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Author Topic:   Quantum Interference
JustinC
Member (Idle past 4858 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 1 of 31 (166174)
12-08-2004 8:21 AM


I'm having trouble understanding a quantum interference experiment, and was wondering if someone more knowledgable on the subject could help me out.
The experiment was designed by Marlan Scully of the University of California, and the set up is my avatar, since I couldn't figure out how to upload an image that wasn't already on a webpage.
In the first step, an incoming photon is split by a crystal (c) into two weaker photons of different trajectories. These entangled photons are then directed by mirrors (m) to a "beam-splitter" (b). A beam splitter is a device that exploits quantum tunneling, such that a photon will tunnel through the splitter with a probability of 50/50. The photons then either get reflected to the detector on their respective sides, or they pass through the beam-splitter an get detected by the detector on the opposite side. It turns out that in this setup, the two photons always arrive at the same detector, so if the bottom photon gets transmitted, the top on gets reflected, and they both go to D1. The opposite is also true.
In the book I am reading, About Time by Paul Davies, he explains this concordance in terms of quantum interference. He writes
quote:
The reason both photons always end up in the same detector, upper or lower, hinges on the fact that, in the setup just described, the experimenter cannot know which photon took which path. It may be that photon 1 took the upper path and photon 2 the lower path, or vice versa, but with the arrangement shown, the experiment cannot reveal that actual routes taken. According to the bizarre rules of quantum physics, this lack of information about the routing implies a schizophrenic world in which both alternatives coexist in a sort of hybrid reality. That is, without knowing which photon took which route, we have to regard the world as made up of both potential realities existing together in a sort of ghostly overlap. This is not merely a way to visualize the wierd goings-on, but leads to real physical effects. For example, we can tell that the two alternatives "photon 1 takes the lower path, photon 2 the upper path" and "photon 1 takes the upper path, photon two takes the lower path" both contribute to the outcome, because these phantom alternatives add together to produce results which are different from either alternative on its own-a process known as "quantum interference." In the present example, it is this interference of the alternative paths that produces the above-mentioned concordance, directing both photons to the same detector.
The interference arises as a consequence of teh wave nature of light, and has to do with the fact that waves arriving in step reinforce, whereas, those that arrive out of step cancel. Here the interference is between waves associated with one alternative reality combining with waves associated with the other alternative. The overlapping of these alternative-world waves can be convincingly demonstrated by slowly increasing the length of of the paths until the waves associated with the alternative realities now arrive exactlly out of step. In this case, interference causes cancellation of the wave, meaning that the two photons now go to different detectors-i.e., the detectors fire simultaneously. A light additional increase in the path length brings the waves back into phase, and the photons go to the same detectors once more. By gradually extending one path in this way, the experimenters can obtain a series of peaks and troughs-characteristic of an interference pattern-for the simultaneous firings of the two detectors.
I'm having a lot of trouble understanding this effect. Why does the cancellation of the waves cause them to go to different detectors and why does the reinforcement cause them to go the same detectors? How should I visualize these waves associated with the alternative realities?
How does that relate to the probability of quantum tunneling through the beam-splitter, which would seem to imply that the probability of them arriving at the same detector is only 50%?

Replies to this message:
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 Message 8 by Peeper, posted 12-15-2004 11:58 AM JustinC has not replied

  
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Message 2 of 31 (166179)
12-08-2004 8:32 AM


Thread moved here from the Proposed New Topics forum.

  
JustinC
Member (Idle past 4858 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 3 of 31 (166519)
12-09-2004 11:25 AM


bump

  
Peeper
Inactive Member


Message 4 of 31 (168345)
12-15-2004 12:36 AM
Reply to: Message 1 by JustinC
12-08-2004 8:21 AM


I am not familiar with this experiment but I’ll give it a shot. Davies explanation makes a lot of sense. First, as he says, you must consider both photons along all possible paths. If the wave functions of the photons along the possible paths are 180 deg out of phase at a point in space they will cancel giving a zero probability for the photons to be at that point.
Here are my assumptions:
1) The photons start in phase and travel the same distance to both detectors.
2) A photon reflected from the beam splitter will undergo a 180 deg shift in phase (yeah my E&M is probably weak, someone correct me if I am wrong).
Therefore:
1)One photon on top, reflected — Cancelled by photon on bottom, not reflected
2) One photon on top, not reflected — Cancelled by photon on bottom, reflected
3) Both photons travel along top; one reflected, one transmitted — (Both) Cancelled by
the two photons from the bottom; one reflected one transmitted.
Which leaves the only non-destructive situations as those where both photons travel along the upper or lower paths and are either both reflected or both transmitted. Both of these situations result in the photons arriving at the same detector.
How should I visualize these waves associated with the alternative realities?
Just think of them as plain old sine waves. Add all the amplitudes. Where the amplitudes vanish, there is zero probability to find the photons.

This message is a reply to:
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RAZD
Member (Idle past 1420 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 5 of 31 (168354)
12-15-2004 12:52 AM
Reply to: Message 1 by JustinC
12-08-2004 8:21 AM


more info?
I would need to know some more first.
if you split and mirror as before but now move in two "beam splitters" and 4 detectors to determine if one is always reflected and one is always transmitted if the path lengths are kept the same.

