Are the Schrodinger equation and expectation values related? For a while, I had difficulty understanding the differences between an expectation value and the state-vector derived from the Schrodinger equation. My understanding is that the Schrodinger equation returns a state-vector $\mid\Psi(t)\rangle$ that can be used to calculate probabilities for events corresponding to possible values, or eigenvalues, of a system. On the other hand, expectation values allow one to determine, with varying accuracy, what the value of an observable is 'most likely' to be, regardless of whether it is a possible state. For example, take a single spin. If the spin is prepared in some direction and rotated in an arbitrary direction, the spin will be either +1 or -1. However, the expectation value will be $\sigma_n = \langle 0 \rangle$. Solving the Schrodinger equation will give a ket $\mid\Psi(t)\rangle$ that can be used to calculate the probabilities that the spin will be +1 or -1.
Both of these are helpful tools used to find what the value of an observable may or may not be. Is there a mathematical or physical relationship between the two? Or are they just two separate ways of obtaining similar but fundamentally different information?
 A: $|\Psi(t)\rangle$ contains the time evolution of a superposition of states describing the probability amplitude of each state of the system - it is not a physical entity but an abstract vector representation of the probability amplitude of measurable states (energy, spin, position etc).
Let's ignore the time dependence of $\Psi$ for now since it's not needed to answer your question.
Suppose the system is in a superposition of observable states $A$, $|\Psi \rangle = A_1|A_1\rangle + A_2|A_2\rangle \:+ \; ...$
$A_i$ is the probability amplitude of state $|A_i\rangle$ where $i$ may be $1,2,3...$
The expectation value of $A$ is:
$|$average probability amplitude of all the superposition states of $A$ that $\Psi$ is in$|^2$ 
and can be calculated using the matrix representation of the operator representing measurable $A$:
$A \;\dot{=} \;\hat{A}$ (matrix form)
$\langle A\rangle\, = \,\langle\Psi|\hat{A}|\Psi\rangle =\sum\limits_{i}$$A_{i}P_{A_{i}}$
Where $P_{A_{i}}$ $= |\langle A_i|\Psi\rangle|^2$ which is the |probability amplitude of specific state $|A_i\rangle|^2$
Two conclusions to answer your question:
$\bullet$$\;\;$$\boldsymbol{|\Psi\rangle}$ is a superposition which means it represents the system being in a combination of "$A$" states simultaneously until we measure the state of the system (causing the collapse of $|\Psi\rangle$ to just one state of $A$ let's say is $|A_i\rangle$).
$\bullet$$\;\;$$\boldsymbol{\langle A\rangle}$ is the mean value of a large number of experiments measuring A. It is not a time average, but an average over many identical experiments.
A: The expectaion value of quantity $A$ is given by
$$\langle A\rangle = \langle\Psi|\hat{A}|\Psi\rangle,$$
where $\hat{A}$ is the operator for this quantity, whereas $|\Psi\rangle$ is a state vector, also called a wave function.
As the system evolves in time, the expectation values also change. In the so-called Schrödinger picture/representation this is accounted for by evolving the state vectors according to the Schrödinger equation:
$$i\hbar\partial_t|\Psi(t)\rangle = \hat{H} |\Psi(t)\rangle,$$
so that the time-dependent expectation values are given by 
$$\langle A(t)\rangle = \langle\Psi(t)|\hat{A}|\Psi(t)\rangle.$$
Extra material
An alternative view is the so-called Heisenberg picture/representation, where the evolution is contained in the operators, which obey the Heisenberg equations
$$\partial_t \hat{A}(t) = \frac{1}{i\hbar}[\hat{A},\hat{H}],$$
where $[...,...]$ designates a commutator and the expectation value is given by
$$\langle A(t)\rangle = \langle\Psi|\hat{A}(t)|\Psi\rangle.$$
