CN105245332A - Four-particle x state-based two-party quantum key agreement protocol - Google Patents

Abstract

The invention discloses a four-particle x state-based two-party quantum key agreement protocol. The protocol includes following steps that: step 1, Alice and Bob negotiate on quantum-state codes; step 2, Alice prepares n x-state|x<00>>ABCD, and divides all the particles into four ordered sequences, and selects out 2m decoy photons and inserts the decoy photons into the sequences and transmits the sequences to Bob; step 3, Bob measures corresponding decoy photons with a correct measurement base, and informs Alice of measurement results; Alice compares the measurement results and the initial states of the decoy photons, and calculates an error rate, and executes step 4 if the error rate is low, otherwise, executes step 2; and step 4, Alice performs Bell measurement on the particles, Bob executes Z convolution Z-base measurement on the sequences, and Alice and Bob both obtain a shared key. The protocol of the invention can resist existing participant attacks, external attacks and Troy Trojan horse attacks; and the protocol is safe in a noise-free quantum channel and a quantum noise channel. The quantum bit efficiency of the protocol of the invention is higher than that of an existing secure multi-particle entangled state-based quantum key agreement protocol.

Description

Based on two side's quantum key agreement protocols of four particle χ states

Technical field

The invention belongs to quantum communications field, be specifically related to a kind of quantum key and consult (Quantumkeyagreement) agreement, particularly a kind of two side's quantum key agreement protocols based on four particle χ states.

Background technology

Quantum cryptography is the new technology of communication and network security, and its fail safe is ensured by fundamental principles of quantum mechanics.Be the different of computationally secure mostly from conventional cipher, quantum cryptography can realize unconditional security, has attracted a large amount of concern thus.It is the new important branch of of quantum cryptography that quantum key consults (QKA) agreement, it allows participant to consult a classical shared secret key by disclosed quantum channel, and the contribution of each participant is identical, the subset that any one participant or participant are formed all can not independently determine this shared key.Utilization sub-key consults the shared secret key of (QKA) agreement foundation and the cryptographic algorithm of one-time pad, and communicating pair can realize the secure communication of unconditional security.

Existing most of quantum key agreement protocol is based on single-particle or Bell state, based on multiparticle Entangled State quantum key agreement can be counted on one's fingers, and they or the external attacks such as Te Luoyi wooden horse can not be resisted, be unsafe, or quantum bit rate is too low.

D.S.Shen, W.P.MaandL.L.Wang utilizes the Cluster State of four particles to propose a both sides QKA agreement in paper " Two-partyquantumkeyagreementwithfour-qubitclusterstates " (QuantumInf.Process.2014:2313-2324), and this agreement has the sub-bit efficiency of higher amount.The concrete steps of agreement are: the first, the Cluster State of some four particles of each self-generating of communicating pair A and B.Communication party A (communication party B) issues communication party B (communication party A) by being inserted after trick photon by the sequence that (first) particle is formed of the 3rd in Cluster State, and retains and its particle sequence.The second, after communicating pair receives corresponding particle sequence, perform eavesdropping monitoring together.3rd, the unitary transformation of the particle sequence execution that communicating pair just receives separately oneself.Then insert after inveigling photon and it is issued the other side mutually.4th, after communicating pair receives corresponding particle sequence, perform eavesdropping monitoring together.5th, communication party A (communication party B) performs respective unitary transformation to by the sequence that (the 3rd) particle is formed of first in Cluster State.Then both sides perform the measurement of cluster base to respective Cluster State, and both sides can obtain identical measurement result.The privacy key shared can be obtained according to coding and the correspondence of measurement result.This agreement Shortcomings part is: because this agreement is a Ping-Pong agreement, namely same particle has been transmitted one back and forth in quantum channel, and therefore this agreement cannot be resisted invisible photon eavesdropping (IPE) Trojan attack and be postponed photon Trojan attack.

