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.


1 Answer 1


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.

  • $\begingroup$ Thanks! This is what I am looking for more or less. Just wondering before I close this question: is this bound more or less optimal, i.e., I cannot do any better unless I give more (possibly strong) promises)? $\endgroup$ Commented Apr 6 at 15:38
  • $\begingroup$ @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. $\endgroup$
    – Tristan Nemoz
    Commented Apr 7 at 11:08

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