# Entanglement and the double slit experiment

## Main Question or Discussion Point

Hello,

I'm wondering how the two slit experiment interacts with entanglement. Here's an ascii art picture of the standard two slit experiment:
Code:
                                       |                  |
|                  | #
|                  | ##
B |                  | #
| ###
@-->            |                  | #######
| ###
A |                  | #
|                  | ##
|                  | #
|                  |

^               ^                  ^
Source          Slits       Diffraction pattern
on screen
Through my simplistic understanding, if we put a detector at slit A or B to measure which slit the photon goes through, then the diffraction pattern will be destroyed, resulting in something like this:

Code:
                                       |                  |
|                  | #
|                  | #
B |                  | ##
| ####
@-->            |                  | ##
| ####
A |                  | ##
|                  | #
|                  | #
|                  |

^               ^                  ^
+ detectors           on screen
Now, lets say that our source produces two entangled photons, one going left, one going right.

Code:
                                       |                  |
|                  |
|                  |
C                             |                  |
|
<--@-->            |                  |
|
D                             |                  |
|                  |
|                  |
|                  |

^               ^                  ^
Entangled         Slits           Which pattern?
Source
So, what happens when we put a detector at C or D? Without the detectors, I'd expect the normal diffraction pattern, but when we add the detectors, we "know" which slit the photon went through, so you'd expect the "shadow" pattern.

Does this mean we can control the pattern shown on the screen based on whether there are detectors present at C and D or not? Doesn't that imply faster than light communication?

Where have I oversimplified here?

Cheers,
Douglas

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olgranpappy
Homework Helper
Hello,

I'm wondering how the two slit experiment interacts with entanglement. Here's an ascii art picture of the standard two slit experiment:
Code:
                                       |                  |
|                  | #
|                  | ##
B |                  | #
| ###
@-->            |                  | #######
| ###
A |                  | #
|                  | ##
|                  | #
|                  |

^               ^                  ^
Source          Slits       Diffraction pattern
on screen
Through my simplistic understanding, if we put a detector at slit A or B to measure which slit the photon goes through, then the diffraction pattern will be destroyed, resulting in something like this:

Code:
                                       |                  |
|                  | #
|                  | #
B |                  | ##
| ####
@-->            |                  | ##
| ####
A |                  | ##
|                  | #
|                  | #
|                  |

^               ^                  ^
+ detectors           on screen
Now, lets say that our source produces two entangled photons, one going left, one going right.

Code:
                                       |                  |
|                  |
|                  |
C                             |                  |
|
<--@-->            |                  |
|
D                             |                  |
|                  |
|                  |
|                  |

^               ^                  ^
Entangled         Slits           Which pattern?
Source
So, what happens when we put a detector at C or D? Without the detectors, I'd expect the normal diffraction pattern, but when we add the detectors, we "know" which slit the photon went through, so you'd expect the "shadow" pattern.

Does this mean we can control the pattern shown on the screen based on whether there are detectors present at C and D or not? Doesn't that imply faster than light communication?

Where have I oversimplified here?

Cheers,
Douglas
Your oversimplification is that you have not yet explained how the measurement at C and D tells you which slit the particle goes through. You seem to imply that "entanglement" will somehow allow you to do this... Exactly, How?

DrChinese
Gold Member
This is a good question. The answer may surprise you a bit. An entangled photon will NOT self-interfere in this situation. This is an example in which an entangled photon acts differently than a normal photon. So it is not possible to send FTL messages this way. You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

Experiment and the foundations of quantum physics

Last edited:
Your oversimplification is that you have not yet explained how the measurement at C and D tells you which slit the particle goes through. You seem to imply that "entanglement" will somehow allow you to do this... Exactly, How?
Thanks to DrChinese, I can just point you to Figure 3 in the article he linked to. The tricky bit is making the entangled photons with correlated beam directions (so that a measurement of the momentum of one particle tells you about the momentum of the other) in the first place, but after that it would just be a case of moving whichever detectors you would use at A and B to positions C and D respectively.

You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.
Thanks for the link, I especially like this phrase wrt Fig 3 and 4: "Note that the photons registered in detector D1 [when put in the focal plane] exhibit a double-slit pattern even though they never pass through a double-slit assembly."

At first glance, it seems that the presence of the interference pattern depends on whether the path information for the left hand photon is erased or not ("Photon 1" in Figure 3, or somewhat confusingly, "Particle 2" in Figure 2, i.e., the one that doesn't go through the double slits), which in turn depends on the position of the detector D1 (Fig 3.). Moving the detector would then control whether an interference pattern is shown on the screen or not; it seems like I'm back to my original question. I'll have a better look at the paper in the morning to see if I can resolve it, I've just skimmed section III so far.

Edit: Hmm, I think I must be miss-reading the article somehow. In the caption for Figure 2, it says that "particle 2" is emitted into beam b or b', but at the bottom of the page it suggests we have to erase the knowledge of whether it took path a' or b'. Surely it means b or b', since it was "particle 1" which went down beam a' in the first place?

Interestingly, looking for the Dopfer 1998 paper leads back to a physicsforums thread, so I'll have to follow that up too.

Thanks,
Douglas

Last edited:
olgranpappy
Homework Helper
Edit: Hmm, I think I must be miss-reading the article somehow. In the caption for Figure 2, it says that "particle 2" is emitted into beam b or b', but at the bottom of the page it suggests we have to erase the knowledge of whether it took path a' or b'. Surely it means b or b', since it was "particle 1" which went down beam a' in the first place?
Yes, it looks like a typo. Typos abound. For example, the paragraph right after Eq. (4) ends with "...states $|b'\rangle_2$ and $|b'\rangle_2$." which is an obvious typo.

It seems like the author is not so good at typing the letter "b".