W.Huang, Q.Su, X.Wu, Y.B.LiandY.Sun utilize the DF state of four particles to propose the both sides QKA agreement of an energy immunity associating noise in paper " Quantumkeyagreementagainstcollectivedecoherence " (Int.J.Theor.Phys.2014:2891-2901).The concrete steps of agreement are: the first, and communication party A generates two random bit strings, and an individual as shared key contributes string, and one as selecting the control string measuring base.The second, communication party A contributes string and selection to measure the sequence of the DF state of control string preparation four particles of base according to individual, and issues communication party B after inserting trick photon.3rd, after communication party B receives the sequence of DF state of four particles, both sides perform eavesdropping monitoring jointly.If by detecting, the individual that communication party B announces his shared key contributes string.4th, communication party A contributes string according to oneself and the individual of communication party B, can calculate the shared secret key of both sides.The control string of base is measured in 5th, the communication party A selection disclosing him.Utilize this to control string, communication party B can measure all DF states, and the individual that can obtain the shared key of communication party A according to measurement result contributes string.Therefore, communication party B also can calculate the shared secret key of both sides.This agreement Shortcomings part is: the quantum bit efficiency of this agreement is too low, and its quantum bit efficiency is only 10%.

Summary of the invention

For the defect existed in above-mentioned prior art or deficiency, the object of the invention is to, a kind of quantum key agreement protocol based on GHZ state is provided.

In order to realize above-mentioned task, the present invention adopts following technical scheme to be solved:

Based on two side's quantum key agreement protocols of four particle χ states, specifically comprise the steps:

Step 1:Alice and Bob consults the coding of following quantum state;

Step 2:Alice prepares n χ state | χ ⁰⁰> _aBCD, and by this n χ state | χ ⁰⁰> _aBCDall particles be divided into four orderly sequences:

S _A＝{A ₁,A ₂,…,A _n}，S _B＝{B ₁,B ₂,…,B _n}，

S _c={ C ₁, C ₂..., C _nand S _d={ D ₁, D ₂..., D _n}

Wherein sequence S _a, S _b, S _c, S _drespectively by each χ state | χ ⁰⁰> _aBCDparticle A, B, C, D form; Alice from set | 0>, | 1>, |+>, | select 2m in->} at random and inveigle photon, and by this 2m trick photon radom insertion sequence S _cand S _d, and ensure that inserting m in each sequence inveigles photon, obtains new sequence S' respectively _cand S' _d; Alice oneself reservation queue S _aand S _b, by sequence S' _cand S' _dsend to Bob; N, m are the positive integer being greater than 1;

Step 3: when Bob receives sequence S' _cand S' _dafter, inform Alice by classical authenticated channel; Alice announces and inveigles photon at sequence S' _cand S' _din position and corresponding measurement base | 0>, | 1>} or |+>, |->}; Bob goes to measure with correct measurement base and inveigles photon accordingly, and measurement result is informed Alice by classical authenticated channel; Alice compares and measures result and inveigles the initial condition of photon, mistake in computation rate; If error rate is lower than prespecified limit gate value, then perform step 4, otherwise, perform step 2;

Step 4:Alice is to sequence S _aand S _bevery two particles that middle sequence number is identical perform Bell and measure, and Bob is to sequence S _cand S _devery two particles that middle sequence number is identical perform base is measured; According to the coding of the quantum state that Alice and Bob in the measurement result of Alice and Bob and step 1 consults, Alice and Bob obtains the shared key of identical 2n bit respectively.

$Z\&CircleTimes;Z=\left\{\right|00>,|01>,|10>,|11>\}$

Further, in described step 1, the coding of the quantum state that Alice and Bob consults: | φ ⁺> _aB→ 00, | ψ ^-> _aB→ 01, | ψ ⁺> _aB→ 10, | φ ^-> _aB→ 11, | 00> _cD→ 00, | 01> _cD→ 01, | 10> _cD→ 10, | 11> _cD→ 11.

Further, in described step 2,

$\begin{array}{c}|{\mathrm{\&chi;}}^{00}{>}_{ABCD}=\frac{1}{2\sqrt{2}}(|0000>+|0011>-|0101>+|0110>+|1001>+|1010>+|1100>\\ -|1111>{)}_{ABCD}\end{array}.$

Further, in described step 3, described limit gate value gets 0.1 ~ 0.2.

