Counterfactual computation

If they were independently random, you'd expect 1/4 of the single-photon splits to fire both detectors and 1/4 to fire neither.

That doesn't happen. Those two cases both violate conservation of energy. In a classical system, you'd get two photons at a longer wavelength every time.

I think that QM and COE are deeply linked.

First of all, you can't shoot photons on demand. If you'd use a pulsed source and adjusted the intensity to get one detection event per shot on average, you *will* get events with both or neither detector firing. I see no need to invoke COE here.

Furthermore, in a classical system, splitting the beam would get you two *waves* with half the power each, that is, sqrt(2) times the amplitude, not twice the wavelength. Waves aren't quantized in the classical model! A change of wavelength requires a non-linear or a moving reflector.

What is typically done is to attenuate the light until detection events are well-separated in time. (I'm assuming the use of single-photon detectors, PMTs, SPADs, TES, whatever. Mind you, a photon, to me, is not a discrete particle that travels from source to detector, but only a quantized interaction between EM waves and matter.) You'll get irregular ticks out of your detectors, like a Geiger counter.

Choose some time interval and count the number of events over many such intervals. The number will vary randomly with some mean and standard deviation. The standard deviation converges on the root of the mean. In other words, both detectors show full shot noise. If you histogram the data, you'll obtain a Poisson distribution. The cross-correlation of the number sequence for both detectors converges on zero.

I think that QM has a somewhat frivolous approach to COE. Some even attribute the very existence of the universe to a QM fluctuation, which is so far out that I can't find a superlative strong enough to express my disbelief.

Jeroen Belleman

Sure. Fresnel coefficients derived from Maxwell's equations.

Cheers

Phil Hobbs

No, it doesn't. It changes the EM boundary conditions, which changes the field amplitude, and the photon detection rate goes as the power density. Jeroen's quite right about the semiclassical model--you have to work really hard to get any other result than full shot noise (or worse, of course).

Photons aren't where that actually bites, because they aren't objects, they're elementary excitations of the EM field in a given set of boundary conditions. Doing interferometry with things like neutrons and even buckyballs is where objects act like waves.

Cheers

Phil Hobbs

They get _detected_ on one side or another, which is another matter.

Cheers

Phil Hobbs

Jeroen's talking about the _detector outputs_, because that's the only place that the quantum character of light can be detected. Everywhere else it just follows Maxwell's equations as one would expect. You really can't reason correctly about light by visualizing little blurry billiard balls.

Nope. You can't match the boundary conditions if the frequency changes--phase matching is the key physics underlying the Fresnel formulae. Maxwell's equations are a classical theory, after all.

Cheers

Phil Hobbs

I can reduce the intensity, or use a shutter, so that if I detect a photon there is effectively zero chance that another is nearby. If both detectors fire, COE has been volated.

Better yet, get one of those entangled photon sources and detect one of them at the source. They you know where and when the other one exists. Fire that into the splitter experiment.

All this sort of thing works better at shorter energies where detectors are 100% efficient, hard xrays or gammas. Some of the poor statictics of things like this are artifacts of noisy detectors. A good gamma is as subtle as a bowling ball.

I don't think there is in principle no reason one couldn't invent a clocked single photon emitter. Something similar has been done for single electrons.

If you'd

I said "longer wavelength."

Where did the universe come from?

Fine if you don't believe in photons. Life is much simpler without annoying things like shot noise.

You can indeed reduce the intensity, reducing the probability that both detectors fire for the same light pulse, but that probability won't go to zero, not even with hypothetical perfect detectors. (100% QE and no dark counts.)

Single electrons, OK. Single photons, I don't believe it. I'd love to be wrong.

If only I knew. I have trouble believing the Big Bang, despite the evidence. I suppose we all have. The cosmological red-shift is strong evidence, but there are too many unphysical ad-hoc patches in the model to make it credible.

Jeroen Belleman

I have abundant data to show that a photon does not behave even slightly like a billiard ball, and that with the exception of photodetection statistics, classical EM theory is a complete description of the propagation of light. ;)

Cheers

Phil Hobbs

QM just redefines waves as the probability of having waves. That is purely a mental game, but it seems to work.

The interesting non-destructive interaction of a photon/wave with matter is the beam splitter.

Sure, they don't.

Waves need a *physical* implementation.

That's pretty much a meaningless statement.

What, exactly do you claim the "EM Wave" is, physically, if not the photons themselves? An Aether for light was rejected quite some time ago....hint MMX experiment.

A standard wave is a disturbance in a medium. A physical wave without a medium is pretty much nonsensical.

"Physical interactions" must be, essentially, by definition, interaction between real physical objects. Declaring a wave is pretty much word salad without identifying what it is.

As far as results go, an individual pea shooter for photons have been demonstrated many times.

"Interference of individual particles section."

The "Wave" either in QM or EM, does not appear to be physically real. It is just way of accounting that it is a probability that particles have observable values, that are not in agreement with values that deterministic Newtonian Mechanics would predict. This is really the route, imo, of the waffle of alleged existence of "waves". "Particles" do not follow newton's laws. Motion has random components.

The universe is statistical. That statistical nature uses an equation. Its unfortunate that this equation is called a "wave function". The reality is, the Quantum Mechanical Wave function does not behave like a classical wave. Indeed if it did, the Schrodinger Equation (non relativistically) would fail to satisfy the Galilei Transformation, which would be problematic. (Quantum Mechanics, A Modern Development - Leslie Ballentine p.103-p104)

-- Kevin Aylward

Not at all. The Quantum state is a probability state function, for example P(x) which should not be confused with a classical state X.

