# How does the Entropy of Entanglement generalize to tripartite and multipartite pure states?

Entropy of Entanglement is a very beautiful measure of entanglement.

I wonder how this concept generalizes to three-parite entangled states, such as $$\left| \psi \right\rangle=\frac{1}{\sqrt{2}}\left(\left| 0,0,0 \right\rangle + \left| 1,1,1 \right\rangle \right)$$ or for higher-dimensional systems $$\left| \psi \right\rangle=\frac{1}{\sqrt{3}}\left(\left| 0,0,0 \right\rangle + \left| 1,1,1 \right\rangle + \left| 2,2,2 \right\rangle \right)$$ or for asymmetrically entangled pure states $$\left| \psi \right\rangle=\frac{1}{\sqrt{3}}\left(\left| 0,0,0 \right\rangle + \left| 1,0,1 \right\rangle + \left| 2,1,0 \right\rangle \right).$$

Question 1: What is the Entropy of Entanglement of these three states?

Question 2: If Entropy of Entanglement cannot be calculated for pure, tripartite states, are there related measures that I can apply to these three-partite pure states?

Thank you very much!

• I disagree with your first sentence: EoF is an important concept for MIXED states. For pure states, EoF, distillable entanglement, etc., are all the same concept: The "entropy of entanglement". (Which is indeed a beautiful thing!) -- So is your question about pure or mixed states? – Norbert Schuch Oct 5 at 8:35
• @NorbertSchuch Thanks, i changed the first sentence to avoid confusion. I am only interested in pure states, and want to see how this generalizes to more than two particles, especially in a higher state-space. Thank you! – NiceDean Oct 6 at 2:40
• I think the short answer is that this concept does not generalize, at least not with the same simplicity and elegance: Three- and multi-partite systems are complicated. – Norbert Schuch Oct 6 at 11:31
• @NorbertSchuch My specific question, as stated in the main text, is the entropy of entanglement of the three states. They are the most simplest examples of multiparty states, i cannot believe that this measure (or a related measure) is not applicable. Can you please provide references for your claims? – NiceDean Oct 6 at 18:12

For a bipartite state $$\vert\psi\rangle_{AB}$$, the entropy of entanglement (EoE) is defined as $$E(\vert\psi\rangle) = S(\mathrm{tr}_B\vert\psi\rangle\langle\psi\vert)\ ,$$ with $$S(\rho)=-\mathrm{tr}(\rho\log\rho)$$ the von Neumann entropy.
Why is the entropy of entanglement such a great measure for the entanglement in a pure bipartite state? The reason is that it uniquely quantifies the entanglement in a bipartite state in an asymptotic setting -- just as in classical information theory the Shannon entropy quantifies the information content (=compressibility) of a data souce. Specifically, given many copies $$N_1$$ of a state $$\vert\psi_1\rangle$$, they can be asymptotically converted with arbitrary accuracy to $$N_2$$ copies of another state $$\vert\psi_2\rangle$$, and back, using only local operations and classical communication (LOCC), as long as the ratio $$\frac{N_1}{N_2}\to \frac{E(\vert\psi_2\rangle)}{E(\vert\psi_1\rangle)}\ ,$$ or equivalently $$N_1E(\vert\psi_1\rangle)\approx N_2E(\vert\psi_2\rangle)\ .$$