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A photon certainly carries a finite amount of energy. So it has a frequency and wavelength. You can do interferometry to a tiny fraction of a fringe over kilometer baselines, so the photons themselves have pretty good discipline about staying on frequency. To get the measurement uncertainty down, you just need a big, expensive instrument.

But the point here was a practical, easily measurable issue: does the quantum uncertainty in a multi-path interferometer, the fact that a photon may, apparently simultaneously, take several different paths from source to detector, change the TOF histogram that one would expect from classical optics?

As an engineer, I want to measure real things. Here's one.

John

Reply to
John Larkin
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It doesn't say that at all. It says that put together you can never know both simultaneously in a way that nails one completely without being totally uncertain about the other. To know the energy exactly you have to wait for an infinitely long measurement time. Squeeze one tightly and the uncertainty runs away in the other conjugate direction.

And by implication that there is somewhere a sweet spot where you know both closely enough for most practical purposes but always inexactly.

Quantum mechanics like relativity is somewhat counter intuitive. Einstein never liked it very much which led to his famous comment "God does not play dice with the universe". It now looks like not only does he do that but he throws the dice in such a way that if *we* attempt to peek we change the outcome. Even hidden variable theories have been experimentally disproved by testing Bell's inequality.

It is a finite amount of energy. But you cannot measure the frequency more accurately than nature will permit and every photon has its as emitted wavefunction amplitude as a function of time. People have been

*very* creative about creating modulated wavefunctions using biphoton processes to torment quantum mechanics with great success. So far QM is behaving pretty much as expected. If there is to be a paradigm shift I reckon it will come from experiments where the wave particle duality is also split by the divider on the input side of the interferometer leg.

But you are still limited by the finite length of the photon.

It has an average energy which is seemingly immutable in terrestrial labs but there is some slop depending on the things wavefunction (not true on cosmological scales or we would all be fried by the plasma at the surface of last scattering - thankfully it is red shifted to 4K).

You surely know that modulating a pure sine wave causes sidebands. It is exactly the same for wavefunctions.

If it obeys nearly classical optics then to a close approximation no. But if you look more closely the photons that arrive at the detector before there is time for the longer path to have been traversed are not subject to interference fringes. To save conservation of energy which really does not hold at the quantum level - exactly why not is open to debate but it doesn't you have traded causality which is generally believed by most physicists to be far more fundamental.

Since you won't take my word for it how about Stephen Hawkings lecture:

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So demonstrate a system that shows Youngs slit fringes on a time gated detector before there has been time for the longer leg to be travelled.

Regards, Martin Brown

Reply to
Martin Brown

Cool. Let's build an interferometer that makes light and dark fringes on a screen, and that has two paths from laser to screen, with different path lengths. We can add an optical switch or some such to gate photons into the system, so we could measure TOF. We'll still get fringes, since any photons that get past the gate are just photons.

Now, at the location of a dark band in the fringe pattern, poke a hole in the screen and put a photodetector behind it. Measure and histogram TOFs. Being a dark band, you won't see many photons.

But if, as you just said, "the photons that arrive at the detector before there is time for the longer path to have been traversed are not subject to interference fringes" then turning on the photodetector electronics will make the interferance cease, so the dark band magically fills with light.

Right?

John

Reply to
John Larkin

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You still have interference in classical optics. That's just a wave thing. So no QM doesn't change the that.

(OK I'm liable to say something stupid or wrong here... hopefully someone will correct me if I do.)

We could set up an interferometer with different path lenghts. (Piece of cake, the hard thing is getting the two paths exactly equal.) If the path length difference is greater than the coherence length of the light, then there are no interference effects. (Now the uncertain part) I believe that the coherence length is something like the 'size' of the photon. So your TOF detector will have a timing uncertainty of the coherence length/ speed of light. To get interference effects you can't see separate blips on your TOF detector and if you get two blips, then there is no interference.

That's my 'best guess' anyway.

George H.

Reply to
George Herold

A single photon doesn't have a coherence length. Coherence length is the measure of the spectral purity of a light source, namely the statistical spread in wavelength of a lot of photons.

