adding time of flight measurements to double slit experiment

So, if you shoot blue light into a beam splitter, why don't you get two beams of red light? And why would two detectors, observing the split beams, always show a blue photon in one leg or the other?

And how can you explain gamma ray pulse-height analysis?

John

Reply to
John Larkin
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On a sunny day (Wed, 08 Jun 2011 06:51:19 -0700) it happened John Larkin wrote in :

Well that this his way of looking for attention, apart from posting political jive.

OK, I will try again. I get the impression that you look at EM waves as particles named 'photons'.

EM waves come in many ways, as WAVES in a medium that your eartlings have not yet understood

To make a 'wave' in the medium earthlings change the orbits of electrons in their atoms.

As there are fixed orbits with fixed energy levels, a fixed amount of energy is transferred to the medium. Or to put it in simpler words, a wave front (some may say packet, but think of a stone thrown in the water perhaps) of specific energy (p = h.v v stands for frequency, and h for Planck's constant) is created.

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it says: In 1905 Einstein f*cked things up by suggesting EM waves could only exist as thse discrete wave packets, he was an earthling too. This mechanism says nothing about particles at that level. In the detector the reverse happens, you need a specific amount of energy to knock an electron out of orbit, it says nothing about the waves and the medium those travel in. The whole 'wave particle duality' is completely classic, and the random element comes from interaction with the medium the wave travels in.
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The theory referred to in that article:
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Reply to
Jan Panteltje

Because an ordinary beam splitter is a linear device? In fact, there

*are* splitters that do this. Look up 'BBO crystals'.

You mean that they would be perfectly anti-correlated? They aren't. In fact, in the limit, they would be perfectly independent. This experiment does not invalidate the idea that light is a wave.

... or the frequency-dependence of the photo-electric effect? I admit I do not know how to explain this in terms of a semi- classical theory. Anyone else?

Jeroen Belleman

Reply to
Jeroen Belleman

Gamma rays ionize a whole bunch of atoms so you get a large number of electrons for each gamma ray. Higher energy gamma rays ionize more atoms... bigger pulse.

For the 'state of the art' in photons, google (photon anti-correlation

+Grangier)

There is a circa 2000 AJP article, that brings this to anyones lab. (You need several $k for the detectors.)

George H.

Reply to
George Herold

No, I'm trying to figure out, in specific and measurable terms, whether TO measurement is consistant with making fringes and, if not, specifically why. I think the answer is that a shutter scatters the wavelengths of single photons that pass through it, which, as a practical matter, has all sorts of interesting sidelines.

All I want to do is understand how the world actually works. But as an engineer, I want to understand in terms that lead to actual, measurable effects. It's like COE: one can generalize that energy must be conserved, bit in a real system, it's the details that are interesting.

You can certainly pass photons through a grating spectrometer and measure their wavelength to arbitrary precision. But looking at photons as particles - which they will soon be, if you detect them - the grating must scatter them in time.

Right. The shutter randomizes photon energies. This suggests a few interesting experiments.

John

Reply to
John Larkin

By Jove! I think he might finally have got it!!!

So far so good. But there are some interesting second order effects that have already been exploited to measure stellar diameters based on the intensity interferometer. Using monochromatic photons from a distant point source and you get fringes - but if the distant point source is too big compared to your baseline difference you do not.

Michelson and Pease did it first with direct periscope optical baseline hung on the 100" Mt Wilson. Hanbury Brown & Twiss at Jodrell went on to develop the controversial (until they made it work) second order theory of intensity interferometry for measuring stellar diameters. It is written up in a book which ISTR Phil pointed out online a whileback.

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These days it carries their names. And it has been applied to studying quantum systems on a miniscule scale in recent years. Glauber got the

2005 Nobel Prize in physics for his contributions to quantum physics and optical coherence. I hate to use an appeal to authority but he describes the history of quantum optics and Diracs Quantum Electrodynamics formalism better than I can in his Nobel Prize lecture. Online at

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(afraid his first joke falls flat and he skips the last slide)

The part around 30 minutes in is relevant to this thread, but you should watch it from the beginning. You will find the final part too mathematical and very sketchy unless you already understand the notation used.

