Closeness of $\rho$ such that $\text{Tr}(|\psi\rangle\langle\psi|\rho)\le1/2^n+{\cal O}(2^{-2n} )$ for all $|\psi\rangle$ to the maximally mixed state

Question

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?

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Note that the maximally mixed state trivially satisfies this requirement.

Answer 1

We can bound the trace distance between the state $\rho$ and the maximally-mixed state using your condition above to bound the operator norm.

Recall that can write the operator norm of a (Hermitian) matrix as $\|X\|_\infty = \max_{|\psi\rangle} \langle\psi|X|\psi\rangle$. It then follows that $$ \|\rho - I/2^n \|_{\infty} = \max_{|\psi\rangle}\,\langle\psi| \rho|\psi\rangle - \frac{1}{2^n} \leq O\left(\frac{1}{2^{2n}}\right), $$ from the bound on tr$(|\psi\rangle\langle\psi|\rho)$ for any state $|\psi\rangle$.

The operator norm upper bounds the 1-norm up to a dimension factor as $\|X\|_1\leq d \|X\|_\infty$ for a $d\times d$ matrix. Conveniently, the above error is enough to overcome the dimension factor and gives that the state $\rho$ must be (extremely) close to the maximally-mixed state $$ \|\rho - I/2^n \|_1 \leq 2^n\|\rho - I/2^n \|_{\infty} \leq O\left(\frac{1}{2^{n}}\right). $$

(note that an error of $O(1/2^n)$ would not have been enough to prove the claim, at least using this norm inequality.)

Answer 2

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 $\|\sigma-I/2^n\|_F\leq 2$.