1 Introduction

Quantum informatics is formed by the intersection of multiple disciplines, and quantum entanglement is an important physical resource in quantum informatics. Using the entangled nature of two-body or multi-body quantum states can achieve many tasks that cannot be accomplished in classical information theory, such as quantum key distribution (QKD) [1], quantum teleportation (QT) [2], quantum superdense coding [3], and so on.

Quantum entanglement means that if we measure one of the physically separated particles belonging to the entangled state (no matter how far apart they are in theory), then it will immediately affect the state of the other particles [4]. The most commonly used Bell state is the maximum entangled state of two particles in a 2-dimensional Hilbert space. In 2019, Shi [4] conducted research on Bell states and derived a large number of Bell state entanglement exchange formulas, and gave the application of Bell states in quantum secret sharing. On this basis, Shi et al. [5] used the Bell state entanglement property to propose a verifiable quantum key exchange with authtication (VQKEA) in 2021.

In addition to Bell states, quantum entanglement also includes multi-body quantum states, such as GHZ states, GHZ-like states, Cluster states, and so on. This paper mainly discusses the three-particle entangled GHZ state. Given the measurement basis, the exchange of some particles in the GHZ state will not affect the original entanglement of the GHZ state, and the GHZ state after the simultaneous measurement exchange is the same as the original GHZ state measurement result. In 2020, Yang et al. [6] gave a maximum of n-particle entangled GHZ state \( \left|{G}_n\right\rangle =\frac{1}{\sqrt{2}}\left(\left|{q}_1,{q}_2,\dots, {q}_n\right\rangle \pm \left|{\overline{q}}_1,{\overline{q}}_2,\dots, {\overline{q}}_n\right\rangle \right) \). Then three-particle entangled GHZ state can be expressed as \( \left|{G}_3\right\rangle =\frac{1}{\sqrt{2}}\left(\left|{q}_1,{q}_2,{q}_3\right\rangle \pm \left|{\overline{q}}_1,{\overline{q}}_2,{\overline{q}}_3\right\rangle \right) \). In 2015, Shima Hassanpour and Monireh Houshmand [7] proposed that the three-particle GHZ-like state can be generated by a single qubit state, an EPR state and a CNOT operation. From this, we derived a three-particle GHZ state composed of a single qubit state and an EPR state. Furthermore, according to the truth table of the three-particle entangled GHZ-like state given by Zheng et al. [8], we obtain the truth table of the three-particle entangled GHZ state.

At present, many researchers use GHZ states as quantum resources and have achieved many research results through entanglement swapping. However, there is no systematic analysis of the relationship between states and measurements [4]. On this basis, we systematically analyzed the properties of the three-particle GHZ state before and after the entanglement exchange, including states and measurements, perfectly established six interesting and flexible equations, and formally proved their correctness. Combined with decoy photons, the practical application of the equation in quantum key agreement (QKA) is given. Key agreement is the basic prototype of many cryptographic protocols in classic ciphers [9]. Both Ref. [10] and Ref. [11] use GHZ entangled states as entanglement resources to achieve QKA.

The rest of the paper is organized as follows: Some equations are given in Section 2. In Section 3, we describe the specific application environment of the equation. This paper is finally concluded in Section 4.

2 Equations

There are eight GHZ entangled state bases in two different basis as follows:

$$ {\displaystyle \begin{array}{l}{\left|{\Psi}_{000}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|000\right\rangle +\left|111\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|+++\right\rangle +\left|--+\right\rangle +\left|+--\right\rangle +\left|-+-\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{001}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|000\right\rangle -\left|111\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|++-\right\rangle +\left|---\right\rangle +\left|+-+\right\rangle +\left|-++\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{010}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|100\right\rangle +\left|011\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|+++\right\rangle -\left|--+\right\rangle +\left|+--\right\rangle -\left|-+-\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{011}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|100\right\rangle -\left|011\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|++-\right\rangle -\left|---\right\rangle +\left|+-+\right\rangle -\left|-++\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{100}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|010\right\rangle +\left|101\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|+++\right\rangle -\left|--+\right\rangle -\left|+--\right\rangle +\left|-+-\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{101}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|010\right\rangle -\left|101\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|++-\right\rangle -\left|---\right\rangle -\left|+-+\right\rangle +\left|-++\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{110}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|110\right\rangle +\left|001\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|+++\right\rangle +\left|--+\right\rangle -\left|+--\right\rangle -\left|-+-\right\rangle \right)}_{123},\end{array}} $$
$$ {\left|{\Psi}_{111}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|110\right\rangle -\left|001\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|++-\right\rangle +\left|---\right\rangle -\left|+-+\right\rangle -\left|-++\right\rangle \right)}_{123}. $$

The main calculation formulas can be expressed as:

$$ \left|0\right\rangle =\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\kern0.5em \left|1\right\rangle =\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right) $$
(1)

Then the specific derivation process of the entangled state base of |Ψ000123 is as follows:

$$ {\displaystyle \begin{array}{l}\begin{array}{l}{\left|{\Psi}_{000}\right\rangle}_{123}=\frac{\sqrt{2}}{2}{\left(\left|000\right\rangle +\left|111\right\rangle \right)}_{123}\\ {}=\frac{\sqrt{2}}{2}{\left(\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)+\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\right)}_{123}\\ {}=\frac{1}{4}{\left(\left(\left|+\right\rangle +\left|-\right\rangle \right)\left(\left|+\right\rangle +\left|-\right\rangle \right)\left(\left|+\right\rangle +\left|-\right\rangle \right)+\left(\left|+\right\rangle -\left|-\right\rangle \right)\left(\left|+\right\rangle -\left|-\right\rangle \right)\left(\left|+\right\rangle -\left|-\right\rangle \right)\right)}_{123}\\ {}=\frac{1}{4}{\left(\left(\left|++\right\rangle +\left|+-\right\rangle +\left|-+\right\rangle +\left|--\right\rangle \right)\left(\left|+\right\rangle +\left|-\right\rangle \right)+\left(\left|++\right\rangle -\left|+-\right\rangle -\left|-+\right\rangle +\left|--\right\rangle \right)\left(\left|+\right\rangle -\left|-\right\rangle \right)\right)}_{123}\\ {}=\frac{1}{4}\Big(\left(\left|+++\right\rangle +\left|++-\right\rangle +\left|+-+\right\rangle +\left|+--\right\rangle +\left|-++\right\rangle +\left|-+-\right\rangle +\left|--+\right\rangle +\left|---\right\rangle \right)+\\ {}\begin{array}{l}\left(\left|+++\right\rangle -\left|++-\right\rangle -\left|+-+\right\rangle +\left|+--\right\rangle -\left|-++\right\rangle +\left|-+-\right\rangle +\left|--+\right\rangle -\left|---\right\rangle \right)\Big){}_{123}\\ {}=\frac{1}{2}{\left(\left|+++\right\rangle +\left|+--\right\rangle +\left|-+-\right\rangle +\left|--+\right\rangle \right)}_{123}\end{array}\end{array}\\ {}\kern3em \end{array}} $$
(2)

“0” represents the measurement result {|0〉, |+〉}, “1” represents the measurement result {|1〉, |−〉}. Taking |Ψ000123 as an example, the first two terms are |+ + +〉 and |+ − −〉 in X-basis, both the first particle is |+〉, the last two particles |++〉 and |−−〉 are the same; the last two terms are |− − +〉 and |− + −〉 in X-basis, both the first particle is |−〉, the last two particles |−+〉 and |+−〉 are opposite. In Z-basis, according to |+ + +〉 and |− − −〉, the three particle measurement results are identical. Table 1 is the truth table of |Ψ000123 measured on different basis.