Beneficial effect of the present invention:

Two side's quantum key agreement protocols based on four particle χ states of the present invention can guarantee that communicating pair sets up the classical privacy key shared between them liberally; Utilize the cryptographic algorithm of this classical privacy key and one-time pad, communicating pair can realize the secure communication of unconditional security, and obvious the present invention can resist existing participant and attack and external attack.Because each particle in agreement of the present invention is only transmitted once, therefore assailant successfully can not perform Trojan horse attack.Find that this agreement is not only safe at noiseless quantum channel by analyzing, and it is also safe on quantum noise channel.In addition, the quantum bit efficiency of agreement of the present invention is higher than the existing safe quantum key agreement protocol based on multiparticle Entangled State.

Embodiment

1, pre-knowledge

As everyone knows, | 0>, | 1>} defines Z base, |+>, |->} defines X base, wherein $|->1/\sqrt{2}\left(\right|0>-|1>).$ Four Bell state are defined as follows:

$|{\mathrm{\&phi;}}^{+}>=1/\sqrt{2}\left(\right|00>+|11>),|{\mathrm{\&phi;}}^{-}>=1/\sqrt{2}\left(\right|00>+|11>),$

$|{\mathrm{\&psi;}}^{+}>=1/\sqrt{2}\left(\right|01>+|10>),|{\mathrm{\&psi;}}^{-}>=1/\sqrt{2}\left(\right|01>-|10>),$

They form one group of complete orthogonal basis in four-dimensional Hilbert space, i.e. Bell base.χ state is the maximal entangled state of four particles, and they can form one group of complete orthogonal basis in 16 dimension Hilbert spaces.In agreement of the present invention, we use a following χ state as quantum information source, namely

$\begin{array}{c}|{\mathrm{\&chi;}}^{00}{>}_{ABCD}=\frac{1}{2\sqrt{2}}(|0000>+|0011>-|0101>+|0110>+|1001>+|1010>+|1100>\\ -|1111>{)}_{ABCD}\\ =\frac{1}{2}\left(\right|{\mathrm{\&phi;}}^{+}{>}_{AB}|00{>}_{CD}-|{\mathrm{\&psi;}}^{-}{>}_{AB}|01{>}_{CD}+|{\mathrm{\&psi;}}^{+}{>}_{AB}|10{>}_{CD}+|{\mathrm{\&phi;}}^{-}{>}_{AB}\left|11{>}_{CD}\right)\end{array}$

According to expression formula, if to χ state | χ ⁰⁰> _aBCDparticle A and B perform Bell measure, particle C and D is performed base measure, then system with 1/4 probability be collapsed to state | φ ⁺> _aB| 00> _cD, | ψ ^-> _aB| 01> _cD, | ψ ⁺> _aB| 10> _cDwith | φ ^-> _aB| 11> _cD.

2, two side's quantum key agreement protocols based on four particle χ states of the present invention

Two side's quantum key agreement protocols based on four particle χ states of the present invention, specifically comprise the steps:

Step 1:Alice and Bob consults the coding of following quantum state: | φ ⁺> _aB→ 00, | ψ ^-> _aB→ 01, | ψ ⁺> _aB→ 10, | φ ^-> _aB→ 11, | 00> _cD→ 00, | 01> _cD→ 01, | 10> _cD→ 10, | 11> _cD→ 11.