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Peeper
Inactive Member


Message 6 of 31 (168360)
12-15-2004 12:59 AM
Reply to: Message 4 by Peeper
12-15-2004 12:36 AM


I almost forgot. The two photons along the top, both reflected, cancel with the two photons from the bottom, not reflected. And the two photons from the top, not reflected, cancel with the two photons from the bottom, reflected.
Therefore, I have a grand total of zero photons anywhere.
bah

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Dr Jack
Member
Posts: 3514
From: Immigrant in the land of Deutsch
Joined: 07-14-2003
Member Rating: 10.0


Message 7 of 31 (168406)
12-15-2004 4:45 AM
Reply to: Message 6 by Peeper
12-15-2004 12:59 AM


Edit: Should read properly before posting.
This message has been edited by Mr Jack, 12-15-2004 04:47 AM

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Peeper
Inactive Member


Message 8 of 31 (168520)
12-15-2004 11:58 AM
Reply to: Message 1 by JustinC
12-08-2004 8:21 AM


Phase wrong among other things
I had, at the least, my phase shift wrong. This derivation is taken from the website of
Frank Rioux
Department of Chemistry
Saint John's University
College of Saint Benedict
Ok, as Davies said you must consider the photons (labeled 1 and 2) as traveling along both the upper and lower paths. Therefore, the state of the entangled photons is give by
|state) = 1/sqrt(2){|1, up)|1, down) + |1, down)|2, up)},
where the wavefunction is symmetric as required by bosons.
A photon traveling along the upper path will be transmitted to detector d2 and reflected to detector d1. According to Dr. Rioux the phase change upon reflection is not 180 deg but 90 deg and is required by conservation of energy. Writing the phase shift of the wavefunction as exp[ix] where x is the change in phase then gives
|up) = 1/sqrt(2) {i|d1) + |d2)},
where i is the imaginary number sqrt(-1).
Similarly |down) = 1/sqrt(2) {|d1) + i|d2)}.
Multiplying it all out,
|state) = {i|1,d1)|2,d1) + i^2 |1,d1)|2,d2) + |2,d1)|1,d2) + i|1,d2)|2,d2) + i|1,d1)|2,d1)
+ i|1,d1)|1,d2) + |1,d1)|2,d2) + i^2 |2,d1)|1,d2) + i|1,d2)|2,d2)}/ 2^(3/2).
This reduces to
|state) = 1/sqrt(2) {i|1,d1)|2,d1) + i|1,d2)|2,d2)}.
Squaring the coefficient of each term and taking the square root of the magnitude gives the various probabilities:
Prob of 1 at d1 and 2 at d1 = 1/2
Prob of 1 at d2 and 2 at d2 = 1/2
Prob of 1 at d1 and 2 at d2 = 0
Prob of 1 at d2 and 2 at d1 = 0.
Therefore as stated, if you consider the photons as traveling along all possible paths you can show that the photons cannot be detected simultaneously at different detectors.

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RAZD
Member (Idle past 1420 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 9 of 31 (168740)
12-15-2004 9:37 PM
Reply to: Message 8 by Peeper
12-15-2004 11:58 AM


Re: Phase wrong among other things
I'm glad we helped you find that answer ... :whew:
(and you wanted help???)
so what happens when you run my version (where they go through the same individual experiences but do not meet at the centered "beam splitter")?

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Peeper
Inactive Member


Message 10 of 31 (168746)
12-15-2004 11:08 PM
Reply to: Message 9 by RAZD
12-15-2004 9:37 PM


Re: Phase wrong among other things
What happens in your version! Heh, I had to go look up what happens in the other version!

This message is a reply to:
 Message 9 by RAZD, posted 12-15-2004 9:37 PM RAZD has replied

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RAZD
Member (Idle past 1420 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 11 of 31 (168754)
12-15-2004 11:35 PM
Reply to: Message 10 by Peeper
12-15-2004 11:08 PM


Re: Phase wrong among other things
Don't you think it should be part of the first experiment? a control?

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Peeper
Inactive Member


Message 12 of 31 (168764)
12-16-2004 12:01 AM
Reply to: Message 11 by RAZD
12-15-2004 11:35 PM


Re: Phase wrong among other things
Looking at my first attempt I can see plenty of mistakes. The initial state is one where one photon travels along the lower path and one photon travels along the upper path. There is not a state where both photons travel the upper or lower path simultaneously. As such, if both photons are to be detected at the same detector, one must be reflected and one must be transmitted. The experiment I saw described actually did use 4 detectors and 2 beam splitters. Two of the set ups described by JustinCy set back to back. Is this what you are describing? Or is yours different?

This message is a reply to:
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RAZD
Member (Idle past 1420 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 13 of 31 (168840)
12-16-2004 7:17 AM
Reply to: Message 12 by Peeper
12-16-2004 12:01 AM


Re: Phase wrong among other things
Mine is defferent as it takes out the recombination possibility by cutting the pathe short -- take the original diagram and put the beam splitters at half the distance from the mirrors and move the detectors up accordingly. you end up with four possible end paths and four detectors to test them.

This message is a reply to:
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Replies to this message:
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Peeper
Inactive Member


Message 14 of 31 (168881)
12-16-2004 10:49 AM
Reply to: Message 13 by RAZD
12-16-2004 7:17 AM


Re: Phase wrong among other things
Ahh, I see. I think to set up at Berkeley had a way to block the up and down paths independently.

This message is a reply to:
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RAZD
Member (Idle past 1420 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 15 of 31 (169163)
12-16-2004 8:54 PM
Reply to: Message 14 by Peeper
12-16-2004 10:49 AM


Re: Phase wrong among other things
that doesn't give the same control on the experiment. I want to see it they always go to the same side even if they do not have the {combination\interference} possibility.

This message is a reply to:
 Message 14 by Peeper, posted 12-16-2004 10:49 AM Peeper has replied

Replies to this message:
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