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Member (Idle past 4858 days) Posts: 624 From: Pittsburgh, PA, USA Joined: 

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Author  Topic: Quantum Interference  
JustinC Member (Idle past 4858 days) Posts: 624 From: Pittsburgh, PA, USA Joined: 
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 "beamsplitter" (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 beamsplitter 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: 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 beamsplitter, which would seem to imply that the probability of them arriving at the same detector is only 50%?


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JustinC Member (Idle past 4858 days) Posts: 624 From: Pittsburgh, PA, USA Joined: 
bump


Peeper Inactive Member 
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 bythe two photons from the bottom; one reflected one transmitted. Which leaves the only nondestructive 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.


RAZD Member (Idle past 1420 days) Posts: 20714 From: the other end of the sidewalk Joined: 
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.


Peeper Inactive Member 
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


Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 10.0 
Edit: Should read properly before posting.
This message has been edited by Mr Jack, 12152004 04:47 AM


Peeper Inactive Member 
I had, at the least, my phase shift wrong. This derivation is taken from the website of
Frank RiouxDepartment 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) {id1) + d2)}, where i is the imaginary number sqrt(1). Similarly down) = 1/sqrt(2) {d1) + id2)}. Multiplying it all out, state) = {i1,d1)2,d1) + i^2 1,d1)2,d2) + 2,d1)1,d2) + i1,d2)2,d2) + i1,d1)2,d1) + i1,d1)1,d2) + 1,d1)2,d2) + i^2 2,d1)1,d2) + i1,d2)2,d2)}/ 2^(3/2). This reduces to state) = 1/sqrt(2) {i1,d1)2,d1) + i1,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 = 1/2Prob 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.


RAZD Member (Idle past 1420 days) Posts: 20714 From: the other end of the sidewalk Joined: 
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")?


Peeper Inactive Member 
What happens in your version! Heh, I had to go look up what happens in the other version!


RAZD Member (Idle past 1420 days) Posts: 20714 From: the other end of the sidewalk Joined: 
Don't you think it should be part of the first experiment? a control?


Peeper Inactive Member 
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?


RAZD Member (Idle past 1420 days) Posts: 20714 From: the other end of the sidewalk Joined: 
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.


Peeper Inactive Member 
Ahh, I see. I think to set up at Berkeley had a way to block the up and down paths independently.


RAZD Member (Idle past 1420 days) Posts: 20714 From: the other end of the sidewalk Joined: 
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.



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