Quantum direct communication protocols using discrete-time quantum walk

Abstract

The unique features of quantum walk, such as the possibility of the walker to be in superposition of the position space and get entangled with the position space, provide inherent advantages that can be captured to design highly secure quantum communication protocols. Here we propose two quantum direct communication protocols, a quantum secure direct communication protocol and a controlled quantum dialogue (CQD) protocol using discrete-time quantum walk on a cycle. The proposed protocols are unconditionally secure against various attacks such as the intercept-resend attack, the denial of service attack, and the man-in-the-middle attack. Additionally, the proposed CQD protocol is shown to be unconditionally secure against an untrusted service provider and both the protocols are shown more secure against the intercept resend attack as compared to the qubit-based LM05/DL04 protocol.

Appendix

1.1 LM05/DL04 protocol

This qubit-based protocol was introduced in [8, 9]. In this protocol, the encoding rules for the message sender are as follows:

To encode the bit 0, do nothing to the incoming qubit.

To encode the bit 1, apply the operator $iY=ZX$ on the incoming qubit. The transformations are as follows:

$$\begin{aligned} iY|0\rangle= & {} -|1\rangle \\ iY|1\rangle= & {} |0\rangle \\ iY|\pm \rangle= & {} \pm |\mp \rangle \end{aligned}$$

The protocol is as follows:

• 1. Alice chooses n random qubits from the set {$|0\rangle ,|1\rangle ,|+\rangle ,|-\rangle$} and sends them to Bob.

• 2. Out of these n qubits received from Alice, Bob randomly chooses n/2 of them and classically sends their coordinates to Alice.

• 3. Alice publicly announces the states of the n/2 qubits which Bob chose in step 2. Bob measures each of the n/2 qubits in their corresponding bases and checks for eavesdropping. If the error is within a tolerable limit, then the protocol continues to step 4. Else, the protocol is discarded and they start all over again.

• 4. Among the remaining n/2 qubits, Bob randomly chooses n/4 of them and encodes the message in them according to the encoding rules above and does nothing to the remaining n/4 qubits. He sends all these n/2 qubits back to Alice.

• 5. After Alice confirms receiving the n/2 qubits, Bob sends the coordinates of the qubits on which he did not encode the message. Alice uses these qubits to check for eavesdropping just like how Bob does it in step 3.

• 6. After confirming no eavesdropping, Alice measures the remaining qubits in their respective bases to obtain the message sent by Bob.

1.2 Mutual information

Let us take two random variables, say x and y. The mutual information $I_{XY}$ between two random variables x and y is the decrease in uncertainty of one random variable when the value of the other random variable is observed, measured, or determined. If x and y are discrete, the formula for $I_{XY}$ is given by [37]

$$\begin{aligned} I_{XY}=\underset{x}{\sum }\underset{y}{\sum }p(x,y)log_{2}\frac{p(x,y)}{p(x)p(y)} \end{aligned}$$

(15)

where p(x, y) is the joint probability mass function and p(x) and p(y) are the individual probability mass functions.

If x and y are continuous, then the formula for $I_{XY}$ is given by

$$\begin{aligned} I_{XY}=\underset{x}{\int }\underset{y}{\int }p(x,y)log_{2}\frac{p(x,y)}{p(x)p(y)}dxdy \end{aligned}$$

(16)

where p(x, y) is the joint probability density function and p(x) and p(y) are the individual probability density functions.

There can also be a case where one of the random variables is discrete and the other is continuous. For example, if x is discrete and y is continuous, then the formula for $I_{XY}$ becomes

$$\begin{aligned} I_{XY}=\underset{x}{\sum }\underset{y}{\int }p(x,y)log_{2}\frac{p(x,y)}{p(x)p(y)}dy \end{aligned}$$

(17)

where p(x) is the probability mass function of x, p(y) is the probability density function of y, and p(x, y) is a function that is a probability density-mass function that is discrete in x and continuous in y.

This concept of mutual information can also be generalised to $r=mn>2$ random variables $\{x_{1},x_{2},...,x_{m}\}$ and $\{y_{1},y_{2},...,y_{n}\}$ where $x_{i}$ are discrete and $y_{i}$ are continuous. The generalised mutual information $I_{mutual}$ is given by [37]

$$\begin{aligned} I_{mutual}= & {} \underset{x_{1},x_{2},...,x_{n}}{\sum }\underset{y_{1},...,y_{n}}{\int }p(x_{1},x_{2},...,x_{m},y_{1},y_{2},...,y_{n})log_{2}\nonumber \\&\frac{p(x_{1},x_{2},...,x_{m},y_{1},y_{2},...,y_{n})}{p(x_{1})p(x_{2})...p(x_{m})p(y_{1})p(y_{2})...p(y_{n})}dy_{1}dy_{2}...dy_{n}. \end{aligned}$$

(18)

1.2.1 Mutual information for the intercept-resend attack for the LM05/DL04 protocol

Let us consider the first transmission from Alice to Bob. In this transmission, Alice first selects either of the four states and prepares them and sends them to Bob. Eve intercepts this channel before the state reaches Bob and randomly chooses a basis for each incoming state and measures the state in that basis. Let $a,e\in \{0,1,+,-\}$. Let the probability of Alice sending the qubit a and Eve receiving the qubit e be p(a, e). For example, the probability p(0, 0) is

$$\begin{aligned} p(0,0)= & {} \overset{probability\,of\,Alice\,choosing\,0}{\frac{1}{4}}\times \overset{probability\,of\,Eve\,choosing\,the\,computational\,Z\,basis}{\frac{1}{2}}\nonumber \\&\times \overset{probability\,of\,Eve\,getting\,0}{1}=\frac{1}{8}. \end{aligned}$$

(19)

Similarly,

$$\begin{aligned} p(0,1)= & {} \frac{1}{4}\times \frac{1}{2}\times 0=0, \end{aligned}$$

(20)

$$\begin{aligned} p(0,+)= & {} \frac{1}{4}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{16}, \end{aligned}$$

(21)

$$\begin{aligned} p(0,-)= & {} \frac{1}{4}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{16}, \end{aligned}$$

(22)

and similar probabilities for $p(1,e),p(+,e)$, and $p(-,e),$ where $e\in \{0,1,+,-\}$.

Hence, the mutual information $I_{AE}$ for the LM05/DL04 protocol is given by

$$\begin{aligned} I_{AE}= & {} \underset{a}{\sum }\underset{e}{\sum }p(a,e)log_{2}\frac{p(a,e)}{p(a)p(e)}\nonumber \\= & {} 4\left( \frac{1}{8}log_{2}\frac{\frac{1}{8}}{\frac{1}{16}}+\frac{1}{16}log_{2}\frac{\frac{1}{16}}{\frac{1}{16}}+\frac{1}{16}log_{2}\frac{\frac{1}{16}}{\frac{1}{16}}\right) =0.5. \end{aligned}$$

(23)

(We can see that for all a and e, $p(a)=p(e)=\frac{1}{4}$. Hence, $p(a)p(e)=\frac{1}{16}$).