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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.