Time measurement in a double slit experiment with single photons

[Marcin]

Assumption: Screen detector is much closer to the slits than in "standard experiment" and the small angle approximation can't be used to determine the interference fringe maxima, but the interference pattern still occurs.
Is it possible to measure the time of detection in such setup accurately? If so, how would the time distribution for fringe maxima look like? Could it be calculated from path-integral formulation? Would it correspond to the averaged path length divided by speed? "Time" is the period of time between the emission and the detection.

 

[PeroK]

Assumption: Screen detector is much closer to the slits than in "standard experiment" and the small angle approximation can't be used to determine the interference fringe maxima, but the interference pattern still occurs.
Is it possible to measure the time of detection in such setup accurately? If so, how would the time distribution for fringe maxima look like? Could it be calculated from path-integral formulation? Would it correspond to the averaged path length divided by speed? "Time" is the period of time between the emission and the detection.

What's the relevance of the screen being much closer? And, what's the relevance of the time of detection?

You could certainly measure the time that an electron hits the screen, but how do you measure the time it started? Or, the time it passed through the slits?

Note descriptions of such experiments often include a mixture of QM and classical concepts. In some ways this is fair enough: an electron, for the most part, follows a near-classical trajectory except where it interacts with a narrow slit or slits (or, at least, can be modeled as such). But, you cannot then be arbitarily precise about this and pretend that QM and the UP do not apply to the electron at these times.

It might be useful for you to try to decsribe the double-slit experiment in purely QM terms, without the concept of a classical trajectory at all.

 

[Marcin]

What's the relevance of the screen being much closer?

The point is to make the difference between path lengths (from slits to the same point on screen) relatively greater so they could not be approximated as parallel, to make this difference more significant in comparison to the distance between the slits, so the inequality: slitsDist<<pathLength is no longer true.

And, what's the relevance of the time of detection?
(...)
It might be useful for you to try to decsribe the double-slit experiment in purely QM terms, without the concept of a classical trajectory at all.

I want to know as much as I can about the difference between QM wave function propagation of photon interfering with itself and CM paths of photon considered as a classical wave or a particle. I think that travel time is an important aspect in this respect.

You could certainly measure the time that an electron hits the screen, but how do you measure the time it started? Or, the time it passed through the slits?

As I reckon, any measurement at the slits would destroy the interference pattern, so I want to measure the time of photon leaving the gun, so I need the detector right there.

Note descriptions of such experiments often include a mixture of QM and classical concepts. In some ways this is fair enough: an electron, for the most part, follows a near-classical trajectory except where it interacts with a narrow slit or slits (or, at least, can be modeled as such). But, you cannot then be arbitarily precise about this and pretend that QM and the UP do not apply to the electron at these times.

If such setup would make the time measurement inaccurate due to the uncertainty principle or because of any other reason, I still want to know if time calculation is possible and what are the results.

 

[PeroK]

The point is to make the difference between path lengths (from slits to the same point on screen) relatively greater so they could not be approximated as parallel, to make this difference more significant in comparison to the distance between the slits, so the inequality: slitsDist<<pathLength is no longer true.

I want to know as much as I can about the difference between QM wave function propagation of photon interfering with itself and CM paths of photon considered as a classical wave or a particle. I think that travel time is an important aspect in this respect.

As I reckon, any measurement at the slits would destroy the interference pattern, so I want to measure the time of photon leaving the gun, so I need the detector right there.

If such setup would make the time measurement inaccurate due to the uncertainty principle or because of any other reason, I still want to know if time calculation is possible and what are the results.

If we are talking about light (photons) then there is no way to detect a photon that does not effectively destroy it. The exception is where pairs of entangled photons are created and you can measure (gain information about) a photon by detecting its entangled partner.

To be precise the photon is, without doubt, a relativistic particle. And, standard QM is strictly non-relativistic. The description of the double-slit experiment usually skirts round this issue and assumes that certain principles of QM must also hold for photons. Which indeed can be shown experimentally.

To make any genuine comparison (in terms of experimental results matching theory) between light as an EM wave and light in QM, you would need QED or the full-blown QFT.

