Quiltlet – Mathematical Connections

In this scheme, two parties (traditionally called Alice and Bob, a bit more personalized than just A and B) have to communicate over an open channel, and they start by establishing a secret binary key to be used for further encrypted communication, beyond the protocol. The problem they have is that to agree on that secret key they also have to use the open channel, on which signals may be intercepted by an eavesdropping third party (typically called Eve).

In the protocol depicted on the quiltlet, Alice sends linearly polarized photons to Bob, of which Bob then measures the polarization. In both the production of polarized photons by Alice, and their detection by Bob, calcite crystals are used.

About the calcite crystals

When a beam of unpolarized light falls on one face of such a crystal, perpendicular to the face, then it splits into two parts at the interface. One part will traverse the crystal, and exit the opposite face in a perfect straight extension of the incident beam, as is usual in refraction experiments. But the other part formed at the incidence point is *at an angle* with the incident beam — most unusual in a refraction set-up for an incident beam perpendicular to the surface, and this second beam is appropriately called the *extraordinary refracted beam*. On exiting the crystal on the other side, this beam bends again, and it will then continue in a path parallel to the other exiting beam. Inside the crystal, and also after they have exited, both beams are linearly polarized. The polarizations of the two beams (ordinary and extraordinary) are perpendicular to each other, with the polarization on the extraordinary beam (or E beam) lying in the plane spanned by the two beams, and the polarization for the ordinary beam (or O beam) perpendicular to this plane. In the 3D figure below (and only in this figure), the incident light ray is shown in green only for clarity, to distinguish it from the others – in reality there is no color change.

For each crystal in her set-up, Alice uses lightrays from two different fireflies, both carefully positioned. The light produced by each firefly is masked by a pinhole-screen, so that only a narrow beam is let through (when Alice opens up the pinhole), directed so that it hits the crystal facet perpendicularly. The different incidence spots for the two fireflies are picked with a precise goal: on refracting, each of the beams splits into an ordinary and an extraordinary refraction beam – and the incident spots are chosen so that the extraordinary E₁ from firefly 1 and the ordinary beam O₂ from firefly 2 exit the crystal in the same spot on the opposite facet. The other two refracted beams (ordinary beam O₁ from fly 1 and extraordinary beam E₂ from fly 2) are of no use to the setup, and are not drawn in the 3D-figure below. (In an experimental setup, a small light absorber could be put at the spot where they would exit the crystal.) Because of the different polarization properties of E and O beams, each crystal-and-two-fireflies setup provides Alice with a source that produces a narrow lightbeam of the polarization she chooses (depending on which pinhole she opens). The next 3D figure shows the geometry of beam-forming for one of Alice’s crystals.

For her full setup, Alice has two crystals, each with its team of two fireflies. The crystals are arranged so that the beams they produce lie in the same plane, and intersect at a 90-degree angle. A half-silvered mirror is placed at the intersection point, perpendicular to the plane of the beams, and at an angle of 45 degrees to each of the two beams; because the mirror is half-silvered, photons in each beam have a 50% chance of being transmitted without change of direction, and a 50% chance of being reflected. As shown in the next 3D figure, the direction of the reflection of one beam coincides with the uninterrupted direction of the other beam.

To distinguish the two “wings” of Alice’s setup, let’s call one Straight, and the other Diagonal. The Straight side produces linearly polarized photons with polarization directions that are either in the plane spanned by the two beams arriving at the half-silvered mirror, or perpendicular to it – we call these polarization directions, respectively, Horizontal (or H) and Vertical (or V). The linearly polarized photons from the Diagonal Side have polarization directions that we distinguish by calling them Left (L) or Right (R).

When she wants to send information to Bob, Alice will send a stream of photons one-by-one, at successive precise time instants that have been pre-arranged with Bob, over the open channel. For each photon she sends, Alice picks at random the side and the polarization direction; she thus sends, in an order that is completely random but while keeping careful track of the choices made, a stream of individual photons each of which has a linear polarization of four possible types: SV, SH, DL or DR. Since the half-silvered mirror sends only half of these photons in the Bob’s direction, there will be time-slots at which B does not receive a photon at all, because there was no photon in the beam coming his way at those times.

On his side, Bob has a set-up that completely mirrors that of Alice, as shown in the figure below. The beam coming his way first strikes a half-silvered mirror, which sends the arriving photon into one of two directions: straight through, or reflected by 90^o .

Bob’s half-silvered mirror sends each arriving photon to either the crystal that is set up exactly like the crustal used by Alice to produce that photon (that is its Correct crystal) or to the other one (the Incorrect crystal), with a probability of 50%. If the photon arrives at its Correct crystal, then its polarization direction is already aligned with one of the special polarization directions of that crystal, and it will just follow the path through the crystal associated to that polarization, and leave the crystal in the corresponding spot. At each of the possible exit spots, a frog is positioned, of a very special species in which the retina can perceive individual photons; the frogs have been moreover conditioned so that they will jump up when they perceive a photon. So if the photon arrives at its Correct crystal, then Bob can deduce, by observing the reactions of his frogs, whether Alice’s photon was V or H photon (if S is the Correct crystal) or (if D is the Correct crystal) whether it was L or R. The photon may also, however, arrive at the Incorrect crystal. Because its own polarization vector makes an angle of 45^o with each of the special polarization directions for that crystal, its polarization will then be reoriented to one of the crystal-specific directions, each with probability 50%, and Bob can then not deduce anything useful from the reactions of the frogs at the exit points.