Step 2:Alice prepares n χ state | χ ⁰⁰> _aBCD, and by this n χ state | χ ⁰⁰> _aBCDall particles be divided into four orderly sequences:

S _A＝{A ₁,A ₂,…,A _n}，S _B＝{B ₁,B ₂,…,B _n}，

S _c={ C ₁, C ₂..., C _nand S _d={ D ₁, D ₂..., D _n}

Wherein sequence S _a, S _b, S _c, S _drespectively by each χ state | χ ⁰⁰> _aBCDparticle A, B, C, D form; Alice from set | 0>, | 1>, |+>, | select 2m in->} at random and inveigle photon, and by this 2m trick photon radom insertion sequence S _cand S _d, and ensure that just in time inserting m in each sequence inveigles photon, obtains new sequence S' respectively _cand S' _d; Alice oneself reservation queue S _aand S _b, by sequence S' _cand S' _dsend to Bob; N, m are the positive integer being greater than 1;

$\begin{array}{c}|{\mathrm{\&chi;}}^{00}{>}_{ABCD}=\frac{1}{2\sqrt{2}}(|0000>+|0011>-|0101>+|0110>+|1001>+|1010>+|1100>\\ -|1111>{)}_{ABCD}\end{array}$

Step 3: when Bob receives sequence S' _cand S' _dafter, inform Alice by classical authenticated channel; Alice announces and inveigles photon at sequence S' _cand S' _din position and corresponding measurement base | 0>, | 1>} or |+>, |->}; Bob goes to measure with correct measurement base and inveigles photon accordingly, and measurement result is informed Alice by classical authenticated channel; Alice compares and measures result and inveigles the initial condition of photon, mistake in computation rate; If error rate is lower than prespecified limit gate value, then perform step 4, otherwise, perform step 2; Described limit gate value gets 0.1 ~ 0.2.

Step 4:Alice is to sequence S _aand S _bevery two particles that middle sequence number is identical perform Bell and measure, and Bob is to sequence S _cand S _devery two particles that middle sequence number is identical perform base is measured; According to the coding of the quantum state that Alice and Bob in the measurement result of Alice and Bob and step 1 consults, Alice and Bob obtains the shared key of identical 2n bit respectively.

$Z\&CircleTimes;Z=\left\{\right|00>,|01>,|10>,|11>\}$

According to $|{\mathrm{\&chi;}}^{00}{>}_{ABCD}=\frac{1}{2}\left(\right|{\mathrm{\&phi;}}^{+}{>}_{AB}|00{>}_{CD}-|{\mathrm{\&psi;}}^{-}{>}_{AB}|01{>}_{CD}+|{\mathrm{\&psi;}}^{+}{>}_{AB}|10{>}_{CD}+|{\mathrm{\&phi;}}^{-}{>}_{AB}\left|11{>}_{CD}\right),$ Known, if to χ state | χ ⁰⁰> _aBCDparticle A and B perform Bell measure, particle C and D is performed base measure, then system with 1/4 probability be collapsed to state | φ ⁺> _aB| 00> _cD, | ψ ^-> _aB| 01> _cD, | ψ ⁺> _aB| 10> _cDwith | φ ^-> _aB| 11> _cD.Therefore, Alice and Bob can obtain the shared key of identical 2n bit.

4 fail safes and efficiency analysis

The QKA agreement of a safety can not only resist external attack, and can resist participant's attack.

4.1 participants attack

Below, explanation dishonest participant can not be obtained alone this shared key by us.Without loss of generality, suppose that Alice is a dishonest participant, she wants to allow the 2l bit in shared key be 0 entirely, and she needs to measure sequence S with Bell base _aand S _bin corresponding l to particle.But according to the characteristic of Quantum Entangled States, the measurement result of every a pair particle is all randomly | φ ⁺> _aB, | ψ ^-> _aB, | ψ ⁺> _aBwith | φ ^-> _aB, namely Alice with 25% probability obtain 00,01,10 or 11.Therefore, every 2 in 2l bit is 00,01,10 or 11 randomly.So Alice cannot any one bit in Independent Decisiveness shared key.So this agreement can be resisted participant and be attacked.

4.2 external attack

Suppose that Eve is a listener-in wanting to steal shared key, the possible method that she attacks has: Trojan horse attack, measurement-multi-sending attack, intercepting and capturing-multi-sending attack and tangle-measure attack.

Trojan horse attack: in this agreement, because each photon in quantum channel is only transmitted once, therefore Eve successfully can not perform invisible photon eavesdropping (IPE) Trojan attack and postpone photon Trojan attack.

