Propagators Homework: Understanding K(x,t;x',0) & More

[ehrenfest]

Homework Statement

I am so confused about propagators:

$$K(x,t;x',0) = \int |E\rangle e^{-iEt/\hbar} \langle E| dE$$

I understand the RHS of that equation perfectly: it just decomposes the time-independent state into its eigenstates and then propagates each of the eigenstates individually.

I would understand the LHS if and only if the ";x'," were removed from it. I simply do not understand why you need to get rid of the prime after you propagate the state? Why can you not propagate a time-independent wave-function of x' and get a time-independent wavefunction of x not x'?

EDIT: here is another equation from the wikipedia site on propagators:

$$\psi(x,t) = \int_{-\infty}^\infty \psi(x',0) K(x,t; x', 0) dx'$$

I think I am starting to understand this better. So, the reason you have an x and an x' is that the x' is summed over (continuously) if we want to think of the propagator just as a huge summation. But still why don't other operators like the Hamiltonian have an (x,x') attached to them? You can think of the Hamiltonian as a matrix operator as well.

Homework Equations

The Attempt at a Solution

 

[CompuChip]

Was just going over this stuff for an exam tomorrow (also, you may consider that a disclaimer: if I write nonsense somewhere, I probably haven't really understood that yet); anyway, here's how I like to look at it:

You can also define the propagator as the overlap between two wave functions at different times, that is
$$K(x, t; x', 0) = \langle x, t \mid x' 0 \rangle,$$
where I put the earlier time on the right. Now let's insert a completeness relation into
$$\psi(x, t) = \langle x t \mid \psi \rangle = \langle x t \mid \left( \int dx' |x' 0\rangle \langle x' 0 \rangle \right) | \psi \rangle = \int dx' \langle x t | x' 0 \rangle \langle x' 0 | \psi \rangle = \int dx' K(x, t; x', 0) \psi(x', 0),$$
which is the equation you cited.
So the propagator can be seen as the function that describes the odds of a system in state [itex]\psi(x', t' = 0)[/tex] ending up in the state $\psi(x, t)$ and by integrating over all possible x', we get the chance of being in state $\psi(x, t)$ at time <i>t</i>, no matter what the state at t = 0 was.<br /> <br /> In addition, the Hamiltonian can indeed be considered a matrix operator, but in the position basis, it's a diagonal matrix. That is, H(x, x') vanishes if $x \neq x'$.[/itex]