Table 1 Measurement results

Three-particle GHZ state can also be represented by the EPR state and the single qubit state, and the EPR state needs to be deformed appropriately. The specific deformation process is as follows:

$$ {\displaystyle \begin{array}{l}{\left|\phi \right.}_{00}\left\rangle =\frac{\sqrt{2}}{2}\left(\left|00\right\rangle +\left|11\right\rangle \right)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)+\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\right)=\frac{\sqrt{2}}{2}\left(\left|++\right\rangle +\left|--\right\rangle \right)={\left|\phi \right.}_{++}\right\rangle, \\ {}\left|{\phi}_{01}\right.\left\rangle =\frac{\sqrt{2}}{2}\left(\left|00\right\rangle -\left|11\right\rangle \right)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)-\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\right)=\frac{\sqrt{2}}{2}\left(\left|++\right\rangle -\left|--\right\rangle \right)=\left|{\phi}_{+-}\right.\right\rangle, \\ {}{\left|\phi \right.}_{10}\left\rangle =\frac{\sqrt{2}}{2}\left(\left|01\right\rangle +\left|10\right\rangle \right)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)+\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\right)=\frac{\sqrt{2}}{2}\left(\left|+-\right\rangle +\left|-+\right\rangle \right)={\left|\phi \right.}_{-+}\right\rangle, \\ {}\left|{\phi}_{11}\right.\left\rangle =\frac{\sqrt{2}}{2}\left(\left|01\right\rangle -\left|10\right\rangle \right)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)-\frac{\sqrt{2}}{2}\left(\left|+\right\rangle -\left|-\right\rangle \right)\frac{\sqrt{2}}{2}\left(\left|+\right\rangle +\left|-\right\rangle \right)\right)=\frac{\sqrt{2}}{2}\left(\left|+-\right\rangle -\left|-+\right\rangle \right)=\left|{\phi}_{--}\right.\right\rangle, \end{array}} $$
(3)

where {|ϕ00〉, |ϕ01〉, |ϕ10〉, |ϕ11〉} indicates Bell states. In Eq. (3), in order to facilitate subsequent calculations, we specify here that {|ϕ00〉, |ϕ01〉, |ϕ10〉, |ϕ11〉} can correspond to {|ϕ++〉, |ϕ+−〉, |ϕ−+〉, |ϕ−−〉} respectively. Any Bell state |ϕij〉 can be rewritten as Eq. (4):

$$ {\left|\phi \right.}_{ij}\Big\rangle =\frac{\sqrt{2}}{2}\left(\left|+\right\rangle \left|i\right\rangle +{\left(-1\right)}^j\left|-\right\rangle \left|\overline{i}\right\rangle \right)=\frac{\sqrt{2}}{2}\left(\left|0i\right\rangle +{\left(-1\right)}^j\left|1\overline{i}\right\rangle \right) $$
(4)

Then eight GHZ entangled states can be expressed as below:

$$ {\displaystyle \begin{array}{l}{\left|{\Psi}_{000}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{++}\right\rangle \left|+\right\rangle +\left|{\phi}_{-+}\right\rangle \left|-\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{00}0\right\rangle +\left|{\phi}_{10}1\right\rangle \right)}_{123},{\left|{\Psi}_{001}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{++}\right\rangle \left|-\right\rangle +\left|{\phi}_{-+}\right\rangle \left|+\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{00}1\right\rangle +\left|{\phi}_{10}0\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{010}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{+-}\right\rangle \left|+\right\rangle +\left|{\phi}_{--}\right\rangle \left|-\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{01}0\right\rangle +\left|{\phi}_{11}1\right\rangle \right)}_{123},{\left|{\Psi}_{011}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{+-}\right\rangle \left|-\right\rangle +\left|{\phi}_{--}\right\rangle \left|+\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{01}1\right\rangle +\left|{\phi}_{11}0\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{100}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{+-}\right\rangle \left|+\right\rangle -\left|{\phi}_{--}\right\rangle \left|-\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{01}0\right\rangle -\left|{\phi}_{11}1\right\rangle \right)}_{123},{\left|{\Psi}_{101}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{+-}\right\rangle \left|-\right\rangle -\left|{\phi}_{--}\right\rangle \left|+\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{01}1\right\rangle -\left|{\phi}_{11}0\right\rangle \right)}_{123},\\ {}{\left|{\Psi}_{110}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{++}\right\rangle \left|+\right\rangle -\left|{\phi}_{-+}\right\rangle \left|-\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{00}0\right\rangle -\left|{\phi}_{10}1\right\rangle \right)}_{123},{\left|{\Psi}_{111}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{++}\right\rangle \left|-\right\rangle -\left|{\phi}_{-+}\right\rangle \left|+\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{00}1\right\rangle -\left|{\phi}_{10}0\right\rangle \right)}_{123},\end{array}} $$
(5)

According to Eq. (4) and Eq. (5), any GHZ states |Ψijk〉 can be expressed as the following general formulas:

$$ {\displaystyle \begin{array}{l}i=0,{\left|{\Psi}_{ij k}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{ij}k\right\rangle +\left|{\phi}_{\overline{i}j}\overline{k}\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\\ {}i=1,{\left|{\Psi}_{ij k}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{\overline{i}\overline{j}}k\right\rangle +{\left(-1\right)}^i\left|{\phi}_{i\overline{j}}\overline{k}\right\rangle \right)}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{j}}k\right\rangle -\left|{\phi}_{1\overline{j}}\overline{k}\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^{\overline{j}}\left|11k\right\rangle -\left|01\overline{k}\right\rangle -{\left(-1\right)}^{\overline{j}}\left|10\overline{k}\right\rangle \right)}_{123}\end{array}} $$
(6)

where i, j, k ∈ {0, 1} and \( \overline{j}=1\oplus j \), \( \overline{k}=1\oplus k \) (i.e., a bar over a bit value indicates its logical negation.). In this paper, ⊕denotes bitwise XOR, the classical bit string ijk is regarded as the characteristic of the GHZ state |Ψijk〉.

2.1 Theorem 1

For any two GHZ states |Ψijk123 and |Ψrst456 exchange particles {3, 6} (see in Fig.1), under the same measurement basis (X-basis or Z-basis), the two GHZ states |Ψijk123 and |Ψrst456 still satisfy the following equations:

$$ {M}_{126}\oplus {M}_{453}={B}_{123}\oplus {B}_{456} $$
(7)
$$ {B}_{123}\oplus {M}_{126}={B}_{456}\oplus {M}_{453} $$
(8)

where {B123, B456} represent the characteristics of the initial states, {M126, M453} represent the characteristics of the possible measurement results. Eq. (7) can be transformed into Eq. (8).

Fig. 1
figure 1

Entanglement swapping of two GHZ states

Remark: The ellipse dotted line denotes the entanglement between three particles. The cuboid dotted line denotes X-basis or Z-basis measurement. After entanglement swapping, the parity of the characteristics of GHZ states does not change, i.e., M126 ⊕ M453 = B123 ⊕ B456.

