3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ i think the claim follows directly from the operator norm, and given how small the error term you've written above is, you can also upper bound the 1-norm distance. if what i said isn't clear i'm happy to write a longer answer $\endgroup$
    – 4xion
    Dec 15, 2021 at 5:49
  • $\begingroup$ Yes, a longer answer would have been much appreciated. $\endgroup$
    – BlackHat18
    Dec 15, 2021 at 6:20
  • $\begingroup$ out of curiosity, what was the motivation for this question? $\endgroup$
    – 4xion
    Dec 15, 2021 at 21:37

2 Answers 2

3
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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