also transform unitarily as $\rho' = U(\Lambda)\rho U^\dag(\Lambda)$.

The von Neumann entropy of a density matrix is $$S \equiv -{\text{Tr}}{(\rho\ln\rho)} = -\sum^N_{n=1} \lambda_n\ln\lambda_n.$$ with $\lambda_n$ being the eigenvalues of $\rho$ (we assume normalization of ${\text{Tr}}\rho = 1$ has been carried out in the definition of the von Neumann entropy). Since $\rho$ is Hermitian it can be diagonalized, and we may write $\rho = V[diag(\lambda_1,..., \lambda_{N})]V^{\dagger}$. It follows that the eigenvalues are [*invariant*]{} under unitary transformations (and, in particular, Lorentz transformations) since $\rho' = U(\Lambda)\rho
U^\dag(\Lambda) = U(\Lambda)V[diag(\lambda_1,...,
\lambda_{N})][{U(\Lambda)V}]^{\dagger}$ obviously has the same eigenvalues as $\rho$. Thus the von Neumann entropy is [*Lorentz invariant*]{}. Moreover the von Neumann entropy of [*all*]{} reduced density matrices defined by $S_m
\equiv -{\text{Tr}}{(\rho_m\ln\rho_m)}$ with $(m=1,...,{{\ensuremath{\mathbb N}}})$ are also invariant for the same reasons. A physical system defined by $\rho$ can thus be parametrized by a set of Lorentz invariant measures $\{S_1,..., S_m,..., S_{{\ensuremath{\mathbb N}}}=S\}$. We shall proceed to show that $S_m$ can be useful measures of entanglement shortly.

Reduced density matrix, Bell states, and Lorentz-invariant entanglement: a worked example
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Consider a system of two identical fermions, and for the first part of the illustration let us follow the discussion of Ref.[@paskauskas]. A two-particle state 