The double-slit is not the be-all and end-all of QM experiments. One key difference in using QED is not what happens in the double-slit, but what happens when light interacts with matter.

 

[PeroK]

I want to know as much as I can about the difference between QM wave function propagation of photon interfering with itself and CM paths of photon considered as a classical wave or a particle. I think that travel time is an important aspect in this respect.

To answer this question as best I can. If you study the theory of EM, by applying Maxwell;s equations you get transmission and reflection coefficients for the behaviour of light at a boundary (where the refractive index changes).

If you study QM, by solving the SDE (Schrodinger equation) you get transmission and reflection coefficients for a particle at a potential barrier. And, remarkably these coefficients take the same mathematical form as you get in the theory of EM.

This leaves you with two ways to model EM radiation: as an EM wave obeying Maxwell's equations; and, as a quantum particle behaving probabilistically. With the assumption that the same QM behaviour you can show for non-relativistic particles extends to photons.

I'm personally not aware of how QED would be used to predict what happens in the double-slit. Maybe someone else can help you there.

 

[Nugatory]

The point is to make the difference between path lengths (from slits to the same point on screen) relatively greater so they could not be approximated as parallel

You can produce the same effect by making the slits farther apart. There will be no interference when the difference in the travel times is greater than the coherence time; the position of the dot on the screen provides a which-way measurement.

 

[Marcin]

The double-slit is not the be-all and end-all of QM experiments. One key difference in using QED is not what happens in the double-slit, but what happens when light interacts with matter.

Wikipedia quote:

Richard Feynman called it "a phenomenon which is impossible […] to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery [of quantum mechanics]."

That's why I'm sticking to it :)

To answer this question as best I can. If you study the theory of EM, by applying Maxwell;s equations you get transmission and reflection coefficients for the behaviour of light at a boundary (where the refractive index changes).

If you study QM, by solving the SDE (Schrodinger equation) you get transmission and reflection coefficients for a particle at a potential barrier. And, remarkably these coefficients take the same mathematical form as you get in the theory of EM.

This leaves you with two ways to model EM radiation: as an EM wave obeying Maxwell's equations; and, as a quantum particle behaving probabilistically. With the assumption that the same QM behaviour you can show for non-relativistic particles extends to photons.

That's quite helpful, thank you.

I'm personally not aware of how QED would be used to predict what happens in the double-slit. Maybe someone else can help you there.

Thank you for all the answers.

 

[Nugatory]

Wikipedia quote:

Careful here - Feynman was speaking metaphorically here, not making a rigorous statement about QM. You’ll find a lot more than just the double slit experiment in his QED textbook.

 

[Marcin]

You can produce the same effect by making the slits farther apart. There will be no interference when the difference in the travel times is greater than the coherence time; the position of the dot on the screen provides a which-way measurement.

The point is to keep the interference. The difference in travel times must still be less than the coherence time - using your words.

 

[Marcin]

Careful here - Feynman was speaking metaphorically here, not making a rigorous statement about QM. You’ll find a lot more than just the double slit experiment in his QED textbook.

I sincerely believe you, but I'm still not afraid of quoting wikipedia. This is a full quote taken from the context specifically about the double slits, with a reference to the source.

 

[Marcin]

I'd like to know if this is a reasonable assumption: Travel time of SINGLE photon from slits to the fringe maxima on screen has to be equal or greater than the longer path divided by speed (c for photon). If the time was shorter, the interference pattern could not occur.

 

[PeterDonis]

Travel time of SINGLE photon

If you aren't measuring the travel time, you can't assume it has any definite value. To measure the travel time, you would need to measure the time the photon passes the slits, and the time the photon reaches the detector. But, as has been pointed out, the first time measurement would destroy the photon.

 

[Marcin]

If you aren't measuring the travel time, you can't assume it has any definite value. To measure the travel time, you would need to measure the time the photon passes the slits, and the time the photon reaches the detector. But, as has been pointed out, the first time measurement would destroy the photon.

I've decided to skip the drawing of the part of the path from the gun to the slits, to make it more clear. Imagine, that the first detector is at the outlet of the gun. I'm also asking is it a reasonable assumption regardless of the measurement.

