The security analysis of E91 protocol in collective-rotation noise channel

Abstract

The bit error in quantum communication is mainly caused by eavesdropping and noise. However, most quantum communication protocols only take eavesdropping into consideration and ignore the result of noise, making the inaccuracy situations in detecting the eavesdropper. To analyze the security of the quantum E91 protocol presented by Ekert in collective-rotation noise channel, an excellent model of noise analysis is proposed. The increment of the qubits error rate (ber) is used to detect eavesdropping. In our analysis, eavesdropper (Eve) can maximally get about 50% of the key from the communication when the noise level approximates to 0.5. The results show that in the collective-rotation noise environment, E91 protocol is secure and the raw key is available just as we have knew and proved. We also presented a new idea in analyzing the protocol security in noise channel.

Introduction

Cryptography is the basis of information security; the task of cryptograph is to ensure that only the legitimate users like Alice and Bob can read the secret message in the secure communication, which the unauthorized users like Eve cannot read. In Cryptography, the secret message is generated by the key and should not be read without the key. In 1926, Vernam ¹ presented the one-time pad (OTP). In 1949, Shannon ² proved that the OTP is perfectly secure with equal length of key and plain text. But there is still a serious problem on how to distribute the key safely.^3–5

Different from the classical secure communication protocol, quantum secure communication (QSC) protocol is based on the fundamental laws of quantum physics rather than the complexity of computing. So QSC has much higher performance of security. Quantum communication and quantum cryptography mainly include quantum key distribution (QKD),^6–8 quantum teleportation (QT),^9,10 quantum secret sharing (QSS),^11,12 quantum secure direct communication (QSDC),^13–15 and so on. In 1984, CH Bennett and Brassard ¹⁶ presented the first QKD protocol, which is called the BB84 protocol. The BB84 protocol uses four quantum single particle state and avoids using quantum entangled states and storing quantum particles, making it easy to operate. Later in 1991, Ekert ¹⁷ proposed that QKD be implemented using the quantum entangled states explored in the Einstein-Podolsky-Rosen (EPR) thought experiment, which is called the E91 protocol.

Compared with the BB84 protocol which Alice sends quantum particles to Bob, the E91 protocol uses an EPR pair during the communication, which divides the EPR pair and sends one particle to Alice and Bob separately. As we know, it is a difficult problem to keep EPR entangled. In other words, the E91 protocol is more difficult to application than the BB84 protocol. But there is a center source that sends one of the EPR particles to Alice and Bob, respectively, with the E91 protocol. Which is similar to the classical cryptographic protocol structure. The E91 has received lots of attention since its inception.^18–22

However, those protocols and analyses almost only take the eavesdropping into consideration and ignore the result of noise which cannot be neglected in practice. For application in the real environment, we must take the noise into account.

With the model of collective-rotation noise, the effect of noise can be described as a sine $\sin \theta$ , where $\theta$ is the noise intensity. If there is a eavesdropper, then the bit error rate $\mathrm{ber}$ will increase and be greater than $\mathrm{be}{r}_0$ which is only caused by the effect of noise.^23,24 Moreover, we also analyze how much information that Eve can get in different levels of noise $\epsilon$ , the amount of information that Eve can get $I$ is from 0.399 to 0.5, and she can get the maximum $I_{max}=0.5$ when $\epsilon \to 0.5$ . It can be concluded that the E91 protocol is secure in the real noise environment, and Eve can only get parts of incomplete key, meaning that she cannot read the secret message in the quantum channel.

Related works

The E91 protocol

Let us give a brief introduction of the quantum E91 protocol. There are two groups of measurement basis: $Z-\mathrm{basis}:B_Z=\{|0〉,|1〉\}$ and $X-\mathrm{basis}:B_X=\{|+〉,|-〉\}$ . Alice and Bob randomly choose $B_Z$ or $B_X$ to measure the received particle each time. The E91 protocol can be described in the following steps:

1. The source center $\mathcal{S}$ chooses the EPR pair $|{\varphi }^{+}〉=1/\sqrt{2}(|00〉+|11〉)$ , sends the first particle $|{\varphi }^{+}{〉}_1$ to Alice and second particle $|{\varphi }^{+}{〉}_2$ to Bob.

2. Alice and Bob randomly choose $B_Z$ or $B_X$ to measure their received particle. They record the measurement result and broadcast the measurement basis which they used through the classical channel.

3. Alice and Bob know the choice of both parties. They divide the measurement result into two groups: one is the decoy qubits $G_d$ where they choose different measurement basis and another is the raw key qubits $G_k$ where they choose the same measurement basis.