Any single photon in an interferometer interferes with itself, and has a probability of hitting various points on a screen that maps a fringe pattern. If the light source is broadband, the fringe probability patterns from different photons are not aligned, so fringes aren't discernable.

All photons travel at exactly c, so a photon's TOF is independent of wavelength or spread of wavelength. Except for any classical geometric path or detector effects, TOF is constant for each leg of the interferometer.

To get

If I have fringes on a screen, and place a photodiode on the screen to measure TOF, do the fringes disappear?

Please note that I'm thinking my way through this, and learning in the process, not just debating to be right. I'm probably not 100% right, but it seems to me that one can do a number of thought experiments that mandate my conclusion: one can measure TOF through the split-beam interferometer, get two distinct spikes in the time histogram, and still have interferance. Each photon interferes with itself over both legs, and, after being detected, appears from TOF to have taken only one leg or the other.

John

Reply to
John Larkin

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Hmm, I can do interferometery with single photons (in principle). If the coherence length of my light source is less than the path difference then no interference. I've done this with 'lots' of photons. (A HeNe laser and unequal arm Michelson interferometer.) So how do you explain this if the coherence lenght of the light source is not the same as that of the individual photons.

But I find photons to be slippery things. You can take a light bulb... broad band, small coherence length, and pass it through an interfence filter (An optical band pass filter.) and have a much longer coherence length.

I can also set up the same Michelson interferometer to see 'white light' fringes. (Real zero path length difference.) Here's a pic,

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Pictures a bit blurry, It's better when you can just eye ball it.

Now if I move one of the mirrors a bit then the fringes goes away. But If I hold a green interference filter to my eye... after the output of the interferometer, then green fringes appear.

OK so this is just my picture of things and most likely wrong when looked at in detail, but there is going to be some spread in time of your detector. Given just one path for the moment, not all pulses are going to come at the same time... there will be a spread in time, and I think that spread will be determined by the coherence length.

(A man could also ask how you are starting your TOF clock, but I think I know how to do that.)

Sure that's fine, My guess is you'll have jitter in your TOF timing that depends on the coherence length.

George H.

Reply to
George Herold

Sure. It just takes longer to build up enough photon impacts to see the bands. Actually, interferance is always done with single photons; a photon interferes with itself, but not with other photons.

If

Well, the interferance patterns are blurred by the fact that different photons have different wavelengths, so their band patterns are confused. The interferance patterns of the single photons, actually probability patterns, are still there.

I've done this with 'lots' of

As I said, one photon doesn't have a coherence length. A laser does, because it spits out a lot of photons of various wavelengths.

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Right. You throw away most of the photons and only keep the ones that are close together in wavelength.

Yes. You can select the narrowband photons before they go into the interferometer, or after they hit the screen. It's all the same.

No; all photons travel at the speed of light. Maybe the detector would have different time response for different wavelengths, or maybe there's some dispersion in the optics, but coherence length can't change the speed of light.

Sure, gate the laser output or pulse the laser.

I don't think so, unless it's an artifact of the detector. The TDCs we make have jitter in the 10s of picoseconds, and no decent laser will have enough wavelength spread to make that much detector jitter. So TOF is mostly an instrumentation issue. It would be easy to resolve the two photon TOFs in an optical-table sized 2-leg interferometer. The measurement jitter might be, say, 30 ps RMS, and the difference in path langth might be, say, 5 ns. The histogram would have two narrrow spikes.

I wonder if a PMT detection speed depends on photon wavelength. Probably a tiny bit.

That would be a cool project-kit thing, an interferometer with TOF. That would push the wave/particle duality right into peoples' faces.

John

Reply to
John Larkin

Not correct. A single photon is finite in time and space and as a result its frequency is necessarily imprecise. Coherence length isn't the right name for this though - wavefunction envelope is better.

All true. And I think the bit that you have been missing is that the fringe minima do not quite go down to zero even when you should be on a perfect null. The only photons that can hit the nominal fringe null are the ones that have come down the shorter leg of the interferometer.

As you make the interferometer path length larger the noise baseline of photons that did not self interfere increases.

