Why does the Schrödinger equation tell us about motion? The Schrodinger equation reads $$-\frac {\hbar^2}{2m} \frac {\partial^2\psi (x)}{\partial x^2} + V(x) \psi(x) = E\psi(x)$$ and $$i\hbar \frac {\partial \psi (x,t)}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V(x,t)\right]\psi(x,t)$$
When we solve this equation for $\psi (x,t)$, my understanding is this tells us where a particle is at a point in time, or rather, the probability that it is there. Furthermore, we can write the solutions to this equation as a linear combination of stationary states, which are restricted to a certain time with the relevant time dependence tacked on. My interpretation is that the total information we have from this equation and it's solutions is if we have identical ensembles of particles, the chances that a particle will be in a given position at a given time and how this value changes over time. However, I have read that the Schrodinger equation is the quantum analog of Newton's Laws and that it governs the motion of particles. I am currently on chapter 3 of Griffiths, and I am wondering why the claim is made that this tells us anything about motion for quantum particles. to me, it only characterizes position, and the beginning of Griffiths makes sure to point out that the derivative of the expectation of position is not velocity in the sense of the moving particle, but is how the average position probabilities are expected to change as time evolves.
To sum up the question: why can we conclude that this equation and its solutions tell us anything about motion?
 A: I wouldn't necessarily say that the Schrodinger equation is specifically about the motion of particles, since the classical ideas of position and velocity are no longer generally meaningful. Instead, the Schrodinger equation tell us how to evolve the state of our system forward in time, and it is in this way that it is analogous to Newton's laws.
In Newtonian mechanics, the state of the system is given by a list of the positions and velocities of each particle. If you know the state of the system at time $t=0$ then Newton's laws will tell you how to determine the state of the system at some later time.
Analogously, in Quantum mechanics the state of the system is (roughly) given by its wavefunction. If you know the wavefunction at a time $t=0$ then the Schrodinger equation will tell you how to determine the wavefunction of the particle at some later time.
A: First of all, concerning the equations: Putting $\psi(t,\mathbf{x})=e^{-\frac{i}{\hbar}Et}\psi(\mathbf{x})$ in the second equation and shortening $e^{-\frac{i}{\hbar}Et}$ from both sides yields the second equation as you probably already know. (By the way, I find the notation $\nabla^2$ a bit awkward, since when applied to a vector field $\mathbf{v}$, it could mean $\nabla\cdot(\nabla\times\mathbf{v})$ or $\Delta\mathbf{v}=(\nabla\cdot\nabla)\mathbf{v}$.) But there is also another condition for the wave function: Due to the Schrödinger equation being linear, for every solution $\psi$, the function $\lambda\psi$ with $\lambda\in\mathbb{C}$ is also a solution. Because of the interpretation as probability, we only consider normed (In mathematics it is called the $L^2$ norm.) solutions with:
\begin{align*}
\langle\psi|\psi\rangle
&=\int_{\mathbb{R}^3}\langle\psi|\mathbf{x}\rangle\langle\mathbf{x}|\psi\rangle\mathrm{d}^3\mathbf{x}
=\int_{\mathbb{R}^3}\langle\mathbf{x}|\psi\rangle^*\langle\mathbf{x}|\psi\rangle\mathrm{d}^3\mathbf{x} \\
&=\int_{\mathbb{R}^3}\psi(\mathbf{x})^*\psi(\mathbf{x})\mathrm{d}^3\mathbf{x}
=\int_{\mathbb{R}^3}|\psi(\mathbf{x})|^2\mathrm{d}^3\mathbf{x}
\stackrel{!}{=}1.
\end{align*}
The average value measured for a physical observable is now given as the expection value in such a quantum state, for example for the Hamilton operator $\widehat{H}(\mathbf{x})=-\frac{\hbar^2}{2m}\Delta+V(\mathbf{x})$, which shortens the first equation to $\widehat{H}(\mathbf{x})|\psi\rangle=E|\psi\rangle$, we have:
\begin{equation}
\langle\widehat{H}(\mathbf{x})\rangle
=\langle\psi|\widehat{H}(\mathbf{x})|\psi\rangle
=\langle\psi|E|\psi\rangle
=E\underbrace{\langle\psi|\psi\rangle}_{=1}=E.
