2
$\begingroup$

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 $\rho$).

$\endgroup$
1
  • $\begingroup$ 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

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.