Proof. According to Eq. (6), any two GHZ states |Ψijk123 and |Ψrst456 are expressed as follows:

$$ {\displaystyle \begin{array}{l}i=0,{\left|{\Psi}_{ijk}\right\rangle}_{123}={\left|{\Psi}_{0 jk}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\\ {}r=1,{\left|{\Psi}_{rst}\right\rangle}_{456}={\left|{\Psi}_{1 st}\right\rangle}_{456}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{s}}t\right\rangle -\left|{\phi}_{1\overline{s}}\overline{t}\right\rangle \right)}_{456}=\frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle -\left|01\overline{t}\right\rangle -{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \right)}_{456}\end{array}} $$

Under the same measurement, we will get:

$$ {\displaystyle \begin{array}{l}{\left|{\Psi}_{ijk}\right\rangle}_{123}\otimes {\left|{\Psi}_{rst}\right\rangle}_{456}={\left|{\Psi}_{0 jk}\right\rangle}_{123}\otimes {\left|{\Psi}_{1 st}\right\rangle}_{456}\\ {}\kern7.25em =\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{s}}t\right\rangle -\left|{\phi}_{1\overline{s}}\overline{t}\right\rangle \right)}_{456}\\ {}\kern7.25em =\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle -\left|01\overline{t}\right\rangle -{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \right)}_{456}\\ {}\kern7.25em =\frac{1}{4}\Big({\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{\overline{s}}{\left|00k\right\rangle}_{123}{\left|11t\right\rangle}_{456}-{\left|00k\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}-{\left(-1\right)}^{\overline{s}}{\left|00k\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern8.75em {\left(-1\right)}^j{\left|11k\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{j\oplus \overline{s}}{\left|11k\right\rangle}_{123}{\left|11t\right\rangle}_{456}-{\left(-1\right)}^j{\left|11k\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|11k\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern8.75em {\left|01\overline{k}\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{\overline{s}}{\left|01\overline{k}\right\rangle}_{123}{\left|11t\right\rangle}_{456}-{\left|01\overline{k}\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}-{\left(-1\right)}^{\overline{s}}{\left|01\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern8.75em {\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{123}{\left|11t\right\rangle}_{456}-{\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}\Big)\\ {}=\frac{1}{4}\Big({\left|00t\right\rangle}_{126}{\left|00k\right\rangle}_{453}+{\left(-1\right)}^{\overline{s}}{\left|00t\right\rangle}_{126}{\left|11k\right\rangle}_{453}-{\left|00t\right\rangle}_{126}{\left|01\overline{k}\right\rangle}_{453}-{\left(-1\right)}^{\overline{s}}{\left|00t\right\rangle}_{126}{\left|10\overline{k}\right\rangle}_{453}+\end{array}} $$
(Step 1)
$$ {\displaystyle \begin{array}{l}\kern1.25em {\left(-1\right)}^j{\left|11t\right\rangle}_{123}{\left|00k\right\rangle}_{456}+{\left(-1\right)}^{j\oplus \overline{s}}{\left|11t\right\rangle}_{123}{\left|11k\right\rangle}_{456}-{\left(-1\right)}^j{\left|11t\right\rangle}_{123}{\left|01\overline{k}\right\rangle}_{456}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|11t\right\rangle}_{123}{\left|10\overline{k}\right\rangle}_{456}+\\ {}\kern1.25em {\left|01\overline{t}\right\rangle}_{123}{\left|00k\right\rangle}_{456}+{\left(-1\right)}^{\overline{s}}{\left|01\overline{t}\right\rangle}_{123}{\left|11k\right\rangle}_{456}-{\left|01\overline{t}\right\rangle}_{123}{\left|01\overline{k}\right\rangle}_{456}-{\left(-1\right)}^{\overline{s}}{\left|01\overline{t}\right\rangle}_{123}{\left|10\overline{k}\right\rangle}_{456}+\\ {}\kern1.25em {\left(-1\right)}^j{\left|10\overline{t}\right\rangle}_{123}{\left|00k\right\rangle}_{456}+{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{t}\right\rangle}_{123}{\left|11k\right\rangle}_{456}-{\left(-1\right)}^j{\left|10\overline{t}\right\rangle}_{123}{\left|01\overline{k}\right\rangle}_{456}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{t}\right\rangle}_{123}{\left|10\overline{k}\right\rangle}_{456}\Big)\\ {}=\frac{1}{2}{\left(\left|00t\Big\rangle +{\left(-1\right)}^j\left|11t\Big\rangle \right.\right.+\left|01\overline{t}\right\rangle +{\left(-1\right)}^j\left|10\overline{t}\right\rangle \right)}_{126}\otimes \frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^{\overline{s}}\left|11k\right\rangle -\left|01\overline{k}\right\rangle -{\left(-1\right)}^{\overline{s}}\left|10\overline{k}\right\rangle \right)}_{453}\\ {}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}t\right\rangle +\left|{\phi}_{1j}\overline{t}\right\rangle \right)}_{126}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{s}}k\right\rangle -\left|{\phi}_{1\overline{s}}\overline{k}\right\rangle \right)}_{453}\\ {}={\left|{\Psi}_{0 jt}\right\rangle}_{126}\otimes {\left|{\Psi}_{1 sk}\right\rangle}_{453}\end{array}} $$
(9)

In Step1, two GHZ states |Ψijk123 and |Ψrst456 exchange particles 3 and 6. It is proved that the calculation result of |Ψijk123 ⊗ |Ψrst456 is equivalent to |Ψijt126 ⊗ |Ψrsk453. In Eq. (9), the subscripts of the initial two GHZ states |Ψijk123 and |Ψrst456 are {0jk, 1st}, and the subscripts after exchanging one particle are {0jt, 1sk} (see in Fig.2).

Fig. 2
figure 2

Relationship flow chart

2.2 Theorem 2

For any two GHZ states |Ψijk123 and |Ψrst456 exchange particles {2, 3} and {5, 6} (see in Fig.3), under the same measurement basis (X-basis or Z-basis), the two GHZ states |Ψijk123 and |Ψrst456 always satisfy the following equations:

$$ {M}_{156}\oplus {M}_{423}={B}_{123}\oplus {B}_{456} $$
(10)
$$ {B}_{123}\oplus {M}_{156}={B}_{456}\oplus {M}_{423} $$
(11)

where {B123, B456} represent the characteristics of the initial states, {M156, M423} represent the characteristics of the possible measurement results. Furthermore, Eq. (10) can be transformed into Eq. (11).

Fig. 3
figure 3

Entanglement swapping of two GHZ states

Proof. Two GHZ states |Ψijk123 and |Ψrst456 are expressed as follows:

$$ {\displaystyle \begin{array}{l}i=0,{\left|{\Psi}_{ijk}\right\rangle}_{123}={\left|{\Psi}_{0 jk}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\\ {}r=0,{\left|{\Psi}_{rst}\right\rangle}_{456}={\left|{\Psi}_{0 st}\right\rangle}_{456}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0s}t\right\rangle +\left|{\phi}_{1s}\overline{t}\right\rangle \right)}_{456}=\frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^s\left|11t\right\rangle +\left|01\overline{t}\right\rangle +{\left(-1\right)}^s\left|10\overline{t}\right\rangle \right)}_{456}\end{array}} $$

Under the same measurement, we will get:

