cos² Φ   which clearly is dependant on Φ.    Specifically, the phase term, or wave character contained in Φ has not been completely lost.

   - - - - - - - -
That's enough about the lecture content.  It went on and said more stuff but overall I finished and thought.... I'm really not sure that was money well spent.   None the less, I'm open minded and considered it.   Do we have any lab detectors that can genuinely detect or react to the oscillatory nature of light or are all detectors really just picking up the energy of a photon?
    Let's take a reply from @Bored chemist  while I'm here:   Thanks for recommending the article about Wiener's experiments.   In that experiment a photographic plate was used to show the wave nature of light.   The trouble is the interaction between the light and photosensitive material is very much a  'photon and atoms' interaction much as described earlier.   At the point of interaction, the material was simply reacting to a deposit of energy by a photon and not conclusively to the wave or oscillation that may have been in the light.   Now don't get me wrong, it's clear that we would have trouble trying to explain the results of that experiment unless we assign some wave properties to light but that's a slightly different thing.
    It was agreed, even in the controversial lecture, that it is a very useful model to consider light as an oscillation in the E and B fields.  There are many experimental results that can only be explained by assuming light has a wave-like nature.  The main issue is only that this wave nature can not be directly observed.  We can draw some immediate parallels with Quantum Mechanics and it will probably help to explain what they (the lecturer and his team) were driving at.   In QM we assume a wave function, ψ(x,t) , exists and describes the state of a particle but we accept that you cannot directly observe or measure the wave function.  It is a model and may not be anything real or tangible that you can measure.   There are observables, things that we can measure or observe, you can find the position or momentum of the particle for example -  but we can't directly observe or measure the wave function.  The best you could hope for is that you measure something that will collapse the wave function to a known state.
    Their (the lecturer and team) main argument is that Light may behave like an oscillation in the E field under certain circumstances but when you go to measure the oscillation in the E field you just can't.  All you can detect is a photon and the properties of a photon which don't seem to contain any information about the wave or oscillation in the E field that you might assume existed.   E.g. the energy of the photon, as outlined above, does not depend on the wave form like E=E₀ Cos ωt   that you have assigned to the E field. 

    Anyway, as I've probably indicated I'm not especially convinced that the online lecture was worth the money but I've got to think about it and try to dismiss some of it systematically and categorically.   The trouble is,  I just do not know of any experiment or piece of equipment that really does detect the oscillation in the E field when some light passes by,   hence the original question.    There is no shortage of experiments that can only be explained by assuming light has wave-like properties but that just isn't good enough.   I'd very much like to establish that the oscillations in the E field are there and can be directly observed and measured.   Then there is no doubt that assigning an oscillation in the E field to describe light is NOT just a model to explain wave-like behaviour in light.   Compare with the wave function in Quantum Mechanics (i.e. something that could be regarded as a useful model but cannot be directly observed).

Best Wishes.

[Edited to remove the error spotted by @paul cotter in post #51 ]

[Eternal Student (OP)]

Hi again,
   I'm sorry to write on the post again when no-one has had any time to respond yet.   I'm just a bit troubled and would like to ask another question.   Which might help to focus attention on what is happening, or claimed to happen.

In the post just above I mentioned this:

Indeed any atom absorbing light is the textbook example of a quantum effect, a photon must have the right energy.  It doesn't matter how many photons you throw at it or how quickly they are thrown one after another, if no photon has the right energy then the electron does not get excited.

   Now suppose that making an oscillation in the E field and B field is all you need to do to create some light.   You create an appropriate oscillation and what you have is some light there.

