Exercise 7.11

The state vector of the near-singlet state is \[ \Psi = \sqrt{0.6}\ket{ud} - \sqrt{0.4}\ket{du} \]

We are asked to compute Alice’s density matrix using the eigenvectors of \(\sigma_z\) as a basis. Remember, the matrix representation of an operator depends on the basis system chosen. In this case, the two basis vectors are the well known vectors \(\ket{u}\) and \(\ket{d}\).

Alice’s density matrix is defined by \[ \rho_{a'a} = \sum_b \psi^*(a, b) \, \psi(a', b) \]

The non-zero values of the wave function are: \[\psi(u, d) = \psi^*(u, d) = \sqrt{0.6}\] \[\psi(d, u) = \psi^*(d, u) = -\sqrt{0.4}\]

The elements of the density matrix are: \[\rho_{uu} = \psi^*(u, u) \psi(u, u) + \psi^*(u, d) \psi(u, d) = 0.6\] \[\rho_{du} = \psi^*(u, u) \psi(d, u) + \psi^*(u, d) \psi(d, d) = 0\] \[\rho_{ud} = \psi^*(d, u) \psi(u, u) + \psi^*(d, d) \psi(u, d) = 0\] \[\rho_{dd} = \psi^*(d, u) \psi(d, u) + \psi^*(d, d) \psi(d, d) = 0.4\]

\[ \rho_A = \begin{pmatrix} 0.6 & 0 \\ 0 & 0.4 \end{pmatrix} \]