$\langle B|A \rangle$ expressed in terms of the Partition Function Say you have an electron departing from point A and reaching poing B after a time t.
According to some helping friend, the Partition Function for that electron going from point A to B can be written as
$$Z = \int_{A \to B} [\mathcal{D}x]~ e^{iS[x]}$$
where $\mathcal{D}x$ is the measure that sums up over all paths going from $A$ to $B$, and $e^{iS[x]}$ is the weight of each path, $S[x]$ is the action.
That friend states then "From this partition function all desirable quantities can be obtained."
Not having much idea about Feynman's Path Integral Formulation of Quantum Mechanics, I have looked around a bit, and I would like someone to confirm the following statement I make:
The amplitude for the electron to go from A to B in the time t can be found in terms of that Partition Function $Z$ given above, as
$$\langle B|e^{-iHt}|A\rangle = Z$$
Did I catch it right?
 A: For simplicity, let's restrict the discussion to that of a single particle moving in one dimension.  Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point.
Let $H$ denote the (time-independent) quantum hamiltonian.  Then the time evolution on the system is governed by the operator $U(t) = e^{-itH/\hbar}$ called the time evolution operator.  If at a time $t_a$ the particle is in the state $|x_a\rangle$, the eigenstate of the position operator corresponding to position $x_a$, then the probability amplitude for finding the particle in the state $|x_b\rangle$ at some later time $t_b$ is given by
$$
  K(t_b,x_b;t_a,x_a) = \langle x_b|U(t_b-t_a)|x_a\rangle
$$
This object is often referred to as the propagator.  It turns out the propagator has the following path integral expression
$$
  K(t_b,x_b;t_a,x_a) = \int \mathcal Dx\mathcal Dp\, e^{\frac{i}{\hbar}S}
$$
where the integral is over all paths $(x(t), p(t))$ in classical phase space satisfying the boundary conditions $x(t_a) = x_a$ and $x(t_b) = x_b$.
Now you could ask, what about if at time $t_a$, the system started in a state $|\psi_a\rangle$ that is not necessarily a position eigenstate, and I want to know what the probability is of finding that particle in the state $|\psi_b\rangle$ at time $t_b$.  Can I write this in terms of the path integral?  Yes!  
What we're looking to compute is the following quantity:
$$
  \langle \psi_b|U(t_b-t_a)|\psi_a\rangle
$$
Now we recall that the we can write the identity operator as
$$
  I = \int dx\,|x\rangle\langle x|
$$
Inserting this expression twice into the amplitude we want to compute, and using integration variables $x_a$ and $x_b$ gives
\begin{align}
  \langle \psi_b|U(t_a-t_a)|\psi_a\rangle
&= \int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_b|U(t_a-t_a)|x_a\rangle\langle x_a|\psi_a\rangle \\
&= \int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_a|\psi_a\rangle K(t_b,x_b;t_a,x_a)
\end{align}
Now we just write the propagator in terms of the path integral as above to obtain
$$
  \langle \psi_b|U(t_a-t_a)|\psi_a\rangle
=\int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_a|\psi_a\rangle \int \mathcal Dx\mathcal Dp\, e^{\frac{i}{\hbar}S}
$$
A: What your friend actually meant is that you can obtain all desirable correlation functions. Assuming you're talking about a non-relativistic electron, consider a source term added to your action
$$
S'[x] = S[x] + \int dt \, J(t) x(t)\,.
$$
Now you can write any correlation function as a derivative of $\ln Z$ calculated at $J = 0$, i.e.
$$
\langle B| \hat{x}(t') | A \rangle = \left.\frac{\partial}{\partial J(t')} \ln Z\right|_{J=0}\,,
$$
$$
\langle B|\hat{T}\left\{ \hat{x}(t') \hat{x}(t'') \right\}| A \rangle = \left.\frac{\partial^2}{\partial J(t')\partial J(t'')} \ln Z\right|_{J=0}\,,
$$
and so on, where $\hat{T}$ is the time ordering operator.
Hope this helps you.