4. The group $G_d$ is used to detect whether there is a eavesdropping. $G_d$ is always entangled without eavesdropper in an ideal environment. If there is bit error in $G_d$ , which means that there is also a eavesdropper, Alice and Bob think that the quantum channel is not safe and they will interrupt this communication and restart a new one.

5. If the quantum channel is safe, $G_k$ can be used as the raw keys because Alice and Bob can receive the same measurements. Both Alice and Bob agree on that the measurement $|0〉$ represents the classical bit 0, while the measurement $|1〉$ represents the classical bit 1.

6. The E91 protocol ends successfully.

The noise in E91 protocol

An explanation should be stated that in the noisy environment, the noise and eavesdropping cannot be distinguished between each other. The qubits error rate $\mathrm{ber}$ may result from Alice to Bob, Alice to Eve, and Eve to Bob.

The $\mathrm{ber}$ is only caused by Eve in an ideal environment, that is to say, there is eavesdropping if $\mathrm{ber}\ne 0$ . However, taking the noise into consideration, $\mathrm{ber}$ is caused by both noise and Eve’s eavesdropping. There will be bit error even if there is not eavesdropper, which means that the original E91 protocol cannot be used in noise environment. The mechanism to judge whether there exists eavesdropping in quantum noise channel needs to be improved for protecting the information.

In the process of quantum communication, the bit error rate only caused by noise can be seen as $\mathrm{be}{r}_0$ , and $\mathrm{be}{r}_0$ is stable and invariable. If there is a eavesdropping, then the bit error rate $\mathrm{be}{r}_i$ will increase and would be greater than than $\mathrm{be}{r}_0$ . No matter what reason causes this situation, Alice and Bob determined that the quantum channel is not secure and exit Eve’s eavesdropping.

The original E91 protocol uses the decoy group $G_d$ to detect the eavesdropping; however, we use the raw key group $G_k$ to analyze the bit error rate caused by Eve and noise, which means that qubit error rate is from the raw keys.

The level of collective-rotation noise

During the process of analyzing the security conveniently, an assumption can be reasonably proposed that the environmental noise is constant. Even if actually noise is variable, the maximum or average value of the noise can still be conducted. To ensure the security of the E91 protocol, we should take the maximum value into account.

The noise will produce the same effect on every particle in an ideal collective-rotation noise environment. The effect is making every particle left or right deflect $\theta$ angle.^25–27 The effect of the collective-rotation noise could read as the following unitary matrix $U$

$$U=\left(\begin{array}{cc}\cos \theta \hfill & -\sin \theta \hfill \\ \multicolumn{1}{c}{\sin \theta }& \cos \theta \hfill \end{array}\right)$$

(1)

Under the effect of collective-rotation noise, the quantum states become

$$\begin{array}{c}|0〉\to \cos \theta |0〉+\sin \theta |1〉\hfill \\ \multicolumn{1}{c}{=\frac{\cos \theta +\sin \theta }{\sqrt{2}}|+〉+\frac{\cos \theta -\sin \theta }{\sqrt{2}}|-〉}\end{array}$$

(2)

$$\begin{array}{c}|1〉\to -\sin \theta |0〉+\cos \theta |1〉\hfill \\ \multicolumn{1}{c}{=\frac{\cos \theta -\sin \theta }{\sqrt{2}}|+〉-\frac{\cos \theta +\sin \theta }{\sqrt{2}}|-〉}\end{array}$$

(3)

$$\begin{array}{c}|+〉\to \frac{\cos \theta -\sin \theta }{\sqrt{2}}|0〉+\frac{\cos \theta +\sin \theta }{\sqrt{2}}|1〉\hfill \\ \multicolumn{1}{c}{=\cos \theta |+〉-\sin \theta |-〉}\end{array}$$

(4)

$$\begin{array}{c}|-〉\to \frac{\cos \theta +\sin \theta }{\sqrt{2}}|0〉-\frac{\cos \theta -\sin \theta }{\sqrt{2}}|1〉\hfill \\ \multicolumn{1}{c}{=\sin \theta |+〉+\cos \theta |-〉}\end{array}$$

(5)

When there is no noise, $\theta =0$ , the four quantum state will not change. The greater the impact of noise, the greater the probability that quantum states change into another state. Therefore, the level of noise $\epsilon$ can be described as the effect only caused by noise, and the value of $\epsilon$ should be carefully analyzed.

Once the level of noise $\epsilon$ is determined, it will not change during the E91 protocol. The Eve’s eavesdropping will cause a different bit error rate $\mathrm{be}{r}_i$ which should satisfy $\mathrm{be}{r}_i\ge \epsilon$ .

