## GRED Answer: Double slit with fast lensless video screen

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### Re: GRED Answer: Double slit with fast lensless video screen

IMHO According to Heisengerg's uncertainty relation, a interference pattern will be seen.

Using a particle interpretation, any photon has to pass one of the to slits; therefore position and moving vector can are limited in that moment. Because position is determined quite exactly due of the dimensons of the slits, there has to be greater uncertainty in the vector of movement. Therefore, first a superimposition of two single-slit interference patterns will be expected.

The x-component of movement at both slits are equal, but the y-component is opposed. Due to the dimensions of the slit the two components are determined by different extend.

The incertainty of the y compontent of movement coupled with the y position will form the two slit interference pattern.

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### Re: GRED Answer: Double slit with fast lensless video screen

Scottopoly wrote: - Measuring the precise time a photon takes to travel is not equivalent to measuring which slit it went through. In principle, a photon could take a path that is not a straight line from the source to the target, travelling at the speed of light and taking the corresponding time to do so, though anything but the most direct path will be highly unlikely.
Okay, I admit I can't stop thinking about this problem. The "perfect" timing equipment, the Uncertainty Principle, the which-way Timing Hypothesis, and the wave-particale duality did not fit well together - the story did not settle well. But there is a new twist to the solution that hasn't been clearly presented yet, and Scottopoly's single idea (above) finally hit me with the missing link:

Bottom Line: Interference pattern doesn't change - This assumes perfect timing with infinite accuracy and precision!

Why?
1. Every photon goes through both slits, ALWAYS - We only measure timing, we don't know where the photon goes, therefore it goes through both slits.
2. Photon arrival times at the EXACT same screen location WILL NOT be identical. - This is the new idea and helps all the puzzle pieces fit together better. This is where I hypothesize the Uncertainty Principle applies. The fact that a collection of point-source photons arrive at the screen showing an interference pattern means that somehow information from both slits is present, and yet a point of light shows up at a specific location! But how can that be? The ONLY answer that works is the photon arrival times (flashes of light) at an identical screen location will vary by ~λ/c.
3. The proposed Timing Hypothesis is wrong - For the previous reasons. The foundation of the model assumed that identical screen locations would yield identical measured arrival times. This I believe is the flaw.

This is a hypothesis, but I think all the pieces go together better now. Though my earlier conclusions are the same, the story is different. And now the simplictiy of how the problem was stated also makes sense. Yes, I think this is the more correct reasoning. It just took some time getting here.
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### The spoiler bit

First of all, one important thing to keep in mind about these double slit experiments: while it's easy to imagine how a huge stream of photons can all be interfering with each other at a screen, it generally defies people's intuition that this still happens with very dim sources. Even if the source is so weak that only one photon ever exists per second, the interference pattern will still take the same form. It implies that a single photon somehow interferes with itself. So you should take that for granted in both the "simple screen" and the "video sensor array" scenarios here.

Not knowing which slit something went through has serious implications in a double slit experiment. When a photon strikes the screen, there are two possible previous versions of reality that might be consistent, one each for the left and right slits. A point on the screen will get lit up only if it doesn't reveal any extra information about left vs right history by doing so. There are places where if a photon lands, you can say definitively, that it went through this slit, or that one. And those places (minima) are dark. Then there are other places where if a photon lands, you would have to enumerate... it could have either traveled there X wavelengths from the left slit, or Y wavelengths from the right slit. And those places (maxima) are where all the photons go. You'll see that the photon distribution arranges itself in such a way as to maximize the ambiguity between the two previous versions of reality, i.e. into an interference pattern with a dependence on wavelength. In this way Nature prevents you from having any "left slit vs right slit" information that you don't specifically measure.

To clarify what we're being given in this problem, a bit more succinctly: the difference between this experiment and the previous one is that we know the flight time, and hence, have a priori knowledge of which slit a photon went through. That is the relevant one single bit of new information that all this business about the flight time is supposed to be hinting at, that we always know which slit now, because that's what we're measuring.

