Theorem: Spectrum, negativity, and separability

Let $p_0=1-q$ and $s={\Vert R_{AB}\Vert }_{\mathrm{F}}$. As a multiset, the complete spectrum is

where

Zeros already present in the first three multisets are retained. The formula holds for $s=0$ as well.

Therefore

Within the state class (eq:reduction),

Also,

Proof. A cross matrix unit transforms according to

The vacuum and the states with one excitation on each side therefore carry the block

The cross-sector basis is orthonormal. For $s>0$, restriction to the vacuum and $v/s$ gives ${\lambda }_{\pm }$; the remaining directions give zeros. For $s=0$, choose any one cross-sector basis vector instead, giving eigenvalues $p_0,0$. Such a vector exists because both parts are nonempty.

The other nonzero blocks are $R_{AA}$ and ${R_{BB}}^{\mathsf{T}}$, on mutually orthogonal local single-excitation spaces. All remaining sectors are zero. Their dimensions give exactly $2^N-N-2$ additional zeros in (eq:PTspec). The positive local blocks have combined trace $q$, giving the trace norm and the negativity formulas.

If $R_{AB}\ne 0$, then ${\lambda }_{-}<0$. Every separable state has positive partial transpose, by transposing the positive second factor in each product term of a separable decomposition. Hence this state is entangled. If $R_{AB}=0$, the vacuum term and the two local excitation blocks are positive product terms. Normalizing the nonzero terms by their traces gives a separable decomposition.

Finally,

Insert this into (eq:EN). The inequality $\sqrt{(1-q{)}^2+q^2}\le 1$ proves the last bound.