Our target is to find the relationship between the bit error rate $\mathrm{be}{r}_i$ and the level of noise $\epsilon$ . In another words, we should find the mathematical relation between $\mathrm{be}{r}_i$ and $\theta$ . In this article, we will also analyze how much information that Eve can get; as we know, Eve cannot read the secret message with incomplete keys. That is to say, if the information that Eve get $I<1$ , which means that Eve can only get parts of keys and she cannot read the secret message, then the E91 protocol is secure.

The security analysis of E91 protocol in the noise environment

The analysis without eavesdropping

In the security analysis of E91 protocol, we ignore the decoy group and only discuss the raw key group. Alice and Bob randomly receive the qubit $|0〉$ and $|1〉$ with the probability of 50% separately. To make the analysis easy, we only discuss the situation that Alice receive qubit. The situation of Bob is the same as Alice; it is symmetrical.

According the formulas (2)–(5), we can easily calculate the probability when $\mathcal{S}$ sends $|0〉$ and Alice receives $|0〉$ , $|1〉$ , |+〉, |−〉. And the situation that $\mathcal{S}$ sends $|1〉$ is the same. Just as Table 1 shows, where $P$ is the probability that Alice received, $A$ is Alice, and $\mathcal{S}$ is the source center.

Table 1.

The result of Alice’s measurement.

$P$	$\mathcal{S}/A$	$|0〉$	$|1〉$	|+〉	|−〉
$\frac{1}{2}$	$|0〉$	$\frac{\stackrel{2}{\cos }\theta }{4}$	$\frac{{\sin }^2\theta }{4}$	$\frac{1-\sin 2\theta }{8}$	$\frac{1+\sin 2\theta }{8}$
$\frac{1}{2}$	$|1〉$	$\frac{{\sin }^2\theta }{4}$	$\frac{\stackrel{2}{\cos }\theta }{4}$	$\frac{1+\sin 2\theta }{8}$	$\frac{1-\sin 2\theta }{8}$

Let us analyze the situation that the source center $\mathcal{S}$ sends qubit $|0〉$ to Alice; when analyzing the bit error rate only caused by noise, we can easily get the level of noise $\epsilon$

$$\begin{array}{c}\epsilon =1-(\frac{\stackrel{2}{\cos }\theta }{4}+\frac{\stackrel{2}{\cos }\theta }{4})\times 2\hfill \\ ={\sin }^2\theta \hfill \end{array}$$

(6)

That is to say, when there is no eavesdropping, the bit error rate $\mathrm{be}{r}_0$ should be set as $\epsilon$

$$\mathrm{be}{r}_0=\epsilon ={\sin }^2\theta$$

(7)

The information that Eve can get

Before we analyze how much bit error rate is caused by eavesdropping, let us analyze how much information that Eve can get.

According to Deng et al., ²⁸ the maximal amount of the information that Eve can eavesdrop from Alice is given

$$I(A,E)=H(A)-H(A|E)$$

(8)

The qubit is sent randomly by Alice, so H(A) = 1 and

$$H(A|E)=-\sum p(A,E)\underset{2}{\log }p(A|E)$$

(9)

According to Table 1 and formulas (6), (8), and (9). To make it easy to express, let $\xi =\stackrel{2}{\cos }\theta =1-\epsilon$ , so the maximal information Eve $I(A,E)$ can be rewritten

$$\begin{array}{c}\multicolumn{1}{c}{I(A,E)=1+\frac{\xi }{2}\underset{2}{\log }\xi +\frac{\epsilon }{2}\underset{2}{\log }\epsilon }\\ +\frac{1-2\sqrt{\epsilon \xi }}{4}\underset{2}{\log }\frac{1-2\sqrt{\epsilon \xi }}{2}\hfill \\ +\frac{1+2\sqrt{\epsilon \xi }}{4}\underset{2}{\log }\frac{1+2\sqrt{\epsilon \xi }}{2}\hfill \end{array}$$

(10)

Formula 10 shows that $I(A,E)$ is only related to the bit error rate, which means that we can analyze it with the level of noise $\epsilon$ and get its maximum value.

The analysis with eavesdropping

Just as Figure 1 shows, as we know, Eve will intercept the qubits, take a measurement, and reprepare them to Alice. The quantum state that Eve receives is one of $\{|0〉,|1〉,|+〉,|-〉\}$ , not $\{|0〉,|1〉\}$ . According to formulas (2)–(5), and Table 1, the situation that Eve resends qubits to Alice, which is described in Table 2, where E is Eve and A is Alice.

