# Compute the negativity of maximally entangled (pure, bipartite) states

The entanglement negativity $$\mathcal N(\rho)$$ of a (bipartite) state $$\rho$$ is defined as the absolute value of the sum of the negative eigenvalues of the partial transpose of a state, or equivalently, $$2\mathcal N(\rho)=\|\rho^{T_B}\|_1-1$$.

Consider a maximally entangled pure state: $$|\psi\rangle \simeq \sum_{k=1}^N |k,k\rangle$$. What is the entanglement negativity of such states? What are good ways to derive this quantity?

## 1 Answer

Let $$\sqrt N|\psi\rangle=\sum_{k=1}^N|k,k\rangle$$ be a maximally entangled state of dimension $$N$$, and $$\rho\equiv|\psi\rangle\!\langle\psi|$$. More generally, we don't need to stick to maximally entangled states: any state with this (or equivalent) Schmidt decomposition will behave identically.

The partial transpose reads $$\rho^{T_B}=\frac{1}{N}\sum_{ij}\lvert ij\rangle\!\langle ji\rvert.$$ Separating this into terms that are symmetric and terms that are not we get $$N\rho^{T_B} = \sum_i \lvert ii\rangle\!\langle ii| + \underbrace{\sum_{i\neq j}\lvert ij\rangle\!\langle ji\rvert}_{A},$$ where here $$A^\dagger =A$$, $$\operatorname{tr}(A)=0$$, and $$A^2=I$$. It follows that $$A$$ has an equal number of eigenvalues equal to $$+1$$ and $$-1$$. Moreover, the rank of $$A$$ is $$N(N-1)$$, so the multiplicity of both eigenvalues is $$N(N-1)/2$$.

The eigenvalues of $$\rho^{T_B}$$ are therefore $$\frac1 N$$ with multiplicity $$N+\frac{N(N-1)}2=\frac{N(N+1)}{2}$$, and $$-\frac1 N$$ with multiplicity $$\frac{N(N-1)}2$$.

We conclude that the negativity equals $$\frac{N(N-1)}{2}\frac{1}{N}=\frac{N-1}{2}$$.