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### Re: GRED Answer: Double slit with fast lensless video screen

alter-ego wrote:2. Photon arrival times at the EXACT same screen location WILL NOT be identical. - This is the new idea and helps all the puzzle pieces fit together better. This is where I hypothesize the Uncertainty Principle applies. The fact that a collection of point-source photons arrive at the screen showing an interference pattern means that somehow information from both slits is present, and yet a point of light shows up at a specific location! But how can that be? The ONLY answer that works is the photon arrival times (flashes of light) at an identical screen location will vary by ~λ/c.
Yes, they have to -- because otherwise its wave packet would be so short that its wavelength was fuzzy, which would destroy the interference pattern.

How do you determine the wavelength? Count the number of crests in the wave packet, divide by its length. However, the crests are not all of the same height (they go towards 0 at the edges of the wave packet), and for two neighboring crests to have different amplitudes requires the waveform to be composed of at least two slightly different frequencies. As a rule of thumb, assume an uncertainty of at least one wave over the entire packet, which means that in order to pinpoint the wavelength within 1%, your wavepacket must take ~100λ/c time to arrive.
Henning Makholm

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### Re: GRED Answer: Double slit with fast lensless video screen

Hi Wavefunction,

Very interesting. This seems like a good homework or test problem for a QM class.
I have some questions, but I already getting feedback.
This was a good problem!

I'd still like to know if there can be intrinsic timing uncertainty imposed by the Uncertainty Principle. Why isn't it showing up in this problem?

Thanks for any further insights you can provide,
ae
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### Re: GRED Answer: Double slit with fast lensless video screen

alter-ego wrote:I'd still like to know if there can be intrinsic timing uncertainty imposed by the Uncertainty Principle.
There is. Time and energy are one of the standard examples of pairs of quantities that the uncertainty principle applies to.
Why isn't it showing up in this problem?
Why do you think it isn't?
Henning Makholm

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### Re: GRED Answer: Double slit with fast lensless video screen

Henning Makholm wrote:
alter-ego wrote:2. Photon arrival times at the EXACT same screen location WILL NOT be identical. - This is the new idea and helps all the puzzle pieces fit together better. This is where I hypothesize the Uncertainty Principle applies. The fact that a collection of point-source photons arrive at the screen showing an interference pattern means that somehow information from both slits is present, and yet a point of light shows up at a specific location! But how can that be? The ONLY answer that works is the photon arrival times (flashes of light) at an identical screen location will vary by ~λ/c.
Yes, they have to -- because otherwise its wave packet would be so short that its wavelength was fuzzy, which would destroy the interference pattern.

How do you determine the wavelength? Count the number of crests in the wave packet, divide by its length. However, the crests are not all of the same height (they go towards 0 at the edges of the wave packet), and for two neighboring crests to have different amplitudes requires the waveform to be composed of at least two slightly different frequencies. As a rule of thumb, assume an uncertainty of at least one wave over the entire packet, which means that in order to pinpoint the wavelength within 1%, your wavepacket must take ~100λ/c time to arrive.
Hi Henning,
It's late for me, so please excuse me if I'm missing something. The timing of the flashes are uncertain and the positions are not. The quantity ~λ/c seem to me to be within the bounds of the Uncertainty Principle (but it does have to be applied legitimately which I'm not sure). But by itself, the timing uncertainty does not imply positional UNcertainty. It arises solely from deciding to pin down the timing of the flashes.
ae
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### Re: GRED Answer: Double slit with fast lensless video screen

Henning Makholm wrote:
alter-ego wrote:I'd still like to know if there can be intrinsic timing uncertainty imposed by the Uncertainty Principle.
There is. Time and energy are one of the standard examples of pairs of quantities that the uncertainty principle applies to.
Why isn't it showing up in this problem?
Why do you think it isn't?
I was resonding to the spoiler. I need to get some rest and maybe I won't have the same questions tomorrow

Thanks Henning.
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### Re: GRED Answer: Double slit with fast lensless video screen

alter-ego wrote:It's late for me, so please excuse me if I'm missing something. The timing of the flashes are uncertain and the positions are not. The quantity ~λ/c seem to me to be within the bounds of the Uncertainty Principle (but it does have to be applied legitimately which I'm not sure). But by itself, the timing uncertainty does not imply positional UNcertainty. It arises solely from deciding to pin down the timing of the flashes.
That sounds about right. Of course the positions are also slightly uncertain, but I don't think that is crucial to resolving the question.

