Polarization of Photons

[JohnWB07]

I have a quick question.

 

If I sent a photon through a fiber-optic cable from point A and measured the polarization of the photon at point B , and then measured it again at point C, would the results between B and C be the same?

 

 

A -----------B-----------C

 

Thanks.

[danny8522003]

id have thought so - undertaking wave behaviour it would just go:

 

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ through the cable. Would there be any reason for the plane of polarisation to change?

[Klaynos]

My understanding is that the polerisation would be the same. If you could measure it at both points without destroying the photon.

[YT2095]

and you can`t Actively "polarise" a photon. you can absorb those that aren`t at the same polarisation, thus leaving the others free to pass, but you can`t polarise them with a filter

[swansont]

I have a quick question.

 

If I sent a photon through a fiber-optic cable from point A and measured the polarization of the photon at point B ' date=' and then measured it again at point C, would the results between B and C be the same?

 

 

A -----------B-----------C

 

Thanks.[/quote']

 

 

There are polarizing fibers and polarization-maintaining fibers that make the answer straightforward. If you know which way to orient your polarizer, you'll get the right answer. But if you align them at some odd angle, you will either measure a photon at that orientation (you will have changed the polarization) or you will measure nothing, according to the law of Malus.

 

With other fibers, though, the polarization will not necessarily be maintained.

[DQW]

1. Since you haven't defined exactly what you mean by "measured the polarization of the photon at point B", it would be hard to answer your question satisfactorily,

 

2. Since you've posted this under QM, I'm guessing it's a prelude to an SG question, but that is merely speculation on my part,

 

Please describe how you measure the polarization.

[JohnWB07]

This is more of a computer science + quantum physics subject. But I've been looking at quantum encryption where Alice sends Bob a photon. Bob determines if the polarization of the photon is | - / \ using filters + and x. Bob then tells Alice which filter he used and Alice responds with which ones are correct. For example:

 

Alice:

| | \ - /

Bob (filter used):

+ x x + +

Correct filter (Y/N):

Y N Y Y N

Binary:

0 N/A 0 1 N/A

 

My question from earlier is in case there is an Eavesdropper "Eve" who can predict what filter to use based on giving Alice a virus of some sort.

 

Thanks,

John

[DQW]

First of all, it is not possible to measure the polarization of a single photon (of unknown polarization) using a polaroid filter. The probability that a photon of polarization angle [imath] \theta [/imath] with respect to the filter orientation, goes through the filter is [imath]cos^2 \theta[/imath]. Even in your scenario with 4 distinct polarizations, it is only possible to rule out one of the 4 polarizations by making a measurement. And once you make the measurement, you lose the knowledge of the original polarization because you have now polarized the photon along the filter orientation.

[DQW]

I'm not sure I clearly understand the scheme outlined in post #7.

 

The first line represents the polarizations transmitted by A. I imagine that 2 of these 4 polarizations represent a 0 bit and the other 2 represent a 1 bit.

 

Now in the second line, you describe the filters that B uses - they can be either up/down/left/right (+) or diagonals (x). Okay, now I'm not sure how B interprets a photon that correctly goes through a filter. For example with the | photon (the second photon in A's message) and the x filter, B will observe 'transmission' or 'blocking' with equal probability. And in half those cases when the photon is transmitted, he will either measure \ or / (which can be determined by a second filter) with equal probability. Now one of these measurements WILL correspond to the correct bit, but the other will not.

 

This brings me to line 3 : B does not know whether he is using the correct filter, so I imagine this line corresponds to knowledge that only A possesses.

 

As for the 4th line, I've got no clue what it means. You haven't even told us the convention used to translate a polarization into a binary bit...so I can't make any sense of that line.

 

In any case, you are asking about something that E does. You state that E has somehow (through a virus or other clever hacking trick) obtained a list of the correct filer sequence. But the correct filter sequence is essentially the key to the ciphertext. With the key, E can correctly determine all the photon polarizations. But this determination does not have to destroy the existing data. For instance, after E deciphers the photon polarizations, she merely prepares the same polarizations and passes the message onto B as though nothing happened along the way.

 

In short, with a knowledge of the key, it is possible for E to decrypt the message and pass on an identical message to B. When B receives the message, it will be exactly what was sent by A.

[JohnWB07]

 

This brings me to line 3 : B does not know whether he is using the correct filter' date=' so I imagine this line corresponds to knowledge that only A possesses.

 

As for the 4th line, I've got no clue what it means. You haven't even told us the convention used to translate a polarization into a binary bit...so I can't make any sense of that line.

[/quote']

 

 

You are correct about line #3, however in a public discussion B tells A what filter he used, and A says if it is correct or not. Suppose A sends | and B measures it with + B tells A what he used and A tells B that it is correct, so both B and A know that the polarization is | which would translate into a 0. Now if A sends a - and B measures with + they compare filters and assign - to be a 1. This is to prevent someone from gaining the key simply by listening to which filters were used.

 

Bob will be incorrect a few of the times, and so will a potential Eve, this means that it is based on probability that Eve will mess up and Bob will be able to detect it. If Bob messes up however, that part of the key is simply thrown out.

 

It's really an exciting technology, but I don't know how useful it will be due to cost and distance limitations, right now they've only transmitted the key up to 120km, which isn't far enough for Satellites. (They have tested wireless transmissions, http://www.magiqtech.com )

[YT2095]

IIRC, there IS a way to detect if a fiber-optic cable has been intercepted and "Listened to", it`s something to do with QM, and that a photon once "Observed" will alter, and will thus be detectable by the receiver. I don`t really know much more about it than that though, it Might even be in use Now or at least sometime soon anyway

[JohnWB07]

There are two companies that I know of that are already selling this system. Magiq http://www.magiqtech.com and ID Quantique http://www.idquantique.com however, there are still downfalls to this system. If there is interest I can post what I have found in my research.

[DQW]

IIRC, there IS a way to detect if a fiber-optic cable has been intercepted and "Listened to", it`s something to do with QM, and that a photon once "Observed" will alter, and will thus be detectable by the receiver.

YT : As I explained in my previous pair of posts. You alter the photon polarization only if you do not know what the polarization is (and hence use an arbitrary filter). And with any given filter, the best you can do is eliminate one polarization - namely the one normal to the filter orientation. But if you are told that the polarization is along one of two normal directions, then you can eliminate one direction and hence determine the polarization direction. This is essentially what JohnWB's evesdropper "Eve" has achieved. So, it is possible for Eve to intercept the message and prepare an identical message. The only difficulty I can see, will be in making up for the time lag. In theory, this difficulty could be overcome by transmitting over air (conventionally) for some distance.

[JohnWB07]

YT :The only difficulty I can see, will be in making up for the time lag. In theory, this difficulty could be overcome by transmitting over air (conventionally) for some distance.

 

The technology that currently exists allows the system to send up to 120km wireless transmission. They have also used fiber optics to transmit, but I don't see a fiber optic network being very successful since it would have to be a star-network. This means that every computer is linked to a server and cannot communicate except through the server. Even a star network like this between 10 computers would become very slow. Just imagine if it gets really popular and it becomes a global interest. I think the only real future this system has is if the system can communicate with satellites and create a wireless global network.

 

Another question this raises is, if satellites can be used in this system, will you be able to send encrypted keys between LA and NY and NY to LA at the same time?

 

For example I want to send something to LA, and someone in LA wants to send something to my region. Will the key be able to get through?

[5614]

but I don't see a fiber optic network being very successful since it would have to be a star-network.

This is not always true. All that needs to be done is give the computer the ADC (analogue (light) digital (electronic) converter) and it can be done.