$$ {\displaystyle \begin{array}{l}{\left|{\Psi}_{ijk}\right\rangle}_{123}\otimes {\left|{\Psi}_{rst}\right\rangle}_{456}={\left|{\Psi}_{0 jk}\right\rangle}_{123}\otimes {\left|{\Psi}_{0 st}\right\rangle}_{456}\\ {}\kern7.5em =\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0s}t\right\rangle +\left|{\phi}_{1s}\overline{t}\right\rangle \right)}_{456}\\ {}\kern7.5em =\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^s\left|11t\right\rangle +\left|01\overline{t}\right\rangle +{\left(-1\right)}^s\left|10\overline{t}\right\rangle \right)}_{456}\\ {}\kern7.5em =\frac{1}{4}\Big({\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^s{\left|00k\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left|00k\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^s{\left|00k\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern9.5em {\left(-1\right)}^j{\left|11k\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|11k\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left(-1\right)}^j{\left|11k\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|11k\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern9.5em {\left|01\overline{k}\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^s{\left|01\overline{k}\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left|01\overline{k}\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^s{\left|01\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\kern1.5em \\ {}{\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{k}\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}\Big)\end{array}} $$
(12)
$$ {\displaystyle \begin{array}{l}{\left|{\Psi}_{ist}\right\rangle}_{156}\otimes {\left|{\Psi}_{rjk}\right\rangle}_{423}={\left|{\Psi}_{0 st}\right\rangle}_{156}\otimes {\left|{\Psi}_{0 jk}\right\rangle}_{423}\\ {}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0s}t\right\rangle +\left|{\phi}_{1s}\overline{t}\right\rangle \right)}_{156}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{423}\\ {}=\frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^s\left|11t\right\rangle +\left|01\overline{t}\right\rangle +{\left(-1\right)}^s\left|10\overline{t}\right\rangle \right)}_{156}\otimes \frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{423}\\ {}=\frac{1}{4}\Big({\left|00t\right\rangle}_{156}{\left|00k\right\rangle}_{423}+{\left(-1\right)}^j{\left|00t\right\rangle}_{156}{\left|11k\right\rangle}_{423}+{\left|00t\right\rangle}_{156}{\left|01\overline{k}\right\rangle}_{423}+{\left(-1\right)}^j{\left|00t\right\rangle}_{156}{\left|10\overline{k}\right\rangle}_{423}+\\ {}\kern1.5em {\left(-1\right)}^s{\left|11t\right\rangle}_{156}{\left|00k\right\rangle}_{423}+{\left(-1\right)}^{j\oplus s}{\left|11t\right\rangle}_{156}{\left|11k\right\rangle}_{423}+{\left(-1\right)}^s{\left|11t\right\rangle}_{156}{\left|01\overline{k}\right\rangle}_{423}+{\left(-1\right)}^{j\oplus s}{\left|11t\right\rangle}_{156}{\left|10\overline{k}\right\rangle}_{423}+\\ {}\kern1.5em {\left|01\overline{t}\right\rangle}_{156}{\left|00k\right\rangle}_{423}+{\left(-1\right)}^j{\left|01\overline{t}\right\rangle}_{156}{\left|11k\right\rangle}_{423}+{\left|01\overline{t}\right\rangle}_{156}{\left|01\overline{k}\right\rangle}_{423}+{\left(-1\right)}^j{\left|01\overline{t}\right\rangle}_{156}{\left|10\overline{k}\right\rangle}_{423}+\\ {}{\left(-1\right)}^s{\left|10\overline{t}\right\rangle}_{156}{\left|00k\right\rangle}_{423}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{t}\right\rangle}_{156}{\left|11k\right\rangle}_{423}+{\left(-1\right)}^s{\left|10\overline{t}\right\rangle}_{156}{\left|01\overline{k}\right\rangle}_{423}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{t}\right\rangle}_{156}{\left|10\overline{k}\right\rangle}_{423}\Big)\end{array}} $$
(13)
$$ {\displaystyle \begin{array}{l}=\frac{1}{4}\Big({\left|00k\right\rangle}_{423}{\left|00t\right\rangle}_{156}+{\left(-1\right)}^s{\left|00k\right\rangle}_{423}{\left|11t\right\rangle}_{156}+{\left|00k\right\rangle}_{423}{\left|01\overline{t}\right\rangle}_{156}+{\left(-1\right)}^s{\left|00k\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{156}+\\ {}\kern1em {\left(-1\right)}^j{\left|11k\right\rangle}_{423}{\left|00t\right\rangle}_{156}+{\left(-1\right)}^{j\oplus s}{\left|11k\right\rangle}_{423}{\left|11t\right\rangle}_{156}+{\left(-1\right)}^j{\left|11k\right\rangle}_{423}{\left|01\overline{t}\right\rangle}_{156}+{\left(-1\right)}^{j\oplus s}{\left|11k\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{156}+\\ {}\kern1em {\left|01\overline{k}\right\rangle}_{423}{\left|00t\right\rangle}_{156}+{\left(-1\right)}^s{\left|01\overline{k}\right\rangle}_{423}{\left|11t\right\rangle}_{156}+{\left|01\overline{k}\right\rangle}_{423}{\left|01\overline{t}\right\rangle}_{156}+{\left(-1\right)}^s{\left|01\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{156}+\\ {}\kern1.25em {\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{423}{\left|00t\right\rangle}_{156}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{k}\right\rangle}_{423}{\left|11t\right\rangle}_{156}+{\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{423}{\left|01\overline{t}\right\rangle}_{156}+{\left(-1\right)}^{j\oplus s}{\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{156}\Big)\\ {}=\frac{1}{4}\Big({\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^s{\left|00k\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left|00k\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^s{\left|00k\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern1.5em {\left(-1\right)}^j{\left|11k\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|11k\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left(-1\right)}^j{\left|11k\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|11k\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern1.5em {\left|01\overline{k}\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^s{\left|01\overline{k}\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left|01\overline{k}\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^s{\left|01\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}+\\ {}\kern1.5em {\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{123}{\left|00t\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{k}\right\rangle}_{123}{\left|11t\right\rangle}_{456}+{\left(-1\right)}^j{\left|10\overline{k}\right\rangle}_{123}{\left|01\overline{t}\right\rangle}_{456}+{\left(-1\right)}^{j\oplus s}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}\Big)\\ {}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^s\left|11t\right\rangle +\left|01\overline{t}\right\rangle +{\left(-1\right)}^s\left|10\overline{t}\right\rangle \right)}_{456}\\ {}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0s}t\right\rangle +\left|{\phi}_{1s}\overline{t}\right\rangle \right)}_{456}\\ {}={\left|{\Psi}_{0 jk}\right\rangle}_{123}\otimes {\left|{\Psi}_{0 st}\right\rangle}_{456}\end{array}} $$
(Step 2)

Eq. (12) shows the result of |Ψijk123 ⊗ |Ψrst456 and Eq. (13) shows the result of |Ψist156 ⊗ |Ψrjk423. In Step2, the positions of GHZ state |Ψist156 and GHZ state |Ψrjk423 are exchanged. Finally, we prove that the result of |Ψist156 ⊗ |Ψrjk423 is equal to |Ψijk123 ⊗ |Ψrst456. The subscripts of the initial two GHZ states |Ψijk123 and |Ψrst456 are {0jk, 0st}, and the subscripts after exchanging two particles are {0st, 0jk} (see in Fig.4). Obviously, the two GHZ states still satisfy Eq. (10) after exchanging two particles with each other.

Fig. 4
figure 4

Relationship flow chart

2.3 Theorem 3

For any three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 exchange particles {3, 6, 9} (see in Fig.5), under the same measurement basis (X-basis or Z-basis), the three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 still satisfy the following equations:

$$ {M}_{129}\oplus {M}_{453}\oplus {M}_{786}={B}_{123}\oplus {B}_{456}\oplus {B}_{789} $$
(14)
$$ {B}_{123}\oplus {M}_{129}=\left({B}_{456}\oplus {M}_{453}\right)\oplus \left({B}_{789}\oplus {M}_{786}\right) $$
(15)

where {B123, B456, B789} represent the characteristics of the initial states, {M129, M453, M786} represent the characteristics of the possible measurement results. Furthermore, Eq. (15) is a variation of Eq. (14).

Fig. 5
figure 5

Entanglement swapping of three GHZ states

Proof. Three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 are expressed as follows:

$$ {\displaystyle \begin{array}{l}i=0,{\left|{\Psi}_{ijk}\right\rangle}_{123}={\left|{\Psi}_{0 jk}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\\ {}r=1,{\left|{\Psi}_{rst}\right\rangle}_{456}={\left|{\Psi}_{1 st}\right\rangle}_{456}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{s}}t\right\rangle -\left|{\phi}_{1\overline{s}}\overline{t}\right\rangle \right)}_{456}=\frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle -\left|01\overline{t}\right\rangle -{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \right)}_{456}\\ {}u=1,{\left|{\Psi}_{uvw}\right\rangle}_{789}={\left|{\Psi}_{1 vw}\right\rangle}_{789}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{v}}w\right\rangle -\left|{\phi}_{1\overline{v}}\overline{w}\right\rangle \right)}_{789}=\frac{1}{2}{\left(\left|00w\right\rangle +{\left(-1\right)}^{\overline{v}}\left|11w\right\rangle -\left|01\overline{w}\right\rangle -{\left(-1\right)}^{\overline{v}}\left|10\overline{w}\right\rangle \right)}_{789}\end{array}} $$

