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Propagator and delta functions

  1. Jun 13, 2010 #1
    http://en.wikipedia.org/wiki/Propagator

    What does this equation mean:

    [tex]
    \Big(H_x - i\hbar\partial_t\Big) K(x,t,x',t') = -i\hbar\delta(x-x')\delta(t-t')
    [/tex]

    Wouldn't it be more relevant to emphasize these equations:

    [tex]
    \Big(H_x - i\hbar\partial_t\Big) K(x,t,x',t') = 0,\quad\quad t\neq t'
    [/tex]
    [tex]
    K(x,t,x',t') = \delta(x-x'),\quad\quad t=t'
    [/tex]
     
  2. jcsd
  3. Jun 14, 2010 #2

    reilly

    User Avatar
    Science Advisor

    <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<,,

    The second formulation, apart from the missing h bar, and the first are really the same. But, the latter form was used prior to Dirac to define Green's functions, and the techniques for solution are somewhat tedious -- see any E&M text written before 1925, and "elementary" discussions of Green's functions.

    Four dimensional Green's functions, that is a source at x' and t', are basic to solutions of Maxwell's equations. Once QM arrived, physicists were comfortable with inverse operators, and your first fomulation became the approach of choice.

    If I recall correctly, Feynman discussed this issue early on in his first QED papers; Schwinger, likewise. Schwinger, however is very difficult to read.

    Regards,
    Reilly
     
  4. Jun 15, 2010 #3
    Once the propagator and the time evolution are defined:

    [tex]
    K(x,t,x',t') = \frac{1}{\sqrt{2\pi\hbar |t-t'|}}\exp\Big(\frac{i(x-x')^2}{2\hbar(t-t')}\Big)
    [/tex]

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

    we are usually interested in two things. One is

    [tex]
    i\hbar\partial_t\psi(x,t) = -\frac{\hbar^2}{2}\partial_x^2\psi(x,t)
    [/tex]

    so that the Schrödinger equation is satisfied, and the other one is

    [tex]
    \lim_{t\to t'} \psi(x,t) = \psi(x,t')
    [/tex]

    so that the correct initial value is satisfied. In order to prove that the Schrödinger equation is satisfied, we need this:

    And in order to prove that the initial value is satisfied, we need this:

    How could this equation:

    replace the previous two? I see that if you set [itex]\delta(t-t')=0[/itex] for [itex]t\neq t'[/itex], then you get the Schrödinger equation satisfied, but how about the initial value? What is the delta function with respect to time supposed to do there?

    Not only do I not understand how to use this delta function with respect to time, I also don't understand where this equation is supposed to come from.

    When [itex]t\neq t'[/itex], the following can be calculated:

    [tex]
    i\hbar\partial_t K(x,t,x',t') \;=\; -\frac{1}{2}\sqrt{\frac{\hbar}{2\pi}} \frac{1}{|t-t'|^{5/2}}
    \Big(i(t-t') \;-\; \frac{(x-x')^2}{\hbar}\Big) \exp\Big(\frac{i(x-x')^2}{2\hbar (t-t')}\Big)
    \;=\; -\frac{\hbar^2}{2}\partial_x^2 K(x,t,x',t')
    [/tex]

    You cannot get the delta function with respect to time with a limit [itex]t\to t'[/itex] at least, because there is no divergent quantity there. It's pure zero:

    [tex]
    \lim_{t\to t'} \Big(-\frac{\hbar^2}{2}\partial_x^2 - i\hbar\partial_t\Big) K(x,t,x',t') = \lim_{t\to t'} 0 = 0
    [/tex]
     
  5. Jun 16, 2010 #4
    I see (roughly) what the identity means now. A following calculation can be carried out:

    [tex]
    \int\limits_{-\infty}^{\infty} dx \int\limits_{-\infty}^{\infty} dt \Big(\big(-\frac{\hbar^2}{2}\partial_x^2
    \;+\;i\hbar\partial\big)\psi(x,t)\Big) K(x,t,x',t')
    \;=\; \lim_{\epsilon\to 0^+} \int\limits_{-\infty}^{\infty} dx
    \int\limits_{]-\infty,t'-\epsilon]\cup [t'+\epsilon,\infty[} dt
    \Big(\psi(x,t)\big(-\frac{\hbar^2}{2}\partial_x^2\big)K(x,t,x',t')
    \;+\; \big(i\hbar\partial_t\psi(x,t)\big) K(x,t,x',t')\Big)
    [/tex]

    [tex]
    = \lim_{\epsilon\to 0^+}
    i\hbar \int\limits_{-\infty}^{\infty} dx\Big(\psi(x,t'-\epsilon)K(x,t'-\epsilon,x',t') \;-\; \psi(x,t'+\epsilon)
    K(x,t'+\epsilon,x',t')\Big)
    [/tex]

    [tex]
    = \lim_{\epsilon\to 0^+}
    \frac{i\hbar}{\sqrt{2\pi\hbar\epsilon}} \int\limits_{-\infty}^{\infty} dx\big(\psi(x,t') + o(\epsilon)\big)
    (-2i)\sin\Big(\frac{(x-x')^2}{2\hbar\epsilon}\Big)
    \;=\; \frac{(i\hbar)(-2i)}{\sqrt{\pi}} \psi(x',t') \int\limits_{-\infty}^{\infty} \sin(y^2)dy \;=\; \hbar\psi(x',t')
    [/tex]

    Which would be the same thing as

    [tex]
    \Big(-\frac{\hbar^2}{2}\partial_x^2 - i\hbar\partial_t\Big)K(x,t,x',t') = \hbar\delta(x-x')\delta(t-t')
    [/tex]

    I see the coefficient [itex]-i[/itex] vanished somewhere. Maybe there's a mistake somewhere...

