Get the transition amplitudes from a wavefunction? Given a wavefunction $\psi(x,t)$ a transition from time $t_1$ to $t_2$ might be written:
$$\psi(x,t_1) = \int \Delta(x,y,t_1-t_2) \psi(y,t_2) d^3y.$$
But can we solve this to get $\Delta$ in terms of $\psi$? i.e. if we knew the wave function for all times.
Edit: A second condition we must impose:
$$\Delta(x,y,0) = \delta(x,y)$$
Is there a way to get an expression for $\Delta$ in terms of $\psi$? (i.e. knowing what the wavefunction is at all times should we be able to derive what the transition amplitude is. It seems like all the information is there?)
Edit: As an example take $\psi(x,t) = e^{-p x + i p^2 t}$ for some constant p. Then we want to find:
$$\Delta(x,y,t) = e^{it\frac{\partial}{\partial x^2}}\delta(x-y) = \frac{e^{-(x-y)^2/t}}{\sqrt{t}}$$
in terms of the original wave functions.
I think this would be impossible because changing $p^2$ to $p$ for example but setting $p=1$ would give the same wave function but then $\Delta(x,y,t)$ would be different. Not sure if this is true for the general case. (It might just be for these plane wave solutions.)
 A: You may only reconstruct the propagator corresponding to the hamiltonian propagating your specific state, and of course not the part of the hamiltonian acting on spaces orthogonal to it.
Let us nondimensionalize $\hbar=1$ for simplicity, and take $t_2=0, ~~t_1=t$, as customary in QM. The propagator for unitary QM evolution is then 
$$
\psi(x,t)=\int dy ~~ \Delta(x,y;t) \psi(y,0),\qquad \Longrightarrow \\
\langle x| \psi_t\rangle= \int dy ~\langle x| e^{-itH} | y\rangle \langle y|\psi_0\rangle,\qquad  \Longrightarrow \\
|\psi_t\rangle  = e^{-itH}|\psi_0\rangle ~~.
$$
You are looking for the relevant part of the unitary evolution operator U, the exponential of the above unknown hamiltonian, given the entire orbit $|\psi_t\rangle$. 
You will, of course, never get to see the parts of the propagator acting on wavefunctions orthogonal to $|\psi_t\rangle$, as they will never touch it. So you are not really looking at the full propagator $\Delta$, but instead, the projection of the propagator on the subspace of  your wavefunction truncated  by $P=|\psi_0\rangle \langle \psi_0|$.
Think of a $|\psi_0\rangle$ being an eigenstate of the hamiltonian. Then it will be simply evolved by multiplying by the obvious energy phase. Next, consider the paradigmatic 2-state system to illustrate and interpret your formulas. 
The formal correct representation of the idea of your deeply garbled last formula (now deleted) is 
$$
\langle x| e^{-itH} P |y\rangle= \langle x| e^{-itH}|\psi_0\rangle \langle \psi_0 | y\rangle = \\ \langle x|\psi_t\rangle \langle \psi_0 | y\rangle = \psi(x,t)\overline{\psi}(y,0).
$$
That is why it works, as you may check by direct plugin. Again, recall the $t\to 0$ limit gives you the projection of the $\delta$-function on your small subspace, and not the full $\delta$-function. 
You may amuse yourself, to appreciate the method in the madness of the construction, by working out  a toy two-state system, for a diagonal hamiltonian with two different eigenvalues. Indeed, products of the corresponding components of your trajectory wavefunctions will produce the phases. Moreover, the Fourier transform $\int dt ~ \exp(i\omega t) \langle \psi_0|\psi_t\rangle $ will give you energy delta functions of $\omega$ at each of the eigenvalues, etc.

NB. Aside on your freaky particular delta-function propagator, (now deleted). You already know propagators, even for the oscillator, are diffusive. Your particular "one solution" corresponds to the freaky case of a hamiltonian diagonal in coordinate space, i.e. one that does not couple states of different locations: each location evolves with its own eigenenergy and stays put! Eigenfunctions are not normalizable. Indeed, in that case, each phase is gotten from that particular solution; but focussing on this type of ultralocal hamiltonian might not well do wonders for one's peace of mind...

Edit in response to question edit of plane wave propagation with fixed momentum  : Your example of a free plane wave of today is (unnormalizable) fixed momentum p plane wave free propagation. (I reinsert the missing i),
$\psi(x,t)\propto \exp (ipx -ip^2t)$. So, now, you have the full orbit for plane waves of fixed momentum p only, ultralocal in p. This orbit will tell you absolutely nothing about the propagation of any other momentum: all other momenta are projected out!  Essentially, your state will propagate through the kernel $\exp(ip(x-y)-ip^2t/2)$, where the now idle integration w.r.t. y will "provide" the infinite normalization of plane waves. This is at the heart of my point about the irrelevance of the $\delta$-function. The above spectral function is $\int dx dt \exp( i\omega t) \exp (-ip^2 t/2)\propto \delta (\omega-p^2/2)$.
Propagation of ultra-local states in p will only tell you about propagation of momenta p and not, magically,  any other momenta. In fact, you may dial any freaky energy function E(p) beyond free propagation: it will not change, visibly, just as in the ultralocal coordinate case above! That is the point of my sanity-preserving warning above.
