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?

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?

• The trace you are referring to is the average value of the observable $O_k$ when the quantum state is $U(\theta)\rho_k U^{\dagger}(\theta)$ Sep 16 '21 at 13:59
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.