# Permutation invariant states have permutation invariant purifications - proof?

I don't remember where I came across the statement but I'm pretty sure it is true and am interested in understanding why it holds. For any $$n-$$ register state $$\rho^n \in H^{\otimes n}$$ that is permutation invariant, there exists a purification that is also permutation invariant.

That is, if

$$\pi \rho^n = \rho^n$$

for any permutation $$\pi$$ among the $$n$$ registers, then there exists a purification $$\vert\Psi^n\rangle\langle\Psi^n\vert \in (H\otimes H)^{\otimes n}$$ of $$\rho^n$$ such that

$$(\pi\otimes \pi)\vert\Psi^n\rangle\langle\Psi^n\vert = \vert\Psi^n\rangle\langle\Psi^n\vert$$

for any permutation $$\pi$$. How does one see that this is true?

• Yes, this is true, and not hard to prove. There' s a proof in Renato Renner's PhD thesis. Jun 2, 2021 at 8:44
• Extended @Mateus 's answer. Here is the thesis link: arxiv.org/pdf/quant-ph/0512258.pdf , see lemma 4.2.2 Jun 2, 2021 at 17:07