Entropy and mode restrictions of finite quasifree CAR states

By [F5], write $C=V\mathrm{diag}({\nu }_1,\dots ,{\nu }_r)V^{\ast }$ with $V$ unitary and ${\nu }_j\in [0,1]$ . By [F2], after conjugation by the induced exterior-power unitary, ${\rho }_C$ is diagonal on the occupation basis with probabilities $p_m={\prod }_j{w}_j(m_j)$ , where $w_j(0)=1-{\nu }_j$ and $w_j(1)={\nu }_j$ . These probabilities sum to one because ${\sum }_{m\in \{0,1{\}}^r}{p}_m={\prod }_j((1-{\nu }_j)+{\nu }_j)=1$ .

Choose the ordered orthonormal bases of $K$ and $K^{\perp }$ using [F6]. The map from the full occupation vector to the tensor product of its $K$ -occupation and $K^{\perp }$ -occupation vectors sends one orthonormal basis onto another, hence extends to a unitary factorization. Because the $K$ modes come first, their creation and annihilation operators act on the first factor as $c_j^K\otimes I$ ; the CAR parity signs from preceding modes are therefore entirely within that first factor. The partial trace over $K^{\perp }$ has the same expectation as ${\rho }_C$ for every word in the $K$ modes. Restricting the determinant rule [F1] to those words gives the compressed covariance $C_K=P_{K}C{\mathrm{\mid }}_K$ and the quasifree moment rule; gauge invariance is inherited under restriction. Thus ${\rho }_K$ is the unique quasifree density for $C_K$ by [F2].

Unitary conjugation preserves the eigenvalues, so the entropy is $-{\sum }_m{p}_m\ln p_m$ , with zero terms omitted under [F3]. When $0<{\nu }_j<1$ , expand $\ln p_m={\sum }_j[m_j\ln {\nu }_j+(1-m_j)\ln (1-{\nu }_j)]$ and sum the finite product distribution; its $j$ th occupation has mean ${\nu }_j$ . The result is $-{\sum }_j[{\nu }_j\ln {\nu }_j+(1-{\nu }_j)\ln (1-{\nu }_j)]$ . If ${\nu }_j=0$ or $1$ , that mode has a deterministic occupation and contributes zero by [F3], which is the same formula with $0\ln 0=0$ . Spectral calculus for $C$ and $I-C$ now identifies this sum with the displayed matrix trace.

Apply the calculation of step 2.1 to the finite quasifree density ${\rho }_K$ and its positive contraction $C_K$ . The same endpoint convention $0\ln 0=0$ handles pure or rank-deficient reduced states. A change of orthonormal basis in $K$ conjugates ${\rho }_K$ by the induced unitary and $C_K$ by the corresponding matrix unitary, so [F3] makes the trace entropy unchanged. A change of basis in $K^{\perp }$ conjugates only the traced factor; for a unitary matrix $U$ on that factor, the partial-trace entry sum in [F3] contains ${\sum }_j{U}_{ja}\overline{U_{jb}}={\delta }_{ab}$ , so ${\rho }_K$ is unchanged. This proves the restriction formula and its basis independence.