# Constructing a maximally entangled qutrit state from $n$ Bell states

I've read that maximally entangled qubit states are a good "unit" of bipartite entanglement since it is possible to create any other entangled state from them using local operations and classical communication (LOCC) provided sufficiently many copies are available. What would the protocol be to construct a maximally entangled qutrit ($\vert \psi \rangle_{AB} = \frac{1}{\sqrt{3}}(\vert 00 \rangle + \vert 11 \rangle + \vert 22 \rangle)$) between two space separated parties from a set of $n$ Bell states ($\vert \Phi^+ \rangle^{\otimes n}_{AB} = 2^{-n/2} (\vert 00 \rangle + \vert 11 \rangle)^{\otimes n}$) initially shared by those parties using only LOCC?

Basically, when you have $n$ Bell-states, you already have a maximally entangled $2^n$ dimensional state. For example, when $n=2$,
\begin{align} \lvert\Phi^+\rangle_{AB}^{\otimes 3} &= \frac12 (\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B)(\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B)\\ &=\frac12 (\lvert00\rangle_A\lvert00\rangle_B + \lvert01\rangle_A\lvert01\rangle_B + \lvert10\rangle_A\lvert10\rangle_B + \lvert11\rangle_A\lvert11\rangle_B\\ &=\frac12 (\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B + \lvert2\rangle_A\lvert2\rangle_B + \lvert3\rangle_A\lvert3\rangle_B ) \end{align}
If you really want to confert thes ququarts to qutrits, then you have to do this probabilistically. If you are not to picky about the success probability, you just have to measure whether the sate written above is in the $\lvert3\rangle$ state or not. If not (which happens with probability 3/4, you have converted 2 qubit maximally entangled pairs into 1 maximally entangled qutrit pairs. You need therefore on average $\frac2{3/4}=8/3=2.667$ qubit pairs per produced qutrit pairs.
Of course, processing more pairs simultaneously allows you to approach the asymptotic rate of $\log_2 3=1.585$ qubit pair per qutrit pairs.