Interestingly the experiment that Jan highlighted (which despite the SciAm title doesn't wreck quantum mechanics at all but does come quite close to doing the experiment we have be discussing). And it looks on the face of it like you can get a nice ensemble result - Deutsch says it is nothing new and I am inclined to believe him but it is an interesting experiment non-the-less. It would appear the Copenhagen interpretation that I was taught might be too hard line about what can happen in quantum systems when you can measure all the parameters.

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It doesn't do anything like what Jan claims. But it might lead to a resurrection of pilot-wave theory or at least some tweaks. I might be wrong about how nature really behaves but I have described here to the best that words will allow what is thought to happen according to the Copenhagen interpretation of quantum mechanics.

But hey that is what experiments are for to find tiny defects in the theories that we think are correct. I am still hoping for an interesting result out of one of the experiments that use quantum tunnelling to force wave particle duality to make a choice at the outset and then combine them again at the end.

No not at all. But if you were able to do it you would see that the photons that could not have had time to traverse the long leg of the interferometer were uniformly distributed on the screen.

You have to be more devious than that to determine the trajectory but according to the SciAm paper someone has done something close to testing this. My classic QM prediction is still that the photon can hit the detector at any time or place where its wavefunction at the screen is non-zero and with a probability determined by the amplitude.

Regards, Martin Brown

Reply to
Martin Brown

Replace coherence length with wavefunction envelope and it will do. The emitted wave has some amplitude modulation starting at zero and ending at zero in between it is oscillatory with a characteristic wavelength but because of the finite length it has some ambiguity.

The only way to be certain is to try it. FWIW I think it *is* worth doing the experiment even though I also think I know the answer.

It would be so much more interesting to be wrong on this one.

It would be interesting to see what could be done in this line with modern time gating devices. It is just possible that such an experiment if it gave the outcome that you expect and showed fringes could at require a modification of how the Copenhagen interpretation is taught.

Regards, Martin Brown

Reply to
Martin Brown

If you know its energy, Planck's relation tells you its wavelength. It may be hard to measure, but that's another issue.

Is a photon finite in space? I thought it was infinite in space, which is why interferance is seen over huge baselines.

Coherence length isn't the right

The term "coherence length" has a commonly accepted definition

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and it's associated with a light source, not a single photon. It's just another way of stating the spectral purity of the source.

A perfect null fringe would have zero width, and there's no way to make a detector that has zero width.

The only photons that can hit the nominal fringe null are

But every photon goes down both legs of the interferometer.

I think that if you actually measure TOF in a beam-splitter interferometer, you'll find two distinct values, one for each path length, independent of fringe position.

Every photon, as a wave, splits in half at the beam splitter, traverses both legs, and interferes with itself. That same photon, as a particle, can't split but has a 50-50 probability of passing straight through the splitter or being deflected. You get the best of both worlds, interferance fringes from every photon and two distinct TOFs.

They all self interfere. But if the path length is longer than the coherence length, you can't *see* the jumbled fringes.

"I think I can safely say that nobody understands quantum mechanics."

Richard Feynman, in The Character of Physical Law (1965)

John

Reply to
John Larkin

I found this neat applet to visualize the double slit experiment:

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I set it up for "double slit" and then it can show the waves progressing from the two slits, towards an imaginary screen at the bottom. You can even change the incidence angle of the light source!

The applet shows that the light hits the screen in two places (one from each slit) before any interference pattern has emerged, and then a short time later, there will be only interference pattern light on the screen. So its pretty simple too see, if you send only "one photon" at a time and this light goes through both slits, then there won't be enough energy to trigger a quantum orbital jump in the screen until the interference pattern emerges (recombines the split light source). The TOF measurements will not be possible to make unless there is an interference pattern because of this (if only shooting "one photon" at a time.

The TOF will be lowest at the point on the screen that is BETWEEN the two slits, and then the TOF will increase either direction outwards on the screen. The exact location between the two slits on the screen where the TOF is lowest depends on the incidence angle of the light source.

cheers, Jamie

Reply to
Jamie

That's cool.

If a photon gets through the slits, it can be detected. And it either gets through, at 100% energy, or it doesn't. You can't split a photon.