Figure 1.

Eve intercepts and resends qubits.

Table 2.

The result of Alice’s measurement.

$E/A$	$|0〉$	$|1〉$	|+〉	|−〉
$|0〉$	$\frac{\stackrel{2}{\cos }\theta }{2}$	$\frac{{\sin }^2\theta }{2}$	$\frac{1-\sin 2\theta }{4}$	$\frac{1+\sin 2\theta }{4}$
$|1〉$	$\frac{{\sin }^2\theta }{2}$	$\frac{\stackrel{2}{\cos }\theta }{2}$	$\frac{1+\sin 2\theta }{4}$	$\frac{1-\sin 2\theta }{4}$
|+〉	$\frac{1+\sin 2\theta }{4}$	$\frac{1-\sin 2\theta }{4}$	$\frac{\stackrel{2}{\cos }\theta }{2}$	$\frac{{\sin }^2\theta }{2}$
|−〉	$\frac{1-\sin 2\theta }{4}$	$\frac{1+\sin 2\theta }{4}$	$\frac{{\sin }^2\theta }{2}$	$\frac{\stackrel{2}{\cos }\theta }{2}$

According to Tables 1 and 2, Alice and Bob always choose the same measurement basis; the probability that $|0〉$ changes into $|1〉$ is ${\sin }^2\theta$ , which has the same probability that $|1〉$ changes into $|0〉$ and so is |+〉 and |−〉. So the result should be locally symmetrical, as shown in Table 3, where $\mathrm{Basis}$ is the measurement basis that Alice and Bob choose, A is Alice, and B is Bob.

Table 3.

Alice’s and Bob’s measurement result.

$\mathrm{Basis}$	$A/B$	$|0〉$	$|1〉$	|+〉	|−〉
$B_Z$	$|0〉$	$p_1$	$p_2$	0	0
$B_Z$	$|1〉$	$p_2$	$p_1$	0	0
$B_X$	|+〉	0	0	$p_3$	$p_4$
$B_X$	|−〉	0	0	$p_4$	$p_3$

From Table 3, it can be seen that when Alice’s measurement is different from Bob’s measurement, there is a bit error. We only need to calculate the value of $p_2$ and $p_4$ , while $p_1$ and $p_3$ can be ignored.

$p_2$ means that Alice gets $|0〉$ and Bob gets $|1〉$ ; this situation can be divided into four parts:

1. Alice gets $|0〉$ , Eve gets $|0〉$ , and Bob gets $|1〉$ , the probability is $[({\cos }^2\theta )/4]\times [({\sin }^2\theta )/2]$ ;

2. Alice gets $|0〉$ , Eve gets $|1〉$ , and Bob gets $|1〉$ , the probability is $[({\sin }^2\theta )/4]\times [({\cos }^2\theta )/2]$ ;

3. Alice gets $|0〉$ , Eve gets |+〉, and Bob gets $|1〉$ , the probability is $[(1-\sin 2\theta )/8]\times [(1-\sin 2\theta )/4]$ ;

4. Alice gets $|0〉$ , Eve gets |−〉, and Bob gets $|1〉$ , the probability is $[(1+\sin 2\theta )/8]\times [(1+\sin 2\theta )/4]$ ;

And so is $p_4$ , consider the symmetry between Alice and Bob, $p_2$ and $p_4$ can be easily calculated

$$\begin{array}{c}\multicolumn{1}{c}{p_2=(\frac{\stackrel{2}{\cos }\theta }{4}\times \frac{{\sin }^2\theta }{2}+\frac{{\sin }^2\theta }{4}\times \frac{\stackrel{2}{\cos }\theta }{2}}\\ +\frac{1-\sin 2\theta }{8}\times \frac{1-\sin 2\theta }{4}\hfill \\ +\frac{1+\sin 2\theta }{8}\times \frac{1+\sin 2\theta }{4})\times \frac{1}{2}\times 2\hfill \\ \multicolumn{1}{c}{=\frac{4{\sin }^2\theta \stackrel{2}{\cos }\theta +1+{\sin }^{2}2\theta }{16}}\end{array}$$

(11)

$$\begin{array}{c}\multicolumn{1}{c}{p_4=(\frac{1-\sin 2\theta }{8}\times \frac{\stackrel{2}{\cos }\theta }{2}+\frac{1+\sin 2\theta }{8}\times \frac{{\sin }^2\theta }{2}}\\ +\frac{1+\sin 2\theta }{8}\times \frac{\stackrel{2}{\cos }\theta }{2}+\frac{1-\sin 2\theta }{8}\times \frac{{\sin }^2\theta }{2})\hfill \\ \times \frac{1}{2}\times 2=\frac{1}{8}\hfill \end{array}$$

