6H30.15 - Quantum Cryptography
The Quantum Crypography demo uses a laser, one fixed polarizer and two rotatable polarizers, and a screen. Basically, Alice and Bob can communicate with message photons, but Eve can intercept the message photons. The long argument is below.
Alice can send light that is polarized 0, 90, 45, or -45 degrees. She makes two lists. In the first list she writes 0 if she sent out a photon polarized by 0 or 90 degrees, and she writes a 1 if the photon was polarized by 45 or -45 degrees. In the second list she writes 0 for a photon polarized by -45 or 90 degrees and a 1 for 45 or 0 degrees. Alice's Lists
Bob only has one chance to measure each photon with a 0 or 45 degree filter. If it’s a single photon then it can either go through Bob’s filter completely or not at all, but if he tries to measure a 90 degree polarized photon with the 45 degree filter there will be a 50/50 chance of it going through.
Chance of a photon going through each filter.
|0 degree filter||100%||0%||50%||50%|
|45 degree filter||50%||50%||100%||0%|
Like Alice, Bob makes two lists. In the first list he records with a 0 or a 1 whether he used the 0 or the 45 degree filter, and in the second list he records whether the photon made it through the filter.
Bob’s lists (agreement with Alice shown in bold)
Now Alice and Bob compare their first lists. They keep the data where they agree and discard everything else. About half should remain. The remaining data on their second lists should now be in perfect agreement, even though they did not directly compare the second lists.
Now suppose Eve intercepts some of the photons before they reach Bob and measures them herself with a 0 degree filter. If the photon goes through she sends Bob a new 0 degree polarized photon, and if the one she intercepts is blocked she sends Bob a 90 degree polarized photon.
But if Alice sends a 45 degree polarized photon that is intercepted by Eve, and if Bob measures Eve’s replacement photon with the 45 degree filter, there is a 50/50 chance it will be blocked. If Bob had received Alice’s original photon it would pass through the 45 degree filter with certainty. Likewise if Alice sent a -45 degree photon, there is a 50/50 chance that Eve’s replacement photon passes through Bob’s 45 degree filter when it shouldn’t. If Eve intercepted every photon, then half of Bob’s measurements with the 45 degree filter would be wrong, and overall 25% of his data would be wrong (if he used both filters equally). Alice and Bob can compare a sample of their second lists, and if there is any disagreement, then they know that Eve has been interfering.
The demo: (diagram) Alice and Bob are rotating filters, with a lamp behind Alice and a screen behind Bob. Show that when Alice and Bob are aligned with each other, “all” of the lamp light passes through (it is actually attenuated quite severely by the old, cheap filters). When Alice and Bob are off by 90 degrees in either direction, no light makes it through. And when they are off by 45 degrees, half of the light passes through. Now introduce Eve, a fixed filter in between Alice and Bob. Show that Eve does not affect the light when Alice is parallel to her. Then show that when Alice and Bob are both at 45 degrees, less light (fewer photons) pass through, and when Alice is at 45 degrees while Bob is at -45 degrees, Eve’s presence causes some light to get through.
The Physics: In the demo, Eve does not send any new photons to Bob if her filter blocks them; they are still Alice’s original photons. So how do some of them end up passing through Bob’s filter in the last arrangement, with Alice at 45 degrees, Bob at -45, and Eve at 0 in between them? Because according to quantum mechanics, measuring a particle can change its state. After passing through Alice’s filter the photons are polarized to 45 degrees, with a 50/50 chance of passing through a filter at 0 degrees, and no chance of passing through a filter at -45 degrees. But the photons that pass through Eve’s filter come out polarized at 0 degrees, just as if she had sent her own photon. A photon polarized at 0 degrees has a 50/50 chance of going through Bob’s -45 degree filter, and it doesn’t “remember” its previous state.
In the cryptography scheme, if Eve wanted to measure Alice’s photons without changing Bob’s measurements and revealing that their communication had been tampered with, she would have to use the same filters as Bob, and have access to his first list to know which ones he was going to use on each photon. That way Eve could always send Bob a photon that he would measure the same way she did, and he would get the same result. Note that in this way Eve still might send Bob a photon with a different polarization from Alice’s original.
Alternately, Eve could copy Alice’s first list to find which filters to use, and then send Bob new photons with the same polarizations as the originals. Either method can be used because Alice and Bob, after comparing their first lists, only use the data from where those lists agree, meaning that Bob’s measurements would not change the state of the photon.
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