of identical fermions may be written as $${\ensuremath{|\Psi\rangle}} = C_{ij}a^\dag_i a^\dag_j{\ensuremath{|0\rangle}}  \qquad i,j=1,2,\cdots,{\cal N}.$$ $C_{ij}$ is anti-symmetric and can be set into block diagonal form[@yang] via $$\label{e:k}
  {\cal U}C{\cal U}^T = \bigoplus_{i=1}^{{\cal N}_f} \begin{pmatrix} 0 & c_i \\ -c_i & 0 \\
  \end{pmatrix}
   \overbrace{\bigoplus(0)}^\text{for ${\cal N}$ odd};$$ with ${\cal N}_f \equiv ({\cal N}/2)$ for even $\cal N$, and for odd ${\cal
N}$, ${\cal N}_f \equiv ({\cal N}-1)/2$; and ${\cal U}$ is a unitary matrix. Considering a redefinition with $a'^\dag_i = {\cal U}^*_{ij}a^\dag_j$, we may also rewrite $${\ensuremath{|\Psi\rangle}} = 2\sum_{i=1}^{{\cal N}_f} c_i a'^\dag_i a'^\dag_{i+1} {\ensuremath{|0\rangle}}.$$ This is the analog of “Schmidt decomposition" for identical fermion systems[@schliemann]. The total system has density matrix $\rho = {\ensuremath{|\Psi\rangle}}{\ensuremath{\langle\Psi|}}$ and entropy $S=0$. Following the prescription for density matrix reduction discussed earlier, the one-particle reduced density matrix is then $${{\ensuremath{\langlei|}}}\rho_1{{\ensuremath{|j\rangle}}} = {\text{Tr}}{\{a_i\rho\, a^\dag_j}\} = 4(CC^\dag)_{ij}.$$ In terms of (${\cal U}C{\cal U}^T$) combination of Eq.(\[e:k\]), we note that $\rho_1 = 4{\cal U}^\dag[({\cal U}C{\cal U}^T)({\cal U}C{\cal U}^T)^\dag]({\cal
U})$; thus $$\label{e:gg}
 {{\cal U}\rho_1{\cal U}^\dag} = 4\bigoplus_{i=1}^{{\cal N}_f}
 \begin{pmatrix} {{\ensuremath{\vert c_i\vert}}}^2 &0\\ 0& {{\ensuremath{\vert c_i\vert}}}^2 \\ \end{pmatrix}
   \overbrace{\bigoplus(0)}^\text{for ${\cal N}$ odd}.$$ The normalization condition is $$\label{e:h}
{\langle \Psi|\Psi\rangle} = 4\sum_{i=1}^{{\cal N}_f}{{\ensuremath{\vert c_i\vert}}}^2=1
\qquad\Rightarrow {\text{Tr}}\rho_1 = 8\sum_{i=1}^{{\cal N}_f}{{\ensuremath{\vert c_i\vert}}}^2 = 2;$$ 