 

[PeterDonis]

Imagine, that the first detector is at the outlet of the gun.

Doesn't matter, since, as noted, if you try to detect the photon at the slits--which you need to if you want to measure the time from the slits to the detector--you will destroy the photon.

I'm also asking is it a reasonable assumption regardless of the measurement.

Read this again:

If you aren't measuring the travel time, you can't assume it has any definite value.

 

[Marcin]

Doesn't matter, since, as noted, if you try to detect the photon at the slits--which you need to if you want to measure the time from the slits to the detector--you will destroy the photon.

I'm not trying to detect the photon at the slits. There are just two detectors: the first is at the outlet of the gun, the second is the screen.

 

[PeterDonis]

I'm not trying to detect the photon at the slits.

Then you can't make any assumptions about the time it takes for the photon to travel from the slits to the detector. Which means you can't make any assumptions about whether or not that time is shorter than the coherence time, which was what you said you were interested in.

 

[Marcin]

If you aren't measuring the travel time, you can't assume it has any definite value.

If I can't measure it, at least I want to calculate it and compare the results of CM calculation and QED or QM calculation.

 

[PeterDonis]

If I can't measure it, at least I want to calculate it

If you aren't measuring it, you cannot make any assumptions about it. Which means you can't calculate anything meaningful either.

compare the result of CM calculation and QED or QM calculation.

What QED or QM calculation? There isn't one.

 

[Marcin]

If you aren't measuring it, you cannot make any assumptions about it. Which means you can't calculate anything meaningful either.

What QED or QM calculation? There isn't one.

Is it impossible to determine the probability of detection in certain point and time from solving the Schrodinger equation or it's modification, or derive it from path-integral formulation? If so, I could say what time and detection position are the most probable.

 

[PeterDonis]

Is it impossible to determine the probability of detection in certain point and time from solving the Schrodinger equation or it's modification, or derive it from path-integral formulation?

You could calculate probabilities for detection at the slits at particular times, but doing that would invalidate any calculations about probabilities at the detector, since detection at the slits would destroy the photon. So you cannot calculate any joint probabilities for detection at the slits at some time $t_1$ and detection at the detector at some later time $t_2$.

 

[Marcin]

You could calculate probabilities for detection at the slits at particular times, but doing that would invalidate any calculations about probabilities at the detector, since detection at the slits would destroy the photon. So you cannot calculate any joint probabilities for detection at the slits at some time $t_1$ and detection at the detector at some later time $t_2$.

But the calculation (even at the slits) can't destroy the photon, right? :) But even if, I can skip it and calculate only the probabilities at the gun outlet and on the screen. That would satisfy me - I could make my comparison. Is it also impossible?

 

[PeterDonis]

the calculation (even at the slits) can't destroy the photon, right?

How is this relevant to anything?

I can skip it and calculate only the probabilities at the gun outlet and on the screen.

Then you can't say anything about how long it takes the photon to travel from the slits to the screen.

That would satisfy me - I could make my comparison.

Comparison with what? You said you were interested in the difference in travel times of the photon from the two slits to the screen. The only way you can say anything meaningful about that is if you measure the time the photon passes the slits--which destroys the photon. If you do not measure the photon's time of arrival at the slits, you cannot say anything meaningful about the travel time from either slit to the screen. No calculation you can make will change that.

 

[Marcin]

Then you can't say anything about how long it takes the photon to travel from the slits to the screen.

I can commit a horrible heresy and subtract the travel time from the gun to the slits (classical distance divided by speed) from the final result - the time of detection.

Comparison with what? You said you were interested in the difference in travel times of the photon from the two slits to the screen.

No. I want to compare a measured or calculated probabilistically time of detection and compare it with the travel time of classical paths through the both slits.

The only way you can say anything meaningful about that is if you measure the time the photon passes the slits--which destroys the photon.

This is my goal - to get addition information without the measurement at the slits.

If you do not measure the photon's time of arrival at the slits, you cannot say anything meaningful about the travel time from either slit to the screen. No calculation you can make will change that.

