What is the relationship between expectation values of two commuting operators? Given a state $|\psi(t)\rangle$, if the operators A and B commute, and the expectation value of A is constant in time, (i.e., $\langle\psi(t)|A|\psi(t)\rangle$ = constant) what can we infer about the expectation value of B ($\langle\psi(t)|B|\psi(t)\rangle$)? The state $|\psi(t)\rangle$ is expressed in the eigenstates of A and B.
Edit: Neither A or B are the identity operator.
 A: We can't infer anything about the expectation value of B, actually. For instance, A could be the identity operator (i.e. the number 1). Then it will commute with any operator and will always have a constant expectation value of 1 for any state, but of course this fact doesn't help us calculate other expectation values.
Even if $A$ and $B$ have no degeneracies, you still may not be able to tell very much from the fact that $\langle A \rangle$ is constant. Suppose we're in a 3 dimensional Hilbert space and, in their joint eigenbasis,
\begin{equation}
A=\left(
\begin{matrix}
1 & 0&0 \\
0& 0 & 0\\
0& 0 &-1
\end{matrix}
\right),\qquad
B=\left(
\begin{matrix}
-2 & 0&0 \\
0& 1 & 0\\
0& 0 &0
\end{matrix}
\right)
\end{equation}
and we write the wavefunction as
$$
|\psi(t)\rangle = 
\left(
\begin{matrix}
c_1(t)\\
c_2(t)\\
c_3(t)
\end{matrix}
\right)
$$
Suppose we have a wavefunction varying in time but always obeying $|c_1(t)|^2 = |c_3(t)|^2$. Then we will always have $\langle A \rangle = 0 $. But $\langle B \rangle$ on the other hand could be oscillating anywhere between $-1$ (in the case $c_2 = 0$) and $1$ (in the case $c_2 = 1$) in a completely uncontrolled manner.
