= Solution
Fix the mode ordering with the $b^\dagger$ operator before the $c^\dagger$ operator, as in the given state. In the sector with one <fermion> in each subsystem, use the ordered basis
$$
(|B_1C_1\rangle,|B_1C_2\rangle,|B_2C_1\rangle,|B_2C_2\rangle).
$$
The <state vector> is $v=(1/2,1/\sqrt2,0,1/2)^T$, whose squared norm is one. Thus the global <density matrix> is the rank-one projector
$$
\boxed{\rho_{BC}=vv^\dagger=
\begin{pmatrix}
\frac14&\frac1{2\sqrt2}&0&\frac14\\
\frac1{2\sqrt2}&\frac12&0&\frac1{2\sqrt2}\\
0&0&0&0\\
\frac14&\frac1{2\sqrt2}&0&\frac14
\end{pmatrix}}.
$$
It has <trace> one and satisfies $\rho_{BC}^2=\rho_{BC}$, so the global state is pure.
Take the <partial trace> over $B$. The conditional vectors in $C$ are $v_1=(1/2,1/\sqrt2)^T$ for $B_1$ and $v_2=(0,1/2)^T$ for $B_2$. Therefore the <reduced density matrix> is
$$
\boxed{\rho_C=v_1v_1^\dagger+v_2v_2^\dagger
=\begin{pmatrix}\frac14&\frac1{2\sqrt2}\\\frac1{2\sqrt2}&\frac34\end{pmatrix}}.
$$
Its determinant is $1/16$, giving <eigenvalues>
$$
\lambda_\pm=\frac{2\pm\sqrt3}{4}.
$$
Both are positive; in particular, $\operatorname{Tr}\rho_C^2=7/8<1$, so the reduced state is mixed despite global purity. The <entanglement entropy> is obtained by diagonalizing this reduced <matrix> and evaluating
$$
\boxed{S_{\mathrm{ent}}=-\lambda_+\ln\lambda_+-\lambda_-\ln\lambda_-}.
$$
For a globally pure bipartite state this equals the entropy of either reduced subsystem. No numerical evaluation is needed.
Back to article page
