# Given an observable $O$, what's the achievable maximum value of $\operatorname{Tr}(O\rho)$?

The maximum value of expectation value of an observable $$O$$ with respect to a density matrix $$\rho$$ can be computed by using Holder's inequality as follows:

$$$$\text{Tr}(O\rho) \leq \Vert O \Vert_p \Vert \rho \Vert_q, ~~ \text{where} ~~ \frac{1}{p} + \frac{1}{q} = 1.$$$$ Consider $$O = \sigma_x$$ (single-qubit Pauli X operator). Choosing $$p = \infty, q = 1$$ gives $$\text{Tr}(O\rho) \leq 1$$, which is correct. But choosing $$p = 1, q = \infty$$ gives $$\text{Tr}(O\rho) \leq \Vert O \Vert_1 = 2$$, which cannot be achieved for any density matrix $$\rho$$. So my question is as follows: given an arbitrary Hermitian observable $$O$$, how do we actually know the (achievable) maximum value of its expectation value? (Perhaps it depends on rank of $$\rho$$).

• You have to realize that the natural norm for density matrices is the $1$-norm. Thus, the right Hölder inequality is the one where $q=1$, and this is the one which can he made tight. Aug 25, 2023 at 15:28

Expectation values are bounded by the spectrum of the observable $$O$$. This follows essentially from the definition of eigenvalues and the fact that Hermitian operators always have real eigenvalues:

$$\lambda_{\mathrm{min}} \leq \langle \psi | O | \psi \rangle \leq \lambda_{\mathrm{max}},$$

where the equalities are saturated whenever $$|\psi\rangle$$ is a minimal or maximal eigenvector, respectively. Note that density operators are just convex combinations of pure states so this statement applies just as well to them:

$$\mathrm{tr}(O\rho) = \sum_i p_i \langle \psi_i | O | \psi_i \rangle \leq \sum_i p_i \lambda_{\mathrm{max}} = \lambda_{\mathrm{max}},$$

and similarly with the lower bound. Indeed, pure states constitute the extremal points of the set of density operators (which is a convex body).

Two further things to note. One, it doesn't really depend on the rank of $$\rho$$. Indeed, suppose $$O$$ has a highly degenerate eigenspace associated with $$\lambda_{\mathrm{max}}$$. Then any convex combination of basis vectors in that eigenspace is a high-rank density operator with the same expectation value as a pure state (rank-1) from that eigenspace. And at any rate, if your question is "What is the maximum expectation value of an observable," then this is essentially a state-independent question (i.e., you are asking to maximize over all possible states).

Second, your observation with Holder's inequality simply is a reflection of the fact that it is not tight if you choose "suboptimal" values of $$p, q$$. Indeed, $$p = \infty$$ and $$q = 1$$ is the optimal choice in this setting because $$\|O\|_\infty$$ is precisely the spectral norm of the observable and $$\|\rho\|_1$$ is the sum of singular values of $$\rho$$, which is always $$1$$ for density operators.