# Bounds on local expectation values for two states close in trace distance

I feel like this should have been recorded somewhere but I could not find any result in the literature (except in very specific cases). Consider two states $$\rho,\sigma$$ such that they are $$\epsilon$$-close in trace distance: \begin{align} T := \frac{1}{2}||\rho-\sigma||_1 = \epsilon. \end{align} In this case I would like to consider these states to describe $$N$$-partite qubits so the full dimension is $$2^N$$. $$A$$ be some $$k$$-local observable, say $$k$$-tensor product of Pauli operators. Is there some good and generic bound (with or without additional promises) on the expectation value of $$A$$, i.e., \begin{align} \text{Tr}(A(\rho-\sigma)) \end{align} is somehow $$\epsilon$$-close?

In the context of things that I have seen, this tends to appear in the context of de Finetti theorem or mean-field applications, where $$\sigma$$ is a product state approximation of $$\rho$$. In those cases, there are situations where it is possible to estimate, say, ground state energy (see, e.g., here), but I could not find a more generic result and I am not sure if it is because it is not possible or something else. The exception I found was in the case when I have bosonic or fermionic systems (that obeys canonical (anti-)commutation relations, where Hudson's theorem (see, e.g., Appendix A here) implies the above (and becomes exact in the infinite-volume/thermodynamic limit).

If $$\sigma$$ is fixed to be product state approximation of $$\rho$$, then what I am asking boils down to the question of whether there is a "finite de Finetti theorem" for expectation value of observables that becomes exact in the thermodynamic limit?

Remark: I am aware of the fact that in general, closeness in trace distance does not imply closeness in some information-theoretic quantities of interest: for example, two states close in trace distance can have very large separation in entropy, as Fannes-Audenaert inequality suggests. It is not clear to me if this also applies to local observables in general.

We first have: $$|\mathrm{tr}(A(\rho-\sigma))|\leqslant\mathrm{tr}(|A(\rho-\sigma)|)=\|A(\rho-\sigma)\|_1$$ We can then use Hölder's inequality: $$\|A(\rho-\sigma)\|_1\leqslant\|\rho-\sigma\|_1\|A\|_{\infty}=\left|\lambda_{\text{max}}\right|\varepsilon$$ where $$\lambda_{\text{max}}$$ is the largest eigenvalue of $$A$$ in absolute value.
• @EvangelineA.K.McDowell I might be wrong on this, but I don't think an additional assumption on $A$ would work, since even if $A$ is a Pauli word we can probably find $\rho$ and $\sigma$ to saturate this bound. Maybe an assumption on $\rho$ and $\sigma$ might help, but I'm not sure about this. Commented Apr 7 at 11:08