Exercise 7.10

The initial composite state of the spin and the apparatus is \[ \ket{\Psi} = \alpha_u \ket{u, b} + \alpha_d \ket{d, b} \]

We compute the density matrix and the corresponding eigenvalues. If there is exactly one non-zero eigenvalue with value +1, then the state is completely unentangled, as shown in chapter 7.7.2.

The density matrix is given by Eq. (7.20): \[ \rho_{a'a} = \psi^*(a, b) \psi(a', b) \]

There is no need for a sumation, since there is only a single state \(|b\}\) for the apparatus. \[ \psi(u, b) = \alpha_u \] \[ \psi(d, b) = \alpha_d \] \[ \psi^*(u, b) = \alpha_u^* \] \[ \psi^*(d, b) = \alpha_d^* \]

The elements of the density matrix are as follows: \[ \rho_{uu} = \alpha_u \alpha_u^* \] \[ \rho_{ud} = \alpha_d^* \alpha_u \] \[ \rho_{du} = \alpha_u^* \alpha_d \] \[ \rho_{dd} = \alpha_d^* \alpha_d \]

Now we solve for the eigenvalue \(\lambda\): \[ \rho \ket{\lambda} = \lambda \ket{\lambda} \]

Using the standard procedure \[ \det (\rho - \lambda \mathrm{I}) = 0 \]

\[ \begin{vmatrix} \alpha_u^* \alpha_u - \lambda & \alpha_d^* \alpha_u \\ \alpha_u^* \alpha_d & \alpha_d^* \alpha_d - \lambda \end{vmatrix} = 0 \]

we obtain the characteristic polynomial: \[ \lambda^2 - \lambda (\alpha_u^* \alpha_u + \alpha_d^* \alpha_d) = \lambda^2 - \lambda = 0 \]

The two eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = +1\). This proves that the state is completley unentangled.