A: No, you can't, because the propagator $\Delta$ depends on your Hamiltonian. It is the thing that does the dynamics, (i.e. the time translations), and so you can't just get it from one wavefunction. Knowing the propagator is equivalent to solving Schrödinger's equation.
A: Trnsition Amplitudes and Wavefunctions
Wave Functions
A good reference book is ‘Modern Quantum Mechanics’ by JJ Sakurai, published by Addison Wesley,1994. 
Wave
For our purposes a wave is a disturbance which moves with constant speed in a given direction. This disturbance is assumed to have a constant profile. 
The Profile of the Wave as it Progresses
A wave, such as a Mexican wave, has a general profile of each person’s arms rising, waving then falling as it moves around the stadium. If you are sitting at location x in the stadium then at various times your arms will be rising, waving then falling. Let P(x,t)  denote the amplitude (rising, waving or falling ) of such a wave travelling around the stadium, at location x at time t. 
The simplest example is the oscillating wave---
If ν is the frequency of this wave, λ the wavelength, then at seat x at time t we can show that;
$$P(x,t) = \exp 2\pi i(\frac{x}{\lambda } - \nu t)
% MathType!MTEF!2!1!+-
% feaaheart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiaacI
% cacaWG4bGaaiilaiaadshacaGGPaGaeyypa0JaciyzaiaacIhacaGG
% WbGaaGOmaiabec8aWjaadMgacaGGOaWaaSaaaeaacaWG4baabaGaeq
% 4UdWgaaiabgkHiTiabe27aUjaadshacaGGPaaaaa!49C4!
$$
Where P(x,t) is at its largest with your arms fully extended. P(x,t) is the 'amplitude' of the wave. 
So far this is just general wave motion. If we want a wave to represent an electron for example then a good place to start is the de Broglie relations for the energy E and momentum p of a free particle.
Substituting these into our wave amplitude function gives---
$$\Psi (x,t) = \exp \frac{{2\pi i}}{h}(\frac{{hx}}{\lambda } - h\nu t) = \exp \frac{{2\pi i}}{h}\left( {px - Et} \right)
% MathType!MTEF!2!1!+-
% feaaheart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiQdKLaai
% ikaiaadIhacaGGSaGaamiDaiaacMcacqGH9aqpciGGLbGaaiiEaiaa
% cchadaWcaaqaaiaaikdacqaHapaCcaWGPbaabaGaamiAaaaacaGGOa
% WaaSaaaeaacaWGObGaamiEaaqaaiabeU7aSbaacqGHsislcaWGObGa
% eqyVd4MaamiDaiaacMcacqGH9aqpciGGLbGaaiiEaiaacchadaWcaa
% qaaiaaikdacqaHapaCcaWGPbaabaGaamiAaaaadaqadaqaaiaadcha
% caWG4bGaeyOeI0IaamyraiaadshaaiaawIcacaGLPaaaaaa!5BC5!
$$
Note the change of notation from the profile amplitude P to the quantum wave amplitude Ψ. This is the most basic example possible of a wave function. It corresponds to a free running particle moving in the x direction with energy E and momentum p
Sums of waves corresponding to different energy values of the particle for example, are also waves, and multiplying by α (a complex number) also yields a wave. Thus the set of all such wave functions for a single particle forms a space called a linear space or vector space. Given two wave functions   in this linear space we can also form the inner product---
$\left\langle {{\Psi _1}(t)|{\Psi _2}(t)} \right\rangle  = \int {\Psi _1^*(x,t){\Psi _2}(x,t)dx}$ 
a quantum state is represented by a normalised wave function and the expression above is interpreted as the amplitude for the transition from the first state to the second. 
The ability to form such inner products makes the linear space of wave functions   of our particle into a special type of linear space known as a Hilbert space.
Time Transition
Going back to our basic wavefunction we have;
$\begin{array}{c}
\frac{\partial }{{\partial t}}\Psi (x,t) = \frac{\partial }{{\partial t}}\left\{ {\exp \frac{{2\pi i}}{h}\left( {px - Et} \right)} \right\}\\
 =  - E\frac{{2\pi i}}{h}\left\{ {\exp \frac{{2\pi i}}{h}\left( {px - Et} \right)} \right\}\\
 =  - E\frac{{2\pi i}}{h}\Psi (x,t)
\end{array}[/$
from which we deduce that $i\rlap{--} h\frac{\partial }{{\partial t}}\Psi (x,t) = E\Psi (x,t)[/$
Since E is an eigenvalue of the quantum Hamiltonian $\hat E$ and any reasonable 
wavefunction is a weighted sum of the basic wavefunctions ${\Psi _j}$ we have;
$i\rlap{--} h\frac{\partial }{{\partial t}}\Psi (x,t) = \hat E\Psi (x,t)$ 
The quantum operator solution to this equation is $U(t) = \exp \left( {\frac{{ - i\hat E}}{{\rlap{--} h}}} \right)$
U(t) is the unitary operator giving rise to state transformations in time for a quantum system with a time invariant quantum Hamiltonian. Other cases are trickier-see book reference.  