\end{equation}
There you also see, why we wanted the solution to be normed, it's so that the expection value of the Hamilton operator is exactly the energy of the quantum state. You can interpret the expectation value, which is an integral, as the possible values of the measure process weightened by their likelihood to appear. For example for the momentum operator $\widehat{p}=-i\hbar\nabla$, we have:
\begin{equation}
\langle\widehat{p}\rangle
=\langle\psi|\widehat{p}|\psi\rangle
=\int_{\mathbb{R}^3}\psi(\mathbf{x})^*(-i\hbar\nabla)\psi(\mathbf{x})\mathrm{d}^3\mathbf{x}.
\end{equation}
This value depends on the respective wave function, which is a solution to the Schrödinger equation. Given an observable $\widehat{O}$, you could now try to look at the time derivative of $\langle\widehat{O}\rangle=\langle\psi|\widehat{O}|\psi\rangle$, which you can do by applying the product rule considering the three time-dependend terms ($\langle\psi|$, $\widehat{O}$ and $|\psi\rangle$) and use the time-dependend Schrödinger equation. This is called the Ehrenfest theorem and shows, you do indeed get classical mechanics as a limit, especially Newton's Third law, for the expection values. In short, we have:
$$m\frac{\mathrm{d}}{\mathrm{d}t}\langle\mathbf{x}\rangle
=\langle\mathbf{p}\rangle$$
$$\frac{\mathrm{d}}{\mathrm{d}t}\langle\mathbf{p}\rangle
=-\langle\nabla V(\mathbf{x})\rangle$$
and therefore:
$$m\frac{\mathrm{d}^2}{\mathrm{d}t^2}\langle\mathbf{x}\rangle
=\langle F(\mathbf{x})\rangle
\approx F(\langle\mathbf{x}\rangle).$$
A: Schrodinger's equation is analogous to Newton's laws because it tells how the particle state responds to the interactions the particle undergoes. In particular, these interactions are embodied in the Hamiltonian operator:
$$\hat{H} = \underbrace{\left(\sum_i \frac{\hat{p}_i^2}{2m_i}\right)}_\text{autokinetic interaction} + \underbrace{\hat{U}}_\text{all 2-, 3-, 4-, etc.-body interactions}.$$
That is, most generally,
$$\hat{H} = \text{(sum of all 1-body interactions)} + \text{(sum of all 2-body interactions)} + \cdots$$
Then Schrodinger's equation tells you the impact that these have upon the state, $|\psi\rangle$, which includes information about the particle position. Thus it tells you about motion.
Also, you are sort of right about the ensemble concept, but I think a better way to think of it, particularly in light of recent work showing how quantum mechanics is a "generalized probability theory", is that you can also take a single-system interpretation as that the state represents an agent's knowledge about the system, and when that state is equivalent to a vector ket, i.e. $|\psi\rangle$ it means the agent's information is extremal. That is, at this point any further gains in information will invariably come with losses - as though there was no more information for the system to give up.
For what it's worth, the connection to Newton's laws actually becomes clearer when we consider this further and expand to the case of possibly submaximal information, in which case we should ditch the ket vector for the more general density state $\hat{\rho}$, which is an operator itself: there, we have
$$\frac{d\hat{\rho}}{dt} = \frac{1}{i\hbar} [\hat{H}, \hat{\rho}]$$
which is directly analogous to the Hamilton's equations
$$\frac{d\rho}{dt} = -\{ \rho, H \}$$
in the Hamiltonian reformulation of Newtonian mechanics, using the Poisson bracket. Schrodinger's equation is then just the special case of this equation for maximal ("pure") states, viz. those where $\hat{\rho}$ has a range which is a ray in the Hilbert space and thus can "essentially" be considered as a vector therein.
Indeed, we can even give a "phase space" formulation in terms of the so-called Wigner function $W$ which is more directly analogous to classical $\rho$ than the density operator $\hat{\rho}$ (the two are actually just a transformation of each other) - there, we have
$$\frac{dW}{dt} = -\{\{ W, H \}\}_\hbar$$
which is almost exactly the same, the only difference being the strange doubling up: we have replaced the Poisson bracket, $\{ \cdot, \cdot \}$, with the Moyal bracket, $\{\{ \cdot, \cdot \}\}$, which might be interpretable as imposing a ceiling on total information content at the maximum phase space resolution, $\hbar$, whose dimensions are of phase-space area.
