What does the propagator in Quantum Mechanics mean? I mean, except from the mathematics behind it, what does it tell us? Is it something that has to do with translations in time?
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asked Mar 7, 2016 at 15:26
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4$\begingroup$ What about e.g. the introduction to the Wikipedia article is unclear to you? $\endgroup$– ACuriousMind ♦Mar 7, 2016 at 15:28
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3$\begingroup$ @ACuriousMind i am asking in here because the site is full of people who actually have insight on things like this and that wikipedia might not include. So, there is no need to act like this. If you do not want to contribute to an answer, then just don't. Trying to "hear" something from someone because you believe that they might know more than what is written on a wikipedia article does not mean that i did not search the internet. In the internet i have only found mathematical descriptions. $\endgroup$– TheQuantumManMar 7, 2016 at 15:32
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4$\begingroup$ I think what ACM is trying to say is that you should really be more specific about what you're interested in. Outlining what exactly you already understand, and what points you'd like to understand better is generally a good idea. As the question looks now, it's hard to tell that you actually did already do some prior research, and it's also hard to know where to start/what to focus on in an answer to this question. $\endgroup$– DanuMar 7, 2016 at 15:35
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2$\begingroup$ I quote the first sentence of the Wikipedia article: "In quantum mechanics and quantum field theory, the propagator gives the probability amplitude for a particle to travel from one place to another in a given time, or to travel with a certain energy and momentum." That's what it "does" in general. This is also standard lore that will be found in good introductions to QFT - it's the two-point function. I firmly believe questions here should be about things that are not easily found on Wikipedia. $\endgroup$– ACuriousMind ♦Mar 7, 2016 at 15:40
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1$\begingroup$ @ACuriousMind Wikipedia states more things about the propagator in QFT than introductory Quantum Mechanics. Also, most answers in here often offer more than one sentence. You could write an answer containing just that. But somebody else might bother and write something that offers more information and intuition. That's the idea behind the question. $\endgroup$– TheQuantumManMar 7, 2016 at 15:43
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2 Answers
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Propagator in quantum mechanics is just a different name for Green's function for time-dependent Schroedinger equation. It is a unique function that enables us to write, for any time $t_0$,
$$ \psi(x,t) = \int G(x,t;x',t_0)\psi(x',t_0)\,dx' $$ for all $x$ and all $t$. This means the $\psi$ function of $x$ (at any time $t$) can be written as a result of linear operator acting on the $\psi$ function at any other time $t_0$ (usually $t_0<t$ but this is not necessary).
When $\psi(x',t_0)$ is put equal to $\delta(x'-x^*)$ for some $x^*$, we obtain $$ \psi(x,t) = G(x,t;x^*,t_0) $$
so propagator is a result of evolution of $\delta$ distribution governed by the same equation as ordinary $\psi$ functions are - time-dependent Schroedinger equation.
However, $\delta$ distribution is not admissible as $\psi$ in the sense of the Born interpretation, so be cautious when giving $G$ physical interpretation.
answered Mar 7, 2016 at 17:08
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The clarity offered by using Feynman diagrams to set up the calculations for a scattering or a decay process also gives an intuitive meaning to the propagator function. The Feynman diagrams are iconal representations with one to one mathematical rules of how to set up, order by order in a perturbative expansion, the integrals which will give a measurable quantity.
Here is a Feynman diagram for estimating the probability of neutron decay , to first order.
The internal lines are represented mathematically by a propagator function which will be integrated over the available phase space.
Virtual particles carry the quantum numbers of the name, in the above case of the W-, but not the mass. Evidently with the mass of the neutron around 1 GeV and the W close to 80GeV the reaction would not go if W were on shell.
So the propagator in this example tells us how the quantum numbers propagate from the initial state to the final state and how the phase space contributes to and kinematically sets bounds to the decay amplitude under study. To be noted, the propagator depresses the value of the integral because of the very much larger mass of W with respect to the energy available for the bounds of the integral .
The same logic holds true for the internal lines for any Feynman diagram, they represent propagators..