As above.

 

[Nugatory]

I can commit a horrible heresy and subtract the travel time from the gun to the slits (classical distance divided by speed) from the final result - the time of detection.

It’s not a heresy, just an irrelevant calculation that gives you a number that doesn’t represent anything.

 

[Marcin]

It’s not a heresy, just an irrelevant calculation that gives you a number that doesn’t represent anything.

If I repeated the experiment and this calculation multiple times for single photons, would the statistic also mean nothing?

 

[Nugatory]

If I repeated the experiment and this calculation multiple times for single photons, would the statistic also mean nothing?

What do you think it might mean? Be precise about how you will prepare the initial state of the system for each iteration and which measurements you will perform.

 

[PeterDonis]

to get addition information without the measurement at the slits

You can't.

 

[Marcin]

What do you think it might mean? Be precise about how you will prepare the initial state of the system for each iteration and which measurements you will perform.

It would be the statistic or distribution of travel times from the gun outlet (first detection) to the specific point on the screen (final detection) minus constant - the same for all photons, subtracted from all the results.

 

[Marcin]

I'd like to know if this is a reasonable assumption: Travel time of SINGLE photon from slits to the fringe maxima on screen has to be equal or greater than the longer path divided by speed (c for photon). If the time was shorter, the interference pattern could not occur.

Updated picture:

Updated question:
Is it possible to check this assumption experimentally: Travel time of SINGLE photon from the gun outlet to the fringe maxima on screen has to be equal or greater than the longer path divided by speed (c for photon). If the time was shorter, the interference pattern could not occur.

If such setup would make the time measurement inaccurate due to the uncertainty principle or because of any other reason, I still want to know is it possible to determine the probability of detection in certain point and time by solving the Schrodinger equation or it's modification, or derive it from path-integral formulation. If so, I could know the most probable detection time for certain points on screen, especially fringe maxima, and verify the assumption.

 

[DrChinese]

Is it possible to check this assumption experimentally: ...

The distance difference (L1-L2) is going to be an integer multiple proportional to the wavelength of the particle. This reference may assist you:

https://arxiv.org/abs/quant-ph/0703126

I think you are asking whether you could measure the time the particle took from when it passed some point at the source until it was detected at the screen. If you resolved that accurately enough, then it would be either L1 or L2. Which it was would then tell you the slit it went through, even though interference had been demonstrated.

Of course, you and I know intuitively already that can't be true. (Precisely because there is interference, the particle could not have traversed a specific slit.) I think your question then is actually: why not? I don't know all the issues that explain that, but here are a few:

1. You would need to know the time of emission of the particle quite accurately. (PS That is no piece of cake. With a laser source, how would you even know how many photons were emitted?)
2. You would need to know the time the particle hit the detector screen quite accurately.
3. You would need to exclude all detections where the travel time was somewhere between L1 and L2; since the result would not indicate unambiguously which slit was traversed.

I am not sure myself of what physical limits exist for the first two. To put this in perspective:

A common 405 nm laser has a frequency of 740228291 MHz, and that requires very high timing resolution. If I did my arithmetic correctly, such a photon travels one wavelength in on the order of 1.3 * 10^-15 seconds (1.3 femtoseconds). Not sure what's out there that could provide such resolution in this particular type of experiment, perhaps others can weigh in.

 

[Marcin]

The distance difference (L1-L2) is going to be an integer multiple proportional to the wavelength of the particle.

I know :) (L2-L1)=m*λ; It's in the picture.

Of course, you and I know intuitively already that can't be true. (Precisely because there is interference, the particle could not have traversed a specific slit.)

I agree. That's why I don't want to prove that it went through the specific slit. Interference disproves it.

I think your question then is actually: why not?

I want to check whether the interference pattern requires the contribution of both slits to the probability wave function after the assumed time and not earlier. I can see that L2 in the assumed inequality is misleading.

1. You would need to know the time of emission of the particle quite accurately. (PS That is no piece of cake. With a laser source, how would you even know how many photons were emitted?)
2. You would need to know the time the particle hit the detector screen quite accurately.

