Does the formula $\sum_k f_k{\rm Tr}(O_k U\rho_k U^\dagger)$ have any physical meaning, in the context of variational quantum algorithms?

Question

I am reading a review about the variational quantum algorithm. And there is a definition of the cost function:

$C(\theta)=\sum_k f_k (Tr[O_k U(\theta)\rho_kU^\dagger(\theta)])$

Where $U(\theta)$ is a parametrized unitary with $\theta$ as the parameters. $\rho_k$ are density matrices, and ${O_k}$ are a set of observables, $f_k$ are functions which encode the task at hand.

I am confused about these two things: (1) Is there any general physical meaning of $Tr[O_k U(\theta)\rho_kU^\dagger(\theta)]$? It feels like I have seen it sometime before but I can't really find it now. (2) Why the cost function is defined this way?

Thank you for your help!

Answer 1

Breaking it down:

1. $\rho' \equiv U \rho U^\dagger$ is the quantum state $\rho$ after applying the parameterized quantum circuit described by $U$.
2. As @StarBucK explained in a comment, ${\rm Tr}[O_k\rho']$ is the density-matrix way of writing the expectation value of the observable $O_k$ (ie. $\langle O_k \rangle$) when the quantum state is $\rho'$. You may be more accustomed to the state-vector notation $\langle\psi|O_k|\psi\rangle$.
3. The cost function $C\equiv \sum_k f_k \langle O_k \rangle$ is simply a weighted average of each expectation value $\langle O_k \rangle$, where $f_k$ serves as the weight.

The system being studied (ie. the "task at hand") determines what each of the $O_k$ are, as well as their corresponding weight $f_k$. Apparently the review you are reading also considers a different initial state $\rho_k$ for each operator $O_k$, although I haven't seen that before.