Japanese abstract
前報は、単一量子ビットを利用する量子鍵配送プロトコルBB84の説明であったが、今回は、「もつれた状態」の２つの量子ビット対のストリームを利用した安全な鍵配送プロトコルEkertを説明する。これは、すでに
本ブログで紹介した「Bellのテスト（Bellの不等式）」に基づいている。このような量子ビットを受け取った２者が、それぞれランダムに選択した正規直交基底を使ってそれを測定する。その結果の両者の古典ビット列の一致度が盗聴の有無を明らかにする。この記事は、Prof. Chris Bernhardtの著書[1]に基づいているが、小生が理解した内容を独自の観点からまとめたものである。
English Abstract
In
my previous article, I described the quantum key distribution protocol BB84, which uses a single qubit. Here I describe the Ekert, a secure key distribution protocol that uses a stream of two qubit pairs in an entangled state. This is based on Bell's test (Bell's inequality) which was already introduced in
this blog. Two parties that receive such qubits each measure them using randomly chosen orthonormal bases. The degree of matching between the two resulting classical bit strings reveals the presence or absence of eavesdropping. This article is based on Prof. Chris Bernhardt's book [1] and explains the Ekert from my own point of view.
●Measure qubits in different orthonormal bases
First, let's see what happens when we measure the received qubits in a different orthonormal basis than the one from the sender. In the following, for simplicity, the orthonormal bases will simply be referred to as bases. Here, one is randomly selected from the three types of bases shown in Fig 1. Bases D000, D120, and D240 correspond to setting the orientation of measurement apparatus to due north, southeast, and southwest, respectively.
Next, Fig. 2 summarizes how each ket vector (vertical vector) of the basis used for measurement on the transmitting side is expressed in a different basis. In general, the state of a qubit will be a linear combination of the ket vectors of the chosen basis. The probability that a measurement result falls to a ket vector is the square of the coefficient of that ket vector.
As already shown in Bell's test, Fig. 2 reveals an important point. If the measurement result in the original basis is 0 (falling to the first ket vector), it will be 0 with a probability of 1/4 when measured in another different basis. Also, if the measurement in the original basis is 1 (falling to the second ket vector), it will be 1 again with a probability of 1/4 when measured in a different basis.
That is, there is a 1/4 probability that a measurement on the original basis agrees with another measurement on a different basis. This is a theoretical value, but if you send enough qubits and measure them, it should approach this probability infinitely.
●How the quantum key distribution protocol Ekert works
With the above preparation, we can describe the Ekert protocol. Suppose we send a stream of two qubit pairs to Alice and Bob. Importantly, paired qubits are always in the entangled state
(1/√2) |↑〉|↑〉+(1/√2) |↓〉|↓〉.
Both will each receive one of the pair. My description will be made below with reference to Fig.3.
Note that the entangled states
(1/√2) |↑〉|↑〉+(1/√2) |↓〉|↓〉
and
(1/√2) |↘︎〉|↘︎〉+(1/√2) |↖︎〉|↖︎〉
are exactly the same. This fact is proven in reference [1], but is omitted here.
Suppose that Alice performs measurements on the selected basis D120 (see Figure 1) before Bob does. For example, if the result is 0 (ie, |↘︎〉|↘︎〉), Bob's qubit will also be |↘︎〉 because of the entangled state. Bob measures it on a randomly chosen basis. If the basis is different from Alice's, he gets 0 with probability 1/4, from the fact already revealed in Fig2. Similarly, if Alice's measurement result is 1, Bob's measurement result will be 1 with probability 1/4.
Now, after all qubits have been processed, Alice and Bob exchange their lists of bases (not measurement results) of their choice. In that case, secure communication is not particularly required. Since there are three types of bases that can be selected by both, the two bases match in 1/3 of the total cases. Qubits always give the same result when measured on the same basis. Therefore, If no eavesdropping has occurred, the measurements with their agreed bases can be the common key.
●Why can eavesdropping be detected?
Now consider the case where Eve eavesdrops on an entangled qubit. As I mentioned in the previous BB84 protocol explanation, Eve always needs to set a certain base and use it to make measurements. If measured in such a way, the entanglement will disappear immediately. As a result, the probability that Alice's and Bob's classical bits match will change. For example, if Eve eavesdrops before Alice's measurements, we find that the entanglement disappears and the probability that Alice and Bob's classical bits match increases to 3/8. In other words, when the bases of Alice and Bob are different, if the probability that the classical bits of both match is 1/4, it can be said that there was no eavesdropping.
Acknowledgments
Thanks to Prof. Chris Bernhardt, author of reference [1], for kindly answering some of my questions about entanglement and Bell's inequality.
I used the illustration (icon) of the Qni quantum circuit simulator[2] to show the qubit vectors..
References
[1] Chris Bernhardt, “Quantum Computing for Everyone”, MIT Press, 2020