And ~λ/c underestimates the time uncertainty; it should be ~1/cΔf, so the more certain you are of the frequency/wavelength (and therefore the more well-defined the interference bands you're expecting will be), the less can you know about arrival times.

Bonus observation: Instead of trying to make our timing measurements more precise, we could just increase the scale of the entire experimental setup, such that the differences in travel time between the two slits become large enough to measure. But in that case we would not expect interference in the first place (no matter whether we measure the times or not), because the nonzero part of the wavetrain-through-slit-A would only interfere with a part of the wavetrain-throgh-slit-B whose amplitude is zero and vice versa.
I was resonding to the spoiler.
Sorry then. (To be clear, I think the spoiler goes wrong in assuming that it is possible in principle to measure the times precisely enough to make the interference go away.)
Henning Makholm

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### Re: GRED Answer: Double slit with fast lensless video screen

alter-ego wrote: I'd still like to know if there can be intrinsic timing uncertainty imposed by the Uncertainty Principle. Why isn't it showing up in this problem?
Energy and time are mutually unascertaintable measurables equivalent to position and momentum being mutually unascertainable measurables. (Just look at the units.) But position-momentum is better to use here. It's probably easier to view this as eliminating one single bit of position uncertainty between the two slits (i.e. one bit to indicate which slit); the question's contrivances about the timing apparatus are hinting at a position measurement. (Speed is well known.)

How have we paid for this extra bit of position information in terms of momentum uncertainty? What the hell, spoiler tag:

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### Re: GRED Answer: Double slit with fast lensless video screen

The time of flight from source to screen detector via slit 1 is $t_1$, for slit 2 is $t_2$; $t_2 - t_1=\delta t$. Pulse the source narrower than $\delta t$. The time-modulation is $\pi(t)$, a pulse function. This should'' give time-of-flight information at the detector that can distinguish the slit path. (The detector must be off-axis, so $t_1 \neq t_2$.)

At the detector, with $|\psi_j>$ the wave functions from each of the slits at the detector, we have
$$|\Psi>= \pi(t - t_1) |\psi_1> + \pi(t - t_2) |\psi_2>$$

Consider the probability operator $|\Psi><\Psi|$, and trace out (time integrate) the modulation, getting
$$\rho = \alpha_{11}|\psi_1><\psi_1|+\alpha_{22}|\psi_2><\psi_2|+ \alpha_{12}|\psi_1><\psi_2| + \alpha_{21}|\psi_2><\psi_1|$$
with $\alpha_{jk}=\int \pi(t - t_j)\pi(t-t_k)dt$. Interference'' is just the cross terms.

Naively assume the pulse travels unchanged. Then at the detector at all times one term or the other is zero; $\alpha_{12}=\alpha_{21}=0$. There can be no interference.

{\vskip1ex}
Note carefully: This has nothing to do with the time-resolution'' of the detector. In interference, it is always the distinguishability of the alternate situations, never the detection of that distinguishing factor, that destroys'' the interference.

And it would really help your understanding to give up such ideas as the photon goes through both slits'' or the photon interferes with itself''. Particles do things (eg, move on paths, hit detectors, ...). Wave functions are the probabilities that these things occur. The probability of the occurance of an event is categorically distinct from the occurance of the event. (Categorically distinct'' means conceptually orthogonal --- it is blind confusion to interchange them.)

{\vskip1ex}
Of course, things are much more complicated than this. First, the pulse will not travel unchanged. It will spread, causing the pulses at the detector to overlap to some extent; now the cross-term $\alpha$'s don't vanish, and this decrease in orthogonality results in an increase in interference.

But also, because the source is pulsed, it is no longer monochromatic. This causes the result at the screen to be an overlay of different patterns, smearing any interference pattern that might be present -- the minima and maxima will be broadened, and no longer fully bright and dark.

We are now well past the point of vague hand waving about uncertainty principles and the rest (what appears in the posts above). It is necessary to do some physics. The whole thing needs to be parametrized (slit separation, distance to screen, off-axis distance to detector, central wavelength, etc). Then a pulse shape needs to be chosen, Fourier analyzed, and the resulting interference patterns constructed. Then the pulse shape spread must be computed and applied. Perhaps then something can be said about the appearance of the interference. (Actually, I guess the whole thing needs to be done in {\bf k}-space, rather than trying to separate the different wavelength patterns from the pulse spreading.) This would make a nice senior physics paper.

