## 3 Analyzing the Voting Process
Representatives from party A need not vote \(\left|yes\right\rangle\), but their vote must be within a distance of \(r_{A}\) from \(\left|yes\right\rangle\), that is on the shaded red portion of the circle. Symmetrically, legislators from party B may not vote \(\left|no\right\rangle\), but their vote must be within a distance of \(r_{B}\) from \(\left|no\right\rangle\), that is on the shaded green portion of the circle. \[\rho_{\psi}=|\psi\rangle\langle\psi|=\left[ \begin{array}{cc}|a|^{2}&ab^{\star}\\ ba^{\star}&|b|^{2}\end{array} \right]. \tag{4}\]  
In this case, \(\rho_{\psi}\) is the _orthogonal projector_ onto the unit ket \(|\psi\rangle\). In the special cases where \(|\psi\rangle=|0\rangle\) and \(|\psi\rangle=|1\rangle\), the corresponding density matrices are:  
\[\rho_{0}=|0\rangle\langle 0|=\left[ \begin{array}{cc}1&0\\ 0&0\end{array} \right] \tag{5}\]  
\[\rho_{1}=|1\rangle\langle 1|=\left[ \begin{array}{cc}0&0\\ 0&1\end{array} \right]. \tag{6}\]  
By combining (5) and (6) with (4), we derive:  
\[\rho_{0}-\rho_{\psi}=\left[ \begin{array}{cc}1-|a|^{2}&-ab^{\star}\\ -ba^{\star}&-|b|^{2}\end{array} \right]=\left(\rho_{0}-\rho_{\psi}\right)^{\dagger} \tag{7}\]  
and  
\[\rho_{1}-\rho_{\psi}=\left[ \begin{array}{cc}-|a|^{2}&-ab^{\star}\\ -ba^{\star}&1-|b|^{2}\end{array} \right]=\left(\rho_{0}-\rho_{\psi}\right)^{\dagger}, \tag{8}\]  
respectively. Taking advantage of the fact that \(|a|^{2}+|b|^{2}=1\), we may simplify (7) and (8) as follows:  
\[\rho_{0}-\rho_{\psi}=\left[ \begin{array}{cc}|b|^{2}&-ab^{\star}\\ -ba^{\star}&-|b|^{2}\end{array} \right] \tag{9}\]  
\[\rho_{1}-\rho_{\psi}=\left[ \begin{array}{cc}-|a|^{2}&-ab^{\star}\\ -ba^{\star}&|a|^{2}\end{array} \right]. \tag{10}\]  
It is now easy to see that:  
\[\left(\rho_{0}-\rho_{\psi}\right)^{\dagger}\left(\rho_{0}-\rho_{\psi}\right)= \left[ \begin{array}{cc}|b|^{4}+|a|^{2}|b|^{2}&-ab^{\star}|b|^{2}+ab^{\star}|b |^{2}\\ -ba^{\star}|b|^{2}+ba^{\star}|b|^{2}&|a|^{2}|b|^{2}+|b|^{4}\end{array} \right]= \left[ \begin{array}{cc}|b|^{2}&0\\ 0&|b|^{2}\end{array} \right] \tag{11}\]  
and  
\[\left(\rho_{1}-\rho_{\psi}\right)^{\dagger}\left(\rho_{1}-\rho_{\psi}\right)= \left[ \begin{array}{cc}|a|^{4}+|a|^{2}|b|^{2}&ab^{\star}|a|^{2}-ab^{\star}|a |^{2}\\ ba^{\star}|a|^{2}-ba^{\star}|a|^{2}&|a|^{2}|b|^{2}+|a|^{4}\end{array} \right]= \left[ \begin{array}{cc}|a|^{2}&0\\ 0&|a|^{2}\end{array} \right]. \tag{12}\]  
There is a general result (consult [18] p.75) that if \(U=\sum_{i}a_{i}|i\rangle\langle i|\) is the spectral decomposition of the Hermitian operator \(U\), then:  
\[f(U)=\sum_{i}f(a_{i})|i\rangle\langle i|. \tag{13}\]  
If \(U\) is also _positive_, meaning its eigenvalues \(a_{i}\) are nonnegative, then, in the special case where \(f(U)=\sqrt{U}\), (13) becomes:  
\[\sqrt{U}=\sum_{i}\sqrt{a_{i}}|i\rangle\langle i|. \tag{14}\]Although density matrices are Hermitian and positive, the important observation in Definition 1 is that for any operator \(U\), \(U^{\dagger}U\) is positive (see for instance [18], p.71). Hence, we may apply (14) and derive that:  
\[\big{|}\rho_{0}-\rho_{\psi}\big{|}=\Big{(}\big{(}\rho_{0}-\rho_{\psi}\big{)}^{ \intercal}\big{(}\rho_{0}-\rho_{\psi}\big{)}\Big{)}^{\frac{1}{2}}\stackrel{{ \eqref{eq:U_0}}}{{=}}\bigg{[} \begin{array}{cc}|b|&0\\ 0&|b|\end{array} \bigg{]}, \tag{15}\]  
and  
\[\big{|}\rho_{1}-\rho_{\psi}\big{|}=\Big{(}\big{(}\rho_{1}-\rho_{\psi}\big{)}^{ \intercal}\big{(}\rho_{1}-\rho_{\psi}\big{)}\Big{)}^{\frac{1}{2}}\stackrel{{ \eqref{eq:U_0}}}{{=}}\bigg{[} \begin{array}{cc}|a|&0\\ 0&|a|\end{array} \bigg{]}.