Measurement-multi-sending attack: Eve can to sequence S' _cand S' _din particle perform measurement-multi-sending attack.But the measurement of Eve will affect sequence S' _cand S' _dthe state of middle trick particle.In the eavesdropping of second step detects, Alice and Bob can with 1-(3/4) ^mthe probability of (m represents the quantity for detecting the trick particle that this is attacked) finds the attack of Eve.

Intercepting and capturing-multi-sending attack: if Eve performs intercepting and capturing-multi-sending attack, first she intercept and capture sequence S' _cand S' _d, then send her forgery sequence to Bob.After agreement terminates, she is again to sequence S' _cand S' _din particle perform corresponding measurement.But Eve does not also know position and the initial state of inveigling particle, and the sequence that therefore she forges can not be monitored by the eavesdropping of second step.When inveigling particle to be used to monitor this eavesdropping attack for m, corresponding eavesdropping verification and measurement ratio is 1-(1/2) ^m.Therefore, the intercepting and capturing-multi-sending attack of Eve also have failed.

Tangle-measure attack: Eve also can go to tangle this transmission particle (sequence S' with oneself pre-prepd auxiliary particle _cand S' _din particle), then transmission particle is issued Bob again.After agreement terminates, she measures corresponding auxiliary particle again, thus extracts the useful information about shared key.But Eve is before eavesdropping detects and do not know to inveigle the position of photon, she tangles operation U and certainly also can be performed on trick photon.Further, state is inveigled to become following two particle Entangled State:

U(|0>|E>)＝a|0>|e ₀₀>+b|1>|e ₀₁>，

U(|1>|E>)＝c|0>|e ₁₀>+d|1>|e ₁₁>，

$\begin{array}{c}U\left(\right|+>|E>)=\frac{1}{2}\&lsqb;|+>\left(a\right|e_{00}>+b|e_{01}>+c|e_{10}>+d|e_{11}>)\\ +|->\left(a\right|e_{00}>-b|e_{01}>+c|e_{10}>-d|e_{11}>)\&rsqb;\end{array}$

$\begin{array}{c}U\left(\right|->|E>)=\frac{1}{2}\&lsqb;|+>\left(a\right|e_{00}>+b|e_{01}>-c|e_{10}>-d|e_{11}>)\\ +|->\left(a\right|e_{00}>-b|e_{01}>-c|e_{10}>+d|e_{11}>)\&rsqb;\end{array}$

Wherein | e ₀₀>, | e ₀₁>, | e ₁₀> and | e ₁₁> is by the well-determined pure state of unitary transformation U, and | a| ²+ | b| ²=1, | c| ²+ | d| ²=1.Obviously, CNOT conversion is the special circumstances of unitary transformation U.If Eve wants to be detected by eavesdropping at second step, the unitary transformation U of Eve must satisfy condition b=c=0 and a|e ₀₀>=d|e ₁₁>.Work as equation during establishment, Eve can not distinguish auxiliary photon a|e ₀₀> and d|e ₁₁>, thus Eve does not obtain the useful information of shared key by observing auxiliary photon.But, if a|e ₀₀> ≠ d|e ₁₁the attack of >, Eve will disturb trick state |+> and |->.Therefore, the attack of Eve will be found by Alice and Bob.The eavesdropping verification and measurement ratio of each trick photon is:

$\frac{1}{4}|b{|}^2+\frac{1}{4}|c{|}^2+\frac{1}{4}\&times;\frac{1}{4}\&times;\left(\right|a{|}^2+|b{|}^2+|c{|}^2+\left|d{|}^2\right)\&times;2=\frac{1}{4}\left(\right|b{|}^2+\left|c{|}^2\right)+\frac{1}{4}\&GreaterEqual;\frac{1}{4}.$

4.3 quantum noise channels

In quantum noise channel, the span of quantum bit error rate (QBER) τ introduced by noise is similar at 2%-8.9%, and it depends on the situation of channel as factors such as distances.If the quantum bit error rate that the attack of Eve is introduced is less than τ, so she just can hide her attack with noise.According to above-mentioned safety analysis, in this agreement, the eavesdropping verification and measurement ratio of each trick photon is at least 25%, and it is much larger than τ.Therefore, suitable eavesdropping detectability gate value is selected can to guarantee that this agreement is also safe on quantum noise channel.

