Double Slit--closing a slit while particle is in flight

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I'd like to understand the wave particle duality and how information traverses space further within the double slit experiment.

Let's assume we have two slits, A and B. A is higher up then B, and an electron moves to the right through the slits and hits a final screen.

While the electron is in flight towards the slits, let's say slit B is closed; and the information of slit B being closed will not reach the electron until it hits the final screen. What would the probability distribution look like for the final screen?
 
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I'd like to understand the wave particle duality and how information traverses space further within the double slit experiment.
There is no such thing as woave-particle duality. The words are a label for a confusion that results from insisting on using outdated models to describe phenomenon on a quantum scale.

Let's assume we have two slits, A and B. A is higher up then B, and an electron moves to the right through the slits and hits a final screen.
The standard 2-slit experiment... with a single particle traversing the apparatus at a time.

While the electron is in flight towards the slits, let's say slit B is closed; and the information of slit B being closed will not reach the electron until it hits the final screen. What would the probability distribution look like for the final screen?
...after the electron has hit the screen, the probability distribution is a spike where the electron hit. The probability of the electron being detected at a particular place (i.e. inside the detector) is 1 and elsewhere it is 0.

Before the electron hits the screen, the probability distribution would be that of a single slit provided there was no chance that the electron could have reached the slits before one slit was closed.

The best place to start properly understanding this, if you are serious, would be the Feynman QED lectures:
http://www.vega.org.uk/video/subseries/8
... see all 8 lectures in the series.
 
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tpeabody said:
Thanks for the reply. I didn't communicate my question clearly. What would the probability curve look like at the end of the experimental result for the back screen after multiple electrons have been fired? At each point of the final screen, there could be computed a histogram of how many electrons that detector counted, and a probability curve could be created from that count.

I'm just wondering what the effect of closing one of the slits would do to the probability curve since the electron could pass through one of the slits before the information of the other slit being closed reached the electron. Also, depending where the electron hit the screen, the electron could hit the screen before the information of the closed slit reached the point where the electron hits the final screen.

Basically, how does the information of available paths for the electron get taken into account. ( There are points on the final screen where the information of the closed will arrive before the electron hits that area, and there are points on the final screen where the electron will arrive not knowing that the other slit was closed, but as far as the information the electron had, the slit was open. )

Thanks for the video links. I will check them out.
 
Thanks for the reply. I didn't communicate my question clearly. What would the probability curve look like at the end of the experimental result for the back screen after multiple electrons have been fired? At each point of the final screen, there could be computed a histogram of how many electrons that detector counted, and a probability curve could be created from that count.

You mean you want to know the resulting interference pattern built up over many repititions of the experiment.

It will follow the pattern for a single slit, provided there was no possibility that the electron could have passed through the other slit before it was closed.

Aside. Off the description, not all the source electrons will arrive at the screen.

I'm just wondering what the effect of closing one of the slits would do to the probability curve since the electron could pass through one of the slits before the information of the other slit being closed reached the electron. Also, depending where the electron hit the screen, the electron could hit the screen before the information of the closed slit reached the point where the electron hits the final screen.
The electron knows nothing of the setup that it does not interact with.

Do you mean that the electron passes through a slit and then the other slit us closed?

Basically, how does the information of available paths for the electron get taken into account.
It doesn't. You can do the experiment with light so now there is no possibility that the screen knows the outcome before it happens and it still comes out the same.

It is unclear how you think that information may arrive at the screen ahead of the electron in some places and not in others. Definitely see the lecture series, preferably before replying: Feynman anticipates many of your questions.
 
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I am looking for experimental data along the lines of the following experiment.

http://advances.sciencemag.org/content/2/2/e1501466

Also, does anyone have any more links to experiments as the above? Basically looking for as much data as possible on pilot-wave ( debroglie-bohm qm theories. ) experiments.

Also, can anyone recommend a good book on pilot-wave or debroglie-bohm theory?

I am posting here as this is an answer to my own question.
 
tpeabody said:
I am looking for experimental data along the lines of the following experiment.

http://advances.sciencemag.org/content/2/2/e1501466

Also, does anyone have any more links to experiments as the above? Basically looking for as much data as possible on pilot-wave ( debroglie-bohm qm theories. ) experiments.

Also, can anyone recommend a good book on pilot-wave or debroglie-bohm theory?

I am posting here as this is an answer to my own question.
Do remember that experiment used weak measurements, https://en.wikipedia.org/wiki/Weak_measurement, so the conclusion reached may not actually be the case.
 
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