Under the same measurement, we will get:

$$ {\displaystyle \begin{array}{l}{\left|{\Psi}_{ijk}\right\rangle}_{123}\otimes {\left|{\Psi}_{rst}\right\rangle}_{456}\otimes {\left|{\Psi}_{uvw}\right\rangle}_{789}={\left|{\Psi}_{0 jk}\right\rangle}_{123}\otimes {\left|{\Psi}_{1 st}\right\rangle}_{456}\otimes {\left|{\Psi}_{1 vw}\right\rangle}_{789}\\ {}\kern12em =\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{s}}t\right\rangle -\left|{\phi}_{1\overline{s}}\overline{t}\right\rangle \right)}_{456}\otimes \frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{v}}w\right\rangle -\left|{\phi}_{1\overline{v}}\overline{w}\right\rangle \right)}_{789}\\ {}\kern12em =\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\otimes \frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle -\left|01\overline{t}\right\rangle -{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \right)}_{456}\otimes \\ {}\kern13em \frac{1}{2}{\left(\left|00w\right\rangle +{\left(-1\right)}^{\overline{v}}\left|11w\right\rangle -\left|01\overline{w}\right\rangle -{\left(-1\right)}^{\overline{v}}\left|10\overline{w}\right\rangle \right)}_{789}\end{array}} $$

The above formula can be expressed in the following matrix form:

$$ {\left|{\Psi}_{ijk}\right\rangle}_{123}\otimes {\left|{\Psi}_{rst}\right\rangle}_{456}\otimes {\left|{\Psi}_{uvw}\right\rangle}_{789}=\frac{1}{2}{\left(\begin{array}{l}\left|00k\right\rangle \\ {}{\left(-1\right)}^j\left|11k\right\rangle \\ {}\left|01\overline{k}\right\rangle \\ {}{\left(-1\right)}^j\left|10\overline{k}\right\rangle \end{array}\right)}_{123}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00t\right\rangle \\ {}{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle \\ {}-\left|01\overline{t}\right\rangle \\ {}-{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \end{array}\right)}_{456}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00w\right\rangle \\ {}{\left(-1\right)}^{\overline{v}}\left|11w\right\rangle \\ {}-\left|01\overline{w}\right\rangle \\ {}-{\left(-1\right)}^{\overline{v}}\left|10\overline{w}\right\rangle \end{array}\right)}_{789}=\frac{1}{8}\left(\begin{array}{l}{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|00w\right\rangle}_{789}\\ {}{\left(-1\right)}^{\overline{v}}{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|11w\right\rangle}_{789}\\ {}-{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|01\overline{w}\right\rangle}_{789}\\ {}-{\left(-1\right)}^{\overline{v}}{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|10\overline{w}\right\rangle}_{789}\\ {}\dots \dots \\ {}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|00w\right\rangle}_{789}\\ {}-{\left(-1\right)}^{j\oplus \overline{s}\oplus \overline{v}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|11w\right\rangle}_{789}\\ {}{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|01\overline{w}\right\rangle}_{789}\\ {}{\left(-1\right)}^{j\oplus \overline{s}\oplus \overline{v}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|10\overline{w}\right\rangle}_{789}\end{array}\right) $$

Swapping particles 3, 6 and 9:

$$ \frac{1}{8}\left(\begin{array}{l}{\left|00w\right\rangle}_{129}{\left|00k\right\rangle}_{453}{\left|00t\right\rangle}_{786}\\ {}{\left(-1\right)}^{\overline{v}}{\left|00w\right\rangle}_{129}{\left|00k\right\rangle}_{453}{\left|11t\right\rangle}_{786}\\ {}-{\left|00w\right\rangle}_{129}{\left|00k\right\rangle}_{453}{\left|01\overline{t}\right\rangle}_{786}\\ {}-{\left(-1\right)}^{\overline{v}}{\left|00w\right\rangle}_{129}{\left|00k\right\rangle}_{453}{\left|10\overline{t}\right\rangle}_{786}\\ {}\dots \dots \\ {}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{w}\right\rangle}_{129}{\left|10\overline{k}\right\rangle}_{453}{\left|00t\right\rangle}_{786}\\ {}-{\left(-1\right)}^{j\oplus \overline{s}\oplus \overline{v}}{\left|10\overline{w}\right\rangle}_{129}{\left|10\overline{k}\right\rangle}_{453}{\left|11t\right\rangle}_{786}\\ {}{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{w}\right\rangle}_{129}{\left|10\overline{k}\right\rangle}_{453}{\left|01\overline{t}\right\rangle}_{786}\\ {}{\left(-1\right)}^{j\oplus \overline{s}\oplus \overline{v}}{\left|10\overline{w}\right\rangle}_{129}{\left|10\overline{k}\right\rangle}_{453}{\left|10\overline{t}\right\rangle}_{786}\end{array}\right)=\frac{1}{2}{\left(\begin{array}{l}\left|00w\right\rangle \\ {}{\left(-1\right)}^j\left|11w\right\rangle \\ {}\left|01\overline{w}\right\rangle \\ {}{\left(-1\right)}^j\left|10\overline{w}\right\rangle \end{array}\right)}_{129}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00k\right\rangle \\ {}{\left(-1\right)}^{\overline{s}}\left|11k\right\rangle \\ {}-\left|01\overline{k}\right\rangle \\ {}-{\left(-1\right)}^{\overline{s}}\left|10\overline{k}\right\rangle \end{array}\right)}_{453}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00t\right\rangle \\ {}{\left(-1\right)}^{\overline{v}}\left|11t\right\rangle \\ {}-\left|01\overline{t}\right\rangle \\ {}-{\left(-1\right)}^{\overline{v}}\left|10\overline{t}\right\rangle \end{array}\right)}_{786}={\left|{\Psi}_{ijw}\right\rangle}_{129}\otimes {\left|{\Psi}_{rsk}\right\rangle}_{453}\otimes {\left|{\Psi}_{uvt}\right\rangle}_{786} $$

Remark: In all subsequent calculations, all the matrices with the coefficient of \( \frac{1}{8} \) are 64-dimensional matrices.

It is proved that the calculation result of |Ψijk123 ⊗ |Ψrst456 ⊗ |Ψuvw789 is equivalent to |Ψijw129 ⊗ |Ψrsk453 ⊗ |Ψuvt786. The subscripts of the initial three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 are {0jk, 1st, 1vw}, and the subscripts after exchanging three particles are {0jw, 1sk, 1vt} (see in Fig.6).

Fig. 6
figure 6

Relationship flow chart

2.4 Theorem 4

For any three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 exchange particles {2, 3}, {5, 6} and {8, 9}(see in Fig.7), under the same measurement basis (X-basis or Z-basis), the three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 always satisfy the following equations:

$$ {M}_{189}\oplus {M}_{423}\oplus {M}_{756}={B}_{123}\oplus {B}_{456}\oplus {B}_{789} $$
(16)
$$ {B}_{123}\oplus {M}_{189}=\left({B}_{456}\oplus {M}_{423}\right)\oplus \left({B}_{789}\oplus {M}_{756}\right) $$
(17)

where {B123, B456, B789} represent the characteristics of the initial states, {M189, M423, M756} represent the characteristics of the possible measurement results. In addition, Eq. (16) can be transformed into Eq. (17).