I don't think so. TOF will measure photon time as if the photon passed through one slit or the other. So, in the center of the screen, you'll see only one value for TOF. Anywhere else, it will be bimodal, namely you'll see two different TOFs, one for each path.

TOF would be hard to measure in a dual-slit system, since the slits are usually very close together. A beam-splitter interferometer can make fringes but provide enough path length difference, nanoseconds, to make TOF measurements practical.

In the beam splitter, a photon interferes with itself even though the path lengths are different by nanoseconds, or microseconds. If a photon takes the fast path, the interferance happens and is detected long before its "other half" could have made it through the slow path. That photon didn't take the long path, but it still knows it's there.

John

Reply to
John Larkin

Hi,

I don't know enough about the beam splitter interferometers to comment except to say maybe the light source isn't really shooting "one photon" at a time. Otherwise that beam splitter experiment is hard to explain with wave theory.

I was thinking about another possibility, if the light source is gradually increased in intensity, starting at "one photon" per pulse and moving up, then at some point the area of the screen with the fastest TOF may change in a discrete step, from an area in between the two slits, to an area directly in line with one of the slits. This would be because enough energy is coming through a single slit to trigger a quantum orbital jump in the screen, when the light intensity is increased enough, and the shortest path for the light to take is a straight line.

cheers, Jamie

Reply to
Jamie

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Geesh, I think you are wrong, but it seems like an experiment that is doable. Imagine sending 10 ns (3 m) pulses of light through an unequal arm Michelson interferometer. If the path length difference is greater than 3 meters then I don't think you will see any interference. That's easy to see without doing any fancy timing.

George H.

Reply to
George Herold

[Jeez, what a long thread and no Godwin incident. A record? For sed?]

Interesting article in the current Science mag where some clever folks are sneaking around Herr Heisenberg's restrictions.

and if one has a login there:

--
Rich Webb     Norfolk, VA
Reply to
Rich Webb

You could send in single photons, average one a second, and still get an interference pattern. Or a CW beam. Or a pulsed beam. It's all just photons.

Photons don't interfere with one another. Each photon interferes with itself.

If you are suggesting that my arguments don't make sense, I agree. QM doesn't make sense.

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John

Reply to
John Larkin

Hi,

Also quantum transitions take a certain amount of time to occur, which is proportional to the wavelength of the light that is absorbed by the atom and the speed of light. The wave is fed into the atom at the speed of light, and if it is the right frequency it will be in resonance (like an antenna or LC tank circuit) and will pop an electron into a higher orbital.

cheers, Jamie

Reply to
Jamie

This may be where you are struggling to understand. It is beyond hard to measure. It is impossible to measure without waiting for an infinite time. That is what the Heisenberg uncertainty principle means for the energy and temporal relationship of a photon.

I know you are wedded to conservation of energy, but at the quantum level it no longer holds instantaneously for measurements of E it only holds for E0 = the expectation value averaged over time as time tends to infinity. There is real unavoidable slop in the value of E and also our ability to measure E0 in any experiment.

In the wave formalism this comes directly from the Fourier transform relationship. In QM particle formalism I suppose it comes from the fine structure of spacetime that is believed to exist on the length scale of the Plank length.

The wavefunction has a group velocity which to a good approximation describes where the photon may be found as a function of time radiating away from the source. Provided that the photons length is long compared to the path length difference you will see interference fringes.

Or the spatial purity - closeness to being a true point source. Coherence length was also used to measure (using monochromatic light) the diameters of stars by looking at the visibility fringes.

That is a mere practical problem.

Goes down it yes. But it cannot escape from the other end until a time of more than L/c has elapsed. So for a path length difference of D there is a time gate window of D/c where any photons reaching the detector must have travelled the short leg and could not have been influenced by the wavefunction in the longer leg which collapses to zero at the instant the photon is detected before it has had chance to form an interference pattern. Any other interpretation requires a violation of causality.

And I think you will find a roughly gaussian distribution that spans a range consistent with the emitters coherence time (or if the wavefunction has been later modulated the wavefunction amplitude).

