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:

\begin{equation} \text{Tr}(O\rho) \leq \Vert O \Vert_p \Vert \rho \Vert_q, ~~ \text{where} ~~ \frac{1}{p} + \frac{1}{q} = 1. \end{equation} 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 $\rho$).

  • $\begingroup$ 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. $\endgroup$ Aug 25, 2023 at 15:28

1 Answer 1


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.


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