# Entanglement of formation of the mixture of maximally entangled states

Suppose we have two spin-$S$ systems. Let $\left| \psi_{a,b} \right\rangle = \frac{1}{\sqrt{2}}(\left| a,b \right\rangle+\left| b,a \right\rangle)$ be the maximally entangled state. ($a\neq b$ and $-S\leq a,b \leq S$.)

What is the Entanglement of formation for $\rho=\frac{1}{C} \sum_{a\neq b} \left| \psi_{a,b} \right\rangle \left\langle \psi_{a,b} \right|$? $C$ is the normalizing constant.

A trivial upper bound is 1. But can we give a nontrivial upper bound or even calculate it explicitly?

• If I understand correctly, your states are a subset of Bell-diagonal for qubit-qubit states. The entanglement of formation of these states is known exactly, see: quantiki.org/wiki/bell-diagonal-state May 9, 2018 at 8:49
• Why do you call these "maximally entangled"? They are not, except for qubits. May 9, 2018 at 22:31
• @KennethGoodenough For qubits, $\rho$ is pure. May 9, 2018 at 22:32
• @NorbertSchuch Just to clarify, suppose $S=1$. Then $a,b$ can take values from $\{-1,0,1\}$. I want to know the entanglement of formation of the mixture of $\frac{1}{\sqrt{2}}(\left| 0,1 \right\rangle+\left| 1,0 \right\rangle)$, $\frac{1}{\sqrt{2}}(\left| 0,2 \right\rangle+\left| 2,0 \right\rangle)$, and $\frac{1}{\sqrt{2}}(\left| 1,2 \right\rangle+\left| 2,1 \right\rangle)$. I am not sure how to name these states properly. May 10, 2018 at 3:32
• @KennethGoodenough See my previous comment for clarification. May 10, 2018 at 5:50

Since you say that any other entanglement measure is fine, let's compute the negativity. Let me denote by $\rho_{ab}:=|\psi_{a,b}\rangle\langle\psi_{a,b}|$.
With $^{T_A}$ the partial transpose, we have $$\rho_{ab}^{T_A} = \frac12 \Big[ |a,a\rangle\langle b,b|+|b,b\rangle\langle a,a|+ |a,b\rangle\langle a,b|+|b,a\rangle\langle b,a|\Big]\ .$$ Thus (denoting by $D:=2S+1$ the number of basis states), $$\rho^{T_A} = \frac{2}{D(D-1)}\sum_{a>b} \rho_{ab}^{T_A}$$ is block-diagonal with two blocks: $\rho^{T_A}_{ab,a'b'}$ for $a\ne b$, $a'\ne b'$ is diagonal with entries $\tfrac{1}{D(D-1)}$ (i.e., $D(D-1)$ entries), and $\rho^{T_A}_{aa,a'a'}$ (a $D\times D$ matrix) equals $\tfrac{1}{D(D-1)}$ everywhere except on the diagonal (which is zero). Since the latter equals $$\tfrac{D}{D(D-1)}|+\rangle\langle +|-\tfrac{1}{D(D-1)}1\!\!1\ ,$$ with $|+\rangle = (\sum |a\rangle)/\sqrt{D}$, it has eigenvalues $-\tfrac{1}{D(D-1)}$ with multiplicity $D-1$ and $\tfrac{1}{D}$ with multiplicity $1$, respectively.
The sum of the absolute value of the eigenvalues of $\rho^{T_A}$ is thus $$\|\rho^{T_A}\|_1={D(D-1)}\frac{1}{D(D-1)}+(D-1)\frac{1}{D(D-1)}+\frac{1}{D} = 1+\frac{2}{D}\ .$$ The negativity is thus $$\mathcal N(\rho) = \frac{\|\rho^{T_A}\|_1-1}{2} = \frac{1}{D}$$ and the log-negativity $$E_N(\rho) = \log(\|\rho^{T_A}\|_1) = \log(1+2/D)\ .$$
• Thank you, Norbert. Another question. We know logarithmic negativity isn't a faithful entanglement measure. Is $\rho$ necessarily close to a separable state for large D? If so, we can treat $\rho$ as if it were a separable state. May 14, 2018 at 3:41
• In trace norm. What concerns me is that $\rho$ may be still far from separable in the trace norm even $E_N(\rho) \to 0$. May 14, 2018 at 8:56
• @user193108 Then I would suggest you ask your actual question (i.e. separating $\rho$ from the Bell state -- BTW, what does that even mean, they live in different Hilbert spaces!), and if you want, you can subsequently give your thoughts which relate it to other question. This way, you ask a question which is probably much harder than what you actually want to know! May 15, 2018 at 11:54