Initially Bob has no idea whether an arriving photon was directed to its Correct or Incorrect crystal. But he can find out by communicating with Alice. At each pre-arranged time instant when he knows Alice is sending a photon, he lets her know whether a photon arrived or not (which happens with equal probability), and if it did arrive, whether it arrived on the S or D side. For each of the photons that made it through, Alice lets Bob know whether his S or D reading corresponds with what she sent. After this exchange, Bob and Alice both know exactly which photons arrived at their Correct crystal and also what the polarization was with which Alice sent each of those photons, since their polarization survives the whole process unscathed. By assigning a value of 0 to one of the two possible polarizations of such a correctly arrived photon, and a value of 1 to the the other, Alice and Bob have thus arranged to both know a sequence of bits without ever having articulated that information over the open channel.

But what if Eve is eavesdropping on the open channel, somewhere between the two half-silvered mirrors? (In practice, these can be much further apart than the distance indicated in the figure, foreshortened to make everything fit.) In her ignorance of whether Alice used the S or the D side to generate a photon she wishes to intercept, Eve would have to guess. She would have a 50% chance of guessing it right; if she guessed wrong, Bob’s half-silvered mirror could, half the time, still end up projecting the now corrupted photon back to the Correct crystal on his side, but the bit of information it carries (namely its polarization direction) would then be as likely to be flipped as to be correct. This means that Bob and Alice can detect the presence of an eavesdropper by comparing a random subset of the secret information bits they believe to have arrived unscathed at Bob’s, sacrificing them in order to check the security of the channel; if they find those bits are all correct, they have a security guarantee for the channel, and the other (non-sacrificed) secret information bits can be used by them as a shared secure encryption key.

About the calcite crystals

Alice has several sources for her polarized photons, which: in the “straight” system S, the polarization is either Vertical or Horizontal, and in the “diagonal” system D the polarization vector is diagonal, pointing up to either the Right or the Left. (This will be illustrated below.)

The full set-up is what is illustrated in the quiltlet.  

Please don’t go out to catch poor unsuspecting fireflies or amphibians to replicate this at home. It would be hard to set up the firefly side so that Alice really has sources that let only one photon imping on a crystal at a time. 

Ingrid Daubechies

Single-photon detecting frogs also haven’t been found in nature yet. It has been demonstrated that frog retinas, at cold temperatures (alluded to by the ice cubes in the frogs’ basins on the quiltlet) can indeed perceive the arrival of a single photon and induce a neuronal firing in about 30% of the cases. 

But this is a far cry from having a stable of trained single-photon detecting frogs to provide to Bob …

Quantum protocol explanation

How then does all this work? How can Alice and Bob use this set-up to agree on a secret binary key sequence, and how can it be secure against interception? 

Alice will send her photons one by one (and not in beams); she sends those at a sequence of times that have been pre-arranged with Bob (and which Eve may well know). Whether or not she sends them from the S or the D side is picked completely randomly, with equal probability for both sides, for each individual photon.

For each side she picks equally randomly, each time, one of the two available polarizations (V or H for the S side, LD or RD for the D side); Bob and she have agreed to give the label 1 to a V photon on the S side or to a LD photon on the D side, and the label 0 to H photons on side S, and RD photon on side D.  

Since there are several stages where the photon may, with some non-zero probablity, continue in a different direction than on the desired path, and also a non-zero probability that even a photon that arrives at Bob’s end in one of the four possible channels is not detected by the frog sitting there,  Bob needs to let Alice know for which of those prearranged times one of his frogs jumped up. He also lets her know (again without shielding it from Eve’s possible eavesdropping), for each of these times, whether the reacting frog was on the S or D side. Alice responds, still openly, by telling Bob for which of those time slots she had picked the same side (S or D) as that where Bob’s frogs detected it.

If Eve didn’t intervene (see below for when she does), then for those particular photons where Alice and Bob have identical sides, the polarization that Alice picked (H or V for the Standard crystal, LD or RD for the Diagonal crystal) will have survived the whole trajectory unscathed and Bob will know the exact polarization choice that Alice made by checking his records of which frog jumped for each instance. By translating the polarizations into binary symbols, using the agreed-upon convention, Bob and Alice have then at their disposal identical *secret* binary sequences that they can start using for communication that requires a secret binary key. 

But what if Eve tried, over the open bit of the channel, to detect the polarizations of the photons coming through? Since she doesn’t know whether they are S or D photons at that point, any measurement she makes will, on average, lead to a rearranging of the photon from S to D, or vice versa, in about 50% of all the cases. The photon of which she has measured the polarization will thus be still “correct” (i.e. have a polarization identical to what Alice sent) in only 50% of the cases. In the 50% of the cases where her interference has changed the polarization, the further processing by Bob’s half-silvered mirror may nevertheless send it, half the time, to the crystal corresponding to Alice’s original choice (S or D); this results in it still being a candidate for a “matched” photon once Alice and Bob have compared the S/D status of Bob’s detected photons with Alice’s records. Its polarization, however, has a 50% chance of being in either of the two possible choices available, regardless of the original polarization with which Alice sent it on its way.

In order to detect whether some interference is happening, Alice and Bob can split the secret bit sequence on which they have “agreed” into two parts — one part to preserve as a secret key, and one that they sacrifice to check the security of their quantum key sharing channel. They communicate the bits in this last part to each other; if these supposedly identical bit sequences differ from each other, then Eve must have interfered, and their attempt at obtaining a common secure secret key has failed. But if they are truly identical, then they have succeeded!

Read more about the Cryptography Quilt