Editing BB84 Quantum Key Distribution
Jump to navigation
Jump to search
The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then publish the changes below to finish undoing the edit.
Latest revision | Your text | ||
Line 49: | Line 49: | ||
\epsilon_{\rm sec}= \epsilon_{\rm PA}+\epsilon_{\rm PE},</math> | \epsilon_{\rm sec}= \epsilon_{\rm PA}+\epsilon_{\rm PE},</math> | ||
and the amount of key <math>\ell</math> that is generated is given by</br> | and the amount of key <math>\ell</math> that is generated is given by</br> | ||
<math> | <math> \centering \begin{align} | ||
\ell \geq & (1-\gamma)^2n (1-h(Q_X+\nu) -h(Q_Z)) \\ &-\sqrt{(1-\gamma)^2n}\big(4\log(2\sqrt{2}+1)(\sqrt{\log\frac{2}{\epsilon_{\rm PE}^2}}+ \sqrt{\log \frac{8}{{\epsilon'}_{\rm EC}^2}})) \\& -\log(\frac{8}{{\epsilon'}_{\rm EC}^2}+\frac{2}{2-\epsilon'_{\rm EC}})-\log (\frac{1}{\epsilon_{\rm EC}})- 2\log(\frac{1}{2\epsilon_{\rm PA}}) | \ell \geq & (1-\gamma)^2n (1-h(Q_X+\nu) -h(Q_Z)) \\ &-\sqrt{(1-\gamma)^2n}\big(4\log(2\sqrt{2}+1)(\sqrt{\log\frac{2}{\epsilon_{\rm PE}^2}}+ \sqrt{\log \frac{8}{{\epsilon'}_{\rm EC}^2}})) \\& -\log(\frac{8}{{\epsilon'}_{\rm EC}^2}+\frac{2}{2-\epsilon'_{\rm EC}})-\log (\frac{1}{\epsilon_{\rm EC}})- 2\log(\frac{1}{2\epsilon_{\rm PA}}) \end{align}</math> | ||
\end{align} | |||
</math> | |||
</br>where <math>\nu = \sqrt{ \frac{(1+\gamma^2n)((1-\gamma)^2+\gamma^2)}{(1-\gamma)^2\gamma^4n^2}\log(\frac{1}{\epsilon_{\rm PE}}})</math> | </br>where <math>\nu = \sqrt{ \frac{(1+\gamma^2n)((1-\gamma)^2+\gamma^2)}{(1-\gamma)^2\gamma^4n^2}\log(\frac{1}{\epsilon_{\rm PE}}})</math> | ||
and <math>h(\cdot)</math> is the [[binary entropy function]]. | and <math>h(\cdot)</math> is the [[binary entropy function]]. |