Now put an atom (or quite a few atoms like a small block of metal because that will be easier) in the middle of some laser emitters, something like the diagram below:


atom-laser.jpg (33.87 kB . 1152x648 - viewed 2073 times)

   The idea is that you surround the atom with lasers at an angle from the atom that is whatever you might need, with the lasers at whatever distance from the atom you might need,  have as many of these lasers as you might need, and finally allow the lasers to generate light of a frequency that is whatever you might need EXCEPT  the frequency for "turquoise" coloured light.     The idea is that the superposition of the E field and B field from all of them combines to give you something that is at least a very good approximation to an e-m wave with the wavelength for "turquoise" at the location of the atom, for at least a short while    (the lasers had different frequencies so we can anticipate that there isn't going to be a good standing wave there, the overall superposition of all the E fields and B fields will may change as some lasers slip out of phase with the others BUT for at least a short while, say "long enough" you had some E and B field oscillation that was a good approximation to what we want).      Obviously the colour "turquoise" was supposed to be the colour at which an electron will transition between orbitals in that atom (I used "turquoise" because the obvious choices like red, green, blue, yellow  were already used on my diagram).

   Now the question is:   Will the electron make a transition to the new orbital?   None of the individual photons from the lasers were of the right frequency for this to happen but will they combine under a superposition of their E and B fields to give you something that will be enough to make the electron jump? 

Best Wishes.

[Colin2B]

Hi again,
   I'm sorry to write on the post again when no-one has had any time to respond yet.     

Now the question is:   Will the electron make a transition to the new orbital?   None of the individual photons from the lasers were of the right frequency for this to happen but will they combine under a superposition of their E and B fields to give you something that will be enough to make the electron jump? 

Best Wishes.

Hi ES
Sorry, still been away with unwell family member so not been following. So excuse if I’m covering something already discussed.

A photon is a measurement phenomenon which is detected when an electron in an atom is given sufficient energy to change state either by changing level or by being ejected (photoelectric effect). So the answer is yes, if the combined energies of the beams hit the sweet spot then the electron will transition and we say we have detected a photon. It’s all a bit circular and incestuous though and misunderstanding of what a photon is can cause a lot of problems, it’s why the phrase travels as a wave and interacts as a photon is often used.
In reality, getting this superposition right is very difficult as the lasers need to be very stable and fixed to large concrete blocks, but there have been some successful experiments on superposition - don’t have references to hand.

[paul cotter]

Eternal Student, I would have THOUGHT no, to your question about lasers of different frequencies. One can throw n+1 radio frequencies at a target and they will remain discrete unless there is some non linear element that allows mixing( a+b, a-b ). I hate to say this, but I fear you are doing what I am regularly accused of, that is "mixing quantum and classical" in the argument( I wrote the last sentence reluctantly, I certainly don't want to be offensive ).

[Colin2B]

One can throw n+1 radio frequencies at a target and they will remain discrete ….

How do you know they remain discrete. Describe the detector that shows them to be discrete or non-discrete.
We are now firmly in the realm of rhe measurement problem.

[yor_on]

I was right ES, it was interesting :)

Seems like an argument for photons, instead of waves? Don't know if anyone remember but we used to have arguments about that here, and I've seen it elsewhere too, where the most reasonable idea still seemed to be waves. Few accepting the duality as a 'real phenomena'. And now the wind has turned the other way, with new generations of fundamental science. It reminds me a lot of this discussion. Check out  juanrga.

https://physics.stackexchange.com/questions/46237/is-the-wave-particle-duality-a-real-duality
=

It's weird though, and I think it says something about how we want the world to be. Because it goes back to Newtonian ideas, both of them. Waves and Particles. One or the other as a simple definition of forces. If we take the example of propagation in a vacuum it doesn't really matter if it is a 'wave' or a 'photon' doing it. Either something propagates a distance in time, or there is something entirely different going on, making us think it's a 'propagation' involved to it.

[yor_on]

And yes and no Paul. I like science a lot better than catastrophes, but if I think we're going to one I will write about it.

[alancalverd]

Can we observe light waves?

No.

We can observe light (and indeed all electromagnetic radiation) doing things that we can model as a wave

We can predict the speed of light from wave solutions to electromagnetic equations

But when we detect light, it behaves like particles

Which is odd because all the equipment we use for detecting lower energy EM radiation is based on its wavy nature.

