and the von Neumann entropy of the [*reduced*]{} density matrix is therefore $$S_1 = -{\text{Tr}}(\frac{\rho_1}{{\text{Tr}}\rho_1}\ln\frac{\rho_1}{{\text{Tr}}\rho_1})
  = -4\sum_{i=1}^{{\cal N}_f}{{\ensuremath{\vert c_i\vert}}}^2\ln(2{{\ensuremath{\vert c_i\vert}}}^2);$$ which is bounded by $\ln 2 \leq S_1 \leq \ln(2{\cal N}_f)$. The upper limit is obtained by maximizing $S_1$ subject to the constraint (\[e:h\]); while the lower bound occurs when there is only a single non-vanishing anti-symmetric block in Eq.(\[e:k\])[@paskauskas; @schliemann]. It may appear disconcerting that this lower bound for fermions is $\ln 2$, rather than zero, whereas the analogous result for a system of two bosons is $0\leq S_1\leq\ln
{\cal N}$[@paskauskas]. However, the lowest entropy is associated with what we would call an “unentangled" system; while any of the “entangled" Bell state discussed earlier will be shown to have $S_1 = 2\ln2$ instead. To see this, let us use, for instance, an unentangled two-fermion state, $$\begin{aligned}
{\ensuremath{|\Psi\rangle}} &=&\frac{1}{\sqrt{2}}[{\ensuremath{|B^0({\ensuremath{\textbf p}}_1,{\ensuremath{\textbf p}}_2)\rangle}}+i{\ensuremath{|B^3({\ensuremath{\textbf p}}_1,{\ensuremath{\textbf p}}_2)\rangle}}]\\
&=& {a^\dag({\ensuremath{\textbf p}}_1,+)a^\dag({\ensuremath{\textbf p}}_2,-)}{\ensuremath{|0\rangle}}\\
&\equiv& C_{ij}a^\dag_i a^\dag_j{\ensuremath{|0\rangle}}; \end{aligned}$$ yielding the non-vanishing part of the C-matrix as $  C= \bordermatrix{&(\p_1,+)&\p_2,-)\cr
          (\p_1,+) & 0  & \frac{1}{2} \cr
          (\p_2,-) & -\frac{1}{2} & 0 \cr}.$ Following earlier derivations, we obtain the normalized one-particle reduced density matrix as $${\rho_1} =\begin{pmatrix} &{\ensuremath{\frac{1}{2}}}& 0 \cr & 0 & {\ensuremath{\frac{1}{2}}}\end{pmatrix};$$ giving $S_1 = -2(\frac{1}{2})\ln\frac{1}{2} = 