# Upper bounding a permutation invariant state

Let $$\rho_{A^n}$$ be a permutation invariant quantum state on $$n$$ registers i.e. $$\pi(A^n)\rho_{A^n}\pi(A^n) = \rho_{A^n}$$ for any permutation $$\pi$$ among the $$n$$ registers.

If we trace out $$n-1$$ registers (doesn't matter which due to permutation invariance), we obtain the reduced state $$\rho_A$$. One knows that $$\text{supp}(\rho_{A^n})\subseteq \text{supp}(\rho_{A}^{\otimes n})$$.

What is the minimal $$\lambda_n$$ such that $$\rho_{A^n} \leq \lambda_n \rho_{A}^{\otimes n}$$ where $$A\leq B$$ denotes that $$B-A$$ is positive semidefinite? In particular, is $$\lambda_n$$ necessarily exponential in $$n$$?

Why not take the example of the GHZ state? $$|GHZ\rangle=(|0\rangle^{\otimes n}+|1\rangle^{\otimes n})/\sqrt{2},$$ such that $$\rho_{A^n}=|GHZ\rangle\langle GHZ|$$. The $$\rho_A=I/2$$ and $$\rho^{\otimes n}_A=I/2^n$$. Then for this specific case $$\lambda \rho^{\otimes n}_A-\rho_{A^n},$$ the eigenvalues are $$\lambda/2^n-1$$ (once) and $$\lambda/2^n$$ ($$2^n-1$$ times). Hence you get the inequality iff $$\lambda\geq 2^n$$. So yes, $$\lambda$$ must be exponential in $$n$$.
• Thank you - that was a good state to try. From what I see, it seems like if I want a positive semidefinite matrix of the form $\sigma^{\otimes n}$ that is an upper bound to $\vert GHZ\rangle\langle GHZ\vert$, the smallest choice is setting $\sigma$ to the identity matrix. So even if I allowed $\lambda\rho_A^{\otimes n}$ to be replaced by an arbitrary tensor product state, this still does not help if I understand correctly. Mar 31 at 9:49