    Anyway, I still don't see what's the point of this. In what kind of situation are we interested in integrating with respect to time like this? In order to get the time evolution, there's no need to integrate anything with respect to time.
     
  6. Jun 16, 2010 #5

    strangerep

    User Avatar
    Science Advisor

    You've got a free t' parameter on the rhs, but not on the lhs.
    The point of this sort of thing is that, given a [itex]\psi(x',t')[/itex] you can
    find [itex]\psi(x,t)[/itex]. To do that, a t' integration is needed.

    K(x,t,x',t') should be considered as an inverse operator to
    [tex]
    \left(H_x - i \hbar \partial_t \right)
    [/tex]
    (in the distributional sense).
     
  7. Jun 17, 2010 #6
    The initial time [itex]t'[/itex] can be chosen arbitrarily. Or if you know the wave function only at some given instant, then you should use that given instant.

    The propagator has the following transitivity property

    [tex]
    \int\limits_{-\infty}^{\infty} K(x,t,x',t') K(x',t',x'',t'')dx' = K(x,t,x'',t'')
    [/tex]

    and it doesn't matter what [itex]t'[/itex] you use in the middle. It follows from this property, that the time-evolution equation gives the same final wave function [itex]\psi(x,t)[/itex] for all initial times [itex]t'[/itex]. The time-evolution equation, in which integration is carried out with respect to [itex]x'[/itex], and not with respect to a fixed variable [itex]t'[/itex], is correct after all, so no integration with respect to [itex]t'[/itex] is needed.
     
  8. Jun 18, 2010 #7

    reilly

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    Science Advisor

    Take the FT of your first eq, and get K(E,p) = 1/(E-p*p) -- I'm dropping constants, except those that are really important.

    Now get K(x,t) by inverting. So, K(x,t) == Integral dE dp{exp(-iEt +ipx)/(E-p*p)}. Do the E integration first, to handle the pole, push it below the E axis -- add -i(epsilon) to E, and close the contour in the lower half plane, Then

    K = Integral dp exp{ -ip*pt +ipx } So, as t->0, K does turn into a delta function.
    (This type of contour integration is common in scattering theory, and in EE as well.)

    Regards,
    Reilly Atkinson
     
  9. Jun 18, 2010 #8
    My final question is that what is this identity good for.

    It seems like an interesting result, but I cannot see any use for it.
     
  10. Jun 18, 2010 #9

    K^2

    User Avatar
    Science Advisor

    And what ever happened to integral over t'?

    If you are talking about classical limit, it should actually look like this:

    [tex]
    \int\limits_{-\infty}^{\infty} dx' \int\limits_{t}^{t''} dt' K(x,t,x',t') K(x',t',x'',t'') = K(x,t,x'',t'')
    [/tex]
     
  11. Jun 18, 2010 #10
    The equation that I wrote is correct, and it has fixed [itex]t'[/itex]. It is a property of propagators which appear in QM and also in the study of the heat equation.
     
  12. Jun 24, 2010 #11

    samalkhaiat

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    Science Advisor

    The meaning and importance of this identity become clear if you realize that the propagator (Green function) [itex]K(x,t : y,t_{0})[/itex]can be interpreted as transition amplitude. That is the probability amplitude for the system prepared at [itex]t_{0}[/itex] with position eigenvalue [itex]y[/itex] to be found at a later time [itex]t[/itex] at [itex]x[/itex] ;

    [tex]K(x,t : y,t_{0}) = \langle x | e^{-iH(t-t_{0})}| y \rangle = \langle x,t | y,t_{0}\rangle \ \ \ (1)[/tex]

    where [itex]\langle x,t|[/itex] and [itex]|y,t_{0}\rangle [/itex] are the eigenket and the eigenbra of the position operator in the Hiesenberg picture. Because at any given time these eigen vectors form a complete set, we can insert the identity operator;

    [tex]\int dz |z,t_{1}\rangle \langle z,t_{1}| = \hat{1}[/tex]

    at any place we want. So, by dividing the evolution interval [itex](t_{0},t_{2})[/itex] into two parts [itex](t_{0},t_{1})[/itex] and [itex](t_{1},t_{2})[/itex], we can decompose the transition amplitude as :

    [tex]\langle x , t_{2} | y , t_{0} \rangle = \int dz \langle x ,t_{2} | z , t_{1} \rangle \langle z , t_{1}| y , t_{0}\rangle \ \ (2)[/tex]

    [tex](t_{2} > t_{1} > t_{0})[/tex]

    This composition property of the transition amplitude is an important consistency requirement underlying the whole formalism of path integration. Indeed, we can use it to show that the transition amplitude [itex]\langle x , t| y , t_{0} \rangle[/itex] satisfies Schrodinger equation in the variables (x,t), just as the propagater [itex]K(x,t : y,t_{0})[/itex] ;

    [tex]i\partial_{t}\langle x , t | y , t_{0} \rangle = H(t_{0})\langle x ,t | y ,t_{0} \rangle \ \ \ (3)[/tex]

    ( argue that you can write [itex]\langle x , t + \epsilon | y , t_{0} \rangle = \delta (x - y) - i \epsilon\delta(x-y)H(t_{0})[/itex]
    and use eq(2) for [itex]t_{1} = t[/itex] and [itex]t_{2} = t + \epsilon[/itex])

    When he path integration method is carried over to Brownian motion, eq(2) is known as the Chapman-Kolmogorov equation, and in diffusion theory, the Smoluchowsky equation.

    regards

    sam
     
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