Good to know!

3. You would need to exclude all detections where the travel time was somewhere between L1 and L2; since the result would not indicate unambiguously which slit was traversed.

I think that many reproducible results of travel time between L1 and L2 would disprove my assumption and that would be the end of discussion for me. Still, as I've said, I don't want to prove that a specific slit was traversed.

A common 405 nm laser has a frequency of 740228291 MHz, and that requires very high timing resolution. If I did my arithmetic correctly, such a photon travels one wavelength in on the order of 1.3 * 10^-15 seconds (1.3 femtoseconds). Not sure what's out there that could provide such resolution in this particular type of experiment, perhaps others can weigh in.

I'm grateful for all the details, they show me the difficulty of this experiment. Thank you.

 

[PeterDonis]

I want to check whether the interference pattern requires the contribution of both slits to the probability wave function after the assumed time and not earlier.

You can't. The formula you use to calculate the probability amplitude for detection at any point on the detector screen does not have "the time the photon passed the slits" in it at all. You can subtract $l_1 / c$ and $l_2 / c$ from the measured time of arrival at the screen and attach the words "the calculated times when the photon passed slit 1 and slit 2" to that pair of numbers, but that is completely irrelevant to the actual prediction of the probability amplitude and its comparison with experiment. It's just a calculated number that, as @Nugatory has said, has no physical meaning.

 

[Marcin]

You can't. The formula you use to calculate the probability amplitude for detection at any point on the detector screen does not have "the time the photon passed the slits" in it at all. You can subtract $l_1 / c$ and $l_2 / c$ from the measured time of arrival at the screen and attach the words "the calculated times when the photon passed slit 1 and slit 2" to that pair of numbers, but that is completely irrelevant to the actual prediction of the probability amplitude and its comparison with experiment. It's just a calculated number that, as @Nugatory has said, has no physical meaning.

Please, forget about subtracting. That's why I've rewritten the question and updated the picture.

 

[PeterDonis]

Is it possible to check this assumption experimentally: Travel time of SINGLE photon from the gun outlet to the fringe maxima on screen has to be equal or greater than the longer path divided by speed (c for photon). If the time was shorter, the interference pattern could not occur.

The "travel time" itself is not the critical parameter; it's the uncertainty in the travel time.

Consider: the two different classical "travel times" are $(a + l_1) / c$ and $(a + l_2) /c$. The difference between them is $(l_2 - l_1) / c$. So if we measure the times of emission at the source and arrival at the detector, we have two possible cases:

Case 1: The time resolution of our measurements is smaller than $(l_2 - l_1)/c$. In that case, as @DrChinese has pointed out, measuring the travel time is equivalent to measuring which slit the photon went through, so there is no interference. But you've said you're not interested in this case; you're interested in the case where there is interference.

Case 2: The time resolution of our measurements is larger than $(l_2 - l_1)/c$. In that case, the time measurements do not distinguish which slit the photon went through, and there is interference. But that has nothing to do with the travel time itself; it's only the large enough uncertainty in the travel time that allows the interference to occur.

Or, to put it in a more provocative way: for interference to occur at the detector, the travel time of the photon must be uncertain enough that it is in a superposition of having gone through both slits; but that in turn means that it is in a superposition of having left the source at two different times, which differ by $(l_2 - l_1)/c$. The time of leaving the source has to be uncertain to that extent for interference to occur.

 

[Marcin]

Case 2: The time resolution of our measurements is larger than $(l_2 - l_1)/c$. In that case, the time measurements do not distinguish which slit the photon went through, and there is interference. But that has nothing to do with the travel time itself; it's only the large enough uncertainty in the travel time that allows the interference to occur.

Or, to put it in a more provocative way: for interference to occur at the detector, the travel time of the photon must be uncertain enough that it is in a superposition of having gone through both slits; but that in turn means that it is in a superposition of having left the source at two different times, which differ by $(l_2 - l_1)/c$. The time of leaving the source has to be uncertain to that extent for interference to occur.

Oh dang :D I get it and I don't like it, but in that case I have to give up and say thank you for the vivid and eloquent explanation.