# Do qutrits violate monogamy of entanglement?

According to monogamy of entanglement, if you have three particles $$A,B\;\&\; C$$, then if $$A$$ is maximally entangled with $$B$$, then $$C$$ is not entangled with either $$A$$ or $$B$$.

However one could imagine a state of qutrits like this

$$|\psi\rangle=\frac{1}{\sqrt{3}}(|012\rangle+|120\rangle+|201\rangle)$$

and clearly measuring any of the particles will tell you the state of the other two. What is wrong here?

In this example $$A$$ and $$B$$ are not entangled at all. Parties $$A$$ and $$B$$ being (maximally) entangled means that the reduced state on $$\mathcal H_A\otimes\mathcal H_B$$ is (maximally) entangled. In this case, you have $$\rho_{AB} \equiv \operatorname{Tr}_C(\mathbb{P}_\psi) = \frac13(\mathbb{P}_{01} + \mathbb{P}_{12} + \mathbb{P}_{20}), \qquad \mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|.$$ Such $$\rho_{AB}$$ corresponds to full correlation between the two parties, but is a separable state.