Double slit probability question

[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)

 

[BvU]

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

 

[entropy1]

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

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?

 

[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' !)

 

[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

 

[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.

 

[PeterDonis]

I seems to me that, then, not measuring at the left slit is a measurement also

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.)

 

[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.

 

[Ostrados]

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)

 

[DrClaude]

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.

 

[entropy1]

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)

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).

 

[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 ?

 

[entropy1]

A screen as a detector ? why not call it a screen, then ?

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).

Can you post a drawing of what you mean ?

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

 

[entropy1]

Picture above refers to: https://www.physicsforums.com/threads/double-slit-question.905869/#post-5704445.

 

[BvU]

We now fire an electron that passes both slits

And how do you know that ?

 

[entropy1]

And how do you know that ?

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?

 

[BvU]

No. Only a small fraction of the electrons make it to the single screen.

This is a simple classical question containing no special quantum behavior.
Most will not pass through either slit.

To get an interference pattern, the slits need to be very narrow and very close together (of the order of the De Broglie wavelength)

 

[entropy1]

No. Only a small fraction of the electrons make it to the single screen.

Am I right about this then?

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

 

[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 ?

 

[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?

 

[StevieTNZ]

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

 

[Nugatory]

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

Yes, that's pretty much it.

 

[BvU]

If a particle is a particle from the start

Yes.

and it travels through only one slit

No.

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

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.

 

[Ostrados]

If you split the screens using a physical barrier then the interference pattern will disappear because there will be no interference between the 2 slits.

But if you are talking about splitting the screens without physical barrier (by putting the screen further away like you did) then that will not make any difference in the interference pattern (why would it make a difference?)

 

[entropy1]

If you split the screens using a physical barrier then the interference pattern will disappear because there will be no interference between the 2 slits.

But if you are talking about splitting the screens without physical barrier (by putting the screen further away like you did) then that will not make any difference in the interference pattern (why would it make a difference?)

In my setup the 'screens' are ment to be physically separated. So then there is no interference pattern on either screen, right? In that case my issue #8 applies.

 

[Ostrados]

In my setup the 'screens' are ment to be physically separated. So then there is no interference pattern on either screen, right? In that case my issue #8 applies.

It is not clear what your point is in #8. Why screen A is special? the particles will go into A or B randomly at 50:50 chance.

Because you are physically separating the screens then there will always be no interference between the slits, so screen A being closer will not make any difference.

 

[entropy1]

It is not clear what your point is in #8. Why screen A is special? the particles will go into A or B randomly at 50:50 chance.

My point is more or less: if the electron hits screen A, then it won't hit screen B, and vice-versa. So they are mutually exclusive events, and I was wondering how the electron makes the choice for screen A or screen B,or what is making it for it.

 

[Ostrados]

P(A) = P(~B) = 1 - P(B)

Nothing makes it chose it is just probability. your question is like saying why a coin chooses header or tale.

 

[entropy1]

P(A) = P(~B) = 1 - P(B)

Nothing makes it chose it is just probability. your question is like saying why a coin chooses header or tale.

The point I was trying to make at #8, is that screen A is first-in-line because it is closer, so the 'particle' (wavefunction) has to 'decide' to impact there or travel some further to screen B. In effect, both screens are hit 50% of the time. So, what makes screen A decide to let 50% of the electrons go? Then screen B obviously decides to keep 100% of the ones let through by screen A, while it is an exactly identical screen. (I realize screen A and B are physically separated - I hope you understand what I mean)

 

[DrClaude]

The point I was trying to make at #8, is that screen A is first-in-line because it is closer, so the 'particle' (wavefunction) has to 'decide' to impact there or travel some further to screen B. In effect, both screens are hit 50% of the time. So, what makes screen A decide to let 50% of the electrons go? Then screen B obviously decides to keep 100% of the ones let through by screen A, while it is an exactly identical screen. (I realize screen A and B are physically separated - I hope you understand what I mean)

Nothing is "deciding."

Imagine that, in your setup, the electrons are fired at precise time intervals, with a determined speed, and that you know immediately when an electron hits the screen. After each firing, you will either detect the electron at screen A after a certain amount of time, or a little bit later at screen B. Nothing weird there. After having done this a million times, you look at where on the two screens the electrons hit, and you will find an interference pattern, which looks as if each electron went through both slits an interfered with itself. The results are consistent with the QM prediction, where the wave function evolves through all of space, and thus there can be interference because of the two slits, but detection always takes place randomly at one particular point on one of the screens. Generally speaking, QM has nothing to say about which path the electron actually took in each case.

 

[entropy1]

and you will find an interference pattern

Are you sure? The screens are physically separated! (boxed)

 

[Nugatory]

The point I was trying to make at #8, is that screen A is first-in-line because it is closer, so the 'particle' (wavefunction) has to 'decide' to impact there or travel some further to screen B. In effect, both screens are hit 50% of the time. So, what makes screen A decide to let 50% of the electrons go? Then screen B obviously decides to keep 100% of the ones let through by screen A, while it as an exactly identical screen. (I realize screen A and B are physically separated - I hope you understand what I mean)

This is no different than when there is one screen and the particle "decides to land" somewhere on the left-hand side of the screen instead of the right. One way of seeing this is to compare your two-screen setup with one in which there is only one screen, but we've angled it so that it is no longer parallel to the barrier and one side is closer than the other. Another way of seeing this is to consider what happens if in your setup we completely remove either screen: the rate at which dots appear on the remaining screen and the pattern on that screen doesn't change.

For every point in space we calculate the probability of a particle leaving a dot if we place a screen at that point. We do that by summing the amplitudes contributed by every available path to that point; the presence or absence of other screens at other points only affects that calculation if they happen to block some paths so we don't get a contribution from them.

 

[DrClaude]

Are you sure? The screens are physically separated! (boxed)

Then I misunderstood the picture in post #13. If there is a divider, then there is no interference.

So, what makes screen A decide to let 50% of the electrons go? Then screen B obviously decides to keep 100% of the ones let through by screen A, while it is an exactly identical screen. (I realize screen A and B are physically separated - I hope you understand what I mean)

Then it is simply a question of probability. Right after the slit, there is a 50% probability of the electron being on either side. The electron is in a superposition of being in the A "box" and the B "box." Once a detection is made, that superposition goes away.

If the distance of the two screens from the slits is so different that not detecting the event at A after a while tells you that the electron is flying towards B before hitting the screen, then you can see A as always making a detection, even if no electron is observed.

 

[Nugatory]

Are you sure? The screens are physically separated! (boxed)

By "physically separated" do you mean that there is no path from slit A to screen B, or from slit B to screen A because the middle dashed line in your picture is a solid barrier? If so, there will be no interference pattern on either screen because for any point on either screen there is only one path available. But that's not how you drew your picture.

 

[entropy1]

By "physically separated" do you mean that there is no path from slit A to screen B, or from slit B to screen A because the middle dashed line in your picture is a solid barrier?

Yes.

But that's not how you drew your picture.

No, I realize my mistake.