There will be a small baseline of photons detected that arrive before the long leg can have been traveled. Those photons will not show interference fringes. If you increase the path length the baseline counts increases in proportion as D/c approaches the approximate length of the photons wavefunction.

Yes.

But only if it can reach the combination point before the photon is detected.

I do think the experiment is worth doing. Just in case what you think happens is actually observed. I reckon the kit should be up to it these days so it just requires an optical bench, a dense filter and some very fast sensitive electronics (and a month or ten to perfect the experimental technique and sort out the gremlins).

This is getting into semantics. There is no substitute for doing the experiment. As far as I can tell it has not been done.

I can't say that I like quantum mechanics and have always found the idea of photons in radio interferometry completely useless. The wave formalism gives the right answers whilst photons tend to mislead.

With photons you can so easily get stuck thinking classically and imagining photons as billiard balls with nice exact properties.

Regards, Martin Brown

Reply to
Martin Brown

Yes. Agreed but it must be given sufficient time to reach the other end and interfere. If the photon has already been detected at the screen then the wavefunction vanishes everywhere else.

Think of it this way you could put a shutter in the long leg right at the end so that at the last instant the long path that was open is closed. The photon either hits that shutter or it hits the screen by having gone down the other path both - outcomes with 50:50 probability.

And I think you will find a range of values representing the length/duration of the photons wavefunction convolved with the two different path delays including any phase factors introduced.

I think that you don't and the only way to be sure is to do the experiment. I reckon it should be doable with modern kit.

I am more familiar with coherence length effects in angular resoltion where what happens is that you get decreasing interference amplitude as the phase error increases upto +/- pi and then it increases again to +/-

3pi/2 (with amplitude 33%) null at +/- 2pi etc in a classic sinc(x) dependence (or Bessel function in cylindrical geometry).

That is still true.

Sorry no that isn't right. Anywhere that the photon wavefunction amplitude is non-zero there is a chance of finding the photon in a measurement. If you do this measurement before both paths have been explored then there is no interference. It is a rarer event but it leads to a raised baseline at the interference minima.

You can do the TOF measurement but you won't get two clean spikes at L1/c and L2/c you will get a range of values from L1/c to L1/c+Tphoton instead (where Tphoton is the effective length in time of the photon).

The average TOF will indeed be lowest at the minima.

But photons that took the short cut and managed to arrive and be detected before the longer path could be traversed don't suffer any interference and are smoothly distributed across the screen.

Detection causes the wavefunction to collapse and so it is as if the other longer path did not exist for the small number of photons where this occurs.

Regards, Martin Brown

Reply to
Martin Brown

No. It only goes through one leg or the other, fast or slow. Once it hits the detector, 100% of its energy is deposited instantly.

What would a photon do if it behaved the way you suggest? Would the beam splitter actually split the photon, sending half into each leg?

If a photon took the short leg and arrived at the detector, would it hover just above it, waiting for the other half of itself to arrive?

If the photon has already been detected at the screen

Yes. And it doesn't wait around for that to happen. Once the photon hits the detector, it's all over. And it can hit it after one of two time delays, corresponding to the path lengths.

Once the fast-leg photon has hit the detector, the electrical signal is already inside the oscilloscope. Closing a shutter in the long leg

5 ns later is too late to undo the signal, even if the long leg is 10 ns.

Think of a laser that makes big, fast light pulses. Imagine a beam splitter interferometer with different leg lengths. Put a detector aon the sscreen and connect an oscilloscope. Light pulses will clearly travel one leg or the other, and you'll see two different prop delays from laser to detector. That's classic physics with big burst of light. You could move the photodetector anywhere in the path and the time delat from the laser will increase linearly with the path length, from zero at the laser to max at the screen through the long path. At the screen, you'll see two pulses, one for each path. That's not hard to do. The pulses of light that you see are just lots of photons.

John

Makes no sense, and doesn't happen. Half of the photons take the short path. What do they do while they are waiting for something to emerge from the long path, which it never will? Do they travel at less then c?

It's just as bad if the photon takes the long path.

Applying classical or intuitive concepts to QM just doesn't work.

John

Reply to
John Larkin

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