[paul cotter]

Colin2b, if you document the incident frequencies accurately one will not get new frequencies, ie addition and subtraction species unless there is a non linear element present to perform frequency mixing. On transmission towers frequently multiple frequencies are simultaneously emitted without a hint of mixing; a rusty bolt on the tower can change this, causing multiple spurious intermodulation products.

[Bored chemist]

But when we detect light, it behaves like particles

Unless we do it while it's behaving like a wave.
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/

[Eternal Student (OP)]

Hi.

Thanks for various replies.   I'm busy thinking about them while catching up on some housework.   I may write some more later.
    @Bored chemist , that looks like the same post you made earlier and was addressed.   It doesn't matter, you can't read everything.   It seems like that experiment only suggests that light has wave-like properties, which we all agree on.   However, the photographic paper was catching light as a particle and not responding to the wave nature of it.

Best Wishes.

[Bored chemist]

It's not that I didn't read it, it's that I don't agree with it.
The image on the film would be in shades of grey, not black and white.

[evan_au]

Describe the detector that shows (radio frequencies) to be discrete or non-discrete.

Adding to Paul Cotter's comment, your mobile phone (or your DSL modem) uses numerous closely-spaced frequencies, which can all be used to transmit signals.
They are separated out by a Fast Fourier transform (and are generated by an Inverse Fourier Transform), usually with some error-correcting code to overcome noise.

Around here, we had problems with VDSL2 broadband and telephone pillars which had "dry" joints or oxidised wire-wrap joints, which can be nonlinear. This would cause intermodulation products across the band, corrupting the signal with the dry joint, and also cross-talking into adjacent copper pairs. Linear crosstalk can be cancelled by (linear) "vectoring", but non-linear crosstalk would take far more processing power to analyse & correct. They resolved it by soldering all of the joints in the noisy pillar.
- This reinforced my view that optical fiber is far superior to copper wires - partly because the fiber (mostly SiO₂) is already fully oxidised.

all the equipment we use for detecting lower energy EM radiation is based on its wavy nature.

The point about lower-frequency EM radiation (eg radio-frequency) is that the individual photons ("radons"?) have ultra-low energy. That means you don't transmit individual photons, but instead a coherent wave consisting of trillions of photons. You can easily modulate or demodulate this wave (eg using FFT/IFFT).

However, as you go to higher frequencies, the energy of individual photons increases, and you are more likely to detect them as individual events, and less likely to detect them as a coherent wave.
- It is possible to produce a coherent wave of light (using a laser), but this gets increasingly difficult at higher frequencies like X-Rays and gamma rays.
- Large astronomical telescopes use photon-counting detectors (cooled to near absolute zero), where you can practically count how many photons struck each pixel of the detector. But the photons are still reflected/refracted as a wave,  the very large mirror localising the photon on an individual pixel of the detector, in a way that a smaller mirror would not.

[Bored chemist]

("radons"?)

It's bad enough that radon has several names (thoron, niton, emanation) without using radon to mean two different things.

generated by an Inverse Fourier Transform

Interesting thing about the inverse FT is that it's an FT (and then inversion, which is why telescopes typically turn the image upside down.)

[alancalverd]

But when we detect light, it behaves like particles

Unless we do it while it's behaving like a wave.
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/

Wiener's detector was a photographic film. We now know that the production of an image depends as an absolute minimum on the absorption of at least two visible photons within a fairly short time on a silver halide grain. There is no wave model explanation for reciprocity failure or latent image fade.

PS I just noticed that ES has made the same observation!

[Bored chemist]

But when we detect light, it behaves like particles

Unless we do it while it's behaving like a wave.
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/

Wiener's detector was a photographic film. We now know that the production of an image depends as an absolute minimum on the absorption of at least two visible photons within a fairly short time on a silver halide grain. There is no wave model explanation for reciprocity failure or latent image fade.

