No genuinely mixed two-qubit Hardy state

Identify the Boolean values false and true with $0$ and $1$, and let $e_0,e_1$ be the standard basis of $\mathbb C^2$. For every choice of complex numbers $\alpha_P,\beta_P$, indexed by $P\in\{0,1\}$, satisfying $|\alpha_P|^2+|\beta_P|^2=1$ and $0<|\alpha_P|<1$ for each $P$, define $u_P=\alpha_Pe_0+\beta_Pe_1$ and $v_P=\overline{\beta_P}e_0-\overline{\alpha_P}e_1$. Here tensor products have coordinates $(r\otimes s)_{a,b}=r_as_b$. For every complex matrix $\rho$ indexed by $\{0,1\}^2$ that is positive semidefinite and has trace $1$, put $q_\rho(w)=\operatorname{Re}\operatorname{tr}((ww^*)\rho)$, where $(ww^*)_{z,z'}=w_z\overline{w_{z'}}$. If $q_\rho(e_0\otimes e_0)>0$, $q_\rho(u_0\otimes e_0)=0$, $q_\rho(e_0\otimes u_1)=0$, and $q_\rho(v_0\otimes v_1)=0$, then there exists a vector $\psi\in\mathbb C^{\{0,1\}^2}$ with Euclidean norm $1$ such that $\rho_{z,z'}=\psi_z\overline{\psi_{z'}}$ for all $z,z'\in\{0,1\}^2$.