A state such as

|psi> = |up> + |dn>

Means that, on measurement the result will either be |up> OR |dn> but not both, as the vectors are orthogonal.

Sure, but not relevant to the interpretation of a superposition of states.

That interpretation, has been proven false. An ignorance interpretation, essentially, means that there are hidden variables. Such a view seems unlikely.

The Kochen?Specker (KS) theorem shows that unique values cannot be assigned to observables prior to measurement.

Sure, but not relevant as to whether there are hidden variables that explain an assumed ignorance.

I actually go over these issues here:

-- Kevin Aylward

Big 'if' there; it isn't possible to both 'recombine them' AND use two detectors, so the QM calculations are very different. If, by 'classical' you mean real old-school light theories, neither single photon nor interference are supported options...

On a sunny day (Wed, 29 May 2019 21:23:36 +0100) it happened "Kevin Aylward" wrote in :

LOL

Good point. I think I was a bit too hasty with my 'definite answer' statement. I'll have to think that over.

Jeroen Belleman

On a sunny day (Wed, 29 May 2019 21:23:36 +0100) it happened "Kevin Aylward" wrote in :

When looking at the Michelson and Morley experiment everybody was amazed. I once asked in sci.physics about it and was given a simple calculation example and that worked out OK.

But then...

What is this experiment? it uses reflected light. What is reflected light? Reflected light is _re_emitted_ light, and thus locally recreated if you like that term. Zero result!

Now could it be 20th century, poor fishsicks in school benches get zero result hammered into their heads. No question asked, do not agree? tortured every day by doing counter-intuitive 'Einsteinian math (also called relatitvitty or something). Not agree? then no paper, more years of torture (or go cooking in restaurant work at Michelson and Donalds ehh Mc Doughnuts, whatever),

So, in my view, I question that M&M experiment (though the M&Ms tasted OK just to get that straight).

And with that the whole card house by that Einstein comes crashing down. AL of it There IS an ether, Time deletion, younger and older twins, it all goes out the window. Just as silly as photon (from the same author Einstein).

For me, looking at these things as well as that kwantuun thingy, is just like the Emperor has no clothes.

Einstein died and his last words apart from likely 'shit' was that the failed to unite graffity and the rest of the world. No wonder if you are stuck in math.

We sell a randomly timed single 'photon' source. Small incandescent light bulb run at lower voltage... so it's mostly red. Green interference filter (10nm BW, I think) And then some slits/ apertures.

George H.

The MMX experiment and its relation to SR is subtle.

Einstein introduced the idea of photons. If light was emitted as any ball would be, relative to its source, then the MMX would obviously be NULL as the source and observer were moving together.

So, the question is why did Einstein take the one key property that the Aether gives, namely independence of the emitted speed from the source velocity, that is bits of Aether vibrate hitting other bits of Aether propagating an effect, at the characteristic speed of the medium, yet throw the Aether itself away?

It is crucial to appreciate what Einstein actually said.

On the Electrodynamics of Moving Bodies:

"These two assumptions are quite sufficient to give us a simple and consistent theory of electrodynamics of moving bodies on the basis of the Maxwellian theory for bodies at rest. The introduction of a "luminiferous

conceptions which will be developed, we shall introduce neither a "space absolutely at rest" endowed with special properties, nor shall we associate a velocity-vector with a point in which electro-magnetic processes take place. "

Its the "superfluous" bit that is important. He doesn't say that it doesn't exist. For the following reason.

Assuming the the equations of SR are correct, then those same equations show that that if the Aether were actually there, they would explain why the failure to detect such an Aether in the MMX experiment would occur. SR says that the SR length contraction would result in a NULL result just as LET would, despite an Aether being there.

I address that here:

e.g

Einstein specifically stated:

"The theory of relativity belongs to a class of "principle-theories...As such, it employs an analytic method, which means that the elements of this theory are not based on hypothesis but on empirical discovery."

That is, physical hypothesis (mechanisms) are ignored from the outset, so clearly makes no statement as to what those processes might be. Einstein is directly declaring here explicitly, that he not not going to offer an explanation, for example, a hypothesis as to why "sources could immediately find a common speed". However, there is no suggestion or implication that such a physical process does not exist.

Einstein, apparently, had a deep reason to hold on to independence of source velocity, however, that bit that is not really the problem, its independence of observer that is, although either one can be used to derive the other on the assumption of all uniform motion being equivalent.

Independence of source velocity is a property of all wave motion, so in itself, no big deal, if there is a medium.

Now consider a sun firing light at you. You travel into the sun and measure its velocity. You now travel faster, yet you still measure the same velocity.

The only way that can happen in the conventional universe is if your measuring instrument properties are dependant on that velocity. This directly implies that there is some sort of reference frame to which that velocity is referenced.

I don't question the physical measurement result of the MMX, the issue is its interpretation.

There is no reasonable doubt that the predictions of SR are correct. The issue is why.

Someone taking a really fast trip will effectively travel into the future. The word salad of standard SR says that the traveller "takes a longer path in space-time". The alternative, is that there is an interaction with a background frame that defines the physical process of clocks, such that traveling at a velocity wrt a frame, truly results in processes slowing down.

SR, to me, appears to be so specifically crafted to hide the physical mechanism that makes it work. My take on "space-time" is also in

-- Kevin Aylward