Quantum Cryptography

Now, let me discuss the BB84 protocol. It is based on the name of the inventors Charles Bennet and Gilles Brassard, and it was invented in 1984. Quantum cryptography follows two steps. The first one is sending the secret key, and the second step is sending the message. Here, Alice and Bob make use of two fundamentally different communication channels: a classical channel and a quantum channel. A classical channel is something that you use on the Internet to transfer data. In a classical channel, Eve can observe the bit-stream without affecting the data. But, a quantum channel is something different. It is capable of sending information in terms of quantum, and Eve can't observe the data without affecting the data. In the BB84 protocol, the secret key is sent through the quantum channel, but the message is sent through the ordinary channel but encrypted by the secret key. The first step is called Quantum Key Distribution (QKD). In this step, Alice and Bob use the quantum channel for communication.

First, let's imagine there is no Eve between Alice and Bob. Let's assume that Alice is using two types of polarizer: one is a diagonal polarizer (X) and one a rectilinear polarizer (+). In a rectilinear basis, a photon with a spin "|" (that is, up to down ) is considered as 1, and a "-" (that is, left to right) is 0. In a diagonal basis, a photon with a spin "/" is considered as 1, and "\" is 0. The diagram shown in Figure 5 should help you understand how I'm representing photons as binary values.

Figure 5. Binary Encoding of Photons in My Examples

Now Alice has a key, and for each bit, she will select a random basis (either diagonal or rectilinear) to encode the bit to send. Nobody, not even Bob, knows what basis Alice is using. Bob will receive the encoded qbits, and Bob will use random basis to decode the qbits. If he uses the same basis, he will get the exact bit that Alice sent; otherwise, there is a 50% chance that he will get a wrong bit. For example, if Alice uses a diagonal basis to encode 1, and Bob also uses diagonal basis to decode that, then he will get a 1. If he uses a rectilinear basis, then there is a 50% chance that he will get a 1 and a 50% chance of getting 0. As Bob is also using random basis, there's a 50% chance that he will use the right basis (that is, he will use the basis that Alice used) and will decode 50% of qbits exactly, and for the 50% wrong basis, he will decode 25% of qbits exactly, and that means Bob will decode 75% of qbits exactly.

Alice and Bob will exchange the basis they used for each bit using the normal channel without revealing their bits. They can check for which bits they both used the same basis, and those bits will be used as the secret key. Consider the example shown in Table 1 where Alice is sending the secret key 100101.

Table 1. Alice Sending the Secret Key 100101

Alice Bob
Basis used +,X,+,+,X,X +,+,+,X,+,X

In this case, Bob will decode the key as 1,0/1,0,0/1,0/1,1. Because Bob has used some wrong basis to measure the qbits, he may get a 0 or 1 randomly on those cases. Then, they will exchange their basis with others, and they will find that in positions 2, 4 and 5, Bob used the wrong basis. So they will use the rest of the bit (1st, 3rd and 6th bit) string as the secret key—that is, 101. The rest is simple, just encrypt the message using that key and send it.

The situation becomes critical when Eve comes into action. As they are connecting using the public channel, it is quite possible that Eve will intercept the communication. In this case, as with the previous case, Alice encodes the bit information using any basis and sends it to Bob, but now Eve intercepts the qbits. Like Bob, Eve also has a decoder of the qbit. But Eve also doesn't know the basis Alice is using, so like Bob, she also randomly uses basis to decode the qbits. There is a 50% chance that Eve will use the right basis, and a 50% chance she will use the wrong basis. For the correct 50%, the photon's spin direction will not be affected, but for the wrong 50%, the photon's spin direction will be changed. For the 50% of qbits for which Eve used the right basis, Bob will use a 25% right basis and 25% wrong basis, and for the right 25% of qbits, he will get a 25% right qbit, and for the wrong 25% basis Bob used, he will get 12.5% of qbits correct just due to probability. That means from the first 50% for which Eve used the right basis, Bob will get 37.5% correct qbits. For the rest of the 50%, again Bob will use 25% right and 25% wrong basis. From this, Bob will get 12.5% and 12.5% due to probability, which means he will get 25% right qbits. So when Eve is between them, Bob will have 37.5 + 25 = 62.5% accuracy. Figure 6 demonstrates this calculation.

Figure 6. Accuracy Calculation for Bob When Eve Is Intercepting

In Figure 6, the node with "**", like C**, represents the nodes where Bob decoded the qbits correctly, and the node with "*", like F*, represents the nodes where Bob decoded the qbits incorrectly. One question that may arise is why does Bob get 12.5% accuracy (in E,L) when he used the wrong basis? Remember that when you use a wrong basis to decode a qbit, there is a 50% chance that you will get a 0, and a 50% chance that you will get a 1. By this logic, Bob will have 12.5% accuracy from D. Similarly, in the case of I, when Bob has used the correct basis (with respect to Alice's basis) but Eve already has changed the polarization of the qbits using the wrong basis, Bob has a 50% chance of being right and a 50% chance of being wrong.

So overall, Bob gets 12.5% right qbits in I and 12.5% wrong qbits in J. Now they will match the basis they used for each qbit, and they will use the bits where Bob used the correct basis, and they will throw out the bits for which Bob used the wrong basis. Now they need to check whether Eve is listening. For that purpose, they will use a subset of the matched key (after throwing out the bits for which Bob used wrong basis) and compare with others using the normal channel. Bob will have 100% accuracy if Eve is not there; otherwise, Bob will have 75% accuracy in the basis comparison. If the accuracy is 100%, they will discard the set of bits they used for matching, and the rest of the bit string will be used as the key to encrypt the message. If 100% accuracy is not observed, they will try again to get a key using QKD.

In Table 2, Alice is sending a key of "01101011" to Bob using two types of polarization as stated above.

Table 2. Alice Sending a Key of 01101011 to Bob Using Two Types of Polarization

Alice's basis + X + + X X X X
Alice's data 0 1 1 0 1 0 1 1
Eve's basis + + X + X X X +
Eve's data 0 1 0 0 1 1 1 0
Bob's basis + + + X + X X X
Bob's data 0 0 0 0 0 1 1 1

Now Alice and Bob will compare their basis, and they will find that Bob has guessed the 1st, 3rd, 7th and 8th basis correctly. So they will throw out the bits for the remaining positions—that is, the 2nd, 4th, 5th and 6th. Now the key is "0011". They will choose the first two bits for matching, and then they will find that their second bit in the key is different, which means Eve is between them. Then they will repeat the same procedure again until they get a 100% key match. When they get a key, they easily can encrypt the message using the key and send it via the public network.

______________________

Subhendu Bera is from West Bengal (India). He completed his Master of Science degree in Computer Science from Banaras Hindu University and his Bachelor of Science degree in Computer Science from University of Calcutta.

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