{\vskip1ex}
Kim Kirkpatrick
\bye

Copy this to a *.tex file and process it to see the formuals. Otherwise you can just read through the TeX.

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### Re: GRED Answer: Double slit with fast lensless video screen

I think the spoiler goes wrong in assuming that it is possible in principle to measure the times precisely enough to make the interference go away.
While true in principle, that's a hand-waving thing.

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### Re: GRED Answer: Double slit with fast lensless video screen

Here's a similar question.

The nicest interference patterns are produced by diffraction gratings (lots of evenly spaced slits); just a few thin beams of light (an odd number) come out, each indicating a precise momentum. That's from throwing most of the position information away; you have no clue which slit at all. But you do know a little about where the photon was, since photons can only go through slits, and the cumulative area of the slits doesn't account for the entire area of the grating.

Let's say for example that these slits are 1.0 micrometers apart, and 0.1 micrometers wide. A photon gets through the grating, at position x, in meters. You know nothing about x at all except its tenth-micrometers digit, which you do know. We pay for that information in terms of p, our information on momentum, roughly trading away the first nonzero momentum digit (but only the first) to get the tenth-micrometers digit. We retaining all the rest of the digits we have for momentum. Stick this grating in the path of a nicely collimated coherent laser beam, where photons have a well-defined momentum), and what was previously a single beam spreads out into... about 9 or 11 nicely collimated beams, each indicating one of 9 or 11 well defined momenta. (Pick suitably colored light for this gedankenexperiment so that we get an odd number that's at least near ten.)

Not knowing which beam means that you know almost all of the momentum variable except its first nonzero digit. (Pick suitable units, base-9 or base-11 if required, keep hand waving constants around as necessary, etc. etc. to get the clean separation of known/unknown digits that should be possible.) If we knew which beam a given photon were in, we would know its momentum as precisely as we did before, when there was nothing in the way and there was just a single beam.

If you screw with the grating by trying to figure out which beam (trying to score that first digit of the momentum), using the fancy gizmo this question describes, do you know what is going to happen now?

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### Re: GRED Answer: Double slit with fast lensless video screen

wavefunction wrote:
alter-ego wrote: I'd still like to know if there can be intrinsic timing uncertainty imposed by the Uncertainty Principle. Why isn't it showing up in this problem?
Energy and time are mutually unascertaintable measurables equivalent to position and momentum being mutually unascertainable measurables. (Just look at the units.) But position-momentum is better to use here. It's probably easier to view this as eliminating one single bit of position uncertainty between the two slits (i.e. one bit to indicate which slit); the question's contrivances about the timing apparatus are hinting at a position measurement. (Speed is well known.)
Now this looks like good stuff! I need to take some time to digest your comments, and I want to try to understand where Kim is comming from. There are two important (maybe philosophical) ideas I want to understand, and not just accept at face value:

First is the uncertainty principle and that it is not violated in arriving at the diffraction pattern answer (you've already responded to this, and I need to go over it). I do find that position-momentum is easier to use, but I don't yet get how event timing can be better than ~λ/c (Implied by Kim?), which is also the arrival timing difference between the slits. Let me see if you've answered this, and I'll follow up if I need to.
Second is to understand what constitutes a timing. I can kind of follow what a "detection" is, but in this problem, interference visibility, or not, does depend on timing, not the means of how the photon detection occurs. But when / where does timing collapse the wave function? If I turn off the timing equipment, it seems the fringes simply come back(?) I have created a couple thought experiments to help me answer this, but I realized that using timing knowledge (as opposed to normal photon energy / momentum perturbations) is not obvious, and I lack the understanding of what "timing" really means and how it remains consistent with the uncertainty principle.

