# On $k$-extendability of bipartite states

Definition of $k$-extendability can be given as follows.

Let $k\in \mathbb{N}$. A state $\rho_{AB}$ on a bipartite Hilbert space $\mathrm{A}\otimes\mathrm{B}$ is $k$-extendible with respect to $\mathrm{B}$ if there exists a state $\rho_{AB^k}$ on $\mathrm{A}\otimes\mathrm{B}^{\otimes k}$ which is invariant under any permutation of the $\mathrm{B}$ subsystems and such that $\rho_{AB}=\mathrm{Tr}_{B^{k-1}}\rho_{AB^k}$.

Further a result of Doherty et al (Phys. Rev. A. 69:022308) gives that,

A state on a bipartite Hilbert space $\mathrm{A}\otimes\mathrm{B}$ is separable if and only if it is $k$-extendible with respect to $\mathrm{B}$ for all $k\in\mathbb{N}$.

If this is the case, then isn't a pure maximally entangled state (say $|00\rangle + |11\rangle$) is also extendable for all $k$? Consider the extension $|00\cdots0\rangle + |11\cdots1\rangle$ for some $k$. I think, I am missing some obvious point. Can someone please help?

• A k-partite maximally entangled state is a pure state (symmetrical to permutations), but its partial trace over any of its components is no longer a pure state. In other words, the partial trace does not recover the pure bipartite maximally entangled state. – udrv Mar 28 '16 at 13:13

isn't a pure maximally entangled state (say $|00\rangle + |11\rangle$) is also extendable for all $k$? Consider the extension $|00\cdots0\rangle + |11\cdots1\rangle$ for some $k$.
Take $k=2$ and $$\rho_{AB^2}=\frac12\left(\vphantom{\sum}|000⟩+|111⟩\right)\left(\vphantom{\sum}⟨000|+⟨111|\right),$$ as it seems that you're proposing, and calculate \begin{align} {\mathrm{Tr}_{B^1}}\!\!\left(\rho_{AB^2}\right) & = \frac12\mathrm{Tr}_{3}\!\!\left(\vphantom{\sum}|000⟩+|111⟩\right)\left(\vphantom{\sum}⟨000|+⟨111|\right) \\& = \frac12\mathrm{Tr}_{3}\!\!\left(\vphantom{\sum} |000⟩⟨000|+|000⟩⟨111|+|111⟩⟨000|+|111⟩⟨111| \right) \\& = \frac12\left(\vphantom{\sum} |00⟩⟨00|+|11⟩⟨11| \right). \end{align} This is a completely mixed state, having lost all coherence to the vanishing traces $\mathrm{Tr}\:|1⟩⟨0|=\mathrm{Tr}\:|0⟩⟨1|=0$ on the cross terms, and it has nothing at all to do with the pure state $$\rho_{AB}=\frac12\left(\vphantom{sum}|00⟩+|11⟩\right)\left(\vphantom{\sum}⟨00|+⟨11|\right)$$ you were hoping for.
Moreover, this fits in perfectly well with the result you quote: the maximally entangled pure state is not separable, so no $k$-extension is possible.