# B Interference fringes, what if you sample them?

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1. Jan 21, 2019

### David Byrden

Imagine a standard Young's Slit experiment using photons. We obtain large, distinct interference fringes on a target screen.

Then, we embed into the target screen a telescope focused on and capable of resolving the slits. The width of the lens must be greater than about one fringe to obtain this resolution. So we use a nice big lens with two or three interference bands falling upon the lens glass.

The telescope's display gradually builds up an image of the two slits.

Now, the interesting part. We paint our lens surface with opaque paint, leaving only some narrow strips unpainted. These have the same spacing as the interference bands. They are too wide to cause significant diffraction in any light passing through them.

We place our telescope in two positions;
1. the unpainted strips on the lens coincide with dark interference fringes
2. the unpainted strips on the lens coincide with bright interference fringes

A naive person might expect the telescope to register a higher rate of photon detection in [2] than in [1]. Personally I expect the rates to be the same.

But I would appreciate if somebody who really knows the subject, would tell us what will result here?

Thank you.

David

Last edited: Jan 21, 2019
2. Jan 21, 2019

### RPinPA

My immediate response was going to be that in single-photon "interference" experiments, when you count the photons and plot the counts vs position, you get the classic double-slit pattern building up over time. So that's definitely "higher rate in [2] than [1]".

But hold on, things are a little more complicated than that.
https://sciencedemonstrations.fas.harvard.edu/presentations/single-photon-interference
"The experiment is designed to also make it possible (in principle) to know which of the two slits the photons are passing through. In this so-called which-path case, the Young's double-slit interference pattern does not manifest itself. This can be followed by a quantum eraser (to erase the which-path information) to recover interference. Thus, the act of measurement and the design of the experiment affect what is being measured. Even if not actually measured, the mere possibility that an observer could determine which slit the photon passed through causes the interference pattern to switch to non-interference."

So you can set it up so that you do not get the classic pattern.

Single-photon interference is an experiment where you have light intensity so low that photons arrive at your detector individually, and you can count the individual photons.

3. Jan 21, 2019

### David Byrden

In my scenario, there is an interference pattern on the target screen, and simultaneously I think there is not an interference pattern detected by the telescope which is embedded in the target screen, sampling photons functionally equivalent to those that hit the screen.

I don't think that you need QM to predict the outcome in my scenario, I expect that classical wave theory will tell us the correct answer. I hope that somebody knows the outcome for sure.

What I am hoping to highlight is that interference does not occur in the quantum particles while they fly from the slits to the target; it occurs in the environment after they hit the target.

David

Last edited: Jan 21, 2019
4. Jan 21, 2019

### BvU

What you describe here is called spatial filtering - which google.

Then you will be severely disappointed.
Anyway, this is not the way it works: experiment decides. A simple laserpointer and a lens on an improvised optical bench can teach you a whole lot.
Any basis for this speculation ? Any ideas for experiments to verify/falsify this ?

5. Jan 21, 2019

Staff Emeritus
Such a thing does not exist.

6. Jan 21, 2019

### David Byrden

There is no such thing as a telescope capable of imaging Young's slits from, say, two meters away?

Is this because the slits would be very close together? If so, why could we not resolve them by using a sufficiently large lens? I'm not seeing the show-stopper here.

David

7. Jan 21, 2019

### BvU

It's called a lens

8. Jan 21, 2019

### David Byrden

This is the basis; the particles, while crossing the gap to the target screen, still bear directional information. Therefore they can in principle be distinguished. Therefore interference (addition of wave functions) is not occurring in that gap.

Yes; the thought experiment that I just described.

David

9. Jan 21, 2019

### BvU

10. Jan 21, 2019

### CWatters

If it helps...

Many people have asked what happens if you set up a double slit experiment then make holes in the screen to see what appears on a second screen located behind the first. The hole being either on a fringe or in the dark area between fringes. Google it.

11. Jan 21, 2019

### DrChinese

Your speculation here is not correct. You could simply move the target screen up to see that there is interference occurring.

12. Jan 21, 2019

### Staff: Mentor

That line of thinking doesn't work (unless you move the detector/telescope so close to one of the slits that there is no appreciable amplitude for the paths to it from the other slit, in which case the lack of interference is to be expected). The problem is that the interaction with the slits leaves the particle with an uncertain transverse momentum so there's no directional information in the sense that you're thinking.

(This is a heuristic argument from Ghirardi. You can arrive at the same conclusion by doing the sum of paths thing).

13. Jan 21, 2019

### David Byrden

I'm sorry if I did not make myself clear.

As I see it, there is no interference in the air gap, but when the particle's wave function interacts with that of the target screen, then momentum, charge and most other properties are preserved but directional information is erased.
The interference, as I see it, occurs only after that interaction, because only when directional info is removed do the wave functions pertaining to the two slits become indistinguishable. They add (as complex amplitudes) only from that point on. Their interactions with the environment then also interfere.

So, if you ask me why there are few photons in the "dark" band, I say it's because there are two indistinguishable versions of you who do see impacts there, but the two "yous" almost cancel out.

Moving the screen up closer to the slits will of course not affect this process.

Am I looking at this wrongly?

David

14. Jan 21, 2019

### David Byrden

Doesn't that imply that no lens, no optical intrument of any kind, placed at the target screen, could possibly resolve the two slits as such?
Which, at first glance, would make astronomy an impossible pursuit.... ?

David

15. Jan 21, 2019

### Staff: Mentor

No (and that confusion may be behind the talking-past-one-another exchange between you and Vanadium50 farther up in this thread). Forming an image of the slits, or of an object on the source side of the barrier, is a different thing than assigning a path to a particular photon detection in the region illuminated by both slits.

16. Jan 21, 2019

### DrChinese

Depending on your perspective: this is either wrong (there IS interference there) or meaningless (because you can't meaningfully discuss what happens when no measurement is made).

17. Jan 21, 2019

### David Byrden

Well, now you have me totally flummoxed. I'll spell it out as I see it, please indicate my mistake;

1.An image of the slits looks like two bright lines.
2. It is obtained by summing many, many impacts of individual photons.
3. Every photon that arrives on the detector screen contributes to either one line or the other.
4 That photon therefore appears to have transited one slit or the other - the slit whose image it contributes to.

No?

David

18. Jan 21, 2019

Staff Emeritus
If you have the conditions for interference, and you stand in the light zone and look back at the slits, you will not see two bright lines. I couldn't find a picture with lines, but with dots (figure from the University of Tennessee at Knoxville) it will look like the figure on the right, not the left:

19. Jan 21, 2019

On the experiment presented by the OP, suggestion is to use a lens as opposed to a telescope. $\\$ If the paint stripes on the lens are placed on the bright stripes of the interference pattern at 2 meters, the energy is basically wiped away, so very little energy will be observed past this region. Any observations done in the region beyond the lens will receive very little energy. $\\$ If the opaque paint stripes are placed in front of the dark regions of the two-slit pattern at 2 meters, they will have little effect, and the results would be similar as if there were no paint stripes at all. $\\$ It's all a completely classical description.