# 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 – Kenneth Goodenough May 9 '18 at 8:49
• Why do you call these "maximally entangled"? They are not, except for qubits. – Norbert Schuch May 9 '18 at 22:31
• @KennethGoodenough For qubits, $\rho$ is pure. – Norbert Schuch May 9 '18 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. – user193108 May 10 '18 at 3:32
• @KennethGoodenough See my previous comment for clarification. – user193108 May 10 '18 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. – user193108 May 14 '18 at 3:41
• In trace norm. What concerns me is that $\rho$ may be still far from separable in the trace norm even $E_N(\rho) \to 0$. – user193108 May 14 '18 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! – Norbert Schuch May 15 '18 at 11:54