3
$\begingroup$

Consider an $n$ qubit state $|\psi\rangle$. Let us say we have the following upper bound on the total variation distance between the distribution generated by measuring $|\psi\rangle$ in the standard basis and the uniform distribution.

$$\sum_{x \in \{0, 1\}^{n}} \left||\langle x | \psi \rangle|^{2} - \frac{1}{2^{n}} \right| \leq a,$$

and, additionally,

$$ \left||\langle x | \psi \rangle|^{2} - \frac{1}{2^{n}} \right| \leq \frac{a}{2^{n}},$$

for every $x$.

Now, let us consider a Pauli observable $O$ as follows:

$$O = \sum_{\sigma_i\in\{I, X, Y, Z\}^{\otimes n}} w_{\sigma_i} \sigma_i. $$

Does an upper bound on the total variation distance imply an upper bound on the following quantity?

$$\left|\langle \psi|O |\psi \rangle - \frac{\text{Tr}(O)}{2^{n}} \right|?$$


My intuition is that a change of basis seems to be at the heart of the question -- we are changing from the standard basis to the Pauli basis. How does the total variation distance change with this change of basis?

$\endgroup$

1 Answer 1

1
$\begingroup$

Probably not in general, no. A useful upper bound needs to beat the state-independent bound \begin{equation} \left| \langle \psi | O | \psi\rangle - \frac{1}{2^n} \text{Tr}(O) \right| \leq \lVert O \rVert_\infty + \frac{1}{2^n}|\text{Tr}(O)| \end{equation}

By counterexample, we can consider the state $|\psi\rangle = H^{\otimes n}|0^n\rangle$ which satisfies the tightest form of your requirement with $a=0$ i.e. $D(\text{diag}(|\psi\rangle \langle \psi|), \mathbb{I}/2^n) = 0$ where $D(p, q)$ denotes total variation distance between two (classical) distributions. But choosing $O = w_{X^{\otimes n}} X^{\otimes n}$, we can saturate the naive bound given above, \begin{equation} \left| \langle \psi | O | \psi\rangle - \frac{1}{2^n} \text{Tr}(O) \right| = \lVert O \rVert_\infty + \frac{1}{2^n}|\text{Tr}(O)| = w_{X^{\otimes n}} \end{equation}

meaning $a$ doesn't play a meaningful role in bounding this quantity in this case, or any case involving this choice of $|\psi\rangle$ and traceless observables composed from $X$ and $I$.

$\endgroup$

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.