Wavefunction, I sure appreciate your patience. At some point I will become a deer in the headlights, but I don't think I'm there yet. I have a BS in physics, and I simply love these discussions. This problem has really forced me to clear out some cob webs in my head.
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### Re: GRED Answer: Double slit with fast lensless video screen

Henning Makholm wrote:Bonus observation: Instead of trying to make our timing measurements more precise, we could just increase the scale of the entire experimental setup, such that the differences in travel time between the two slits become large enough to measure. But in that case we would not expect interference in the first place (no matter whether we measure the times or not), because the nonzero part of the wavetrain-through-slit-A would only interfere with a part of the wavetrain-throgh-slit-B whose amplitude is zero and vice versa.
Upon further thought, I'll promote this to be my principal answer. It is simple, elementary, and works for all scales. No need to do Fourier transforms or muck around with uncertainty principles. So:

1. Assume, perhaps generously, that we can measure the exact shape and timing of the wave packet that gets emitted at the left end of the experiment.

1a. It is not meaningful to ask when during that wave packet the "actual" photon leaves the emitter. If you try to measure it anyway, all you'll get for your effort is a shorter wave packet. And shorter wavepackets are, as argued below, inherently less effective at showing interference patterns. (This is not because of the uncertainty principle; it is how the uncertainty principle works in the first place!)

2. Also assume, perhaps generously, that we can measure exactly the time and place where the photon interacts with the detector.

3. How interference works in the usual one-photon experiment is that the wavetrain can arrive at a given point on the detector via two paths of different length, so it will arrive there in two copies of which one is delayed with respect to the other. We add the copies and integrate the square of the sum over time to find the probability of detecting the photon at that particular point.

4. If the wavetrain is shorter than the difference in travel times, integrating the sum will give the same result as adding two separate integrals. Therefore we see no interference, whether or not we take note on the detection time. However, this is also precisely the case where knowing the detection time will allow us to infer which slit the photon must have passed through.

5. If the two wavetrains overlap partially, we can consider the time in three periods. Call period L the interval when the only contribution to the amplitude sum comes from the copy of the wave train that passed through the left slit. If we detect a photon during this period, we can infer that it went through the left slit. Mutatis mutandis, if we detect a photon during period R, it must have come through the right slit. What is left is period B, during which the sum wave contains contributions from both copies of the wave train. This is where interference happens. But when we detect a photon during periond B, we cannot infer which slit the photon must have passed through. It could be either, by the very definition of period B.

6. The probabilities from periods L, B, and R simply add. If we do the experiment without measuring arrival times, we see a strong interference pattern if B is long compared to L and R, and very little interference if B is short compared to L and R. We see the exact same compound pattern when measuring time, and the strength of the interference is directly related to the fraction of photons we detect in the "could-be-either" period B. If we measure times and create three separate scatterplots according to whether we detected the photon in periods L, B, or R, the plots L and R will show no interference, and plot B will show 100% interference.
kirkpatrick wrote:We are now well past the point of vague hand waving about uncertainty principles and the rest (what appears in the posts above). It is necessary to do some physics. The whole thing needs to be parametrized (slit separation, distance to screen, off-axis distance to detector, central wavelength, etc). Then a pulse shape needs to be chosen, Fourier analyzed, and the resulting interference patterns constructed. Then the pulse shape spread must be computed and applied. Perhaps then something can be said about the appearance of the interference. (Actually, I guess the whole thing needs to be done in {\bf k}-space, rather than trying to separate the different wavelength patterns from the pulse spreading.) This would make a nice senior physics paper.
We would need to do all that if we wanted quantitative predictions that could be compared with an actual experiment. But my understanding is that that is not what this thread asks for. It asks a qualitative question: Why is this simple gedankenexperiment not a paradox? And that calls for a qualitative understanding of how it works, with as few nitty-gritty assumptions about particular pulse shapes and exact dimensions as possible. (But no fewer, obviously).
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### Re: GRED Answer: Double slit with fast lensless video screen

...but I don't yet get how event timing can be better than ~λ/c (Implied by Kim?), which is also the arrival timing difference between the slits.
The event timing accuracy is a property of the pulse and detector timers; it doesn't depend on the photon wavelength. Our timers don't suddenly get crappier if we move from blue to red light.

We can generate an arbitrarily precise measurement of a photon's time of flight to within X microseconds (where X is a property of the timer), even if the probability of detecting it doesn't vary significantly during any arbitrary given time window of X microseconds in duration. That's the distinction of the event vs the probability of the event.

