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 $n$?


1 Answer 1


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 $n$.

  • $\begingroup$ 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. $\endgroup$ Mar 31, 2021 at 9:49
  • 1
    $\begingroup$ Yes, that sounds right. $\endgroup$
    – DaftWullie
    Mar 31, 2021 at 10:02

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