Double slit experiment using entangled photons

[a1call]

Hello all,
I hope that all is well. I was chatting with the Copilot app abut the quantum effects and according to it, the following exact experiment may not have ever been conducted.

The question is:
What happens if entangled photons are created and each half is sent through 2 different double slit apparatuses one with one slit blocked and the other with both slits unblocked. Will the interference pattern form on both, one or neither apparatuses?
Normally you would expect the interference pattern present in the both-slits-unblocked apparatus and not present in the one-slit-blocked apparatus. On the other hand you would expect that if one half of entangled photons is found to behave as a particle (one slit-blocked observation) you would expect its entangled photon to behave the same way and vice versa.

Thanks in advance to any insights you may provide.

 

[renormalize]

Hello all,
I hope that all is well. I was chatting with the Copilot app abut the quantum effects and according to it, the following exact experiment may not have ever been conducted.

Microsoft Copilot bills itself as "Your everyday AI companion" and is therefore not a suitable basis for questions on Physics Forums. Have you done any non-AI (manual) searching for the experiment you're asking about?

 

[a1call]

Well, not lately, but I have researched the subject in the past and watched lectures from Feynman. There are experiment which I don't understand such as quantum eraser. So either I have not came upon the exact experiment that I am asking about or I have not understood it, hence this thread.

 

[PeterDonis]

the Copilot app

...is not an acceptable source for PF discussion.

I have researched the subject in the past and watched lectures from Feynman.

What QM textbooks have you read? Ballentine has a similar scenario in problem 9.6.

 

[a1call]

Thanks for the reply. I will see if I can dig out info on Ballentine problem 9.6.

FWIW (my bold):

I'm sorry for any confusion, but I'm unable to directly access or download files from the internet, including the PDF you provided. However, I found a discussion on Physics Stack Exchange that seems to be about Problem 9.6 in Ballentine's textbook¹.

The problem presents an experiment with entangled electrons¹. The solution given is that the probability density on the screen for the whole ensemble will be [|ψ1 (x)|² + |ψ2 (x)|²]/2 which exhibits no interference¹. This means that there's no interference because the result is the same as if you had sent individual electrons, half of them with wave function ψ1 (so they make a pattern on the screen in the shape of |ψ1|2) and half with wavefunction ψ2 (shape of |ψ2|2) - giving a total of 1/2(|ψ1|2 +|ψ2|2)¹. The fact that the electrons are in a superposition doesn't affect the result¹.

Interference is, in some sense, defined by the result being different from the result you would get without making a superposition¹. So a result like 1/2(|ψ1 +ψ2|2) = 1/2(|ψ1|2 +|ψ2|2 +ψ∗1ψ2 +ψ∗2ψ1) shows interference as long as ψ∗1ψ2 +ψ∗2ψ1 isn't zero¹.

Please note that this is a high-level summary and may not cover all the nuances of the problem and its solution. For a detailed understanding, I recommend referring to the textbook and the discussion on the forum¹. If you have specific questions about the problem or its solution, feel free to ask! I'll do my best to help based on the principles of quantum mechanics.

Source: Conversation with Bing, 2024-04-21
(1) quantum mechanics - Entangled electrons: no interference (Ballentine .... https://physics.stackexchange.com/questions/756344/entangled-electrons-no-interference-ballentine.
(2) This Page Intentionally Left Blank - UW Faculty Web Server. https://faculty.washington.edu/seattle/physics441/online/Ballentine.pdf.
(3) SOLUTIONMANUAL QuantumMechanicsforEngineers - Florida State University. https://web1.eng.famu.fsu.edu/~dommelen/quantum/solman/index.pdf.

Thanks again Peter.

 

[Nugatory]

FWIW (my bold):

For what it's worth, yes... But that's not much. All the chatbot has done here is pull up a StackExchange post that you could have found for yourself, but in doing that its algorithm has also not shown many other relevant and informative publications on the subject. Reading that reply is unlikely to produce any real understanding of the the question.

There's a reason why the forum rules do not allow using chatbots as references; please respect that rule.

 

[DrChinese]

… hence this thread.

The simple answer has already been presented. Entangled particle pairs - whether electrons or photons- generally do not exhibit single particle self-interference. For them to do that, the entanglement must cease on the appropriate basis.

I’m sure you realize that were that not true, FTL communication would be possible.

 

[a1call]

Thank you for the reply Dr,
Actually I am not quite sure what you mean. Am I correct to conclude that I was wrong to assume that if single/multiple half/halves of entangled particles are observed to behave as particle (or wave) then the other entangled half/halves will (have to) be observed to behave the same as their first half/halves?
Thanks again for all the insights.

ETA, to be clear I don’t understand your emphasis on single particle vs continuous stream. Otherwise, I can see how this could lead to FTL.

ETA II, on the other hand the same obstacles that exist against FTL, using entanglement in general, would be present in this concept/scenario as well (except for random occurrences). The statistical dependence would be present here as well.

 

[DrChinese]

Thank you for the reply Dr,
Actually I am not quite sure what you mean. Am I correct to conclude that I was wrong to assume that if single/multiple half/halves of entangled particles are observed to behave as particle (or wave) then the other entangled half/halves will (have to) be observed to behave the same as their first half/halves?
Thanks again for all the insights.

ETA, to be clear I don’t understand your emphasis on single particle vs continuous stream. Otherwise, I can see how this could lead to FTL.

ETA II, on the other hand the same obstacles that exist against FTL, using entanglement in general, would be present in this concept/scenario as well (except for random occurrences). The statistical dependence would be present here as well.

Imagine you send an entangled particle though a double slit. It simply will not produce the expected interference pattern. The reasons are a bit complicated, but entangled particles are not coherent.

See figure 2 of the following, along with the related text. Written by one of the premier scientists in the field, I’d recommend reading it all.

http://courses.washington.edu/ega/more_papers/zeilinger.pdf

 

[a1call]

Thanks very much again for the reply and the referenced article. It is very similar to the experiment in OP, except that instead of the one-slit-blocked apparatus it uses 2 detectors to get the which slit info for the both-slit-unblocked apparatus. Because of this no interference fringes are formed. So it should follow that for the OP, no interference fringes will form in any of the two apparatuses (or so I figure). :)
If so that is more strange than I could possibly imagine.

ETA, then again there is this difference that in the OP scenario without the 0-Total momentum aspect the which slit-through info is not known for both-slits-unblocked apparatuses and the info would not have to correlate to the other apparatus in this regard. In the end I think it doesn’t really matter. The experiment in that article is more interesting.

 

[DrChinese]

ETA, then again there is this difference that in the OP scenario without the 0-Total momentum aspect the which slit-through info is not known for both-slits-unblocked apparatuses and the info would not have to correlate to the other apparatus in this regard.

Actually you would see the 2 outcomes as correlated, momentum conserved if they went in opposite directions. There would simply be no interference pattern formed as more and more were observed.