I find it amazing that the Nobel foundation cannot do simulcasts of the lecturer in one corner at lower bandwidth and the presentation slides behind him. I don't need that bandwidth for a talking head, and it still isn't enough to convey the screen capture adequately.

Sorry but it says no such thing. The photons keep on going until they are either absorbed or reflected and there are a lot of them.

Regards, Martin Brown

Reply to
Martin Brown

I was never horribly confused, at least any more than QM confuses everyone else. I was trying to decide if measuring single-photon TOF was consistant with fringing, and if not, *specifically* what in my proposed experiment stopped them from being present simultaneously.

For the record, in the course of the discussion, I answered the specific question myself: a fast shutter will scatter single photon energies enough to smear statistical fringe patterns. I think.

What's wrong with discussing mental experiments to see what specifically happens?

I'm thinking now about the implications of this to fast fiberoptic systems.

John

Reply to
John Larkin

On a sunny day (Wed, 08 Jun 2011 17:05:34 +0200) it happened Jeroen Belleman wrote in :

Look up tghe thread in sci.physics: explaining the photo electric effect from the wave perspective, by Jan Panteltje:

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Reply to
Jan Panteltje

Nothing, but it would help if you picked up a basic grounding in the physics terminology first. I think the experiment is still possibly worth doing because with modern kit it should be just about doable and it doesn't seem to have been done yet at least not that I can see.

There is always an outside chance that it could show something unexpected, but I think it highly unlikely.

You want ideally want to use solitons in that application so that the pulses are unmolested by the effects of the dispersive medium.

Regards, Martin Brown

Reply to
Martin Brown

I took a year of physics in college, but it was all classical stuff, and not much optics. Hey, I'm a circuit designer.

I think the experiment is still possibly

Unlikely to discover new physics, but possibly a novel twist could be found.

Since a shutter scatters the wavelengths of photons, and energy is conserved, the shutter must deliver and absorb energy, which might show up as induced electrical or mechanical noise in the shutter drive. Not much, I admit.

One could also imagine driving a linear EO-modulator sort of shutter with a waveform, a sawtooth maybe, and producing some sort of klystron-like photon bunching. You'd see a similar effect in a medium where the refractive index could be modulated with a signal.

The stuff I do doesn't have the sorts of power levels that would make the medium nonlinear. But we get lots of interferance effects and squirmies from interference, and I suppose they must be caused by very short-distance reflections, not by fiber-end/end reflections, because the VCSEL lasers have short coherence lengths. Some of this stuff might actually be useful.

John

Reply to
John Larkin

Hi,

I guess the question is why will there only be an emission at or above a certain frequency irregardless of the amplitude of the incoming light? The photoelectric effect is a special case of electron orbital increments, in that it is the "final" increment that an electron can make before it becomes free of the atom and flys off. For all other electron orbital increments (atom bound before and after) the frequency absorbed by the electron is narrow band, as the before and after state of the electron are predefined by the orbital states. The big difference for the photoelectric effect is that the end state of the electron is not predefined, only the initial state is. This initial orbital state is what determines the minimum frequency required from the incident light to increment the orbital state of the electron to free it from the atom, however higher frequencies will free the electron to a higher velocity. All electron orbital state changes are frequency keyed, except the last one that only has a minimum frequency value.

