\ln 2$, which is the [*minimum*]{} von Neumann entropy of the reduced 1-particle density matrix for two identical fermions. We chose to compute $C$ explicitly to illustrate the consistency of the approach, but it is not necessary to go through this step. Given $\rho$, computation of $\rho_1$ can also be done directly through Eq.(\[e:vev\]).

Consider instead the Bell state $$\begin{aligned}
{\ensuremath{|\Psi\rangle}} &=&{\ensuremath{|B^0({\ensuremath{\textbf p}}_1,{\ensuremath{\textbf p}}_2)\rangle}}\\
&=&\frac{1}{\sqrt{2}}[{a^\dag({\ensuremath{\textbf p}}_1,+)a^\dag({\ensuremath{\textbf p}}_2,-)}-{a^\dag({\ensuremath{\textbf p}}_1,-)a^\dag({\ensuremath{\textbf p}}_2,+)}]{\ensuremath{|0\rangle}}\\
&\equiv& C_{ij}a^\dag_i a^\dag_j{\ensuremath{|0\rangle}}; \end{aligned}$$ now yielding the non-vanishing part of the $C$-matrix as $C= \bordermatrix{&(\p_1,+)&(\p_2,-)&(\p_2,+)&(\p_1,-)\cr
          (\p_1,+)& 0 & \frac{1}{2\sqrt{2}}  & 0  & 0 \cr
          (\p_2,-)& -\frac{1}{2\sqrt{2}} & 0  & 0  & 0  \cr
          (\p_2,+) & 0 & 0  & 0  & \frac{1}{2\sqrt{2}}  \cr
          (\p_1,-) & 0  & 0  & -\frac{1}{2\sqrt{2}}  & 0 \cr},$ for which the normalized reduced density matrix is $${\rho_1} =\begin{pmatrix} \frac{1}{4} & 0 & 0\\
0 & \frac{1}{4} & 0 & 0\\
0 & 0 & \frac{1}{4} & 0\\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}.$$ The corresponding entropy is now $S_1 = -4(\frac{1}{4}\ln \frac{1}{4}) =
2\ln2$ instead. This is a von Neumann entanglement entropy (of the reduced density matrix) which is $\ln2$ units [*higher*]{} than the lowest value thus justifying that the Bell state ${\ensuremath{|B^0({\ensuremath{\textbf p}}_1,{\ensuremath{\textbf 