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### Re: GRED Answer: Double slit with fast lensless video screen

Henning Makholm has given the correct naive answer in his point 4. It is identical to that which I posted a few hours earlier. (It appears that, by posting a quantitative development in TeX, I made it invisible to readers on this list.)

But, as I said, this is a matter of naively assuming that the pulse travels without change. It does not. So to get an answer to this question, it is necessary to consider this spatial spreading, which will create an overlap of the sort Henning seems to be discussing in his next points. There is also the effect of the spectral spreading due to the pulsing of the wave, which does not destroy interference but will make it less detectable.

I am quite certain that there are no qualitative principles to resolve this. Only a quantitative result (not numbers, but mathematical expressions relating the parameters of the problem) can let us determine the possibility of a combination of parameters for which the spread of the pulse is not great enough to allow coherence to be reestablished.

If you actually want to know whether pulsing the double slit source will destroy the interference pattern, you will have to go through the mathematical calculations I described. There's no free lunch.

jmac

### Re: The spoiler bit

wavefunction wrote:How can we make our old interference pattern come back? By throwing away the timing information that this new video sensor array is giving us
How do the photons know whether or not we are going to throw away the timing data?

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### Re: GRED Answer: Double slit with fast lensless video screen

How do the photons know whether or not we are going to throw away the timing data?
One might ask how (in the simple screen case, with one or two slits) a photon knows that the slit it didn't go through was covered or not.

Measuring the flight time identifies the slit, essentially "covering" the other slit, because a trip through that one would be inconsistent with new observations.

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### Re: GRED Answer: Double slit with fast lensless video screen

kirkpatrick wrote:Henning Makholm has given the correct naive answer in his point 4. It is identical to that which I posted a few hours earlier. (It appears that, by posting a quantitative development in TeX, I made it invisible to readers on this list.)

But, as I said, this is a matter of naively assuming that the pulse travels without change. It does not. So to get an answer to this question, it is necessary to consider this spatial spreading, which will create an overlap of the sort Henning seems to be discussing in his next points. There is also the effect of the spectral spreading due to the pulsing of the wave, which does not destroy interference but will make it less detectable.
I'm assuming we're in vacuum such that light pulses do not spread.

But I don't really need that assumption. Instead of my point 1 "Assume that we can measure the shape and timing of the pulse when we emit it", we can just cut to the chase and say: Assume that we can measure stuff and do whichever calculations are necessary to find out how the two relevant contributions to the wave look like when they arrive at the detector. These calculations can be arbitrarily complex if we fill the space the photon must pass through with dielectrics, plasmas, things with color-dependent optical indices and so forth, but the only thing that matters for my argument is that their answer exists. Two wavetrains arrive at the detector. They have fancy shapes, I don't care exactly which, I'm just assuming that with enough work we could know them. They either overlap or they don't. If they don't overlap, there is no interference whether we measure arrival times or not. If they do overlap, there is interference whether we measure arrival times or not, but measuring arrival times will not allow us to conclude anything about which slit the photons used.
If you actually want to know whether pulsing the double slit source will destroy the interference pattern, you will have to go through the mathematical calculations I described. There's no free lunch.
But that is not the question that was asked.

We don't need to compute from first principles whether or not we will see interference, because the question as asked explicitly stipulates as a boundary condition that we start from an experiment that does produce interference and the only change we make is to replace the "image screen" with a "fast lensless video screan". The question is whether that change alone can make the interference go away. And the answer is that it can't.

The question does say "The time of release of single source photons is also recorded precisely." I read this as meaning that the apparatus to record the release time has been there all the time, even in the initial experiment where we're told that there is interference. If it were not so, the title of the question should not have been "Double slit with fast lensless video screen", but rather "Double slit with precise recording of photon release times". It appears to be quite possible that adding a release-time recorder to an experiment that had none might destroy interference, and deciding that would indeed require the calculations you allude to (and the correct answer would then be "insufficient data"). But it is, to the best of my reading ability, a different question from the one we were asked.
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### Re: GRED Answer: Double slit with fast lensless video screen

wavefunction wrote:
...but I don't yet get how event timing can be better than ~λ/c (Implied by Kim?), which is also the arrival timing difference between the slits.
The event timing accuracy is a property of the pulse and detector timers; it doesn't depend on the photon wavelength. Our timers don't suddenly get crappier if we move from blue to red light.

