# B Double slit probability question

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1. Feb 28, 2017

### entropy1

Consider the double slit experiment; if we position a detector at, say, the left slit, will a single particle, say, an electron, when fired at the slits, always be detected at the left slit, or will it be detected at the left slit 50% of the time? (so that it is 50% of the time at the right slit)

2. Feb 28, 2017

### BvU

If the electrons are fired at the two sits in a symmetrical manner, 50%
Is that really what you wanted to ask ?

3. Feb 28, 2017

### entropy1

Yes, I think so. I seems to me that, then, not measuring at the left slit is a measurement also, for we now know that the electron has to be on the right slit. It that correct?

4. Feb 28, 2017

### BvU

If you know it's passed a slit and hasn't passed the left slit, then yes. Am I painting myself in a corner now ?

(PS there are more ways than one to interpret 'not measuring' !)

5. Feb 28, 2017

### BvU

The electron doesn't know. There's a sequence in time here: the electron does whatever it pleases and the detector detects -- or not.

But now I sense we aren't communicating -- I wrongly assumed your detector sits after the slit (*). You have something else in mind

(*) Feynman influence: to see through which slit the electron passes

6. Feb 28, 2017

### mikeyork

This is a simple classical question containing no special quantum behavior.

Most will not pass through either slit. The slits merely select two trajectories. The detector at the slit merely marks the electron state as having passed through that slit. The result will be the same as throwing golf balls randomly at a fence with two missing planks. A screen on the other side which notices all electrons that have passed through both slits will see 50% from the left slit. But the detector at the left slit or any other apparatus on the other side that ignores all electrons from the right slit, will record 100% of the electrons it notices as having have passed through that left slit.

7. Feb 28, 2017

### Staff: Mentor

It depends on what you mean by "not measuring". If there is no detector at either slit, then you don't know which slit the electron went through. If there is a detector at the right slit but not the left slit, then you do know which slit the electron went through. Note that both of these statements assume that there is a detector well after the slits, where all electrons are detected (and where we can see if their points of impact build up an interference pattern or not). (And it also assumes that we do not consider any runs of the experiment where no electron is detected at all, because it hit something else and never reached any detectors that are present--as mikeyork pointed out, in a real version of this experiment, many electrons will end up this way.)

8. Feb 28, 2017

### entropy1

The following occurred to me:

If we have slit A with a screen A, and slit B with screen B, both A en B setting isolated from each other. Now, screen B is twice as far behind its slit as screen A from its.

We now fire an electron that passes boh slits. Then after a while, the the electron could hit screen A. The probability is 50% that it actually does.

If the electron does not impact screen A, the probability it will impact screen B is 100%!

So, if the electron does not impact screen A, it always impacts screen B.

On the other hand, if screen B is passed, then screen A must have been hit.

So, what does select which screen, A or B, will be hit? If it is collapse, why does screen A collapse the wavefunction 50% of the time, and when it doesn't, why does it collapse at screen B 100% of the time left? Or do both screens collapse 50% of the time and is there retrocausality from B to A? Or is there no collapse? Is there a non-local effect? Et cetera. Or is it just the way it is (do the math and so forth )

It does resemble entanglement to me. Is that correct?

Sorry if I misunderstand completely.

Last edited: Feb 28, 2017
9. Feb 28, 2017

The which path info will cause a collapse regardless if you measure only 1 slit. Any attempt to get the which path info will lead to the same thing even if no direct measurment was made (ex: delayed choice experiement)

10. Feb 28, 2017

### Staff: Mentor

Have a look at this paper, especially fig. 2, looking experimentally at multi-slit interferometry with C70 molecules.

What you see in fig. 2 is what happens as the proportion of molecules that gets excited increases: the contrast of the fringes diminishes, until it vanishes completely. This is because an excited molecule has a significant probability of emitting a photon, which would result in which-way information (even if the photon is not actually detected). Only the molecules that do not emit contribute to the fringe pattern. The others simply add to the constant background.

11. Mar 1, 2017

### entropy1

In my example setup there are no extra detectors. The screens are the detectors. I am wondering why:
• Screen A absorbs only 50% of the particles, while screen B always absorbs (all of the leftover) particles.
A view on this is that screen B equally absorbs only 50% of its particles, and screen A is the one absorbing the leftover 100%. A symmetrical situation, but mutually exclusive. So this resembles to me the collapsing of two entangles particles, where neither is determining the other's measured value.

The local interaction between particle and screen doesn't seem to explain what is happening (of course).

12. Mar 1, 2017

### BvU

Can you post a drawing of what you mean ? Again, it sounds like you have a totally different setup as thte one I read from your description. A screen as a detector ? why not call it a screen, then ?

13. Mar 1, 2017

### entropy1

I do call it a screen. It was ment to explain to Ostrados, who talked about detectors. (He talked about measurements, I assumed wrongly detectors I see now).

You can imagine the left side as a screen and the right side too.

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Last edited: Mar 1, 2017
14. Mar 2, 2017

### entropy1

15. Mar 2, 2017

### BvU

And how do you know that ?

16. Mar 2, 2017

### entropy1

I think I reason this way: the setup between the slits and the two (in this case) separated screens is identical to a setup that could have only one screen. With only one screen there would be an interference pattern, which means that the particle would have passed both slits. (one should do violence to the picture I posted, but I hope you know what I mean )

And besides that, the particle is only that at the moment it is detected, right?

17. Mar 2, 2017

### BvU

No. Only a small fraction of the electrons make it to the single screen.
To get an interference pattern, the slits need to be very narrow and very close together (of the order of the De Broglie wavelength)

Last edited: Mar 2, 2017
18. Mar 2, 2017

### entropy1

19. Mar 2, 2017

### BvU

You mean to say that the particle is not a particle until it is hitting a screen ? What would the source of the particle have to say about that ?

20. Mar 2, 2017

### entropy1

If a particle is a particle from the start, and it travels through only one slit, how can there exist something as an interference pattern?

• In case of two separated screens, the particle travels through one of the two slits;
• In case of one screen, the particle travels through both slits (interference)?
Does the number of screens determine how the particle travels?

I guess what I ask is: what determines which path(s) the particle will take?

Last edited: Mar 2, 2017
21. Mar 2, 2017

### StevieTNZ

Wouldn't you simply see an interference pattern across screens A and B (a bit out of alignment between the screens, though)?

22. Mar 2, 2017

### Staff: Mentor

Yes, that's pretty much it.

23. Mar 3, 2017

### BvU

Yes.
No.
Can't ask that. The probability the particle ends up somewhere is the outcome of a calculation that involves both slits. Same as a water wave with comparable geometry. It is a particle but it obeys a wave equation.

24. Mar 3, 2017