# Shared Quantum Key Distribution Based on Asymmetric Double Quantum Teleportation
## Appendix D Conditional QBER in the Protocol  
Similarly to the [PERSON] success probability, taking the teleported state after of the [PERSON]\({}_{1}\) processing but before to the [PERSON]\({}_{0}\) processing and the [PERSON]'s measurement of the control system (100), then we consider the sifting and measurement from [PERSON], reaching the outcome \(|k^{\prime}\rangle_{\theta^{\prime},p}\), with \(\theta^{\prime}\
eq\theta\), the basis planned by [PERSON]. \(k^{\prime}\) is also not necessarily equal to \(k\oplus 1\) (the outcome finally obtained by [PERSON]\({}_{1}\)). Following the expressions (100)-(101): \({}_{\theta^{\prime},p}\langle k^{\prime}|0\rangle=(-1)^{pk^{\prime}}f_{k^{ \prime}}(\frac{\theta}{2})\) and \({}_{\theta^{\prime},p}\langle k^{\prime}|\psi_{0}^{0}\rangle=(-1)^{pk^{\prime }}W_{1}^{\prime}\), where, in this case, we introduced the quantity:  
\[W_{i}^{\prime} \equiv f_{k^{\prime}}(\frac{\theta^{\prime}}{2})\cos\bigg{(}\omega_{i}-( -1)^{p}\frac{\theta}{2}\bigg{)}+(-1)^{p+k^{\prime}+1}f_{1\oplus k^{\prime}}( \frac{\theta^{\prime}}{2})\sin\bigg{(}\omega_{i}-(-1)^{p}\frac{\theta}{2} \bigg{)} \tag{103}\]  
As before, [PERSON] performs the sifting, measuring, and returning on the Bob\({}_{1}\) state. Then, Bob\({}_{0}\) processing is followed similarly as in (102), giving:  
\[|\psi_{2}^{\prime}\rangle\equiv C^{C}\mathcal{U}_{0_{2}}\cdot|k^{\prime} \rangle_{\theta^{\prime},p_{4}}\langle k^{\prime}|\cdot|\psi_{1}\rangle=(-1)^ {pk^{\prime}}|k^{\prime}\rangle_{\theta^{\prime},p_{4}}\bigg{[}\sqrt{p_{0}}f_{ k^{\prime}}(\frac{\theta^{\prime}}{2})|0\rangle_{C}|\psi_{0}^{0}\rangle_{2}+ \sqrt{p_{1}}W_{1}^{\prime}|1\rangle_{C}|0\rangle_{2}\bigg{]} \tag{104}\]  
Observe that, in any case, [PERSON] obtains \(|k^{\prime}\rangle_{\theta^{\prime},p}\) as the outcome (by selecting the basis defined by \(\theta^{\prime},p\)). Now, the \(C_{SWAP_{2,4}}\equiv\mathbf{X}_{C}\cdot C_{SWAP_{2,4}}^{C}\cdot\mathbf{X}_{C}\) is applied between the Bob's, obtaining:  
\[|\psi_{3}^{\prime}\rangle\equiv C_{SWAP_{2,4}}\cdot|\psi_{2}^{\prime}\rangle=( -1)^{pk^{\prime}}\bigg{[}\sqrt{p_{0}}f_{k^{\prime}}(\frac{\theta^{\prime}}{2})|0 \rangle_{C}|k^{\prime}\rangle_{\theta^{\prime},p_{2}}|\psi_{0}^{0}\rangle_{4}+ \sqrt{p_{1}}W_{1}^{\prime}|1\rangle_{C}|0\rangle_{2}|k^{\prime}\rangle_{\theta^{ \prime},p_{4}}\bigg{]} \tag{105}\]Now, [PERSON] performs the measurement of the control state on the basis stated by \(K\), hitting \(|b_{j}\rangle\) and generating the state \(|K^{\prime\prime}\rangle_{\theta,p}\) on qubit 4, thus:  
\[|\psi_{4}^{\prime}\rangle \equiv {}_{\theta_{A}}\langle K^{\prime}|\psi_{3}^{\prime}\rangle \tag{101}\] \[= (-1)^{\mu^{\prime}}\Big{[}\sqrt{p_{0}}f_{x}(\frac{\theta^{\prime }}{2})|0\rangle_{C}|k^{\prime}\rangle_{\theta^{\prime},p_{2}}\ {}_{\theta_{A}}\langle k^{\prime\prime}|\psi_{0}^{0}\rangle_{4}+\sqrt{p_{1}}W_ {1}^{\prime}|1\rangle_{C}|0\rangle_{2}\ {}_{\theta_{A}}\langle k^{\prime\prime}|k^{\prime} \rangle_{\theta^{\prime},p_{2}}\Big{]}\]  
where it has been assumed that he hits on a different basis selection than [PERSON], and a different outcome, but still in the same basis than [PERSON] planned. It will let, under the reconciliation, notice the presence of [PERSON]. Then, finally performing [PERSON]'s measurement with outcome \(|b_{j}\rangle\):  
\[|\psi_{5}^{\prime}\rangle \equiv {}_{C}\langle b_{j}|\cdot{}_{\theta,p_{k}}\langle k^{\prime\prime }|\psi_{4}^{\prime}\rangle=(-1)^{p(k^{\prime}+k^{\prime})}\Big{[}\sqrt{p_{0} }\hat{\theta}_{0\oplus j}W_{0}f_{x}(\frac{\theta^{\prime}}{2})e^{i\theta_{A} \cdot|k^{\prime}\rangle_{\theta^{\prime},p_{2}}} \tag{102}\] \[+(-1)^{j+pk^{\prime}+k^{\prime}(k^{\prime}+k^{\prime\prime})} \sqrt{p_{1}}\hat{\theta}_{1\oplus j}W_{1}^{\prime}f_{x^{\prime}\oplus k^{ \prime\prime}}(\frac{\Delta\theta}{2})e^{-i\phi_{n}(1+j)}|0\rangle_{2}\Big{]}\]  
where \(W_{0}\) is the same expression as in (102) but changing \(k\) by \(k^{\prime\prime}\) and \(\Delta\theta=\frac{\theta-\theta^{\prime}}{2}\). As in the Appendix C, we set the prescriptions there.