Can the expectation value behave like a function? If $\langle\hat x\rangle_{\Psi} = 0$ can we conclude by the Ehrenfest theorem that $\langle\hat p\rangle_{\Psi}=0$ ?
I've seen that in this video but I'd rather think that an expectation value is like a function (it's derivative can be non-zero when the function itself is zero). What do you think?
 A: Consider a harmonic oscillator with mass $m$ and frequency $\omega$,
and the state :
$$|\psi\rangle = \frac 1{\sqrt{2}}\left(|0\rangle + i|1\rangle\right)$$
In this state we have :
\begin{align}
\langle\hat x\rangle_\psi &= \sqrt{\frac{\hbar}{2m\omega}}\langle\psi|a+a^\dagger|\psi\rangle\\
&= \frac 12 \sqrt{\frac{\hbar}{2m\omega}} (\langle 0 |-i\langle1|)(a+a^\dagger)(|0\rangle + i|1\rangle) \\
&= \frac 12 \sqrt{\frac{\hbar}{2m\omega}} (\langle 0 |-i\langle1|)(i|0\rangle + |1\rangle + i\sqrt 2|2\rangle)\\
&=\frac 12 \sqrt{\frac{\hbar}{2m\omega}}(i-i)\\
&= 0
\end{align}
and :
$$|\psi\rangle = \frac 1{\sqrt{2}}\left(|0\rangle + i|1\rangle\right)$$
In this state we have :
\begin{align}
\langle\hat p\rangle_\psi &=-i \sqrt{\frac{\hbar m\omega}{2}}\langle\psi|a-a^\dagger|\psi\rangle\\
&= \frac 1{2i} \sqrt{\frac{\hbar m\omega}{2}}(\langle 0 |-i\langle1|)(a-a^\dagger)(|0\rangle + i|1\rangle) \\
&= \frac 1{2i} \sqrt{\frac{\hbar m\omega}{2}}(\langle 0 |-i\langle1|)(i|0\rangle - |1\rangle -i\sqrt 2|2\rangle)\\
&=\frac 1{2i}\sqrt{\frac{\hbar m\omega}{2}}(i+i)\\
&\neq 0
\end{align}
