## 2 Description of [PERSON]'s attack and its properties
### The probability of error in the conjugate basis  
We are now interested in \(p_{e}^{\overline{b}}\) where \(\overline{b}=1-b\) (i.e. \(\overline{0}=1\) and \(\overline{1}=0\)) corresponds to the basis conjugate to that given by \(b\). The attack \(U\) is always the one described by (1) and (2) in the \(b\) basis but, in order to calculate the probability or error when [PERSON] encodes \(i_{j}\) as \(|i_{j}^{\overline{b}}\rangle\) instead of \(|i_{j}^{b}\rangle\), we now need to express \(U\) in the \(\overline{b}\) basis. From (10), we know that the probability of error for this situation is given by  
\[p_{e}^{\bar{b}}=\frac{1}{2}\left[\langle E_{01}^{\overline{b}}|E_{01}^{ \overline{b}}\rangle+\langle E_{10}^{\overline{b}}|E_{10}^{\overline{b}} \rangle\right].\]  
Using the fact that  
\[|0\rangle^{\overline{b}}=\frac{1}{\sqrt{2}}[|0^{b}\rangle+|1^{b}\rangle], \hskip 28.452756 pt|1\rangle^{\overline{b}}=\frac{1}{\sqrt{2}}[|0^{b}\rangle-|1 ^{b}\rangle]\]  
and using the linearity of \(U\), we deduce directly from (1) and (2) that  
\[U|0^{E}\rangle|0^{\overline{b}}\rangle =\frac{1}{\sqrt{2}}\left(|E_{00}^{b}\rangle+|E_{10}^{b}\rangle \right)|0^{b}\rangle+\frac{1}{\sqrt{2}}\left(|E_{01}^{b}\rangle+|E_{11}^{b} \rangle\right)|1^{b}\rangle, \tag{11}\] \[U|0^{E}\rangle|1^{\overline{b}}\rangle =\frac{1}{\sqrt{2}}\left(|E_{00}^{b}\rangle-|E_{10}^{b}\rangle \right)|0^{b}\rangle+\frac{1}{\sqrt{2}}\left(|E_{01}^{b}\rangle-|E_{11}^{b} \rangle\right)|1^{b}\rangle. \tag{12}\]  
Replacing \(|0^{b}\rangle\) and \(|1^{b}\rangle\) on the right-hand sides with their values in terms of \(|0^{\overline{b}}\rangle\) and \(|1^{\overline{b}}\rangle\) i.e. \(|0^{b}\rangle=\frac{1}{\sqrt{2}}[|0^{\overline{b}}\rangle+|1^{\overline{b}}\rangle]\) and \(|1^{b}\rangle=\frac{1}{\sqrt{2}}[|0^{\overline{b}}\rangle-|1^{\overline{b}}\rangle]\) we obtain  
\[U|0^{E}\rangle|0^{\overline{b}}\rangle =\phantom{-}\frac{1}{2}\left[|E_{00}^{b}\rangle+|E_{10}^{b}\rangle +|E_{01}^{b}\rangle+|E_{11}^{b}\rangle\right]|0^{\overline{b}}\rangle\] \[\phantom{-}+\frac{1}{2}\left[\left(|E_{00}^{b}\rangle-|E_{11}^{b} \rangle\right)+\left(|E_{10}^{b}\rangle-|E_{01}^{b}\rangle\right)\right]|1^{ \overline{b}}\rangle \tag{13}\] \[U|0^{E}\rangle|1^{\overline{b}}\rangle =\phantom{-}\frac{1}{2}\left[\left(|E_{00}^{b}\rangle-|E_{11}^{b} \rangle\right)-\left(|E_{10}^{b}\rangle-|E_{01}^{b}\rangle\right)\right]|0^{ \overline{b}}\rangle\] \[\phantom{-}+\frac{1}{2}\left[|E_{00}^{b}\rangle-|E_{10}^{b} \rangle-|E_{01}^{b}\rangle+|E_{11}^{b}\rangle\right]|1^{\overline{b}}\rangle \tag{14}\]where the terms for \(|E_{01}^{\overline{b}}\rangle\) and \(|E_{10}^{\overline{b}}\rangle\) are parenthesized so that we can easily see that  
\[p_{e}^{\overline{b}} =\frac{1}{2}\left[\langle E_{01}^{\overline{b}}|E_{01}^{\overline {b}}\rangle+\langle E_{10}^{\overline{b}}|E_{10}^{\overline{b}}\rangle\right]\] \[=\frac{1}{4}\left[(\langle E_{00}^{b}|-\langle E_{11}^{b}|)(|E_{0 0}^{b}\rangle-|E_{11}^{b}\rangle)+(\langle E_{10}^{b}|-\langle E_{01}^{b}|)(|E _{10}^{b}\rangle-|E_{01}^{b}\rangle)\right].\]  
We expand this result by using the identities \(\langle\phi|\psi\rangle=\overline{\langle\psi|\phi\rangle}\) and \(z+\overline{z}=2\text{Re}(z)\) for \(z\in\mathbf{C}\) (here the overline indicates the complex conjugate). Using equalities (3) and (4) we get  
\[|0\rangle^{\overline{b}}=\frac{1}{\sqrt{2}}[|0^{b}\rangle+|1^{b}\rangle], \hskip 28.452756 pt|1\rangle^{\overline{b}}=\frac{1}{\sqrt{2}}[|0^{b}\rangle-|1 ^{b}\rangle]\]0 \[|0\rangle^{\overline{b}}=\frac{1}{\sqrt{2}}[|0^{b}\rangle+|1^{b}\rangle], \hskip 28.452756 pt|1\rangle^{\overline{b}}=\frac{1}{\sqrt{2}}[|0^{b}\rangle-|1 ^{b}\rangle]\]1  
This Formula will be used to connect the disturbance induced by [PERSON] when [PERSON] encodes in the \(\overline{b}_{j}\) basis bits \(i_{j}\) such that \(s_{j}=1\) to the information [PERSON] can get when [PERSON] encodes them in the \(b_{j}\) basis.