(12)

So the bit error rate caused by both eavesdropping and noise can be easily calculated

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{be}{r}_1=p_2+p_2+p_4+p_4}\\ =\frac{8{\sin }^2\theta \stackrel{2}{\cos }\theta +3}{8}\hfill \\ =-{\epsilon }^2+\epsilon +\frac{3}{8}\hfill \end{array}$$

(13)

Figure 2 shows the relationship between the bit error rate caused by both eavesdropping and noise $\mathrm{be}{r}_1$ and the level of noise $\epsilon$ .

Figure 2.

The relationship between $\mathrm{be}{r}_1$ and $\epsilon$ .

From Figure 2, it can be seen that when $\epsilon \le 0.61$ , $\mathrm{be}{r}_1\ge \mathrm{be}{r}_0$ , Eve’s eavesdropping will cause a larger bit error rate and she will be detected. In an ideal environment (when $\epsilon =0$ ), Eve will still cause a bit error rate $\mathrm{be}{r}_{1_{\epsilon =0}}=0.375$ and Eve’s eavesdropping will also be detected; from another point of view, our analysis is proven correct. When $\epsilon >0.61$ , the environment of quantum channel is too bad to communicate, Alice and Bob should intercept this communication and restart a new one. This situation should be avoided and we will not take it into consideration.

Let us define $\Delta =\mathrm{be}{r}_1-\mathrm{be}{r}_0$ , and we can easily get its maximum

$$\begin{array}{c}\Delta =-{\epsilon }^2+\epsilon +\frac{3}{8}-\epsilon \hfill \\ \multicolumn{1}{c}{=-{\epsilon }^2+\frac{3}{8}\le \frac{3}{8}}\end{array}$$

(14)

It is easy to find that $\Delta$ is a continuous monotone decreasing function when $\epsilon \in [0,0.61]$ , which means that the less the level of noise $\epsilon$ , the more effect of detecting Eve’s eavesdropping in E91 protocol. To get an effect E91 protocol, we should try our best to reduce the level of noise $\epsilon$ .

Now, let us analyze how much information that Eve can get from her eavesdropping. The maximal and minimal information $I_{max}$ and $I_{min}$ can be depicted, as shown in Figure 3.

Figure 3.

The relationship between $I$ and $\epsilon$ .

As it shown in Figure 3, the minimum of information that Eve can get is $I_{min}=0.399$ when $\epsilon =0.144$ , and when $\epsilon \to 0.5$ , the maximum that Eve can get is $I_{max}=0.5<1$ . When $\epsilon >0.61$ , the quantum channel is too bad, Alice and Bob should intercept this communication and restart a new one,; we will not take this situation into consideration.

From Figure 3, Eve can most get only 50% information, since each qubits’ transmission is mutual independence; Eve can only get randomly 50% raw keys, but she does not know which parts she has gotten. This means that Eve cannot read the secret message with the raw keys she have obtained, and the quantum E91 protocol has been proven to be safe.

Conclusion

In this article, we have analyzed the relationship between the bit error rate $\mathrm{be}{r}_i$ and the level of noise $\epsilon$ . When the level of noise $\epsilon$ is determined, the eavesdropping from Eve will cause an increment in the bit error rate; Alice and Bob can detect and intercept this communication and restart a new one.

From Figure 2, it can be concluded that the quantum E91 protocol is safe when $\epsilon \in [0,0.61]$ . The less the level of noise $\epsilon$ , the more effective the E91 protocol will be. When applying the quantum E91 protocol into real environment, we should reduce the level of noise $\epsilon$ as much as possible.

From Figure 3, it can be concluded that the maximal information Eve can get is $I_{max}=0.5$ , which means that Eve cannot get the whole raw keys. She does not know which part she has got. And she cannot read the secret message with incomplete raw keys. On the other hand, the E91 protocol has been proved safe.

In conclusion, the QKD protocol named E91’s security has been analyzed. When the level of noise $\epsilon \in [0,0.61]$ , Alice and Bob can detect the eavesdropping, and the maximal amount of information that Eve can get is $I_{max}=0.5$ , and she cannot read the secret message. This result will give new idea for the application of the quantum E91 protocol in communication and information processing with the environmental noise. What is more, our method also presented a new idea in analyzing the other protocol security in collective-rotation noise channel.