Fig. 7
figure 7

Entanglement swapping of three GHZ states

Proof. Three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 are expressed as follows:

$$ {\displaystyle \begin{array}{l}i=0,{\left|{\Psi}_{ijk}\right\rangle}_{123}={\left|{\Psi}_{0 jk}\right\rangle}_{123}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0j}k\right\rangle +\left|{\phi}_{1j}\overline{k}\right\rangle \right)}_{123}=\frac{1}{2}{\left(\left|00k\right\rangle +{\left(-1\right)}^j\left|11k\right\rangle +\left|01\overline{k}\right\rangle +{\left(-1\right)}^j\left|10\overline{k}\right\rangle \right)}_{123}\\ {}r=1,{\left|{\Psi}_{rst}\right\rangle}_{456}={\left|{\Psi}_{1 st}\right\rangle}_{456}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{s}}t\right\rangle -\left|{\phi}_{1\overline{s}}\overline{t}\right\rangle \right)}_{456}=\frac{1}{2}{\left(\left|00t\right\rangle +{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle -\left|01\overline{t}\right\rangle -{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \right)}_{456}\\ {}u=1,{\left|{\Psi}_{uvw}\right\rangle}_{789}={\left|{\Psi}_{1 vw}\right\rangle}_{789}=\frac{1}{\sqrt{2}}{\left(\left|{\phi}_{0\overline{v}}w\right\rangle -\left|{\phi}_{1\overline{v}}\overline{w}\right\rangle \right)}_{789}=\frac{1}{2}{\left(\left|00w\right\rangle +{\left(-1\right)}^{\overline{v}}\left|11w\right\rangle -\left|01\overline{w}\right\rangle -{\left(-1\right)}^{\overline{v}}\left|10\overline{w}\right\rangle \right)}_{789}\end{array}} $$

Under the same measurement, we will get the following matrix form:

$$ {\left|{\Psi}_{ijk}\right\rangle}_{123}\otimes {\left|{\Psi}_{rst}\right\rangle}_{456}\otimes {\left|{\Psi}_{uvw}\right\rangle}_{789}=\frac{1}{2}{\left(\begin{array}{l}\left|00k\right\rangle \\ {}{\left(-1\right)}^j\left|11k\right\rangle \\ {}\left|01\overline{k}\right\rangle \\ {}{\left(-1\right)}^j\left|10\overline{k}\right\rangle \end{array}\right)}_{123}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00t\right\rangle \\ {}{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle \\ {}-\left|01\overline{t}\right\rangle \\ {}-{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \end{array}\right)}_{456}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00w\right\rangle \\ {}{\left(-1\right)}^{\overline{v}}\left|11w\right\rangle \\ {}-\left|01\overline{w}\right\rangle \\ {}-{\left(-1\right)}^{\overline{v}}\left|10\overline{w}\right\rangle \end{array}\right)}_{789}=\frac{1}{8}\left(\begin{array}{l}{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|00w\right\rangle}_{789}\\ {}{\left(-1\right)}^{\overline{v}}{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|11w\right\rangle}_{789}\\ {}-{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|01\overline{w}\right\rangle}_{789}\\ {}-{\left(-1\right)}^{\overline{v}}{\left|00k\right\rangle}_{123}{\left|00t\right\rangle}_{456}{\left|10\overline{w}\right\rangle}_{789}\\ {}\dots \dots \\ {}-{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|00w\right\rangle}_{789}\\ {}-{\left(-1\right)}^{j\oplus \overline{s}\oplus \overline{v}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|11w\right\rangle}_{789}\\ {}{\left(-1\right)}^{j\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|01\overline{w}\right\rangle}_{789}\\ {}{\left(-1\right)}^{j\oplus \overline{s}\oplus \overline{v}}{\left|10\overline{k}\right\rangle}_{123}{\left|10\overline{t}\right\rangle}_{456}{\left|10\overline{w}\right\rangle}_{789}\end{array}\right) $$
(18)
$$ {\left|{\Psi}_{ivw}\right\rangle}_{189}\otimes {\left|{\Psi}_{rjk}\right\rangle}_{423}\otimes {\left|{\Psi}_{ust}\right\rangle}_{756}=\frac{1}{2}{\left(\begin{array}{l}\left|00w\right\rangle \\ {}{\left(-1\right)}^v\left|11w\right\rangle \\ {}\left|01\overline{w}\right\rangle \\ {}{\left(-1\right)}^v\left|10\overline{w}\right\rangle \end{array}\right)}_{189}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00k\right\rangle \\ {}{\left(-1\right)}^{\overline{j}}\left|11k\right\rangle \\ {}-\left|01\overline{k}\right\rangle \\ {}-{\left(-1\right)}^{\overline{j}}\left|10\overline{k}\right\rangle \end{array}\right)}_{423}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00t\right\rangle \\ {}{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle \\ {}-\left|01\overline{t}\right\rangle \\ {}-{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \end{array}\right)}_{756}=\frac{1}{8}\left(\begin{array}{l}{\left|00w\right\rangle}_{189}{\left|00k\right\rangle}_{423}{\left|00t\right\rangle}_{756}\\ {}{\left(-1\right)}^{\overline{s}}{\left|00w\right\rangle}_{189}{\left|00k\right\rangle}_{423}{\left|11t\right\rangle}_{756}\\ {}-{\left|00w\right\rangle}_{189}{\left|00k\right\rangle}_{423}{\left|01\overline{t}\right\rangle}_{756}\\ {}-{\left(-1\right)}^{\overline{s}}{\left|00w\right\rangle}_{189}{\left|00k\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{756}\\ {}\dots \dots \\ {}-{\left(-1\right)}^{v\oplus \overline{j}}{\left|10\overline{w}\right\rangle}_{189}{\left|10\overline{k}\right\rangle}_{423}{\left|00t\right\rangle}_{756}\\ {}-{\left(-1\right)}^{v\oplus \overline{j}\oplus \overline{s}}{\left|10\overline{w}\right\rangle}_{189}{\left|10\overline{k}\right\rangle}_{423}{\left|11t\right\rangle}_{756}\\ {}{\left(-1\right)}^{v\oplus \overline{j}}{\left|10\overline{w}\right\rangle}_{189}{\left|10\overline{k}\right\rangle}_{423}{\left|01\overline{t}\right\rangle}_{756}\\ {}{\left(-1\right)}^{v\oplus \overline{j}\oplus \overline{s}}{\left|10\overline{w}\right\rangle}_{189}{\left|10\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{756}\end{array}\right) $$
(19)

Eq. (19) also can be expressed as follows:

$$ \frac{1}{8}\left(\begin{array}{l}{\left|00k\right\rangle}_{423}{\left|00t\right\rangle}_{756}{\left|00w\right\rangle}_{189}\\ {}{\left(-1\right)}^v{\left|00k\right\rangle}_{423}{\left|00t\right\rangle}_{756}{\left|11w\right\rangle}_{189}\\ {}{\left|00k\right\rangle}_{423}{\left|00t\right\rangle}_{756}{\left|01\overline{w}\right\rangle}_{189}\\ {}{\left(-1\right)}^v{\left|00k\right\rangle}_{423}{\left|00t\right\rangle}_{756}{\left|10\overline{w}\right\rangle}_{189}\\ {}\dots \dots \\ {}-{\left(-1\right)}^{\overline{j}\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{756}{\left|00w\right\rangle}_{189}\\ {}-{\left(-1\right)}^{\overline{j}\oplus \overline{s}\oplus v}{\left|10\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{756}{\left|11w\right\rangle}_{189}\\ {}-{\left(-1\right)}^{\overline{j}\oplus \overline{s}}{\left|10\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{756}{\left|01\overline{w}\right\rangle}_{189}\\ {}-{\left(-1\right)}^{\overline{j}\oplus \overline{s}\oplus v}{\left|10\overline{k}\right\rangle}_{423}{\left|10\overline{t}\right\rangle}_{756}{\left|10\overline{w}\right\rangle}_{189}\end{array}\right)=\frac{1}{2}{\left(\begin{array}{l}\left|00k\right\rangle \\ {}{\left(-1\right)}^{\overline{j}}\left|11k\right\rangle \\ {}-\left|01\overline{k}\right\rangle \\ {}-{\left(-1\right)}^{\overline{j}}\left|10\overline{k}\right\rangle \end{array}\right)}_{423}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00t\right\rangle \\ {}{\left(-1\right)}^{\overline{s}}\left|11t\right\rangle \\ {}-\left|01\overline{t}\right\rangle \\ {}-{\left(-1\right)}^{\overline{s}}\left|10\overline{t}\right\rangle \end{array}\right)}_{756}\otimes \frac{1}{2}{\left(\begin{array}{l}\left|00w\right\rangle \\ {}{\left(-1\right)}^v\left|11w\right\rangle \\ {}\left|01\overline{w}\right\rangle \\ {}{\left(-1\right)}^v\left|10\overline{w}\right\rangle \end{array}\right)}_{189}={\left|{\Psi}_{rjk}\right\rangle}_{423}\otimes {\left|{\Psi}_{ust}\right\rangle}_{756}\otimes {\left|{\Psi}_{ivw}\right\rangle}_{189} $$
(20)