We can generate an arbitrarily precise measurement of a photon's time of flight to within X microseconds (where X is a property of the timer), even if the probability of detecting it doesn't vary significantly during any arbitrary given time window of X microseconds in duration. That's the distinction of the event vs the probability of the event.
I am following these discussions fairly with a bit of hobbling. I like that both kirkpatrick and Henning are presenting explanations keeping the uncertainty principle at bay, but if you don't mind, I'd like clarification regarding proper use and interpretation of the ΔEΔt relation. Per your comment, the timers aren't (significantly?) affected by wavelength. Does that mean that the timers can measure photon flight time to an arbitrary small fraction of its wavelength and not violate the uncertainty principle? If so, is that intuitively explained? Is it that we can effectively determine the "center" of the time window? In your words, you used "X microseconds" as the example which I have no problem with, but I do have a problem with X attoseconds. Again, this is asked in principle. If you did answer this (above), then I didn't get it.
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jmac

### Re: GRED Answer: Double slit with fast lensless video screen

Suppose we run the experiment. After it is over, two people go in at different times. One of them looks at the timing data, the other doesn't. If I understand what you say correctly, the one that looked at the timing data would not see the interference pattern but the other one would, right?

Also, what if one person looked at the results without looking at the timing data, then looked at the timing data, and then looked at the results again. What would he see?

alter-ego
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### Re: GRED Answer: Double slit with fast lensless video screen

Henning,
For whatever it's worth, you description is exactly what I was struggling to put together. I've been extremely bothered why/how timing, per se, could change an observed outcome. Yes, I've assumed all along that the problem, as stated with a pre-existing, pre-timing interference pattern was critical in a deterministic way (if you will) of the post-timing resultant intensity distrubution. It remains to be seen if your description is correct, but you've expressed the qualitative description to the tee, in my opinion.
A pessimist is nothing more than an experienced optimist

wavefunction
Asternaut
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### Re: GRED Answer: Double slit with fast lensless video screen

I am quite certain that there are no qualitative principles to resolve this. Only a quantitative result (not numbers, but mathematical expressions relating the parameters of the problem) can let us determine the possibility of a combination of parameters for which the spread of the pulse is not great enough to allow coherence to be reestablished. If you actually want to know whether pulsing the double slit source will destroy the interference pattern, you will have to go through the mathematical calculations I described. There's no free lunch.
Yes there is- lunch came free with the question.
Aspect to consider: if the precise time-of-flight time for a photon is known, won't it be possible to compare this with the time-of-flight it would take for this photon to go through each slit, and therefore determine which slit this photon went through?
Unless lunch was poisoned, clearly the questioner is under the impression that such a combination of parameters does exist, and the difference to ponder here is the effect of knowing the slit identity. Not whether the premise of the question is really invalid and we're not going to be able to identify which slit at all in the first place because of other factors relating to pulse spreading that will interfere with the precision in a real experiment. I don't think this is a trick question.

A quantitative analytical approach would be appropriate if we were actually designing an apparatus for this experiment costing a lot of money, because in that case we would "actually want to know" whether we really have the slit identity, but for a thread about a gedankenexperiment, that's bringing a gun to a knife fight. This is a Guess the Result of the Experiment of the Day, not a thesis defense. The purpose of gedankenexperiments is to illustrate qualitative principles that do manifest themselves in nature to an extent, even if they end up dominated in reality by unrelated factors that go over everyone's head to the concern of no one. These aren't real experiments.

wavefunction
Asternaut
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Joined: Sun Jun 20, 2010 2:27 am

### Re: GRED Answer: Double slit with fast lensless video screen

Suppose we run the experiment. After it is over, two people go in at different times. One of them looks at the timing data, the other doesn't. If I understand what you say correctly, the one that looked at the timing data would not see the interference pattern but the other one would, right?
If someone is able to go in and look at the timing data afterwards, then it was measured. Nobody will see the interference pattern.