Is it possible to get $\langle AB \rangle$ from $\langle A \rangle$ and $\langle B \rangle$, and vice versa?

Question

Assuming that $A$ and $B$ are operators (not necessarily observables) which do not commute and that the quantum system in an arbitrary state $| \psi \rangle$, then ist it possible to get $\langle AB \rangle$ from $\langle A \rangle$ and $\langle B \rangle$, and vice versa ?

Take the example of a quantum harmonic oscillator where the operators involved are ladder operators. I'm using Ehrenfest theorem $$ \frac{d}{dt} \langle A \rangle = \frac{i}{\hbar}\langle [H,A] \rangle + \langle \frac{d}{dt} A \rangle $$ which yields ordinary differential equations for the expectation values of the operator $A$, no states are mentioned. So $A$ in the last equation above is to be $a^\dagger, a, a^\dagger a$ and $a^2$. So for example if one has that the initial values for $\langle Q \rangle$ and $\langle P \rangle$, it is easy to find the those for $\langle a \rangle$ and $\langle a^\dagger \rangle$. But is it possible from those initial conditions to get the corresponding initial expectation value $\langle T \rangle$, where $T = \frac{1}{2m}P^2$ is the kinetic energy, to get the time evolution of $\langle T \rangle$?

Answer 1

No, it is not possible for an arbitrary state: Let $\{|n\rangle\}_{n\in \mathbb N_0}$ denote the set of (normalized) eigenvectors of $N\equiv a^\dagger a$. From the ladder operator algebra we find that

$$\langle n|a|n\rangle =\langle n|a^\dagger|n\rangle = 0 \quad , $$

for all $n$. On the other hand, it holds that $$\langle n|N|n\rangle = n \quad . $$

Consequently, it is not possible to find the expectation value of $N$ from the knowledge of the expectation values of $a$ and $a^\dagger$ with respect to these states, since the latter are always zero, while the former can be any natural number.

However, note that here in this special case (of being in an eigenstate) the converse is true: Given the expectation value of $N$ with respect to its eigenstates, we trivially know the corresponding expectation values of $a$ and $a^\dagger$.