Eq. (18) and Eq. (19) show the results of |Ψijk123 ⊗ |Ψrst456 ⊗ |Ψuvw789 and |Ψivw189 ⊗ |Ψrjk423 ⊗ |Ψust756 respectively. Eq. (20) is the result of |Ψrjk423 ⊗ |Ψust756 ⊗ |Ψivw189 exchanging the positions of the particles {(1, 8, 9), (4, 2, 3), (7, 5, 6)}. By comparison, we find that the 64-dimensional matrix of Eq. (18) is equivalent to the 64-dimensional matrix of Eq. (20), which means that the calculation results of Eq. (18) and Eq. (19) are the same. The subscripts of the initial three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789are {0jk, 1st, 1vw}, and the subscripts after exchanging three groups of particles are {0vw, 1jk, 1st} (see in Fig.8). Obviously, the three GHZ states still satisfy Eq. (16) after exchanging particles.

Fig. 8
figure 8

Relationship flow chart

2.5 Theorem 5

For m GHZ states {(1, 2, 3), (4, 5, 6), …, (3m − 2, 3m − 1, 3m)}, after entanglement swapping as Fig.9, the possible measurement results and the initial states always meet the following equations:

$$ {M}_{1,2,3m}\oplus {M}_{4,5,3}\oplus \dots \oplus {M}_{3m-2,3m-1,3m-3}={B}_{1,2,3}\oplus {B}_{4,5,6}\oplus \dots \oplus {B}_{3m-2,3m-1,3m} $$
(21)
$$ {B}_{1,2,3}\oplus {M}_{1,2,3m}=\left({B}_{4,5,6}\oplus {M}_{4,5,3}\right)\oplus \left({B}_{7,8,9}\oplus {M}_{7,8,6}\right)\oplus \dots \oplus \left({B}_{3m-2,3m-1,3m}\oplus {M}_{3m-2,3m-1,3m-3}\right) $$
(22)

where Bijk and Mrst denote the characteristics of the initial states of the particles i, j and k, and the possible measurement results of the particles r, s and t, respectively. Eq. (22) is a variation of Eq. (21).

Fig. 9
figure 9

Entanglement swapping of m GHZ states

Proof. When m = 2 and m = 3, Eq. (21) and Eq. (22) are obtained directly from Theorem 1 and Theorem 3. According to Theorem 1, the entanglement swapping of two GHZ states |Ψijk123 and |Ψrst456 satisfies the following formulas:

$$ {M}_{126}\oplus {M}_{453}={B}_{123}\oplus {B}_{456} $$
(7)
$$ {B}_{123}\oplus {M}_{126}={B}_{456}\oplus {M}_{453} $$
(8)

According to Theorem 3, the entanglement swapping of three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 satisfies the following formulas:

$$ {M}_{129}\oplus {M}_{453}\oplus {M}_{786}={B}_{123}\oplus {B}_{456}\oplus {B}_{789} $$
(14)
$$ {B}_{123}\oplus {M}_{129}=\left({B}_{456}\oplus {M}_{453}\right)\oplus \left({B}_{789}\oplus {M}_{786}\right) $$
(15)

When m = 4, the entanglement swapping of four GHZ states |Ψijk123, |Ψrst456, |Ψuvw789 and |Ψxyz10, 11, 12 is shown in Fig.10 and satisfies the following formulas:

$$ {M}_{1,2,12}\oplus {M}_{4,5,3}\oplus {M}_{7,8,6}\oplus {M}_{10,11,9}={B}_{1,2,3}\oplus {B}_{4,5,6}\oplus {B}_{7,8,9}\oplus {B}_{10,11,12} $$
(23)
$$ {B}_{1,2,3}\oplus {M}_{1,2,12}=\left({B}_{4,5,6}\oplus {M}_{4,5,3}\right)\oplus \left({B}_{7,8,9}\oplus {M}_{7,8,6}\right)\oplus \left({B}_{10,11,12}\oplus {M}_{10,11,9}\right) $$
(24)

According to Eq. (7), Eq. (14) and Eq. (23), it is not hard to generalize the equation to the case of arbitrary m GHZ states (i.e., M1, 2, 3m ⊕ M4, 5, 3 ⊕ … ⊕ M3m − 2, 3m − 1, 3m − 3 = B1, 2, 3 ⊕ B4, 5, 6 ⊕ … ⊕ B3m − 2, 3m − 1, 3m). Similarly, when Eq. (21) holds, Eq. (22) can be obtained from Eq. (8), Eq. (15) and Eq. (24).

Fig. 10
figure 10

Entanglement swapping of four GHZ states

2.6 Theorem 6

For m GHZ states {(1, 2, 3), (4, 5, 6), …, (3m − 2, 3m − 1, 3m)}, after entanglement swapping as Fig.11, the possible measurement results and the initial states always meet the following equation:

$$ {M}_{1,3\mathrm{m}-1,3m}\oplus {M}_{4,2,3}\oplus \dots \oplus {M}_{3m-2,3m-4,3m-3}={B}_{1,2,3}\oplus {B}_{4,5,6}\oplus \dots \oplus {B}_{3m-2,3m-1,3m} $$
(25)
$$ {B}_{1,2,3}\oplus {M}_{1,3\mathrm{m}-1,3m}=\left({B}_{4,5,6}\oplus {M}_{4,2,3}\right)\oplus \left({B}_{7,8,9}\oplus {M}_{7,5,6}\right)\oplus \dots \oplus \left({B}_{3m-2,3m-1,3m}\oplus {M}_{3m-2,3m-4,3m-3}\right) $$
(26)

where Bijk and Mrst denote the characteristics of the initial states of the particles i, j and k, and the possible measurement results of the particles r, s and t, respectively. Eq. (25) can be transformed into Eq. (26).

Fig. 11
figure 11

Entanglement swapping of m GHZ states

Proof. When m = 2 and m = 3, Eq. (25) and Eq. (26) are obtained directly from Theorem 2 and Theorem 4. According to Theorem 2, the entanglement swapping of two GHZ states |Ψijk123 and |Ψrst456 satisfies the following formulas:

$$ {M}_{156}\oplus {M}_{423}={B}_{123}\oplus {B}_{456} $$
(10)
$$ {B}_{123}\oplus {M}_{156}={B}_{456}\oplus {M}_{423} $$
(11)

According to Theorem 4, the entanglement swapping of three GHZ states |Ψijk123, |Ψrst456 and |Ψuvw789 satisfies the following formulas:

$$ {M}_{189}\oplus {M}_{423}\oplus {M}_{756}={B}_{123}\oplus {B}_{456}\oplus {B}_{789} $$
(16)
$$ {B}_{123}\oplus {M}_{189}=\left({B}_{456}\oplus {M}_{423}\right)\oplus \left({B}_{789}\oplus {M}_{756}\right) $$
(17)

When m = 4, the entanglement swapping of four GHZ states |Ψijk123, |Ψrst456, |Ψuvw789 and |Ψxyz10, 11, 12 is shown in Fig.12 and satisfies the following formulas:

$$ {M}_{1,11,12}\oplus {M}_{4,2,3}\oplus {M}_{7,5,6}\oplus {M}_{10,8,9}={B}_{1,2,3}\oplus {B}_{4,5,6}\oplus {B}_{7,8,9}\oplus {B}_{10,11,12} $$
(27)
$$ {B}_{1,2,3}\oplus {M}_{1,11,12}=\left({B}_{4,5,6}\oplus {M}_{4,2,3}\right)\oplus \left({B}_{7,8,9}\oplus {M}_{7,5,6}\right)\oplus \left({B}_{10,11,12}\oplus {M}_{10,8,9}\right) $$
(28)
Fig. 12
figure 12

Entanglement swapping of four GHZ states

According to Eq. (10), Eq. (16) and Eq. (27), it is not hard to generalize the equation to the case of arbitrary m GHZ states (i.e., M1, 3m − 1, 3m ⊕ M4, 2, 3 ⊕ … ⊕ M3m − 2, 3m − 4, 3m − 3 = B1, 2, 3 ⊕ B4, 5, 6 ⊕ … ⊕ B3m − 2, 3m − 1, 3m). At the same time, Eq. (26) can be obtained from Eq. (11), Eq. (17) and Eq. (28).

