# Shared Quantum Key Distribution Based on Asymmetric Double Quantum Teleportation
## Appendix B _Swap_ Operations between Faraway Non-Local Parties  
As in the Appendix A, we will show how to perform the \(C_{SWAP_{2}a}\) operation between the faraway parties 2, 4 (assuming they cannot be moved close together). Again, we will use the entangled resource \(|\beta_{00}\rangle_{ab}\) where qubit \(a\) is in possession of [PERSON]\({}_{0}\) and \(b\) is sent to [PERSON]\({}_{1}\):  
\[|\psi\rangle = (\sqrt{p_{0}}|0\rangle_{C}|\psi_{0}^{0}\rangle_{2}|0\rangle_{4}+ \sqrt{p_{1}}|1\rangle_{C}|0\rangle_{2}|\psi_{0}^{1}\rangle_{4})\otimes|\beta_ {00}\rangle_{ab} \tag{28}\]  
As before, by rearranging the qubits 2 and \(a\) in the first term of (28), and expressing it in terms of Bell stats basis:  
\[|\psi\rangle = \frac{\sqrt{p_{0}}}{2}|0\rangle_{C}|0\rangle_{4}\sum_{x,y\in\{0, 1\}}|\beta_{xy}\rangle_{2a}\mathbf{Z}_{b}^{x}\mathbf{X}_{b}^{y}|\psi_{0}^{0} \rangle_{b}+\sqrt{p_{1}}|1\rangle_{C}|0\rangle_{2}|\psi_{0}^{1}\rangle_{4}| \beta_{00}\rangle_{ab} \tag{29}\]  
then, [PERSON] and [PERSON]\({}_{0}\) apply the controlled operation \(\tilde{C}^{C}(H_{2}C^{2}NOT_{a})\) on their qubits (where, \(\tilde{C}^{a}G_{b}\equiv\mathbf{X}_{a}(C^{a}G_{b})\mathbf{X}_{a}\)). Then, it becomes:  
\[|\psi^{(1)}\rangle = \frac{\sqrt{p_{0}}}{2}|0\rangle_{C}|0\rangle_{4}\sum_{x,y\in\{0, 1\}}|xy\rangle_{2a}\mathbf{Z}_{b}^{x}\mathbf{X}_{b}^{y}|\psi_{0}^{0}\rangle_{b }+\sqrt{p_{1}}|1\rangle_{C}|0\rangle_{2}|\psi_{0}^{1}\rangle_{4}|\beta_{00} \rangle_{ab} \tag{30}\]  
then, [PERSON]\({}_{0}\) measures qubits 2 and \(a\) if the control register is \(|0\rangle_{C}\), getting \(|x\rangle_{2}\) and \(|y\rangle_{a}\) using controlled quantum measurements [36; 37]. Thus:  
\[|\psi^{(2)}\rangle = \sqrt{p_{0}}|0\rangle_{C}|0\rangle_{4}|xy\rangle_{2a}\mathbf{Z}_ {b}^{x}\mathbf{X}_{b}^{y}|\psi_{0}^{0}\rangle_{b}+\sqrt{p_{1}}|1\rangle_{C}|0 \rangle_{2}|\psi_{0}^{1}\rangle_{4}|\beta_{00}\rangle_{ab} \tag{31}\]Using controlled classical communication, the measurement outcomes are shared with \(\text{Bob}_{1}\) just if the control register is \(|0\rangle_{C}\) to perform the operation \(\mathbf{X}_{b}^{y}\mathbf{Z}_{b}^{x}\) and then \(SWAP_{ab}\), all of them on his qubits. Similarly, \(\text{Bob}_{0}\) applies \(C^{C}\mathbf{X}_{2}^{x}\) and \(C^{C}\mathbf{X}_{a}^{y}\). It gives:  
\[|\psi^{(3)}\rangle = |0\rangle_{2}(\sqrt{p_{0}}|0\rangle_{C}|\psi_{0}^{0}\rangle_{4}|0 \rangle_{ab}+\sqrt{p_{1}}|1\rangle_{C}|\psi_{0}^{0}\rangle_{4}|\beta_{0}\rangle_ {ab}) \tag{111}\]  
Finally, \(\text{Bob}_{0}\) uses controlled quantum measurements again when the control register is \(|1\rangle_{C}\) to measure the qubit \(a\) getting \(|z\rangle_{a}\) as outcome. He performs \(C^{C}\mathbf{X}_{a}^{z}\) and uses controlled classical communication to share the outcome to \(\text{Bob}_{1}\) who performs \(\mathbf{X}_{b}^{z}\). It gives the state:  
\[|\psi^{(4)}\rangle = |000\rangle_{2 ab}(\sqrt{p_{0}}|0\rangle_{C}|\psi_{0}^{0}\rangle_ {4}+\sqrt{p_{1}}|1\rangle_{C}|\psi_{0}^{1}\rangle_{4}) \tag{112}\]  
which, disregarding \(|000\rangle_{2 ab}\), fits with \(|\psi_{final}\rangle\) in (10) upon the application of \(C_{SWAP_{2,4}}\).