Consider an $n$ qubit density matrix $\rho$ such that

$$\text{Tr}(|\psi\rangle\langle \psi| ~\rho) \leq \frac{1}{2^{n}} + \mathcal{O}\left(\frac{1}{2^{2n}} \right), $$

for every $n$ qubit pure state $|\psi \rangle$.

Is there some way to argue that the density matrix is close to the maximally mixed state, with respect to some distance measure?

Note that the maximally mixed state trivially satisfies this requirement.


I'm not sure that this is a very good bound, but it might be a start.

Let $\rho=\sum_v \alpha_v|v\rangle\langle v|$ be the spectral decomposition. We then remark that from the assumption, we get a bound on the matrix 2-norm or (Frobenius norm)$$\|\rho\|_F^2=Tr(\rho^*\rho)=Tr(\rho^2)=\sum_v \alpha_v\langle v|\rho|v\rangle\leq \frac{rank(\rho)}{2^n}+O(1/2^{2n}),$$ by noting each $|v\rangle$ is a pure state.

Then the triangle inequality says that $$\|I/2^n-\rho\|_F\leq\|I/2^n\|_F+ \|\rho\|_F\leq 1+\sqrt{rank(\rho)/2^n+O(1/2^{2n})}.$$

However, this bound isn't really that great, since for any density matrix $\sigma$ we have $\|\sigma\|_F\leq 1$ and thus, for any density matrix, we have that $\|\sigma-I/2^n\|_F\leq 2$.


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