Also I think the "emission process" occurs over time not and not instantaneously, so that there is time for the electron orbital to absorb the incoming radiation. It would be interesting to think of the wave nature of an electron orbital to see why certain frequencies are required to change their states, I think that would be an interesting thing to try to add to an iterative wave simulator program.

cheers, Jamie

Reply to
Jamie

Hi,

Thanks for the link, I am watching that.. but to me quantum mechanics is like trying to decipher assembly code for a wave simulator instead of running the program to visualize the output.

cheers, Jamie

Reply to
Jamie

Hi,

For measuring time of flight couldn't the required shutter speed be reduced by making the experiment with longer path lengths? Also if a higher frequency light source is used that may allow fringes to still appear even with the lower frequency shutter scattering the light, since the fringes would be higher frequency that the shutter.

cheers, Jamie

Reply to
Jamie

The argument made here is that it's fundamentally impossible to simultaneously have interferance fringes and know which leg of the interferometer the photon took. The rest are details.

It's fairly easy to measure time of flight, and to resolve whether individual photons took the short leg or the long leg. But if you have or get enough information to measure the start time of a light pulse or of single photons, it becomes impossible to get fringes.

I think that's true.

One argument made here was that, if you have fringes, the TOF timings of the differing legs are smeared by the quantum uncertainty. That is not true. It's the path length measurement that destroys the fringes. Specifically, the problem is acquiring the start times. At the instant the photon hits the detector, it becomes an official, non-ambiguous particle, so that end is easy to measure.

What's interesting to me is that, apparently, the wavelength/energy of a single photon is altered by its passing through an open shutter. And the faster the shutter opens and closes around the photon, the bigger the change. It's like the shutter chops off bits of the extended head or tail of the wave.

John

Reply to
John Larkin

On a sunny day (Wed, 08 Jun 2011 16:38:32 -0700) it happened John Larkin wrote in :

The Copenhagen interpretation has been disproven.

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Reply to
Jan Panteltje

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No it hasn't - you have to *read* the whole article rather than the sensationalised headline. The ensemble properties of large numbers of particles trace out something like the pilot ray solution. This is not unexpected and is in line with modern theoretical physics as authors have pointed out. The prohibition is on knowing constraining two cojugate variables for a single particle at the same time.

It also leads to the (N+1/2)hv zero point energy which gets around the problem that otherwise when N=0 you know both exactly.

Regards, Martin Brown

Reply to
Martin Brown

It has been done by using weak measurements - that is what that garbled announcement that Jan is sounding off about is in reality.

A fairly decent technical description of what they have done and the sort of Youngs slit result you are interested in has actually been done by exploiting the birefringence of calcite. It is a very cunning experiment but it doesn't really alter quantum mechanics appreciably - it just uses a new way to do an experiment to horse trade for the sweet spot of knowing both conjugate properties of the wavefunction averaged over many photons well enough to be useful for engineering puposes.

The reconstructed fringes from their observations are very nice. See:

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I don't think anyone was particularly wedded to the either or hardline complimentarity interpretation for particles vs waves it was just a case of it being experimentally very technically difficult to do a weak measurement that tells you something sufficiently useful and doesn't in the process scramble the wavefunction.

No. It is gating the vacuum fluctuations for a very short time that does it - you are effectively borrowing energy from time sampling fluctuations in the zero point energy of the vacuum. And the shorter delta_T is the more energy you can borrow. You have to give it back again when the photon gets absorbed or in the case of virtual particle antiparticle pairs when they recombine and annihilate.

They do that RI trick to pulse shape the very fast lasers.

Regards, Martin Brown

Reply to
Martin Brown

Not true, of course. A photon can be detected and the energy analyzed, but the Heisenberg uncertainty says the absolute energy value is only available if you take infinite time to absorb the photon. The universe isn't old enough for that ever to have happened.

More to the point, a wave is only monochromatic if it extends to infinity (and finite experiments only deal with relatively-monochromatic light).

Reply to
whit3rd

That doesn't mean the energy is unmeasurable. In Mössbauer spectroscopy, gamma ray absorption is changed by doppler effect, with velocities measured in mm/sec. It doesn't take very long to measure insanely narrow line widths, parts in 1e14 or so.

John

Reply to
John Larkin

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