How is $phi$ an expectation value and not an operator in this AdS/CFT discussion?

I’m getting started with AdS/CFT and there is one particular point that I’m struggling to understand. Consider the example of a scalar field $phi$ in ${rm AdS}_{d+1}$. In global coordinates $(rho,tau,hat{x})$ with metric $$ds^2=dfrac{L^2}{cos^2rho}(-dtau^2+drho^2+sin^2rho dOmega_{d-1}^2)$$ $phi$ will have asymptotic expansion of the form $$phi(rho,tau,hat{x})=(cosrho)^{Delta_-}alpha(tau, hat x)+(cosrho)^{Delta_+}beta(tau,hat x)+cdots$$

The functions $alpha$ and $beta$ on the boundary are not determined by the equations of motion and are fixed by boundary conditions. The standard one is to fix $alpha$.

Now, what is often said is that this bulk theory is dual to a ${rm CFT}_d$ on the boundary with scalar primary operator ${cal O}(tau,hat{x})$ such that $alpha(tau,hat{x})$ is a source for ${cal O}(tau,hat{x})$ and $beta(tau,hat{x})$ is its expectation value.

This confuses me for the following reason: shouldn’t we have both a quantum theory in ${rm AdS}_{d+1}$ and in its boundary? In that case both ${cal O}(tau,hat{x})$ should be an operator in the ${rm CFT}_d$ Hilbert space and $phi(rho,tau,hat x)$ should be an operator in the bulk Hilbert space.

But when we say that $alpha$ is a source and $beta sim langle {cal O}rangle$ we are effectively saying that $phi$ is just a classical field configuration (not an operator). Shouldn’t we have in some sense $beta sim {cal O}$ as an operator statement, i.e., true inside expectation values?

What is the point of saying that $alpha$ is a source and $beta$ is an expectation value? How this does not conflict with $phi$ being an operator-valued field?