1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Propagators homework

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data
    I am so confused about propagators:

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

    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:

    [tex]\psi(x,t) = \int_{-\infty}^\infty \psi(x',0) K(x,t; x', 0) dx'[/tex]

    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.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 14, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    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
    [tex]K(x, t; x', 0) = \langle x, t \mid x' 0 \rangle,[/tex]
    where I put the earlier time on the right. Now let's insert a completeness relation into
    [tex] \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), [/tex]
    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 [itex]\psi(x, t)[/itex] and by integrating over all possible x', we get the chance of being in state [itex]\psi(x, t)[/itex] at time t, no matter what the state at t = 0 was.

    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 [itex]x \neq x'[/itex].
    Last edited: Nov 14, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook