Upper bounding a permutation invariant state

Question

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 $Aleq B$ denotes that $B-A$ is positive semidefinite? In particular, is $lambda_n$ necessarily exponential in $n$?

DaftWullie · Accepted Answer

Why not take the example of the GHZ state?
$$
|GHZrangle=(|0rangle^{otimes n}+|1rangle^{otimes n})/sqrt{2},
$$
such that $rho_{A^n}=|GHZranglelangle GHZ|$. 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 $lambdageq 2^n$. So yes, $lambda$ must be exponential in $n$.

Upper bounding a permutation invariant state

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