I Time measurement in a double slit experiment with single photons

[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.

 

[Marcin]

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.

@PeterDonis
Let's assume, that we can get 3 possible, statistical results with $Δt=(l_2 - l_1)/c$:
$t=(a+l_1)/c±Δt$
$t=(a+(l_1 + l_2)/2)/c±Δt$
$t=(a+l_2)/c±Δt$
Would they be noticeably different? For the sake of argument, let's assume that $Δt$ is slightly greater.

 

[Nugatory]

Let's assume, that we can get 3 possible, statistical results...

It sounds as if you are still thinking in terms of the particle leaving the source at some time $t$, and that will mislead you. When we say that the uncertainty in the time of departure is $(l_2-l_1)/c$ that does not mean that the particle was emitted at some point in time within that range but we are uncertain as to exactly when. It means that all times within that range contribute to the probability of the particle being detected at any given point on the screen.

If you're going to form a mental picture of what's going on (it's natural to want to try, but sooner or later you have to give up looking for classical analogies for quantum phenomena) you would be better off thinking of $(l_2-l_1)/c$ as the minimum "width in time" of the undetected photon if there is to be interference. You can't narrow it down any more than that.

 

[Marcin]

If you're going to form a mental picture of what's going on (it's natural to want to try, but sooner or later you have to give up looking for classical analogies for quantum phenomena) you would be better off thinking of $(l_2-l_1)/c$ as the minimum "width in time" of the undetected photon if there is to be interference. You can't narrow it down any more than that.

I think that @PeterDonis explained this uncertainty or resolution very well and I think I get it. My question still stands. I'm just asking about the correctness of the assumption about the statistical results. I've made a small amendment: "For the sake of argument, let's assume that $Δt$ is slightly greater."

 

[PeterDonis]

My question still stands.

@Nugatory answered it:

When we say that the uncertainty in the time of departure is $(l_2-l_1)/c$ that does not mean that the particle was emitted at some point in time within that range but we are uncertain as to exactly when. It means that all times within that range contribute to the probability of the particle being detected at any given point on the screen.

Emphasis mine. This means that this statement of yours...

Let's assume, that we can get 3 possible, statistical results

...is wrong; you can't make this assumption because it is false. If you run this experiment many times, you won't get "3 possible statistical results". You will get one distribution of results, with a standard deviation larger than $(l_2-l_1)/c$.

 

[Marcin]

It sounds as if you are still thinking in terms of the particle leaving the source at some time $t$, and that will mislead you. When we say that the uncertainty in the time of departure is $(l_2-l_1)/c$ that does not mean that the particle was emitted at some point in time within that range but we are uncertain as to exactly when. It means that all times within that range contribute to the probability of the particle being detected at any given point on the screen.

I apologize for omitting this explanation. It's very good and I think I really get it, not just because @PeterDonis said so :)

This means that this statement of yours... ...is wrong; you can't make this assumption because it is false. If you run this experiment many times, you won't get "3 possible statistical results". You will get one distribution of results, with a standard deviation larger than $(l_2-l_1)/c$.

I'm fully aware, that I can get just one statistical result. I want to check which of the three proposed (assumed) results will be the closest to the actual experiment result. I've also made the amendment: "For the sake of argument, let's assume that $Δt$ is slightly greater."

 

[Marcin]

@PeterDonis I'm rephrasing it:
Let's assume, that we can get just one the following, statistical results with $Δt=(l_2 - l_1)/c$:
$t=(a+l_1)/c±Δt$
$t=(a+(l_1 + l_2)/2)/c±Δt$
$t=(a+l_2)/c±Δt$
Would they be noticeably different? For the sake of argument, let's assume that $Δt$ is slightly greater.

 

[PeterDonis]

I'm fully aware, that I can get just one statistical result.

No, you aren't, because your question...

I want to check which of the three proposed (assumed) results will be the closest to the actual experiment result.

...is based on a false premise. A "statistical result" means that the result for a single run of the experiment is not predictable. It could be anywhere in the uncertainty range. That means any of your three proposed results could be the closest to it; there is no way to tell in advance which.

I'm rephrasing it

Doesn't change anything I said above.

 

[Marcin]

No, you aren't, because your question... ...is based on a false premise. A "statistical result" means that the result for a single run of the experiment is not predictable. It could be anywhere in the uncertainty range.

I agree. I don't care about the single run in this case, just about the statistic.

That means any of your three proposed results could be the closest to it; there is no way to tell in advance which.

And I want to check which one, being fully aware, that it tells me nothing about the individual paths. I don't want to guess and tell in advance, I want to check.

 

[PeterDonis]

want to check which one

How would you check this for the results from a large number of runs of the experiment? I have no idea what you want to "check". All three of your proposed values are within the uncertainty range. What else is there to check?

Note that what you quoted from my post just above this statement of yours was talking about the case of a single run, which you said you weren't interested in. For a single run, I suppose you could check which of your three proposed values was closest to the single value given by the single run, but (a) you say you're not interested in that, and (b) that check wouldn't mean anything anyway.

 

[Marcin]

How would you check this for the results from a large number of runs of the experiment? I have no idea what you want to "check". All three of your proposed values are within the uncertainty range. What else is there to check?

These 3 statistical results are "statistically" distinct. Of course they are overlapping. I want to check which of them is the closest to the experiment. However, if the experiment gives me the wider range, overlapping all of them:

it that case I have no additional information at all. Is this the case?

 

[PeterDonis]

Is this the case?

Yes, that's what I've been saying for several posts now.

 

[Marcin]

Yes, that's what I've been saying for several posts now.

How about this: the middle of the widest possible range has to be somewhere. I want to check where it is. Why is this also meaningless?

 

[Marcin]

@PeterDonis
Would you say, that once again we have even wider range, that overlaps them all?

I could ask this question and update this picture recursively.

 

[PeterDonis]

I could ask this question and update this picture recursively.

Yes, you could, which means continuing this discussion is pointless since it could go on forever. You apparently will never be satisfied with any answer we give you. Your question has been answered as well as it can be.

Thread closed.