3 Application

QKA is a session key established by two-party or multi-party making the same contribution. Taking Theorem 1 (Eq. (8)) as an example, Alice and Bob use their GHZ states to obtain the same session key KAB. Assume that Alice and Bob have safely shared the initial key K through face-to-face (or other secure methods). K is a string of binary numbers with length N, K[i] ∈ {0, 1} (K[i] denotes the ith bit of K, i = 1, 2, 3, …, N), the value of K determines the measurement basis (see in Table 2).

Table 2 Measuring basis

StepA Alice and Bob prepare N groups of GHZ state randomly from eight three-particle GHZ entangled states, expressed as AN and BN, respectively.

$$ {\displaystyle \begin{array}{l}{A}_N:\left[{P}_1\left({a}_1\right),{P}_2\left({a}_2\right),{P}_3\left({a}_3\right);\dots; {P}_N\left({a}_1\right),{P}_N\left({a}_2\right),{P}_N\left({a}_3\right)\right]\\ {}{B}_N:\left[{P}_1\left({b}_1\right),{P}_2\left({b}_2\right),{P}_3\left({b}_3\right);\dots; {P}_N\left({b}_1\right),{P}_N\left({b}_2\right),{P}_N\left({b}_3\right)\right]\end{array}} $$

Both Alice and Bob know what kind of GHZ entangled states they prepare belong to. After GHZ state of N is prepared, Alice collects particles from each group of GHZ state to form three groups of particle sequences {Sa1, Sa2, Sa3}. Similarly, Bob gets three groups of particle sequences {Sb1, Sb2, Sb3}. In the same time, Alice and Bob each prepare m single-photon states |D〉 with certain amount of randomly scattered photons in the range of {|0〉, |1〉, |+〉, |−〉}. Alice and Bob randomly insert m single-photon states |D〉 as decoy photons into the particle sequences Sa3 and Sb3 respectively, and record the insertion position and initial state of the decoy photon |D〉. Particle sequences are expressed as follows:

$$ {\displaystyle \begin{array}{l}{S}_{a1}:\left[{P}_1\left({a}_1\right),{P}_2\left({a}_1\right),{P}_3\left({a}_1\right),\dots, {P}_N\left({a}_1\right)\right]\\ {}{S}_{a2}:\left[{P}_1\left({a}_2\right),{P}_2\left({a}_2\right),{P}_3\left({a}_2\right),\dots, {P}_N\left({a}_2\right)\right]\\ {}{S}_{a3}:\left[{P}_1\left({a}_3\right),D,{P}_2\left({a}_3\right),D,{P}_3\left({a}_3\right),\dots, {P}_N\left({a}_3\right)\right]\\ {}{S}_{b1}:\left[{P}_1\left({b}_1\right),{P}_2\left({b}_1\right),{P}_3\left({b}_1\right),\dots, {P}_N\left({b}_1\right)\right]\\ {}{S}_{b2}:\left[{P}_1\left({b}_2\right),{P}_2\left({b}_2\right),{P}_3\left({b}_2\right),\dots, {P}_N\left({b}_2\right)\right]\\ {}{S}_{b3}:\left[{P}_1\left({b}_3\right),{P}_2\left({b}_3\right),D,{P}_3\left({b}_3\right),\dots, D,{P}_N\left({b}_3\right)\right]\end{array}} $$

Lengths of {Sa1, Sa2, Sb1, Sb2} and {Sa3, Sb3} are n and n + m respectively.

StepB Alice sends Sa3 to Bob through the quantum channel and Bob sends Sb3 to Alice through the quantum channel. Alice informs Bob about position and measuring base of decoy photons through classical channel. Similarly, Bob also tells Alice the position and measuring base of the decoy photons. Then Alice and Bob have particle sequences {Sa1, Sa2, Sb3} and {Sb1, Sb2, Sa3} respectively.

At this point, Alice and Bob begin to perform the security check after receiving the particle sequence. Alice selects decoy photons from particle sequence Sb3 according to Bob’s information for single-particle measurement. Bob selects decoy photons from particle sequence Sa3 according to Alice’s information for single-particle measurement. Then, Alice and Bob release their measurement results through the classical channel. Alice calculates the error rate based on the results published by Bob, and Bob calculates the error rate based on the results published by Alice. If the error rate is lower than a threshold, it indicates that the transmitted sequences are complete and there is no external eavesdropper.

StepC After Alice and Bob complete the security check, the length of all particle sequences is n. Alice and Bob start to measure the particle sequences at the same time according to Table 2. K[i] = 0, Alice and Bob use X-basis measurement on particle groups (Pi(a1), Pi(a2), Pi(b3)) and (Pi(b1), Pi(b2), Pi(a3)), respectively. K[i] = 1, Alice and Bob use Z-basis measurement on particle groups (Pi(a1), Pi(a2), Pi(b3)) and (Pi(b1), Pi(b2), Pi(a3)), respectively. Here we agree on that the characteristics of the measurement results are Mi(a1, a2, b3) and Mi(b1, b2, a3) for i = 1, 2, 3, …, N. Bi(a1, a2, a3) and Bi(b1, b2, b3) represent the characteristics of the initial GHZ states of the particle groups (Pi(a1), Pi(a2), Pi(a3)) and (Pi(b1), Pi(b2), Pi(b3)), respectively. In the ith group of GHZ state, Alice infers the value of Pi(a3) based on the measured value of (Pi(a1), Pi(a2)), and thus obtain Bi(a1, a2, a3). Similarly, Bob can infer the value of Pi(b3) based on the measured value of (Pi(b1), Pi(b2)), and get Bi(b1, b2, b3). For i = 1, 2, 3, …, N, Alice computes KA[i] = Mi(a1, a2, b3) ⊕ Bi(a1, a2, a3), Bob computes KB[i] = Mi(b1, b2, a3) ⊕ Bi(b1, b2, b3). Then Alice and Bob get their keys KA = KA[1] ∥ KA[2] ∥ … ∥ KA[N] and KB = KB[1] ∥ KB[2] ∥ … ∥ KB[N], respectively. And ∥ denotes string concatenation. Finally, Alice and Bob obtain their session key KAB = KA = KB.

4 Conclusion

Through the investigate of the general expression |Gn〉, construction and truth table of the GHZ state, it is found that the three-particle entangled GHZ state satisfies certain relations after exchanging partial particles, and further establish the six equations of the three-particle GHZ state between the initial state and the exchanged state. In addition, these six equations are highly scalable and give the deformed equations. In this paper, four GHZ states and m GHZ states are derived from two GHZ states and three GHZ states exchanging one group and two groups of particles respectively, which satisfy some expressions. We applied these interesting and flexible equations to the QKA field, and further proved the correctness of these equations in the actual application environment.