PS I just noticed that ES has made the same observation!

And yet.

It's not that I didn't read it, it's that I don't agree with it.
The image on the film would be in shades of grey, not black and white.

Light isn't waves and it isn't particles.
It's light.

[Colin2B]

Colin2b, if you document the incident frequencies accurately one will not get new frequencies, ie addition and subtraction species unless there is a non linear element present to perform frequency mixing.

I understand what you are saying, but additional frequencies do occur in linear systems.
Alan & I regularly use beat frequencies to tune musical instruments. I agree that the individual component frequencies still exist, but it doesn’t change the fact that the beat frequency exists and has a physical effect, you can for example get it to excite a tuned resonator.
ES was very specific in his hypothetical example. If 2 optical waves did superpose to create a beat frequency, is it conceivable that if that frequency coincided with the absorption band of an atom then would an electron in that atom would absorb energy and make the transition.
The set up in ES’s description is difficult to achieve, but straight laser interference of 2 beams has been demonstrated in free space.
Look at a simpler demonstration in the paper below. Two beams are directed along the same path through a beam splitter and although the beat frequency is outside the optical band it’s effect is real and can be measured and monitored electronically. If the frequency happened to be at the right frequency is it not feasible that the atom might also act as a suitable detector.
I think that an atom is too narrow a filter to respond to the components, but in the spirit of ES’s question, if by some fluke the beat frequency were in the optical range would we see the beat as a colour, ES’s turquoise?

Unfortunately, none of this really helps answer ES’s basic question.


A3AAE3CA-7811-4741-965C-1BE7271872E0.jpeg (673.82 kB . 804x2192 - viewed 4418 times)

[Bored chemist]

Most "green lasers" you see are actually the beat frequency of an IR beam with itself i.e frequency doubling.

The photodetectors use in experiments like that depicted, typically measure the power, rather than the field strength of the incident light.
Since the power is proportional to the square of the field, (which is why it's always positive), the detector is inherently non linear.

[alancalverd]

Quote from: Bored chemist on Yesterday at 19:26:30
It's not that I didn't read it, it's that I don't agree with it.
The image on the film would be in shades of grey, not black and white.
Light isn't waves and it isn't particles.
It's light.

The "shades of grey" actually resolve into "density of dots" at very low intensities.

Photographic film isn't particularly sensitive to x-rays, surprisingly, so Edison invented the fluorescent "intensifying screen" that absorbs incoming x-ray photons and emits large numbers of visible photons which are more efficiently captured by the film. At very low dose rates you can see "quantum mottle" on an x-ray image, each spot corresponding to a single x-ray photon hitting  the screen. So at least the quantum behavior of x-radiation is easy to see.

We can also count the random arrival of individual visible photons striking the cathode of a photomultiplier - a phenomemon that can't logically be modelled by continuous wave emission.

But I completely agree (and have often stated) that light isn't waves or particles: the often-stated "duality" is of our models, not the phenomenon itself.

Sadly, pseudoscientists have become increasingly insistent on modelling rather than observation to promote their strange beliefs. I've been engaged in a ridiculous discussion with a local "development partnership" funded at your expense, which insists that a dual carriageway needs a major bypass and bus lane (at your expense) to reduce congestion. I have pointed out that there is no evidence of congestion: the average speed of traffic in both directions throughout the rush hour is 55 mph, but the Powers that Be say "but the computer model says it is congested, so we need to resolve the problem [at your expense]". Thus it is with wave-particle duality - generations of students have been inducted into a mystery with as much validity as the Holy Trinity or Transubstantiation (i.e. none) through the vanity of people who cannot admit that their model(s) are not reality itself.

End of rant (for now).

[paul cotter]

Colin2B, I too have used beats to tune a guitar-unfortunately my resultant output remained as unmusical as ever!. I will have to have further thoughts on this, I don't want to derail ES's question.