# Quantum Mechanics: Time dependence of an expectation value

In Griffiths's Introduction to Quantum Mechanics, he says that the time dependence of an expectation value is $$\frac{d\langle Q\rangle}{dt}=\frac{i}{\hbar}\langle [H,Q]\rangle+\langle \frac{\partial Q}{\partial t}\rangle$$ And I also saw a lecture note saying that time dependence of an expectation value is: $$\frac{\partial}{\partial t}\langle \hat{Q}(t)\rangle=-\frac{i}{h}\langle [\hat{Q}, \hat{H}]\rangle$$

I know they are saying the same thing. For me the second one is quite easy to understand. But how can I understand the first one?

• possible duplicate of Does Heisenberg equation of motion imply the Schrodinger equation for evolution operator?, where your question is answered in the first part of the answer by @joshphysics. In particular, your first equation is just equation $(\star \star)$ multiplied on the left and on the right by some ket (bra) $|\psi \rangle$ ($\langle \psi |$) – Danu Jul 15 '14 at 19:10
• I have outlined this derivation (by joshphysics) in more detail, as I wasn't sure whether the OP would be able to follow otherwise. – Danu Jul 15 '14 at 19:24

The first equation is quite elementary to derive. First, we define the expectation value of an operator: $$\langle O \rangle=\langle \psi(t)| O | \psi (t)\rangle = \langle \psi(t=0)|U^\dagger O \ U |\psi (t=0)\rangle \tag{1}$$ where $U$ is the time evolution operator: $U=\exp\left( -i\frac{Ht}{\hbar}\right)$ if $H$ is independent of time.
Now, we can take the time derivative of (1), which will clearly have three terms (by the product rule). For the time evolution operator, we have the Schrodinger equation: $$\frac{d}{dt} U=\frac{1}{i\hbar}HU$$ and its hermitian conjugate $$\frac{d}{dt} U^\dagger=-\frac{1}{i\hbar}U^\dagger H$$ where we used that $H=H^\dagger$. Applying these two equations we find: $$\frac{d}{dt} \langle O \rangle =\frac{i}{\hbar}\langle [H,O]\rangle+\left\langle \frac{\partial O}{\partial t}\right\rangle$$ where we have defined $$\left\langle \frac{\partial O}{\partial t}\right\rangle=\langle \psi(t=0)|U^\dagger \frac{d O}{dt} \ U |\psi (t=0)\rangle$$
$$\hat{Q} = \hat{Q}(\hat{x}(t),\hat{p}(t),t)$$