According to above-mentioned analysis, this agreement is all safe on noiseless quantum channel and quantum noise channel.

5 efficiency analysiss

For a QKA agreement, Cabello quantum bit efficiency is defined as: wherein c represents the quantity of the classical bit of negotiation, the quantity of the quantum bit used in q presentation protocol.Therefore, our the quantum bit efficiency of agreement is: wherein χ state in n presentation protocol | χ ⁰⁰> _aBCDquantity, m represents and inveigles the quantity of particle.Make m=n, Wo Menyou $\mathrm{\&eta;}=\frac{1}{3}=33.3\%.$

Claims (4)

1., based on two side's quantum key agreement protocols of four particle χ states, it is characterized in that, specifically comprise the steps:

Step 1:Alice and Bob consults the coding of following quantum state;

Step 2:Alice prepares n χ state | χ ⁰⁰> _aBCD, and by this n χ state | χ ⁰⁰> _aBCDall particles be divided into four orderly sequences:

S _A＝{A ₁,A ₂,…,A _n}，S _B＝{B ₁,B ₂,…,B _n}，

S _c={ C ₁, C ₂..., C _nand S _d={ D ₁, D ₂..., D _n}

Wherein sequence S _a, S _b, S _c, S _drespectively by each χ state | χ ⁰⁰> _aBCDparticle A, B, C, D form; Alice from set | 0>, | 1>, |+>, | select 2m in->} at random and inveigle photon, and by this 2m trick photon radom insertion sequence S _cand S _d, and ensure that inserting m in each sequence inveigles photon, obtains new sequence S' respectively _cand S' _d; Alice oneself reservation queue S _aand S _b, by sequence S' _cand S' _dsend to Bob; N, m are the positive integer being greater than 1;

Step 3: when Bob receives sequence S' _cand S' _dafter, inform Alice by classical authenticated channel; Alice announces and inveigles photon at sequence S' _cand S' _din position and corresponding measurement base | 0>, | 1>} or |+>, |->}; Bob goes to measure with correct measurement base and inveigles photon accordingly, and measurement result is informed Alice by classical authenticated channel; Alice compares and measures result and inveigles the initial condition of photon, mistake in computation rate; If error rate is lower than prespecified limit gate value, then perform step 4, otherwise, perform step 2;

Step 4:Alice is to sequence S _aand S _bevery two particles that middle sequence number is identical perform Bell and measure, and Bob is to sequence S _cand S _devery two particles that middle sequence number is identical perform base is measured; According to the coding of the quantum state that Alice and Bob in the measurement result of Alice and Bob and step 1 consults, Alice and Bob obtains the shared key of identical 2n bit respectively.

$Z\&CircleTimes;Z=\left\{\right|00>,|01>,|10>,|11>\}$

2., as claimed in claim 1 based on two side's quantum key agreement protocols of four particle χ states, it is characterized in that, in described step 1, the coding of the quantum state that Alice and Bob consults: | φ ⁺> _aB→ 00, | ψ ^-> _aB→ 01, | ψ ⁺> _aB→ 10, | φ ^-> _aB→ 11, | 00> _cD→ 00, | 01> _cD→ 01, | 10> _cD→ 10, | 11> _cD→ 11.

3., as claimed in claim 1 based on two side's quantum key agreement protocols of four particle χ states, it is characterized in that, in described step 2,

$\begin{array}{c}|{\mathrm{\&chi;}}^{00}{>}_{ABCD}=\frac{1}{2\sqrt{2}}(|0000>+|0011>-|0101>+|0110>+|1001>+|1010>+|1100>\\ -|1111>{)}_{ABCD}\end{array}.$

4., as claimed in claim 1 based on two side's quantum key agreement protocols of four particle χ states, it is characterized in that, in described step 3, described limit gate value gets 0.1 ~ 0.2.